Abstract Algebra

Cut an equilateral triangle out of stiff card. Number the three corners 1, 2, 3, writing the same number on both faces so a number stays with its corner when you flip the card. Draw the outline of the triangle on a sheet of paper and set the card inside it. Now ask a narrow question: in how many ways can you pick the card up and put it back so that it fills the outline exactly?

Count them. You can leave it alone. You can turn it 120 degrees counterclockwise, or 240 degrees. You can flip it over about the line through corner 1, or about the line through corner 2, or about the line through corner 3. That is six, and there is no seventh. Those six motions, with the operation "do one, then do the other," are the first group in this course, and almost everything in the next seventeen lessons is a generalisation of what you can already see on the table in front of you.

Six motions, and how they multiply

Name them. Write e for leaving the card alone, r for the 120 degree counterclockwise rotation, and r^2 for the 240 degree rotation. Write s for the flip about the line through corner 1. The remaining two flips will turn out not to need new names.

To multiply two motions, perform them in sequence. We will use the convention that ab means "do b first, then a," matching the way composition of functions is written. Track what happens to the corner positions rather than to the card, and record each motion as a permutation of the labels. The rotation r sends the corner in position 1 to position 2, position 2 to position 3, and position 3 to position 1, which we write in cycle notation as (1 2 3). The flip s holds corner 1 and exchanges 2 and 3, so s = (2 3).

Now compute rs. Apply s first: 1 stays at 1, 2 goes to 3, 3 goes to 2. Then apply r: 1 goes to 2, 3 goes to 1, 2 goes to 3. Follow each label through both steps. Label 1 goes to 1 and then to 2. Label 2 goes to 3 and then to 1. Label 3 goes to 2 and then to 3, staying put. So rs = (1 2), the flip about the line through corner 3.

Compute sr the same way. Apply r first, then s. Label 1 goes to 2 and then to 3. Label 2 goes to 3 and then to 2, staying put. Label 3 goes to 1 and then to 1. So sr = (1 3), the flip about the line through corner 2. Two flips, two different answers. The point: rs and sr are not the same motion, so this operation is not commutative, and any theorem you prove about groups in general must survive that fact.

We have also just accounted for all six elements without inventing new letters. The list is e, r, r^2, s, rs, r^2 s, and you can check that r^2 s = (1 3) = sr. That last identity, sr = r^2 s, is the whole multiplication rule in one line: it tells you how to push a rotation past a flip, and with r^3 = e and s^2 = e it determines every product in the table. This group is called the dihedral group D3, and it has order 6.

The four axioms

Strip away the triangle and keep only what the six motions actually did. A group is a set G together with an operation that combines any two elements a and b of G into an element ab of G, subject to four requirements.

• Closure. For all a and b in G, the product ab lies in G. Doing two symmetries in a row is again a symmetry.
• Associativity. For all a, b, c in G, (ab)c = a(bc). Composition of functions is associative for free, which is why so many groups are groups of functions.
• Identity. There is an element e in G with ea = ae = a for every a. Leaving the card alone.
• Inverses. For each a in G there is an element a^(-1) in G with a a^(-1) = a^(-1) a = e. Undoing a motion.

Commutativity is not on the list. A group in which ab = ba always holds is called abelian, after Niels Henrik Abel, and abelian groups are a special case, not the default. The integers under addition are abelian; D3 is not.

Notice how little is being assumed. There is no requirement that elements be numbers, that the operation resemble arithmetic, or that the set be finite. That poverty is deliberate. Every theorem we prove from these four lines will hold for the symmetries of a triangle, the invertible matrices, the permutations of a deck of cards, and the legal moves of a Rubik's cube simultaneously, because all of them satisfy the same four lines.

What the axioms force

The axioms say "there is an identity," not "there is exactly one." Uniqueness has to be earned.

Proposition. A group has exactly one identity element.

Proof. Suppose e and f are both identities. Since e is an identity, e f = f. Since f is an identity, e f = e. The left sides are the same element, so e = f. This completes the proof.

The trick is worth naming, because it recurs: evaluate one product two ways, using a different hypothesis each time, and read the equality off. The same move settles inverses.

Proposition. Each element of a group has exactly one inverse.

Proof. Suppose b and c are both inverses of a, so ab = ba = e and ac = ca = e. Then b = be = b(ac) = (ba)c = ec = c, where the third equality is associativity. This completes the proof.

Every step of that chain cites an axiom, and the middle one is where associativity earns its place on the list. Drop associativity and the argument collapses. Two more consequences follow quickly, and both are used constantly.

Proposition (cancellation). In a group, ab = ac implies b = c, and ba = ca implies b = c.

Proof. Multiply ab = ac on the left by a^(-1). The left side becomes a^(-1)(ab) = (a^(-1)a)b = eb = b and the right side becomes c by the same computation. The right-cancellation statement is the mirror image, multiplying on the right instead. This completes the proof.

Proposition. (ab)^(-1) = b^(-1) a^(-1).

Proof. Compute (ab)(b^(-1)a^(-1)) = a(b b^(-1))a^(-1) = a e a^(-1) = a a^(-1) = e, and symmetrically (b^(-1)a^(-1))(ab) = e. So b^(-1)a^(-1) is an inverse of ab, and by the previous proposition it is the inverse. This completes the proof.

The order reversal is not a formality. Put on socks, then shoes; to undo, remove shoes, then socks. In D3, (rs)^(-1) = s^(-1) r^(-1) = s r^2, and you can check on the card that this is the same flip as rs, which is consistent because a flip is its own inverse.

Groups that are not numbers

Here is a spread of examples chosen so that no two of them look alike. Working through the verification of the axioms in each one is the fastest way to internalise what the definition is asking for.

The cube deserves a comment, because it is the example most likely to convince a sceptic that groups are not a notational habit. A move is any sequence of face turns; two sequences count as the same element when they leave every sticker in the same place. Closure holds because doing one sequence after another is a sequence. Associativity holds because concatenation of sequences is associative. The identity is the empty sequence. The inverse of a sequence is the same turns performed backwards, each reversed. The order of that group, computed by exactly the counting techniques this course develops, is 43252003274489856000.

Now the near misses, which teach as much as the examples. The natural numbers under addition fail the inverse axiom: 3 has no natural number partner summing to 0. The integers under multiplication fail it too, since 2 has no integer reciprocal. The real numbers under multiplication fail only at a single point, 0, and deleting that point repairs everything: the nonzero reals under multiplication are a group. Subtraction on Z fails associativity, since (5 - 3) - 1 = 1 while 5 - (3 - 1) = 3. Worth holding on to: when you meet a candidate group, check the inverse axiom first. It is the one that fails most often, and it usually fails at exactly one element.

Reading a Cayley table

For a small finite group you can write down the whole operation as a table, with the row label multiplied on the left of the column label. Here are the two groups of order 4, side by side.

Both tables have a structural property worth proving, because it is a free consequence of cancellation.

Proposition. In the Cayley table of a finite group, every element of the group appears exactly once in each row and exactly once in each column.

Proof. Fix a in G and consider the map that sends x to ax, which produces the row labelled a. It is injective, because ax = ay forces x = y by left cancellation. It is surjective, because given any g in G the element x = a^(-1)g satisfies ax = g. A map from a finite set to itself that is injective and surjective is a bijection, so the row is a rearrangement of G. Columns are handled by the same argument with right multiplication. This completes the proof.

A table with that property is called a Latin square, and it gives you a cheap sanity check on any table you construct by hand. It is not, however, a sufficient condition: plenty of Latin squares fail associativity and so define no group. Associativity is the axiom that a table hides, since checking it directly on a group of order n means verifying n^3 equations.

Compare the two tables above. In Z_4 the element 1 has the property that 1 + 1 = 2, 1 + 1 + 1 = 3, 1 + 1 + 1 + 1 = 0, so repeatedly adding 1 sweeps out the whole group. In V, the Klein four-group, every non-identity element satisfies x^2 = e, so nothing sweeps out the group. These are the only two groups of order 4, and they are genuinely different. Recognising when two tables describe the same group in different clothing is the problem the isomorphism theorems of Module 2 exist to solve.

Common misconceptions

• "A group is a set of numbers with a kind of arithmetic." Not one of the four axioms mentions numbers. The symmetries of a triangle are motions, not quantities, and the cube group's elements are physical rearrangements. What the axioms describe is reversible combination, and that pattern appears wherever things can be done and undone.
• "Groups are commutative unless something goes wrong." Commutativity was never assumed. The very first group in this lesson fails it, and so do the permutation groups, the matrix groups, and the cube. The abelian ones are the exception you must earn.
• "Closure is automatic, so it is not really an axiom." Closure is a statement about the pair (set, operation), and it is exactly what fails when you take a good operation and a badly chosen set. The odd integers are closed under multiplication but not addition; the rotations of the triangle are closed, but the three flips alone are not, since two flips compose to a rotation.
• "The inverse of ab is a^(-1)b^(-1)." Only in an abelian group. In general the order reverses, and in D3 the difference is visible on the card.

Try it: a group hiding in modular arithmetic

Exercise. Let U(8) be the set of residues mod 8 that are coprime to 8, namely 1, 3, 5, 7, with multiplication mod 8. Show it is a group and identify which of the two order-4 groups it is.

Solution. Closure: the products mod 8 are 3 * 3 = 1, 3 * 5 = 7, 3 * 7 = 5, 5 * 5 = 1, 5 * 7 = 3, 7 * 7 = 1, and every result is again in the set. Associativity is inherited from multiplication of integers. The identity is 1. For inverses, the computations just listed show each of 3, 5 and 7 is its own inverse. So U(8) is a group of order 4 in which every non-identity element squares to the identity, which is the signature of the Klein four-group V, not of Z_4.

Change one input and the answer flips. Take U(5) = {1, 2, 3, 4} under multiplication mod 5. Then 2^2 = 4, 2^3 = 3, 2^4 = 1, so 2 sweeps out the whole group and U(5) behaves like Z_4. Two sets of four residues, two multiplication rules that look equally innocent, and two different groups.

What you now know

Four axioms, and the uniqueness results they force: one identity, one inverse per element, cancellation on both sides, and the reversing rule for the inverse of a product. A catalogue of groups in which only two are number systems. The Latin square property of a Cayley table, proved from cancellation alone, and the warning that it does not certify associativity. If you can verify the axioms for an unfamiliar set and operation, and can say in one sentence why Z_4 and V are not the same group, the lesson has done its work.

Looking ahead

The next lesson looks inside a group rather than at whole groups. A subgroup is a subset that is a group in its own right, and the cleanest way to produce one is to pick a single element and take all its powers. That construction leads to the order of an element, to the cyclic groups, and to a complete description of every subgroup a cyclic group can have. It also sets up the counting argument that Lagrange's theorem turns into the most useful constraint in finite group theory.

Sources

1. Weisstein, E. W. (n.d.). Group. Wolfram MathWorld. mathworld.wolfram.com
2. Judson, T. W. (n.d.). Chapter 3: Groups. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
3. Weisstein, E. W. (n.d.). Dihedral group. Wolfram MathWorld. mathworld.wolfram.com
4. O'Connor, J. J., & Robertson, E. F. (1996). The abstract group concept. MacTutor History of Mathematics Archive. mathshistory.st-andrews.ac.uk
5. Wikipedia contributors. (n.d.). Rubik's Cube group. Wikipedia. en.wikipedia.org
6. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Chapter 1. John Wiley and Sons.

Put your finger on the 12 of a clock face and move it forward four hours at a time. You land on 4, then 8, then back on 12, and from then on you circle those three positions forever. You will never touch 1, or 5, or 7. Move forward three hours at a time instead and you visit 3, 6, 9, 12: four positions, again a closed cycle, again with nine positions permanently out of reach.

Something is being generated here. Each step size produces a set that is closed under the clock's own addition, contains the starting point, and lets you back out of any move. That is the group structure of the whole clock face, restricted to part of it, and the object it produces is called a subgroup. This lesson is about how much of a group's architecture is already determined by what one element does when you keep applying it.

What counts as a subgroup, and how to check it cheaply

A subset H of a group G is a subgroup, written H <= G, if H is itself a group under the operation G already carries. Written out that way it is four conditions, but two of them come free. Associativity is inherited: if (ab)c = a(bc) holds for all elements of G, it certainly holds for those that happen to lie in H. And the identity of H, if H has one, must be the identity of G: if f is an identity for H then ff = f, and cancelling f in G gives f = e. So only closure and inverses need checking, and even those can be bundled.

Proposition (one-step subgroup test). A nonempty subset H of a group G is a subgroup if and only if a b^(-1) lies in H whenever a and b lie in H.

Proof. If H is a subgroup the condition is immediate, since b^(-1) is in H and H is closed. Conversely, assume the condition. Since H is nonempty, choose some a in H; taking b = a gives a a^(-1) = e in H. Now for any b in H, taking the pair e and b gives e b^(-1) = b^(-1) in H, so H is closed under inverses. Finally, for a and b in H, we know b^(-1) is in H, so applying the condition to the pair a and b^(-1) gives a (b^(-1))^(-1) = ab in H. This completes the proof.

Read the order of that argument. Nonemptiness buys the identity, the identity buys inverses, and inverses buy closure. Each step uses the one condition we were given, applied to a cleverly chosen pair. When H is finite there is an even cheaper test, and it is the one that saves the most labour in practice.

Proposition (finite subgroup test). A nonempty finite subset H of a group G that is closed under the operation is a subgroup.

Proof. Let a be in H. By closure the powers a, a^2, a^3, ... all lie in H, and since H is finite they cannot all be distinct, so a^i = a^j for some i < j. Cancelling gives a^(j-i) = e, and j - i is a positive integer, so e lies in H. If j - i = 1 then a = e and a is its own inverse; otherwise a^(j-i-1) is a positive power of a, so it lies in H, and a a^(j-i-1) = a^(j-i) = e shows it is a^(-1). So inverses are present, and H is a subgroup. This completes the proof.

Notice the hypothesis doing the work: finiteness. On the infinite side the test fails at once. The positive integers form a nonempty subset of Z closed under addition, and they are nowhere near a subgroup, because no inverse of 3 is present. Why this matters: in a finite group, closure alone is enough; in an infinite group, closure alone is almost nothing.

The order of an element

For an element a of a group, the order of a, written |a|, is the least positive integer n with a^n = e, and is called infinite if no such n exists. In the clock group Z_12, written additively, the order of 4 is 3 because 4 + 4 + 4 = 12 = 0 and no smaller multiple vanishes. The order of 3 is 4. The order of 5 is 12, since the multiples of 5 mod 12 run 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 0 and hit every position before returning.

The argument inside the finite subgroup test already proved something general: in a finite group every element has finite order, because the powers of a must repeat. The next proposition is the one you will use most.

Proposition. Let a have finite order n. Then a^m = e if and only if n divides m.

Proof. If n divides m, write m = nq; then a^m = (a^n)^q = e^q = e. Conversely suppose a^m = e. By the division algorithm write m = qn + r with 0 <= r < n. Then e = a^m = (a^n)^q a^r = a^r. But n is the least positive integer with a^n = e, and r < n, so r cannot be positive. Hence r = 0 and n divides m. This completes the proof.

Worked example. In the symmetric group S_4 take the permutation t = (1 2 3)(4), a three-cycle. Then t^3 is the identity and t has order 3, so t^100 = t^(3 * 33 + 1) = t. You never have to multiply a hundred permutations; you divide 100 by 3 and read the remainder. The same trick computes 5^(2026) in U(12): since 5^2 = 25 = 1 mod 12, the order of 5 is 2, and 2026 is even, so the answer is 1.

The subgroup a single element generates

Given a in G, define <a> to be the set of all integer powers {..., a^(-2), a^(-1), e, a, a^2, ...}. This set is closed, since a^i a^j = a^(i+j), and contains the inverse a^(-i) of each of its members, so by the one-step test it is a subgroup: the cyclic subgroup generated by a. It is the smallest subgroup of G containing a, because any subgroup containing a must, by closure and inverses, contain every power of a.

A group G is cyclic if G = <a> for some element a, called a generator. Two families exhaust the possibilities. If a has infinite order, the powers a^i are pairwise distinct and <a> behaves exactly like Z under addition, with a^i playing the role of the integer i. If a has order n, then <a> = {e, a, a^2, ..., a^(n-1)} has exactly n elements and behaves exactly like Z_n. That the list has no repeats is worth checking: if a^i = a^j with 0 <= i < j < n then a^(j-i) = e with 0 < j - i < n, contradicting minimality of n. So |<a>| = |a|: the size of the subgroup an element generates is the order of that element, which is why the same word is used for both.

Cyclic does not mean small or trivial. The clock face is cyclic. So is the group of 12th roots of unity in the complex plane, generated by cos(30 degrees) + i sin(30 degrees). So is U(5), generated by 2. And so is the additive group of integers, generated by 1, which is infinite. What cyclic does mean is abelian, always: a^i a^j = a^(i+j) = a^j a^i. That single line rules out ever finding a cyclic copy of D3 or of any other non-commutative group.

Every subgroup of a cyclic group is cyclic

Here is the structural theorem of the lesson. It says that the cyclic property is inherited downwards with no exceptions, which is unusual and very useful.

Theorem. Let G = <a> be cyclic and let H <= G. Then H is cyclic.

Proof. If H = {e} then H = <e> and we are done. Otherwise H contains some a^k with k not zero, and since H also contains a^(-k), it contains a power of a with positive exponent. Let d be the least positive integer such that a^d lies in H; this exists because every nonempty set of positive integers has a least element. We claim H = <a^d>. One inclusion is free: a^d is in H, so all its powers are. For the other, take any element a^k of H and divide: k = qd + r with 0 <= r < d. Then a^r = a^(k - qd) = a^k (a^d)^(-q), a product of two elements of H, so a^r lies in H. Since r < d and d was least positive with that property, r = 0. Hence a^k = (a^d)^q lies in <a^d>. This completes the proof.

The engine is the division algorithm, deployed exactly as it was in the order proposition: divide, produce a remainder that lands inside the set, and let minimality kill the remainder. You will see this pattern a third time in Module 5, when we prove that every ideal of the integers is generated by a single element. It is the same proof wearing a ring-theoretic hat.

Two computational corollaries follow. First, in a cyclic group of order n the order of a^k is n / gcd(n, k). Second, a^k generates the whole group exactly when gcd(n, k) = 1, so Z_n has precisely as many generators as there are integers between 1 and n coprime to n. For n = 12 those are 1, 5, 7 and 11, which is why moving four hours at a time trapped your finger and moving five hours at a time did not.

The subgroups of the clock, listed

Combining the theorem with the corollaries gives a complete answer for Z_n: for every divisor d of n there is exactly one subgroup of order d, generated by n/d, and there are no others. Here is Z_12 in full.

Six subgroups, one for each of the six divisors of 12, and every subgroup order divides 12. That last observation is not an accident of this example, and proving it in general is the business of the next lesson.

Non-cyclic groups behave quite differently. In D3 the subgroups are the trivial one, three subgroups of order 2 generated by the three reflections, one subgroup of order 3 generated by r, and D3 itself: six subgroups, but three of them have the same order, so the tidy one-per-divisor statement is specific to cyclic groups. In the Klein four-group V there are three distinct subgroups of order 2, and no element of order 4, so V is not cyclic at all.

Common misconceptions

• "A subgroup just has to be closed." True for finite subsets, false in general, and the positive integers inside Z are the standard counterexample. Get in the habit of asking whether the subset is finite before you use the cheap test.
• "A subgroup could have a different identity element from the group." It cannot. If f is an identity for H then ff = f = fe, and cancellation in G gives f = e.
• "Cyclic means the group is a circle of elements with no internal structure." Cyclic groups have a very rich internal structure, completely described: one subgroup per divisor of the order. What cyclic rules out is non-commutativity.
• "The order of a^k is n divided by k." It is n / gcd(n, k). In Z_12 the order of 8 is not 12/8 but 12/gcd(12, 8) = 12/4 = 3, and indeed 8, 16 = 4, 24 = 0.

Try it: an element of large order

Exercise. In U(7) = {1, 2, 3, 4, 5, 6} under multiplication mod 7, find the order of 3 and decide whether the group is cyclic.

Solution. Compute powers of 3 mod 7: 3^1 = 3, 3^2 = 2, 3^3 = 6, 3^4 = 4, 3^5 = 5, 3^6 = 1. No smaller positive power gives 1, so the order of 3 is 6. Since |U(7)| = 6, the subgroup <3> has six elements and must be all of U(7), so the group is cyclic with generator 3.

Change one input. Try the same question in U(8) = {1, 3, 5, 7}. Here 3^2 = 1, 5^2 = 1 and 7^2 = 1, so every element has order 1 or 2, no element has order 4, and U(8) is not cyclic. Both groups are built the same way from a modulus, and one is cyclic while the other is not. Which moduli give cyclic unit groups is a genuine theorem of number theory, and the answer is 1, 2, 4, powers of an odd prime, and twice a power of an odd prime.

Pulling it together

Two tests for subgroups, one of which needs only closure and only when the subset is finite. The order of an element, the fact that it equals the size of the subgroup that element generates, and the divisibility criterion that turns a huge exponent into a remainder. The theorem that subgroups of cyclic groups are cyclic, proved by dividing an exponent and letting minimality kill the remainder. And a complete inventory of the subgroups of Z_n, one for each divisor of n.

Looking ahead

Every subgroup order in the Z_12 table divided 12, and the same held in D3. The next lesson proves that this always happens, in any finite group, cyclic or not. The proof does not compute anything: it shows that a subgroup slices the group into equally sized pieces called cosets, and then counts the pieces. From that one counting argument come Fermat's little theorem, the fact that every group of prime order is cyclic, and the constraint that eliminates most candidate structures before you have to examine them.

Sources

1. Weisstein, E. W. (n.d.). Subgroup. Wolfram MathWorld. mathworld.wolfram.com
2. Weisstein, E. W. (n.d.). Cyclic group. Wolfram MathWorld. mathworld.wolfram.com
3. Judson, T. W. (n.d.). Chapter 4: Cyclic groups. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
4. Wikipedia contributors. (n.d.). Cyclic group. Wikipedia. en.wikipedia.org
5. Artin, M. (2011). Algebra (2nd ed.), Chapter 2. Pearson.

In his 1771 memoir on the algebraic resolution of equations, Joseph-Louis Lagrange asked what happens to a polynomial when you shuffle its variables. Take x + y - z and permute the three letters in all six possible ways. You do not get six polynomials. You get three: x + y - z, x + z - y, and y + z - x. Lagrange noticed that the count of distinct results always divides the number of permutations, and he used that fact to organise everything then known about solving cubics and quartics.

He never stated the theorem that now carries his name; the language of abstract groups did not exist for another century. Gauss proved the special case for the nonzero residues modulo a prime in the Disquisitiones Arithmeticae of 1801, Cauchy extended it to symmetric groups in 1844, and Camille Jordan proved it for permutation groups in 1861. But the counting idea in that 1771 paper is exactly the one we are about to make precise, and it is the single most useful constraint in finite group theory: the order of a subgroup divides the order of the group.

Cosets: the group cut into slices

Let H be a subgroup of G and let a be any element of G. The left coset aH is the set {ah : h in H}, the whole subgroup shifted by a. The right coset Ha is {ha : h in H}. These are subsets of G, not usually subgroups: the only left coset that contains the identity is H itself.

Take G = Z_12, written additively so that aH becomes a + H, and H = {0, 4, 8}. Then 0 + H = {0, 4, 8}, 1 + H = {1, 5, 9}, 2 + H = {2, 6, 10}, 3 + H = {3, 7, 11}, and 4 + H = {4, 8, 0}, which is H again. Four distinct cosets, each with three elements, and together they use every element of Z_12 exactly once. Twelve is four times three, and the arithmetic is not a coincidence.

Now a case where the sides differ. In D3 take H = {e, s}, the subgroup generated by one reflection. Using sr = r^2 s and s r^2 = rs, work out both families.

Both columns partition the six elements into three blocks of two, but they are different partitions. The left coset of r and the right coset of r share only r. Hold on to this: the whole theory of quotient groups in the next lesson is about the subgroups for which the two columns agree. For counting, though, either column will do, and that is the point of the next result.

Why the slices are equal and disjoint

Two facts do all the work. Both are short, and both are worth reconstructing rather than memorising.

Lemma 1. For a subgroup H and elements a, b of G, the cosets aH and bH are either identical or disjoint. Moreover aH = bH if and only if b^(-1)a lies in H.

Proof. Suppose the cosets are not disjoint, so some element x lies in both: x = a h_1 = b h_2 with h_1, h_2 in H. Then a = b h_2 h_1^(-1), so b^(-1) a = h_2 h_1^(-1), which lies in H by closure. Now take any element ah of aH. Writing a = b(b^(-1)a) gives ah = b (b^(-1)a) h, and (b^(-1)a)h is a product of two elements of H, so ah lies in bH. Hence aH is contained in bH, and the symmetric argument gives the reverse inclusion. So overlapping cosets coincide. The stated criterion falls out of the same computation: if b^(-1)a is in H then the displayed inclusion argument runs, and if aH = bH then a = a e lies in bH, giving b^(-1)a in H. This completes the proof.

Lemma 2. Every left coset of H has exactly |H| elements.

Proof. Fix a and consider the map from H to aH sending h to ah. It is onto by the definition of aH. It is one to one because ah_1 = ah_2 forces h_1 = h_2 by left cancellation. A bijection preserves cardinality. This completes the proof.

Every element a of G lies in at least one coset, namely aH, since a = ae. Combined with Lemma 1, this says the left cosets are a partition of G: every element is in exactly one. Combined with Lemma 2, each block has the same size. A finite set cut into equal blocks has size equal to the number of blocks times the block size, and that sentence is the theorem.

Lagrange's theorem

Theorem (Lagrange). Let G be a finite group and H a subgroup. Then |H| divides |G|, and the number of distinct left cosets of H in G is |G| / |H|.

Proof. The left cosets of H partition G by Lemma 1, and each contains exactly |H| elements by Lemma 2. Let k be the number of distinct cosets, which is finite because G is. Counting the elements of G block by block gives |G| = k |H|. Hence |H| divides |G| and k = |G| / |H|. This completes the proof.

The number k is called the index of H in G and written [G : H]. Note that no property of G beyond finiteness was used, and no property of H beyond being a subgroup. The theorem holds for abelian and non-abelian groups alike, for cyclic and non-cyclic, and the proof never computes a single product beyond the two lemmas.

Worked example. D3 has order 6. Its subgroups therefore have order 1, 2, 3 or 6, and no others are possible. The subgroup {e, s} has index 3, matching the three cosets in the table above. The subgroup {e, r, r^2} has index 2. There is no subgroup of order 4, and you do not have to search for one.

What the theorem buys

Four corollaries follow in a few lines each, and together they are why Lagrange's theorem is quoted more often than it is proved.

Corollary 1. In a finite group, the order of every element divides the order of the group.

Proof. The order of a equals |<a>|, as shown in the previous lesson, and <a> is a subgroup. Apply Lagrange. This completes the proof.

Corollary 2. For every element a of a finite group G, a^(|G|) = e.

Proof. Write n = |a|. By Corollary 1, n divides |G|, say |G| = nm. Then a^(|G|) = (a^n)^m = e^m = e. This completes the proof.

Corollary 3. Every group of prime order p is cyclic, and its only subgroups are the trivial one and itself.

Proof. A subgroup has order dividing p, so order 1 or p. Take any element a other than e. Then <a> has order greater than 1, so it has order p, so it is all of G. This completes the proof.

Corollary 3 is a classification: for each prime p there is exactly one group of order p up to relabelling, namely Z_p. Groups of order 2, 3, 5, 7, 11 and 13 hold no surprises whatsoever, and that is a theorem, not an observation.

Corollary 4 (Fermat's little theorem). If p is prime and a is an integer not divisible by p, then a^(p-1) is congruent to 1 modulo p.

Proof. The nonzero residues modulo p form a group U(p) under multiplication, of order p - 1. The residue of a is one of its elements, so by Corollary 2 its (p-1)-st power is the identity, which is the residue 1. This completes the proof.

Check it once: p = 7 and a = 3 give 3^6 = 729 = 104 * 7 + 1, so 3^6 is 1 mod 7, as the previous lesson's power table already showed. The same argument in U(n) for composite n gives Euler's theorem, that a^(phi(n)) is 1 mod n whenever a and n are coprime, where phi(n) counts the residues coprime to n. The RSA cryptosystem is built on that corollary of a counting argument about cosets.

Why the converse fails: the tetrahedron

Lagrange says a subgroup order must divide the group order. It does not say every divisor is realised, and the smallest counterexample is worth meeting properly because it is a physical object.

Take a regular tetrahedron and list its rotations. There is the identity. About each of the four axes through a vertex and the centre of the opposite face there are two nontrivial rotations, by 120 and 240 degrees, giving eight. About each of the three axes through the midpoints of opposite edges there is one rotation by 180 degrees, giving three. Total: 1 + 8 + 3 = 12. Labelling the vertices 1 to 4, the vertex rotations act as three-cycles such as (1 2 3) and the edge rotations act as double transpositions such as (1 2)(3 4). This group is the alternating group A_4, of order 12.

Claim. A_4 has no subgroup of order 6, even though 6 divides 12.

Proof. Suppose H were such a subgroup. Its index is 12/6 = 2, so there are exactly two left cosets, H and gH for any g outside H. First observe that x^2 lies in H for every x in A_4: if x is in H this is closure, and if x is not in H then x lies in the other coset xH, and were x^2 also outside H we could write x^2 = xh for some h in H, giving x = h and placing x in H after all, a contradiction. So every square lands in H. Now let t be any of the eight three-cycles. Since t^3 = e, we have t = t^4 = (t^2)^2, so t is a square and therefore lies in H. That puts at least the identity and eight three-cycles inside H, so |H| >= 9 > 6. This contradiction shows no such subgroup exists. This completes the proof.

Bottom line: Lagrange is a one-way street. It rules candidates out and never rules them in. When you want subgroups of a specified order to exist, you need a different theorem, and the Sylow theorems of Module 3 are exactly that theorem for prime power divisors.

Common misconceptions

• "If d divides |G| then G has a subgroup of order d." False, as A_4 shows for d = 6. The partial converses that do hold are Cauchy's theorem for prime divisors and the Sylow theorems for prime powers.
• "Cosets are subgroups." Only one of them is. A coset aH contains the identity exactly when a lies in H, in which case aH = H. The other cosets are shifted copies with no identity in them.
• "Left and right cosets are the same thing." They agree for every subgroup of an abelian group and for some subgroups of non-abelian groups, but in D3 the subgroup {e, s} gives different partitions on the two sides. Which subgroups have the two partitions agree is the question the next lesson answers.
• "Lagrange's theorem needs the group to be abelian, or cyclic, or finite in some stronger sense." It needs only that the group is finite. The proof uses closure, inverses and cancellation, nothing else.

Try it: what Lagrange rules out

Exercise. A group G has order 35. Show that every element other than the identity has order 5 or 7, and deduce that G contains an element of order 5.

Solution. By Corollary 1 the order of any element divides 35, so it is 1, 5, 7 or 35. Only the identity has order 1. If some element had order 35 then G would be cyclic, and a cyclic group of order 35 contains an element of order 5, namely the seventh power of a generator, so we would be done. Otherwise every non-identity element has order 5 or 7. Suppose, for contradiction, that all 34 non-identity elements had order 7. Each generates a subgroup of order 7, and two such subgroups meet only in the identity, because their intersection is a subgroup of a group of prime order 7 and cannot have order 7 without the two coinciding. So the non-identity elements fall into disjoint families of size 6, and 6 does not divide 34. The contradiction shows some element has order 5.

Notice what the argument never did: it never named a single element of G, and it never assumed the group was abelian. All of the leverage came from divisibility.

The short version

A subgroup cuts its parent group into left cosets that are pairwise disjoint and all the same size, so the subgroup order divides the group order and the number of cosets is the index. From that: element orders divide the group order, the |G|-th power of anything is the identity, groups of prime order are cyclic, and Fermat's little theorem is a two-line corollary. The converse fails, and the rotation group of the tetrahedron is the standard witness.

Looking ahead

The table of left and right cosets in D3 left an obvious loose end. For the subgroup of order 3 the two partitions agreed; for the subgroup of order 2 they did not. Subgroups of the first kind are called normal, and they are exactly the subgroups for which the set of cosets can itself be made into a group. That construction, the quotient group, is the subject of the next lesson, and it is the single most productive idea in the first half of this course.

Sources

1. Wikipedia contributors. (n.d.). Lagrange's theorem (group theory). Wikipedia. en.wikipedia.org
2. Bray, N. (n.d.). Lagrange's group theorem. Wolfram MathWorld. mathworld.wolfram.com
3. Judson, T. W. (n.d.). Chapter 6: Cosets and Lagrange's theorem. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
4. O'Connor, J. J., & Robertson, E. F. (1999). Joseph-Louis Lagrange. MacTutor History of Mathematics Archive. mathshistory.st-andrews.ac.uk
5. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Section 3.2. John Wiley and Sons.

A five-digit mechanical odometer reads 99999 miles, you drive one more mile, and it reads 00000. Nothing was destroyed; the wheels simply carried past the end of their range. What the odometer computes is not addition of integers but addition of integers with multiples of 100000 declared invisible. Every mileage that differs from yours by a whole number of 100000 shows the same five digits, and the display cannot tell those mileages apart.

That is a quotient group. The full group is Z under addition, the subgroup being made invisible is the set of multiples of 100000, and the elements of the new group are the classes of integers that agree modulo 100000. The construction is worth taking seriously in general, because most of the structure theory in this course consists of building a quotient and then recognising it. But the construction does not always work, and the first job is to find out exactly when it does.

The obvious definition, and where it breaks

Let H be a subgroup of G. The set of left cosets of H is a perfectly good set, and it has an obvious candidate operation: define the product of aH and bH to be abH. The trouble is that aH does not come with a written on it. The coset {1, 5, 9} in Z_12 is equally 1 + H, 5 + H and 9 + H. If the recipe is to define anything at all, the answer must not depend on which representative you happen to pick.

In Z_12 with H = {0, 4, 8}, that is fine. Take 1 + H and 2 + H. Using the representatives 1 and 2 gives 3 + H. Using 5 and 10 instead gives 15 + H = 3 + H, the same class. Try any other pair of representatives and you get the same answer, because the group is abelian and the difference of two representatives lies in H.

Now watch it fail. In D3 take H = {e, s}. The cosets are H, rH = {r, rs} and r^2 H = {r^2, r^2 s}. Compute the product of H and rH two ways. Representing the first coset by e gives (e)(r)H = rH = {r, rs}. Representing the same first coset by s gives (s)(r)H = (sr)H = (r^2 s)H, and since s H = H, this is r^2 H = {r^2, r^2 s}. Two representatives of the same coset, two different answers. The recipe defines nothing.

What matters here: the question is not whether cosets exist but whether the multiplication rule survives a change of representative. The subgroups that survive it have a name.

Three ways to say normal, and why they agree

A subgroup N of G is normal, written N normal in G, if any of the following equivalent conditions holds.

1. gng^(-1) lies in N for every g in G and every n in N.
2. gNg^(-1) = N for every g in G, where gNg^(-1) means the set of all gng^(-1).
3. gN = Ng for every g in G: the left and right coset partitions coincide.

Proposition. Conditions 1, 2 and 3 are equivalent.

Proof. Condition 2 says gNg^(-1) = N, which certainly gives condition 1, the containment of gNg^(-1) in N. Conversely, assume 1. Then gNg^(-1) is contained in N for every g, and applying that statement with g^(-1) in place of g gives g^(-1)Ng contained in N. Multiplying the second containment on the left by g and on the right by g^(-1) turns it into N contained in gNg^(-1). The two containments give equality, which is condition 2. Finally, multiplying gNg^(-1) = N on the right by g gives gN = Ng, which is condition 3; and multiplying gN = Ng on the right by g^(-1) reverses the step. This completes the proof.

Condition 1 is the one to check in practice, because it is a statement about single elements. Condition 3 is the one that explains the earlier failure: {e, s} in D3 is not normal because r s r^(-1) = r s r^2, and using s r^2 = rs this equals r(rs) = r^2 s, which is not in {e, s}. One computation settles it. Meanwhile K = {e, r, r^2} is normal: conjugating a rotation by a rotation gives a rotation, and conjugating by a reflection gives s r s^(-1) = s r s = r^2, still a rotation.

Two families of normal subgroups come free. In an abelian group every subgroup is normal, since gng^(-1) = n g g^(-1) = n. And in any group the centre Z(G), the set of elements commuting with everything, is normal, for the same reason applied elementwise. A third family is more useful than either.

Proposition. Every subgroup of index 2 is normal.

Proof. Let [G : N] = 2. If g lies in N then gN = N = Ng. If g does not lie in N, then since there are exactly two left cosets and they partition G, the coset gN must be the complement of N in G. The same argument on the right shows Ng is also the complement of N. So gN = Ng in both cases, which is condition 3. This completes the proof.

That is why {e, r, r^2} is normal in D3 without any computation, and why the even permutations are normal inside the symmetric group, and why the rotations of any regular polygon are normal inside its full symmetry group.

Building the quotient group

Theorem. Let N be a normal subgroup of G. Then the set G/N of left cosets of N, with the operation (aN)(bN) = abN, is a group. Its identity is N itself, the inverse of aN is a^(-1)N, and if G is finite then |G/N| = |G| / |N|.

Proof. The only real work is showing the operation is well defined. Suppose aN = a'N and bN = b'N. By the coset criterion of the previous lesson, a' = a n_1 and b' = b n_2 for some n_1, n_2 in N. Then

a' b' = a n_1 b n_2 = a b (b^(-1) n_1 b) n_2,

where we have inserted b b^(-1) between n_1 and b. Normality, in the form b^(-1) n_1 b lies in N, makes the bracketed factor an element of N, so a'b' is ab times an element of N. Hence a'b'N = abN, and the product does not depend on the representatives.

The remaining axioms are inherited. Associativity: ((aN)(bN))(cN) = (ab)cN = a(bc)N = (aN)((bN)(cN)), using associativity in G. Identity: (eN)(aN) = aN and (aN)(eN) = aN, and eN = N. Inverses: (aN)(a^(-1)N) = a a^(-1) N = N. The order statement is Lagrange's theorem, since |G/N| is the index [G : N]. This completes the proof.

Read the proof again and notice where normality entered: exactly once, to move n_1 past b. That single manoeuvre is the whole content of the hypothesis. Without it, a'b' lands in some other coset, which is what happened with {e, s} in D3.

Four quotients, worked

The odometer. G = Z, N = 100000Z. Since Z is abelian, N is normal. The quotient Z/100000Z has 100000 elements, the classes of 0 through 99999, and it is cyclic, generated by the class of 1. The odometer is a physical realisation of a quotient group.

Collapsing the rotations. G = D3, N = {e, r, r^2}, index 2 and so normal. The quotient has two elements: the coset of rotations, which is the identity of the quotient, and the coset of reflections. Multiplying two reflections gives a rotation, which in the quotient reads as "reflection times reflection equals identity." So D3/N is a group of order 2, and it is remembering exactly one bit about each symmetry: whether the card was flipped.

Determinant. G = GL(2,R), the invertible 2 by 2 real matrices, and N = SL(2,R), those with determinant 1. Normality: if det(A) = 1 then det(GAG^(-1)) = det(G) det(A) det(G)^(-1) = 1. The quotient GL(2,R)/SL(2,R) behaves exactly like the nonzero reals under multiplication, with the coset of A corresponding to det(A). Here the quotient is infinite, and it is not a subset of GL(2,R) in any natural way.

A quotient of the clock. G = Z_12, N = {0, 4, 8}. The four cosets are the classes of 0, 1, 2 and 3, and adding them adds the representatives mod 4, so Z_12 / N behaves exactly like Z_4. Notice that Z_4 is also a subgroup-like object inside Z_12, namely {0, 3, 6, 9}. That coincidence is special to abelian groups, and you should not expect the quotient to sit inside the original group in general.

What a quotient forgets

The right way to think about G/N is as G with the elements of N renamed to the identity. Everything the group knew that could be detected after that renaming survives; everything else is gone. In the D3 example, the quotient remembers orientation and forgets which of the three axes a flip used. In the determinant example, the quotient remembers scaling and forgets rotation and shear.

This gives a useful test. The quotient G/N is abelian precisely when N contains every element of the form aba^(-1)b^(-1), called a commutator. The reason is direct: (aN)(bN) = (bN)(aN) says abN = baN, which by the coset criterion says (ba)^(-1)(ab) = a^(-1)b^(-1)ab lies in N. So a quotient is commutative exactly when the failures of commutativity have been thrown into the subgroup being collapsed. Remember: a quotient is an act of deliberate forgetting, and the normality condition is precisely the condition that the forgetting is consistent.

Common misconceptions

• "Normal means every element of N commutes with every element of G." That would be the centre, which is much smaller. Normality asks only that gNg^(-1) equals N as a set; individual elements are allowed to move, as long as they stay inside. In D3, conjugating r by s gives r^2, a different element of the same normal subgroup.
• "G/N is a subgroup of G." It is a group whose elements are subsets of G, not elements of G. Z/100000Z is not a set of integers; it is a set of classes of integers. The occasional appearance of a matching subgroup inside G is a coincidence of abelian examples.
• "Normality is transitive." It is not. If K is normal in N and N is normal in G, it does not follow that K is normal in G. The standard counterexample lives inside the symmetry group of the square, where a reflection subgroup is normal in a Klein four subgroup which is normal in the whole group, and yet is not normal in the whole group.
• "If N is normal, coset multiplication is just multiplication of sets." It happens to be, and that is a pleasant fact: for normal N, the setwise product {xy : x in aN, y in bN} really does equal abN. But the definition is by representatives, and the well-definedness proof is what licenses either reading.

Try it: a quotient of the integers by an integer

Exercise. In Z, let N = 6Z and consider Z/6Z. Show that the class of 4 has order 3, and find all subgroups of the quotient.

Solution. The class of 4 added to itself gives the class of 8, which is the class of 2; added again gives the class of 12, which is the class of 0, the identity. No smaller positive multiple vanishes, since the class of 4 is not zero and the class of 8 is the class of 2, also not zero. So the order is 3. Since Z/6Z is cyclic of order 6, the previous module's classification applies: one subgroup per divisor of 6, so four subgroups, of orders 1, 2, 3 and 6, generated by the classes of 0, 3, 2 and 1.

Change one input. Replace Z by D3 and N by {e, s}. Now there is no quotient group at all, because the operation is not well defined, and the three cosets carry no group structure that respects the multiplication of D3. Same shape of question, and the answer depends entirely on a conjugation condition you can check in one line.

Where this leaves us

Coset multiplication is well defined exactly when the subgroup is normal, and normality has three equivalent formulations of which the elementwise conjugation test is the practical one. Index-2 subgroups, subgroups of abelian groups, and the centre are always normal. When N is normal, G/N is a group of order [G : N], and it is best read as G with N renamed to the identity. Quotients are not subgroups, and normality is not transitivity of anything.

Looking ahead

Every worked example above ended with a phrase like "behaves exactly like." The next lesson makes that phrase precise. A homomorphism is a map between groups that respects multiplication, an isomorphism is a homomorphism that is also a bijection, and the first isomorphism theorem says that every quotient you can build is the image of some homomorphism, and every image of a homomorphism is a quotient. Once you have that, "the determinant map has kernel SL(2,R)" and "GL(2,R)/SL(2,R) is the nonzero reals" become the same sentence.

Sources

1. Weisstein, E. W. (n.d.). Normal subgroup. Wolfram MathWorld. mathworld.wolfram.com
2. Weisstein, E. W. (n.d.). Quotient group. Wolfram MathWorld. mathworld.wolfram.com
3. Judson, T. W. (n.d.). Chapter 10: Normal subgroups and factor groups. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
4. Wikipedia contributors. (n.d.). Normal subgroup. Wikipedia. en.wikipedia.org
5. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Section 3.1. John Wiley and Sons.

An engineer in 1950 multiplying 3.7 by 42 does not multiply. She slides one logarithmic scale along another until the 1 on the sliding scale sits over 3.7, reads across to 42, and finds 155 under the cursor. The slide rule, which William Oughtred built in the 1620s on the strength of John Napier's 1614 tables, performs multiplication by adding lengths. It works because the logarithm converts one operation into another: log(xy) = log(x) + log(y).

That identity is the definition of the central object in this lesson. The logarithm is a map from the positive reals under multiplication to all reals under addition, and it carries the first operation to the second faithfully. Structure-preserving maps like this are how groups are compared, and the theorem at the end of this lesson says that every quotient group in the previous lesson is secretly the image of such a map.

Maps that respect the operation

Let G and H be groups. A homomorphism is a function f from G to H with f(ab) = f(a) f(b) for all a, b in G, where the product on the left is taken in G and the product on the right in H. An isomorphism is a homomorphism that is also a bijection; when one exists we say the groups are isomorphic and write G is isomorphic to H, meaning they are the same group with different names on the elements.

Two near misses sharpen the definition. The map from Z to Z sending n to n + 1 is a bijection but not a homomorphism, since f(a + b) = a + b + 1 while f(a) + f(b) = a + b + 2. And the map sending each matrix to its trace is not a homomorphism from GL(2,R) to anything multiplicative, because the trace of a product is not the product of traces. Being a nice function is not the point; carrying the operation is.

What comes free

The definition mentions only products, yet identities and inverses come along automatically. This is the first sign that the definition was chosen well.

Proposition. Let f be a homomorphism from G to H. Then f(e_G) = e_H, and f(a^(-1)) = f(a)^(-1) for every a, and f(a^n) = f(a)^n for every integer n.

Proof. From e_G e_G = e_G we get f(e_G) f(e_G) = f(e_G) = f(e_G) e_H, and cancelling f(e_G) in H gives f(e_G) = e_H. Then f(a) f(a^(-1)) = f(a a^(-1)) = f(e_G) = e_H, so f(a^(-1)) is the inverse of f(a) by uniqueness of inverses. The power statement follows for positive n by induction on the defining property, for n = 0 from the identity statement, and for negative n by combining the two. This completes the proof.

A useful corollary: the order of f(a) divides the order of a. If a^n = e_G then f(a)^n = f(a^n) = e_H, and the divisibility criterion from Module 1 finishes it. So a homomorphism can shrink orders but never enlarge them, which immediately rules out many maps. There is no homomorphism from Z_3 to Z_4 other than the trivial one, because the image of a generator would have order dividing both 3 and 4.

Proposition. The image f(G) is a subgroup of H.

Proof. It is nonempty, since e_H = f(e_G) lies in it. Given two elements f(a) and f(b) of the image, f(a) f(b)^(-1) = f(a) f(b^(-1)) = f(a b^(-1)), again in the image. The one-step subgroup test applies. This completes the proof.

The kernel, and why it is normal

The kernel of f, written ker(f), is the set of elements of G that f sends to e_H. For the logarithm, the kernel is {1}. For the determinant on GL(2,R), the kernel is SL(2,R). For reduction mod n, the kernel is the multiples of n. For the sign map on S_n, the kernel is the even permutations. Every one of those was a normal subgroup in the previous lesson, and that is not a coincidence.

Theorem. The kernel of a homomorphism is a normal subgroup of its domain.

Proof. First, ker(f) is a subgroup: it contains e_G, and if f(a) = f(b) = e_H then f(a b^(-1)) = f(a) f(b)^(-1) = e_H e_H = e_H, so the one-step test applies. For normality, take k in ker(f) and g in G. Then f(g k g^(-1)) = f(g) f(k) f(g)^(-1) = f(g) e_H f(g)^(-1) = e_H, so g k g^(-1) lies in ker(f). This completes the proof.

The kernel also measures injectivity exactly, which is a considerable convenience: instead of comparing every pair of elements you inspect one subgroup.

Proposition. A homomorphism f is injective if and only if ker(f) = {e_G}.

Proof. If f is injective then only e_G can map to e_H, since f(e_G) = e_H already. Conversely suppose the kernel is trivial and f(a) = f(b). Then f(a b^(-1)) = f(a) f(b)^(-1) = e_H, so a b^(-1) lies in the kernel and therefore equals e_G, giving a = b. This completes the proof.

More is true and worth seeing, because it is the seed of the main theorem: for any a, the set of elements mapping to f(a) is exactly the coset a ker(f). Indeed f(x) = f(a) holds precisely when f(a^(-1)x) = e_H, that is, when a^(-1)x lies in the kernel, that is, when x lies in a ker(f). So the fibres of a homomorphism are the cosets of its kernel. Key idea: a homomorphism collapses each coset of its kernel to a single point, and it separates different cosets.

The first isomorphism theorem

Theorem. Let f be a homomorphism from G to H, with kernel K. Then G/K is isomorphic to the image f(G), by the map sending the coset aK to f(a).

Proof. Call the proposed map F, so F(aK) = f(a).

Well defined. Suppose aK = bK. Then b^(-1)a lies in K, so f(b^(-1)a) = e_H, so f(b)^(-1) f(a) = e_H and therefore f(a) = f(b). The recipe does not depend on the representative.

Homomorphism. F((aK)(bK)) = F(abK) = f(ab) = f(a) f(b) = F(aK) F(bK), using the definition of the quotient operation and then the homomorphism property of f.

Injective. Suppose F(aK) = e_H. Then f(a) = e_H, so a lies in K, so aK = K, which is the identity of G/K. The kernel of F is trivial, so F is injective by the previous proposition.

Surjective onto the image. Every element of f(G) has the form f(a) for some a, and F(aK) = f(a).

So F is a bijective homomorphism from G/K onto f(G). This completes the proof.

The theorem should be read in two directions. Left to right it says: to identify a quotient G/N, find a homomorphism out of G whose kernel is N, and the quotient is its image. Right to left it says: every homomorphic image of G is a quotient of G. Images and quotients are the same list of groups, indexed by normal subgroups.

Four identifications, done in one line each

Determinant. det maps GL(2,R) onto the nonzero reals, with kernel SL(2,R). Hence GL(2,R)/SL(2,R) is isomorphic to the nonzero reals under multiplication. Last lesson this took a paragraph of hand-waving; now it is a citation.

Reduction. The map from Z onto Z_n sending an integer to its residue has kernel nZ. Hence Z/nZ is isomorphic to Z_n. The odometer is now a theorem.

Sign. The sign map from S_n onto {1, -1} has kernel A_n, the alternating group. Hence S_n / A_n has order 2, which forces |A_n| = n! / 2. That is how the order of the alternating group is computed, and we will use it in Module 4.

Winding. The map from the reals under addition to the unit circle under multiplication sending t to cos t + i sin t is onto, with kernel the multiples of 2 pi. Hence the circle group is isomorphic to R / 2 pi Z. The quotient of a line by a lattice is a circle, and the theorem says so in one line.

When are two groups really the same?

Isomorphic groups are indistinguishable by any property phrased in terms of the operation. So to prove two groups are not isomorphic, exhibit such a property that differs. The most useful ones are the order of the group, whether it is abelian, the number of elements of each order, and the number of subgroups of each order.

Z_4 and the Klein four-group both have order 4 and are both abelian, so those two tests fail to separate them. The element orders do not: Z_4 has an element of order 4 and V has none. Similarly Z_6 and S_3 both have order 6, but one is abelian and the other is not. And Z_2 combined with Z_2 combined with Z_2, of order 8, is not isomorphic to Z_8, again on element orders.

The reverse direction, proving two groups are isomorphic, usually means writing down a map and checking it, or applying the first isomorphism theorem to a well-chosen homomorphism. The second route is almost always shorter.

Common misconceptions

• "A homomorphism has to be injective or surjective to be useful." The most useful ones are neither. The determinant is far from injective and the inclusion of a subgroup is far from surjective, and both carry real information.
• "The kernel is where the map is undefined." The kernel is where the map is defined and equal to the identity. It is a subgroup of the domain, never a subset of the codomain.
• "If two groups have the same order they are isomorphic." There are two groups of order 4, two of order 6, five of order 8, and 49487365422 of order 1024. Order is one invariant among many.
• "An isomorphism has to preserve the names of elements." It preserves only the multiplication pattern. The map from Z_4 to the fourth roots of unity sending k to i^k matches an additive group with a multiplicative one, and nothing about the notation survives.

Try it: identify a quotient without listing cosets

Exercise. Let G = Z_12 and N = {0, 6}. Identify G/N using the first isomorphism theorem rather than by writing out cosets.

Solution. Define f from Z_12 to Z_6 by sending the class of a mod 12 to the class of a mod 6. This is well defined because 12 is a multiple of 6, and it is a homomorphism because reduction respects addition. It is surjective, since every residue mod 6 is hit. Its kernel is the set of classes mod 12 that are multiples of 6, namely {0, 6}, which is N. So Z_12 / {0, 6} is isomorphic to Z_6.

Change one input. Try N = {0, 4, 8} instead. Now the natural map is reduction mod 4, from Z_12 onto Z_4, whose kernel is the multiples of 4. So the quotient is Z_4, matching the answer we obtained last lesson by listing cosets. In both cases the theorem replaced a table with a map.

Putting it together

A homomorphism is a map that carries the operation, and that single requirement forces it to preserve the identity, inverses and powers, to make its image a subgroup, and to shrink element orders divisibly. Its kernel is always normal, is trivial exactly when the map is injective, and has cosets that are exactly the fibres of the map. The first isomorphism theorem then identifies G/ker(f) with the image, which turns questions about quotients into questions about well-chosen maps.

Looking ahead

One theorem in, two to go. The second isomorphism theorem describes what happens when you multiply a subgroup by a normal subgroup, and the third describes quotients of quotients, the situation where you collapse in two stages. Both are proved by building the right homomorphism and appealing to the theorem you now have, which is how most of the remaining structure theory in this course will go.

Sources

1. Weisstein, E. W. (n.d.). Group homomorphism. Wolfram MathWorld. mathworld.wolfram.com
2. Weisstein, E. W. (n.d.). First group isomorphism theorem. Wolfram MathWorld. mathworld.wolfram.com
3. Judson, T. W. (n.d.). Chapter 11: Homomorphisms. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
4. O'Connor, J. J., & Robertson, E. F. (1998). John Napier. MacTutor History of Mathematics Archive. mathshistory.st-andrews.ac.uk
5. Wikipedia contributors. (n.d.). Isomorphism theorems. Wikipedia. en.wikipedia.org

Inside the integers, take the multiples of 4 and the multiples of 6. Add them together in every possible way and you get every multiple of 2, because 2 = 6 - 4 is already available and everything you build is even. Intersect them instead and you get the multiples of 12. Two numbers came out of that: 2, the greatest common divisor of 4 and 6, and 12, their least common multiple. And 2 * 12 = 24 = 4 * 6.

That identity is usually filed under elementary number theory. It is in fact a special case of a theorem about subgroups that holds in every group, abelian or not, and the theorem is proved in six lines from the tool built in the previous lesson. This lesson finishes the isomorphism theorems and adds the correspondence theorem, which together let you compute inside a quotient without ever writing down a coset.

Multiplying two subgroups

Given subgroups H and K of G, write HK for the set of all products hk. This is generally not a subgroup: in S_3, taking H = {e, (1 2)} and K = {e, (1 3)} gives a set of four elements, and 4 does not divide 6, so Lagrange forbids it from being a subgroup. Normality repairs the situation.

Proposition. If H is a subgroup of G and N is a normal subgroup of G, then HN is a subgroup of G containing both H and N.

Proof. The set is nonempty, since e = ee lies in it, and it contains H (take n = e) and N (take h = e). Apply the one-step test to two elements h_1 n_1 and h_2 n_2. Their combination is

(h_1 n_1)(h_2 n_2)^(-1) = h_1 n_1 n_2^(-1) h_2^(-1) = (h_1 h_2^(-1)) (h_2 n_1 n_2^(-1) h_2^(-1)),

where we inserted h_2^(-1) h_2 in the middle. The first factor lies in H. The second is a conjugate of the element n_1 n_2^(-1) of N, so it lies in N by normality. Hence the product lies in HN. This completes the proof.

For finite groups there is a counting formula, and it is the group-theoretic form of the identity in the opening paragraph.

Proposition. For finite subgroups H and K of G, the set HK has |H| |K| / |H intersect K| elements, whether or not it is a subgroup.

Proof. Consider the map from the set of pairs (h, k) in H times K to the set HK sending (h, k) to hk. It is onto by definition. Two pairs give the same product exactly when h_1 k_1 = h_2 k_2, that is, when h_2^(-1)h_1 = k_2 k_1^(-1). Call this common value d; it lies in H and in K, so in the intersection. Conversely, for each d in the intersection the pair (h_1 d^(-1), d k_1) has the same product as (h_1, k_1), and distinct d give distinct pairs. So every element of HK has exactly |H intersect K| preimages, and counting gives |H| |K| = |HK| |H intersect K|. This completes the proof.

With H = 4Z and K = 6Z inside Z, written additively, this is the statement that the index of 4Z + 6Z times the index of 4Z intersect 6Z matches, which is the gcd and lcm identity in disguise.

The second isomorphism theorem

Theorem. Let H be a subgroup of G and N a normal subgroup of G. Then H intersect N is normal in H, N is normal in HN, and

H / (H intersect N) is isomorphic to HN / N.

Proof. Define f from H to G/N by f(h) = hN. This is a homomorphism, since f(h_1 h_2) = h_1 h_2 N = (h_1 N)(h_2 N) = f(h_1) f(h_2), using the quotient operation. Its image is the set of cosets hN with h in H, which is exactly the subgroup HN/N of G/N. Its kernel is the set of h in H with hN = N, that is, with h in N, which is H intersect N. Kernels are normal, so H intersect N is normal in H. Since N is normal in all of G it is normal in the subgroup HN, so the quotient HN/N makes sense. Now apply the first isomorphism theorem to f. This completes the proof.

The name "diamond theorem" comes from drawing HN at the top, H and N below it on either side, and H intersect N at the bottom. The two sides of the diamond, from H up to HN and from H intersect N up to N, are the same size, and the theorem says the corresponding quotients are actually the same group.

Worked example. Let G = D4, the eight symmetries of a square, with r the quarter turn and s a reflection. Take N = {e, r, r^2, r^3}, the rotations, normal because its index is 2. Take H = {e, r^2, s, r^2 s}, which you can check is a subgroup isomorphic to the Klein four-group. Then H intersect N = {e, r^2}, so |HN| = 4 * 4 / 2 = 8 and HN = D4. The theorem says H/{e, r^2}, of order 2, is isomorphic to D4/N, also of order 2. Both quotients answer the same question about a symmetry: was the square turned over?

Worked example, integers. Take H = 4Z and N = 6Z inside Z. Then H + N = 2Z and H intersect N = 12Z. The theorem says 4Z/12Z is isomorphic to 2Z/6Z. The left side has three elements, the classes of 0, 4 and 8; the right side has three, the classes of 0, 2 and 4. Both are cyclic of order 3, and the isomorphism doubles: 4 corresponds to 2, 8 to 4.

The third isomorphism theorem

The second theorem is about a subgroup beside a normal subgroup. The third is about two normal subgroups stacked, one inside the other, and it says that collapsing in two stages is the same as collapsing once.

Theorem. Let K and N be normal subgroups of G with K contained in N. Then N/K is a normal subgroup of G/K, and

(G/K) / (N/K) is isomorphic to G/N.

Proof. Define f from G/K to G/N by f(gK) = gN. This is well defined: if gK = g'K then g^(-1)g' lies in K, hence in N since K is contained in N, hence gN = g'N. It is a homomorphism, since f((gK)(hK)) = f(ghK) = ghN = (gN)(hN). It is onto, since every coset gN is the image of gK. Its kernel is the set of gK with gN = N, that is, with g in N, which is precisely N/K. Kernels are normal, so N/K is normal in G/K, and the first isomorphism theorem gives the isomorphism. This completes the proof.

Worked example. Take G = Z, K = 12Z and N = 4Z. Then G/K = Z_12, and N/K is the subgroup {0, 4, 8} of Z_12. The theorem says Z_12 / {0, 4, 8} is isomorphic to Z/4Z = Z_4. That matches the answer computed by listing cosets two lessons ago, and this time nothing was listed. The upshot: reducing mod 12 and then mod 4 is the same as reducing mod 4, and the third isomorphism theorem is that sentence made general.

The correspondence theorem

One more result completes the toolkit. It says that the subgroup structure of a quotient is visible from inside the original group.

Theorem. Let N be normal in G. The map sending a subgroup H of G containing N to the subgroup H/N of G/N is a bijection between the subgroups of G that contain N and all subgroups of G/N. It preserves inclusion, sends normal subgroups to normal subgroups, and multiplies indexes correctly: [G : H] = [G/N : H/N].

Proof sketch with the key steps. Write p for the map from G onto G/N sending g to gN. If H contains N then p(H) = H/N is a subgroup of G/N, since images of subgroups are subgroups. In the other direction, if B is a subgroup of G/N, then the preimage p^(-1)(B) is a subgroup of G containing N, because the preimage of a subgroup under a homomorphism is a subgroup and N maps to the identity. The two constructions undo each other: p(p^(-1)(B)) = B because p is onto, and p^(-1)(p(H)) = H because H already contains N, so any element mapping into p(H) differs from an element of H by something in N, which is in H. Inclusion is clearly preserved both ways, and normality transfers because conjugation commutes with p. This completes the argument.

Worked example. Take G = Z_12 and N = {0, 6}. The quotient is cyclic of order 6, so by the classification of cyclic subgroups it has exactly four subgroups. The correspondence theorem says those four match the subgroups of Z_12 containing {0, 6}, and here they are: {0, 6} itself, {0, 3, 6, 9}, {0, 2, 4, 6, 8, 10}, and Z_12. Four, as promised, and none of the other two subgroups of Z_12 qualify, because neither {0, 4, 8} nor {0} contains 6.

Common misconceptions

• "HK is a subgroup whenever H and K are." It is not. In S_3, the product of two order-2 subgroups has four elements, which Lagrange forbids from being a subgroup of a group of order 6. Normality of one factor is what fixes this.
• "The order formula |HK| = |H||K|/|H intersect K| requires HK to be a subgroup." It does not. The counting argument uses only that H and K are subgroups, so the formula holds even when the product set is not closed.
• "The correspondence theorem lists all subgroups of G." It lists only those containing N. Subgroups that miss part of N are invisible in the quotient, which is exactly the information the quotient discarded.
• "H/(H intersect N) and HN/N are merely the same size." They are the same group. The isomorphism is explicit: the class of h on the left corresponds to the coset hN on the right, and it is the map produced by the proof, not an accident of counting.

Try it: a quotient of a quotient

Exercise. In G = Z_36, let K = {0, 18} and let N be the subgroup of order 6 generated by 6. Identify (G/K)/(N/K).

Solution. First check the hypothesis: N = {0, 6, 12, 18, 24, 30} contains 18, so K is inside N, and both are normal since Z_36 is abelian. The third isomorphism theorem says the answer is G/N, which has order 36/6 = 6 and is cyclic because a quotient of a cyclic group is cyclic. So the answer is Z_6.

Change one input. Take K = {0, 12, 24} instead. Now K is not contained in N, since 12 is in both but the theorem needs full containment, and here it does hold: 12 and 24 are multiples of 6, so K is inside N after all, and the answer is again Z_6. Take K = {0, 9, 18, 27} and the containment genuinely fails, since 9 is not a multiple of 6, and the third isomorphism theorem simply does not apply. The right move then is the second theorem, applied to K and N, which computes K + N as the subgroup generated by gcd(9, 6) = 3.

What to carry forward

A product HN is a subgroup when N is normal, and its size obeys the gcd-lcm pattern |H||N| = |HN||H intersect N| whether or not it is a subgroup. The second isomorphism theorem identifies H/(H intersect N) with HN/N, and the third says collapsing by K and then by N/K equals collapsing by N. The correspondence theorem makes the subgroup lattice of a quotient readable inside the original group. All three came from one construction: build a homomorphism, spot its kernel and image, and quote the first isomorphism theorem.

Looking ahead

Everything so far has treated a group as an object with an internal structure. The next module treats a group as something that does things: permuting the faces of a cube, rotating a necklace, conjugating its own elements. That shift of viewpoint, called a group action, converts structural questions into counting questions, and the counting turns out to be sharp enough to prove that a group of order 15 must be cyclic and that a group whose order is a prime power always has a nontrivial centre.

Sources

1. Wikipedia contributors. (n.d.). Isomorphism theorems. Wikipedia. en.wikipedia.org
2. Judson, T. W. (n.d.). Chapter 13: The structure of groups. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
3. Massachusetts Institute of Technology. (2010). 18.701 Algebra I [Course materials]. MIT OpenCourseWare. ocw.mit.edu
4. Weisstein, E. W. (n.d.). Quotient group. Wolfram MathWorld. mathworld.wolfram.com
5. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Section 3.3. John Wiley and Sons.

Put a die on a table and roll it around without lifting it. Any of the six faces can be brought to the top, and once a face is on top you can still spin the die four ways about the vertical axis. Six choices, then four: the rotation group of a cube has 24 elements, and you have just counted them by splitting the count in two.

That split is a theorem. The six is the size of the set of positions a single face can reach, and the four is the number of rotations that leave that face where it is. The theorem says the product is always the order of the group, and it is the most efficient counting device in finite group theory. Getting to it requires one definition, which turns a group from a thing with an internal structure into a thing that does something to a set.

What it means for a group to act

Let G be a group and X a set. An action of G on X assigns to each g in G and each x in X an element g . x of X, subject to two rules:

• e . x = x for every x, and
• g . (h . x) = (gh) . x for all g, h in G and all x in X.

An equivalent formulation is often more illuminating: an action is a homomorphism from G into Sym(X), the group of all bijections of X. Given an action, send g to the map x goes to g . x; the second rule says this assignment is multiplicative, and the first, together with the second applied to g and g^(-1), shows each such map is a bijection with inverse supplied by g^(-1). Read the other way, any homomorphism into Sym(X) defines an action. Key idea: an action is a way of realising abstract group elements as concrete permutations.

Check one of these against the axioms so the definition stops being abstract. For conjugation: e x e^(-1) = x, and g . (h . x) = g(hxh^(-1))g^(-1) = (gh) x (gh)^(-1) = (gh) . x. Both rules hold, and the second is the reason conjugation is written with the inverse on the right rather than the left. The last row of the table is the action that will prove the Sylow theorems two lessons from now.

Orbits partition, stabilisers are subgroups

Fix an action of G on X. The orbit of a point x, written Orb(x), is the set of all g . x as g ranges over G: everywhere x can be sent. The stabiliser of x, written Stab(x), is the set of g with g . x = x: everything that leaves x alone. Notice that orbits live in X and stabilisers live in G; confusing the two is the most common early error.

Proposition. The orbits partition X.

Proof. Each x lies in its own orbit, since e . x = x, so the orbits cover X. Suppose two orbits meet, say z lies in Orb(x) and in Orb(y), so z = g . x = h . y. Then x = g^(-1) . z = g^(-1) . (h . y) = (g^(-1)h) . y, so x lies in Orb(y), and therefore every point k . x = (k g^(-1) h) . y of Orb(x) lies in Orb(y). The symmetric argument gives the reverse containment. So two orbits are equal or disjoint. This completes the proof.

Proposition. Stab(x) is a subgroup of G.

Proof. It contains e. If g and h fix x, then h^(-1) . x = h^(-1) . (h . x) = (h^(-1)h) . x = x, so h^(-1) fixes x as well, and then (gh^(-1)) . x = g . (h^(-1) . x) = g . x = x. The one-step test applies. This completes the proof.

An action is called transitive when there is only one orbit, that is, when any point can be moved to any other. The cube rotations act transitively on faces, on vertices and on edges; S_n acts transitively on the symbols; conjugation on a non-trivial group is never transitive, because the identity is fixed by everything and so sits alone in its orbit.

The orbit-stabiliser theorem

Theorem. Let G act on X and let x be in X. Then there is a bijection between the left cosets of Stab(x) in G and the points of Orb(x). In particular, if G is finite,

|Orb(x)| * |Stab(x)| = |G|,

so every orbit size divides the order of the group.

Proof. Write S = Stab(x) and define a map from the set of left cosets of S to Orb(x) by sending gS to g . x.

Well defined. If gS = hS then h^(-1)g lies in S, so (h^(-1)g) . x = x, so g . x = h . ((h^(-1)g) . x) = h . x.

Injective. Reverse the computation: if g . x = h . x then (h^(-1)g) . x = x, so h^(-1)g lies in S, so gS = hS.

Surjective. Every element of the orbit has the form g . x, which is the image of gS.

So the number of cosets, which is the index [G : S], equals |Orb(x)|. For finite G, Lagrange's theorem turns the index into |G| / |S|, giving the displayed product. This completes the proof.

Look at where the argument came from. The well-definedness and injectivity steps are the same computation run in two directions, and both are the coset criterion from Module 1. Orbit-stabiliser is Lagrange's theorem wearing a different hat, which is why it inherits the same divisibility conclusion.

The cube, counted three ways

Let G be the rotation group of a cube, whose order we are trying to establish. Take the action on the six faces. Any face can be rotated to any other, so the orbit has six elements. The rotations fixing a given face are the four turns about the axis through that face and the one opposite: by 0, 90, 180 and 270 degrees. So |G| = 6 * 4 = 24.

Three different actions, three different factorisations, one answer. That consistency is a useful check on your reasoning about any symmetric object: if two actions disagree about |G|, you have miscounted an orbit or a stabiliser.

A second worked example, smaller. Let D4 act on the four vertices of a square. The action is transitive, so the orbit of a vertex has size 4. The symmetries fixing a chosen vertex are the identity and the reflection in the diagonal through that vertex, so the stabiliser has order 2. The theorem gives |D4| = 4 * 2 = 8, which matches the count of four rotations and four reflections.

Cayley's theorem: every group is a permutation group

Arthur Cayley observed in 1854 that the left multiplication action makes any group concrete.

Theorem (Cayley). Every group G is isomorphic to a subgroup of Sym(G). In particular a finite group of order n is isomorphic to a subgroup of S_n.

Proof. Let G act on itself by left multiplication, g . x = gx. The axioms hold: ex = x, and g(hx) = (gh)x by associativity. This action corresponds to a homomorphism L from G into Sym(G), sending g to the map x goes to gx. Its kernel consists of those g with gx = x for every x; taking x = e forces g = e, so the kernel is trivial and L is injective. By the first isomorphism theorem, G is isomorphic to its image, a subgroup of Sym(G). When |G| = n, Sym(G) is S_n after numbering the elements. This completes the proof.

Worked instance. Take G = Z_3 = {0, 1, 2} and number the elements 1, 2, 3 in that order. Left addition by 0 is the identity permutation. Left addition by 1 sends 0 to 1, 1 to 2, 2 to 0, which is the cycle (1 2 3) in the numbering. Left addition by 2 is (1 3 2). So Z_3 appears inside S_3 as the subgroup of three-cycles, which is exactly A_3.

Cayley's theorem is often described as saying permutation groups are the only groups. Treat that with care. It is true that every group embeds in a symmetric group, but the embedding it produces is enormous and inefficient: a group of order 100 lands inside S_100, a group with about 10^157 elements. Its value is conceptual, and it also supplies the technique for the Sylow theorems: to learn something about G, let G act on something and read the resulting permutation representation.

Acting on cosets

The most productive action for proofs is the one in the last row of the table. Let H be a subgroup of index k and let G act on the k left cosets by g . (aH) = (ga)H. The action is transitive, since (ba^(-1)) . (aH) = bH. The stabiliser of the coset aH is aHa^(-1), which you can check directly: g . (aH) = aH says a^(-1)ga lies in H. The resulting homomorphism from G into S_k has kernel equal to the intersection of all the conjugates aHa^(-1), the largest normal subgroup of G contained in H.

One immediate payoff: if G is finite and H has index 2, the homomorphism lands in S_2, a group of order 2, so the kernel has index at most 2 and must equal H, recovering the fact that index-2 subgroups are normal. The same technique with index 3 or 4 constrains groups sharply, and it is how many small groups are shown not to be simple.

Common misconceptions

• "The orbit and the stabiliser are both subsets of the group." The orbit is a subset of the set being acted on; the stabiliser is a subgroup of the group. The theorem multiplies a count in X by a count in G.
• "Orbit-stabiliser needs the action to be transitive." It does not. It applies to one point at a time, and when the action is not transitive, different points may have orbits of different sizes. Transitivity only means there is a single orbit, so its size is |X|.
• "Stabilisers of different points are the same subgroup." They are conjugate, not equal: Stab(g . x) = g Stab(x) g^(-1). For the cube acting on faces, the stabiliser of the top face and the stabiliser of a side face are different subgroups of order 4.
• "Cayley's theorem makes abstract group theory unnecessary." It embeds a group of order n into a group of order n factorial. Knowing that a needle is somewhere in a very large haystack is not the same as holding the needle.

Try it: counting with a stabiliser

Exercise. The rotation group of a regular tetrahedron acts on its four faces. Use orbit-stabiliser to compute the order of that group, then check the answer using its action on the six edges.

Solution. Any face can be rotated to any other, so the orbit of a face has size 4. The rotations fixing a face are the turns about the axis through the opposite vertex and the centre of that face: by 0, 120 and 240 degrees, so the stabiliser has order 3. Hence the group has order 4 * 3 = 12. Checking with edges: the group acts transitively on the six edges, and the rotations fixing a given edge are the identity and the half turn about the axis through the midpoints of that edge and the opposite edge, so the stabiliser has order 2 and the product is 6 * 2 = 12. The two counts agree, and 12 is the order of A_4, as the tetrahedron discussion in Module 1 predicted.

The takeaway

An action is a homomorphism into a symmetric group, written in the language of moving points. Orbits partition the set, stabilisers are subgroups, and the orbit-stabiliser theorem multiplies the two to recover the order of the group, which forces every orbit size to divide |G|. Left multiplication gives Cayley's embedding; action on cosets gives a permutation representation whose kernel is the largest normal subgroup inside the chosen subgroup.

Looking ahead

The next lesson takes the one action we have not yet exploited, conjugation of a group on itself. Its orbits are the conjugacy classes, its stabilisers are the centralisers, and orbit-stabiliser applied to it yields the class equation. That single equation proves that any group of prime power order has a non-trivial centre, gives Cauchy's theorem for the prime p, and provides the counting on which the Sylow theorems rest.

Sources

1. Weisstein, E. W. (n.d.). Group action. Wolfram MathWorld. mathworld.wolfram.com
2. Weisstein, E. W. (n.d.). Group orbit. Wolfram MathWorld. mathworld.wolfram.com
3. Judson, T. W. (n.d.). Chapter 14: Group actions. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
4. O'Connor, J. J., & Robertson, E. F. (1997). Arthur Cayley. MacTutor History of Mathematics Archive. mathshistory.st-andrews.ac.uk
5. Weisstein, E. W. (n.d.). Cayley's group theorem. Wolfram MathWorld. mathworld.wolfram.com

A square has four mirror lines, but they are not four of a kind. Two run through opposite corners; two run through the midpoints of opposite edges. Turn the square a quarter turn and a corner mirror becomes the other corner mirror, while an edge mirror becomes the other edge mirror. No symmetry of the square ever turns a corner mirror into an edge mirror, because a corner mirror fixes two vertices and an edge mirror fixes none.

The eight symmetries of the square therefore fall into groups that symmetry itself cannot tell apart: {e}, {r^2}, {r, r^3}, {s, r^2 s}, and {rs, r^3 s}. Sizes 1, 1, 2, 2, 2, adding to 8. That decomposition is the class equation, and this lesson proves it in general and then extracts three results from it that no amount of direct computation would give you.

Conjugacy: the same element seen from elsewhere

Two elements x and y of a group are conjugate if y = gxg^(-1) for some g. Reading right to left, g^(-1) moves you into a new frame, x acts, and g moves you back: gxg^(-1) is what x looks like from g's point of view. This is exactly the conjugation action of the previous lesson, so the orbits are the conjugacy classes and the stabiliser of x is the centraliser

C(x) = {g in G : gx = xg},

the set of elements that commute with x. Since orbits partition, the conjugacy classes partition the group.

Conjugate elements are structurally identical. In particular they have the same order, since (gxg^(-1))^n = g x^n g^(-1) is the identity exactly when x^n is. In symmetric groups the invariant is even sharper, and the computation behind it is worth doing once. For any permutation g and any cycle,

g (a_1 a_2 ... a_k) g^(-1) = (g(a_1) g(a_2) ... g(a_k)).

To see it, apply the left side to g(a_i): first g^(-1) sends it to a_i, then the cycle sends that to a_(i+1), then g sends that to g(a_(i+1)), which is what the right side does. So conjugating a permutation just relabels the symbols inside its cycles, leaving the pattern of cycle lengths untouched. In S_n, two permutations are conjugate precisely when they have the same cycle type.

The point: conjugacy is the equivalence relation that ignores labels and keeps shape.

Class sizes come from orbit-stabiliser

Applying the orbit-stabiliser theorem to the conjugation action gives, for finite G,

size of the class of x = [G : C(x)] = |G| / |C(x)|.

Three consequences are immediate. Every class size divides |G|. A class has one element exactly when C(x) = G, that is, when x commutes with everything, that is, when x lies in the centre Z(G). And in an abelian group every class is a singleton, which is why conjugacy carries no information there.

Check it on the square. The centraliser of r contains all four rotations, and it cannot contain a reflection because s r s^(-1) = r^3; so |C(r)| = 4 and the class of r has size 8/4 = 2, namely {r, r^3}, matching the picture. The element r^2, a half turn, commutes with everything, so its class is a singleton and it lies in the centre.

The class equation

Theorem. Let G be a finite group and let x_1, ..., x_k be representatives of the conjugacy classes with more than one element. Then

|G| = |Z(G)| + [G : C(x_1)] + ... + [G : C(x_k)],

and each index in the sum is greater than 1 and divides |G|.

Proof. The conjugacy classes partition G, so |G| is the sum of the class sizes. Split that sum into singleton classes and the rest. A class is a singleton exactly when its element is central, so the singletons contribute |Z(G)| in total. Each remaining class has size [G : C(x_i)] by orbit-stabiliser, and that index exceeds 1 precisely because the class is not a singleton. Divisibility is Lagrange. This completes the proof.

Worked equations, three of them.

In S_4 the summands are, in order, the identity, the six transpositions, the three products of two disjoint transpositions, the eight three-cycles and the six four-cycles. Every one of 1, 6, 3, 8, 6 divides 24, as it must.

What the equation proves

The class equation is a constraint on the arithmetic of a group order, and constraints of that kind rule out structures wholesale.

Theorem. If |G| = p^n for a prime p and n at least 1, then Z(G) is not trivial.

Proof. Each index [G : C(x_i)] in the class equation divides p^n and is greater than 1, so it is a positive power of p and in particular divisible by p. The left side |G| = p^n is divisible by p. Rearranging, |Z(G)| = |G| - ([G : C(x_1)] + ... + [G : C(x_k)]) is a difference of multiples of p, hence divisible by p. Since |Z(G)| is at least 1 (the identity is central), it is at least p. This completes the proof.

That is a strong statement obtained from pure arithmetic: no group of order 8, 27, 125 or 2^10 can have a trivial centre, and we never looked at a single multiplication table. It has a well-known consequence.

Corollary. Every group of order p^2 is abelian.

Proof. By the theorem, |Z(G)| is p or p^2. Suppose it were p. Then G/Z(G) has order p, so it is cyclic by the prime-order classification. Now use the lemma that G/Z(G) cyclic forces G abelian: if gZ generates the quotient, every element of G has the form g^a z with z central, and two such elements commute because g^a z_1 g^b z_2 = g^(a+b) z_1 z_2 = g^b z_2 g^a z_1. But then G is abelian and Z(G) = G has order p^2, contradicting the assumption. So |Z(G)| = p^2 and G is abelian. This completes the proof.

Cauchy's theorem, in one page

Lagrange says a subgroup order divides the group order. The partial converse for a prime divisor is due to Cauchy, and the proof below, found by James McKay in 1959, is the shortest known.

Theorem (Cauchy). If a prime p divides |G|, then G contains an element of order p.

Proof. Let X be the set of ordered p-tuples (x_1, ..., x_p) of elements of G whose product is the identity. Choosing the first p - 1 entries freely determines the last, so |X| = |G|^(p-1), which is divisible by p.

Let the cyclic group Z_p act on X by shifting a tuple one place: the generator sends (x_1, x_2, ..., x_p) to (x_2, ..., x_p, x_1). This lands back in X, because if x_1 x_2 ... x_p = e then x_2 ... x_p x_1 = x_1^(-1)(x_1 x_2 ... x_p) x_1 = e. Every orbit has size dividing p, so size 1 or p. A tuple is fixed exactly when all its entries agree, that is, when it has the form (x, x, ..., x) with x^p = e.

Let F be the number of fixed tuples. Counting X orbit by orbit gives |X| = F + p m for some non-negative integer m, so F is congruent to |X| modulo p, hence divisible by p. The tuple of all identities is fixed, so F is at least 1, hence at least p, hence at least 2. So some x other than the identity satisfies x^p = e, and since p is prime the order of that x is exactly p. This completes the proof.

Worked instance. A group of order 15 must contain an element of order 3 and an element of order 5, and Lagrange leaves no room for elements of any other non-identity order. In fact one can push this to show every group of order 15 is cyclic, using the Sylow counts of the next lesson.

Counting things up to symmetry

Orbits also answer questions that have nothing to do with group structure. How many ways are there to paint the six faces of a cube red or blue, if two paintings count as the same when a rotation carries one to the other? There are 2^6 = 64 paintings and 24 rotations, but 64/24 is not an integer, so simple division fails: different paintings have different amounts of symmetry.

Lemma (orbit counting). If a finite group G acts on a finite set X, the number of orbits equals the average number of fixed points, that is (1/|G|) times the sum over g in G of |Fix(g)|, where Fix(g) is the set of points g leaves alone.

Proof. Count the pairs (g, x) with g . x = x in two ways. Grouping by g gives the sum of |Fix(g)|. Grouping by x gives the sum of |Stab(x)|, which by orbit-stabiliser is the sum of |G| / |Orb(x)|. Within a single orbit of size m there are m points each contributing |G|/m, for a total of |G| per orbit. So the second count is |G| times the number of orbits, and equating the two counts gives the lemma. This completes the proof.

This result is usually called Burnside's lemma, although Burnside quoted it from Frobenius and it was known to Cauchy; the honest names are the Cauchy-Frobenius lemma or the orbit counting theorem. Apply it to the cube.

The total is 64 + 6(8) + 3(16) + 8(4) + 6(8) = 64 + 48 + 48 + 32 + 48 = 240, and 240 / 24 = 10. There are exactly ten distinguishable two-colourings of a cube. A colouring is fixed by a rotation exactly when every face in a cycle gets the same colour, which is why each entry in the last column is 2 raised to the number of cycles.

Common misconceptions

• "Conjugate elements are equal in some weaker sense." They are genuinely different elements, related by a change of viewpoint. r and r^3 are different quarter turns of the square, and no relabelling makes them the same rotation; conjugacy says a symmetry of the square exchanges them.
• "The centraliser and the centre are the same thing." The centraliser C(x) depends on x and is usually large; the centre is the intersection of all centralisers and is often trivial. In S_4 every centraliser is a proper subgroup and the centre is {e}.
• "You can count colourings by dividing by the size of the group." Only when the action is free, meaning no nontrivial element fixes anything. For the cube, 64/24 is not even an integer, and the orbit counting lemma is the correction.
• "Cauchy's theorem gives subgroups of every order dividing |G|." It gives elements of prime order only. The tetrahedral group of order 12 has elements of order 2 and 3 but no subgroup of order 6, as Module 1 showed.

Try it: necklaces

Exercise. How many distinct necklaces of six beads can be made from two bead colours, if two necklaces are the same when one can be rotated into the other? Use the cyclic group Z_6 acting by rotation.

Solution. A rotation by k positions decomposes the six beads into gcd(k, 6) cycles, and a colouring is fixed exactly when each cycle is monochromatic, so it fixes 2^(gcd(k,6)) colourings. Summing over k = 0, 1, 2, 3, 4, 5 gives 2^6 + 2^1 + 2^2 + 2^3 + 2^2 + 2^1 = 64 + 2 + 4 + 8 + 4 + 2 = 84, and dividing by 6 gives 14 necklaces.

Change one input. Allow the necklace to be flipped over as well, so the group becomes D6 of order 12. The six reflections contribute as follows: the three through opposite beads fix 2^4 colourings each, and the three through opposite gaps fix 2^3 each, adding 3(16) + 3(8) = 72. The total becomes 84 + 72 = 156, and 156/12 = 13. One pair of the fourteen was a mirror image of another.

Summing up

Conjugation is the action of a group on itself whose orbits are conjugacy classes and whose stabilisers are centralisers, so class sizes are indexes and therefore divide the group order. Splitting the partition into central and non-central parts gives the class equation, which forces a non-trivial centre in any group of prime power order and hence makes every group of order p^2 abelian. A different action, cyclic shifting of tuples, gives Cauchy's theorem. And counting fixed points on average counts orbits, which is how a cube gets exactly ten two-colourings.

Looking ahead

Cauchy's theorem produces an element of order p whenever p divides |G|. The Sylow theorems, next, produce a subgroup of order p^k for the largest power p^k dividing |G|, say how many such subgroups there can be, and say that they are all conjugate. That package is strong enough to prove that no group of order 15, 20 or 255 is simple, and to classify the groups of several small orders outright.

Sources

1. Weisstein, E. W. (n.d.). Class equation. Wolfram MathWorld. mathworld.wolfram.com
2. Wikipedia contributors. (n.d.). Burnside's lemma. Wikipedia. en.wikipedia.org
3. Wikipedia contributors. (n.d.). Cauchy's theorem (group theory). Wikipedia. en.wikipedia.org
4. Weisstein, E. W. (n.d.). Conjugacy class. Wolfram MathWorld. mathworld.wolfram.com
5. Judson, T. W. (n.d.). Chapter 14: Group actions. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org

Here is a question with a startling answer. How many different groups of order 15 are there, counting two as the same when they are isomorphic? Not five, not three. Exactly one, and it is Z_15. Every conceivable multiplication table on fifteen elements that satisfies the four axioms is a relabelling of addition modulo 15.

Lagrange cannot prove that. Neither can Cauchy, which gives elements of order 3 and 5 but says nothing about how they interact. The result comes from three theorems that Ludwig Sylow published in a ten-page paper in 1872, Théorèmes sur les groupes de substitutions, in volume 5 of Mathematische Annalen. They are the sharpest general facts known about finite groups, and this lesson proves them and then puts them to work.

The statement

Let G be finite of order p^k m, where p is prime, k is at least 1, and p does not divide m. A Sylow p-subgroup of G is a subgroup of order p^k: as large a p-power subgroup as Lagrange permits. Write n_p for the number of Sylow p-subgroups.

• Sylow I (existence). G has at least one Sylow p-subgroup. In fact G has a subgroup of order p^i for every i from 0 to k.
• Sylow II (conjugacy). Any two Sylow p-subgroups are conjugate: if P and Q are both Sylow p-subgroups then Q = gPg^(-1) for some g.
• Sylow III (counting). n_p is congruent to 1 modulo p, and n_p divides m. Moreover n_p = [G : N(P)], the index of the normaliser of any one of them.

The normaliser N(H) of a subgroup H is the set of g with gHg^(-1) = H, the largest subgroup of G in which H is normal. Note the immediate corollary of Sylow II: n_p = 1 if and only if the unique Sylow p-subgroup is normal, since being alone in its conjugacy class means being fixed by all conjugation.

Sylow I, by induction

Theorem. If p^i divides |G| then G has a subgroup of order p^i.

Proof. Induct on |G|. For |G| = 1 there is nothing to prove, and for i = 0 the trivial subgroup works. So assume i is at least 1 and the statement holds for all groups smaller than G. Split into two cases according to the centre.

Case 1: p divides |Z(G)|. By Cauchy's theorem applied to the abelian group Z(G), there is a central element z of order p. Let N = <z>, a subgroup of order p, normal in G because it sits inside the centre. Then |G/N| = |G|/p, which is smaller than |G| and still divisible by p^(i-1). By the inductive hypothesis G/N has a subgroup B of order p^(i-1). By the correspondence theorem its preimage H in G is a subgroup containing N with H/N = B, so |H| = |B| |N| = p^(i-1) p = p^i.

Case 2: p does not divide |Z(G)|. Read the class equation |G| = |Z(G)| + [G : C(x_1)] + ... + [G : C(x_k)] modulo p. The left side is divisible by p and |Z(G)| is not, so at least one index [G : C(x_j)] is not divisible by p. Fix that x_j and write C = C(x_j). Since |G| = [G : C] |C| and p^i divides |G| while p misses the index entirely, all of p^i must divide |C|. And C is a proper subgroup, because x_j is not central, so |C| < |G|. By the inductive hypothesis C has a subgroup of order p^i, and that is a subgroup of G. This completes the proof.

Both cases used the previous lesson: Cauchy's theorem in the first, the class equation in the second. Why this matters: Sylow I is the converse to Lagrange that actually holds. Lagrange forbids; Sylow, for prime powers, supplies.

A lemma that does the rest of the work

Lemma. Let P and Q be Sylow p-subgroups of G. If P is contained in N(Q), then P = Q.

Proof. Inside N(Q), the subgroup Q is normal, so PQ is a subgroup, and by the order formula from Module 2, |PQ| = |P||Q| / |P intersect Q|. Every factor there is a power of p, so |PQ| is a power of p. But PQ contains Q, which already has the largest possible p-power order in G, namely p^k. So |PQ| = p^k and therefore PQ = Q. Since P is contained in PQ, we get P inside Q, and equal orders force P = Q. This completes the proof.

Now let X be the set of all Sylow p-subgroups and let any p-subgroup act on X by conjugation, which is legitimate because a conjugate of a Sylow p-subgroup is again one. Under the action of a Sylow p-subgroup P, every orbit has size dividing |P| = p^k, hence size 1 or a positive power of p. A member Q of X is a fixed point exactly when P normalises Q, that is, when P is inside N(Q), which by the lemma happens only when Q = P. So P fixes exactly one element of X, namely itself.

Sylow II and III

Theorem (conjugacy). All Sylow p-subgroups of G are conjugate.

Proof. Fix P in X and let O be its orbit under conjugation by all of G, that is, the set of conjugates of P. Suppose some Q in X is not in O. Let Q act on O by conjugation. Orbits have p-power size, and a fixed point would be a member R of O normalised by Q, which by the lemma forces Q = R and puts Q in O, contrary to assumption. So every Q-orbit on O has size divisible by p, and |O| is divisible by p. But now let P act on O. The same size argument applies, except that P does have a fixed point, itself, and by the lemma exactly one. So |O| is congruent to 1 modulo p. The two conclusions contradict each other, so no such Q exists and X = O. This completes the proof.

Theorem (counting). n_p is congruent to 1 modulo p, and n_p divides m.

Proof. By Sylow II, X is a single orbit under conjugation by G, and the stabiliser of P in that action is by definition N(P). Orbit-stabiliser gives n_p = |X| = [G : N(P)], which divides |G|. Since P is contained in N(P), the order of N(P) is a multiple of p^k, so its index divides |G|/p^k = m. For the congruence, let P act on X by conjugation: all orbits have p-power size and exactly one is a singleton, so |X| is congruent to 1 modulo p. This completes the proof.

Order 15, settled

Theorem. Every group of order 15 is cyclic.

Proof. Write |G| = 15 = 3 * 5. For p = 3 we have m = 5, so n_3 divides 5 and is congruent to 1 modulo 3: the candidates 1 and 5 reduce to 1 and 2 modulo 3, so n_3 = 1. For p = 5 we have m = 3, so n_5 divides 3 and is congruent to 1 modulo 5: the candidates 1 and 3 reduce to 1 and 3, so n_5 = 1.

Let P be the unique subgroup of order 3 and Q the unique subgroup of order 5; both are normal, since uniqueness plus Sylow II makes each fixed by conjugation. Their intersection is a subgroup whose order divides both 3 and 5, hence is trivial. For x in P and y in Q, the commutator x y x^(-1) y^(-1) can be read two ways: as (x y x^(-1)) y^(-1), a product of two elements of Q by normality of Q, and as x (y x^(-1) y^(-1)), a product of two elements of P. So it lies in the trivial intersection and equals the identity, meaning xy = yx. Now take generators x of P and y of Q. They commute, so (xy)^n = x^n y^n, and this is the identity only when 3 divides n and 5 divides n, so xy has order 15. A group of order 15 with an element of order 15 is cyclic. This completes the proof.

The same argument works for any order pq with primes p < q where p does not divide q - 1: the congruence and divisibility conditions leave only n_p = n_q = 1. When p does divide q - 1, a second group appears, non-abelian; the smallest instance is order 6 = 2 * 3 with 2 dividing 3 - 1, giving Z_6 and S_3.

Ruling out simple groups

A group is simple if its only normal subgroups are the trivial one and itself. Simple groups are the atoms of finite group theory, and a standard use of Sylow is to prove that a group of a given order cannot be one.

Order 20. Here 20 = 2^2 * 5, so for p = 5 we get m = 4 and n_5 divides 4 with n_5 congruent to 1 modulo 5. The divisors of 4 are 1, 2 and 4, of which only 1 is congruent to 1 modulo 5. So n_5 = 1, the Sylow 5-subgroup is normal, and no group of order 20 is simple.

Order 30. Here n_5 divides 6 and is 1 or 6; n_3 divides 10 and is 1 or 10. If both were maximal we would have 6 * 4 = 24 elements of order 5 and 10 * 2 = 20 elements of order 3, totalling 44 elements in a group of 30. So at least one of n_3 and n_5 is 1, and no group of order 30 is simple.

Order 12, where the count is not forced. For p = 3, n_3 divides 4 and is congruent to 1 modulo 3, so n_3 is 1 or 4. Both occur: Z_12 has n_3 = 1 and A_4 has n_3 = 4, its four subgroups generated by the eight three-cycles in pairs. One can still show no group of order 12 is simple, by arguing that n_3 = 4 forces the Sylow 2-subgroup to be unique, but Sylow alone does not decide the count.

Common misconceptions

• "Sylow gives a subgroup of every order dividing |G|." Only prime power orders. A_4 has order 12 and subgroups of orders 1, 2, 3 and 4, but none of order 6, and Sylow never claimed otherwise.
• "n_p = 1 means the group is abelian." It means one Sylow subgroup is normal. D4 has a unique Sylow 2-subgroup, namely all of D4, and is not abelian; the group of order 20 with a normal Sylow 5-subgroup need not be abelian either.
• "Two Sylow p-subgroups of the same order must be equal." They are conjugate, not equal. S_3 has three Sylow 2-subgroups, each generated by one transposition, all conjugate and all distinct.
• "n_p is the number of elements of order p." It counts subgroups, not elements. In S_3, n_2 = 3 and there happen to be three elements of order 2, but in a group with a Sylow subgroup of order 9 a single subgroup already contains eight elements of order dividing 9.

Try it: order 56

Exercise. Show that no group of order 56 is simple.

Solution. Write 56 = 2^3 * 7. For p = 7, m = 8, so n_7 divides 8 and is congruent to 1 modulo 7, leaving n_7 = 1 or n_7 = 8. If n_7 = 1 the Sylow 7-subgroup is normal and we are finished. So suppose n_7 = 8. Each Sylow 7-subgroup has prime order 7, so two distinct ones intersect trivially, and each contributes 6 elements of order 7. That accounts for 8 * 6 = 48 elements of order 7. The remaining 56 - 48 = 8 elements, together with nothing else, must accommodate every Sylow 2-subgroup, and a Sylow 2-subgroup has exactly 8 elements none of which has order 7. So there is room for only one, giving n_2 = 1 and a normal Sylow 2-subgroup. Either way a proper non-trivial normal subgroup exists.

Notice the shape of the argument: when the count is not forced, count elements instead. That combination, congruence conditions plus an element census, settles most small orders.

Where this leaves us

Sylow I supplies a subgroup of order p^i for every prime power dividing |G|, proved by induction with the class equation. Sylow II makes all the maximal ones conjugate, so uniqueness and normality are the same condition. Sylow III pins the count between a congruence modulo p and a divisibility into m, which is often narrow enough to force n_p = 1. Together they classify groups of order 15, forbid simple groups of order 20, 30 and 56, and give the standard opening move on any question about a group of specified order.

Looking ahead

Simple groups keep appearing as the thing being ruled out. The next module meets the smallest non-abelian simple group face on. It lives inside the permutations of five symbols, has order 60, and its simplicity is the reason the general quintic equation cannot be solved by radicals, which is where this course ends.

Sources

1. Weisstein, E. W. (n.d.). Sylow theorems. Wolfram MathWorld. mathworld.wolfram.com
2. Wikipedia contributors. (n.d.). Sylow theorems. Wikipedia. en.wikipedia.org
3. Judson, T. W. (n.d.). Chapter 15: The Sylow theorems. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
4. O'Connor, J. J., & Robertson, E. F. (2000). Peter Ludwig Mejdell Sylow. MacTutor History of Mathematics Archive. mathshistory.st-andrews.ac.uk
5. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Section 4.5. John Wiley and Sons.

Split a 52-card deck exactly in half and riffle the two halves together so that the cards alternate perfectly, with the original top card staying on top. Do it again. Do it eight times in total and the deck is back exactly as it started. This is not a card trick; it is arithmetic. The perfect out-shuffle moves the card in position i to a position determined by doubling modulo 51, and 2^8 = 256 = 5 * 51 + 1, so eight doublings return every card home.

What that shuffle is, mathematically, is a single element of S_52, the group of all permutations of 52 objects, and the number 8 is its order. This lesson is about how to read such information off a permutation without ever performing it 52 times: cycle notation, order from cycle type, and the parity invariant that splits every symmetric group cleanly in half.

Cycle notation

A permutation of {1, ..., n} is a bijection from that set to itself, and S_n is the group of all of them under composition, of order n factorial. To record one, follow an element until it returns. Start at 1, see where it goes, follow that, and keep going; you must eventually return to 1, because the map is injective and the set is finite. That closed loop is a cycle, written (1 a b c), meaning 1 goes to a, a goes to b, b goes to c, and c goes back to 1. Pick an unused symbol and repeat.

Take the permutation of {1, ..., 7} sending 1 to 4, 2 to 6, 3 to 3, 4 to 7, 5 to 2, 6 to 5, 7 to 1. Following 1: 1 to 4 to 7 to 1, giving (1 4 7). Following 2: 2 to 6 to 5 to 2, giving (2 6 5). The symbol 3 is fixed. So the permutation is (1 4 7)(2 6 5), and by convention fixed points are omitted. Its cycle type is the list of cycle lengths, here 3 and 3, plus the fixed point 1, and it is customary to record it as the partition 3 + 3 + 1 = 7.

Two facts about this notation are worth stating carefully. First, every permutation factors into disjoint cycles, and that factorisation is unique up to the order in which the cycles are written and the choice of starting symbol inside each. The construction above proves existence, and uniqueness holds because the cycle containing a symbol is determined by the permutation alone: it is the orbit of that symbol under the cyclic group generated by the permutation. Second, disjoint cycles commute, because they move disjoint sets of symbols and neither can interfere with the other. Non-disjoint cycles generally do not: (1 2)(1 3) = (1 3 2) while (1 3)(1 2) = (1 2 3), using the convention that the right-hand factor acts first.

The order of a permutation

Proposition. The order of a permutation is the least common multiple of the lengths of its disjoint cycles.

Proof. Write the permutation as t = c_1 c_2 ... c_r with the c_i disjoint of lengths l_1, ..., l_r. Since disjoint cycles commute, t^k = c_1^k c_2^k ... c_r^k. A cycle of length l has order l, since applying it j times shifts each of its symbols j places around the loop and returns them home exactly when l divides j. Because the c_i move disjoint symbols, the product c_1^k ... c_r^k is the identity only when every factor is, that is, only when every l_i divides k. The least such positive k is the least common multiple. This completes the proof.

Worked example. In S_7, the permutation (1 4 7)(2 6 5) has order lcm(3, 3) = 3. The permutation (1 2)(3 4 5) has order lcm(2, 3) = 6. The permutation (1 2 3 4)(5 6 7) has order 12, which is the largest possible order of an element of S_7. Notice how much cheaper this is than raising a permutation to successive powers: you read the cycle lengths and take one least common multiple.

Transpositions and the parity invariant

A transposition is a cycle of length 2, a single swap. Every permutation is a product of transpositions, because a cycle can be unpacked directly:

(a_1 a_2 ... a_k) = (a_1 a_k)(a_1 a_(k-1)) ... (a_1 a_2),

which you can verify by tracking one symbol at a time. So a k-cycle is a product of k - 1 transpositions, and a general permutation is a product of transpositions by handling its cycles one at a time.

The factorisation is far from unique: (1 2 3) = (1 3)(1 2) = (2 1)(2 3) = (1 3)(1 2)(1 2)(1 2). But the parity of the number of factors never changes, and proving that is the point of the next theorem.

Theorem. If a permutation can be written as a product of an even number of transpositions, it cannot also be written as a product of an odd number.

Proof. Work with the polynomial in n variables

D = product over all pairs i < j of (x_i - x_j).

Let a permutation t act on D by permuting the subscripts, producing the polynomial t(D) whose factors are (x_(t(i)) - x_(t(j))). Each unordered pair {i, j} still appears exactly once, possibly with its two entries in the reversed order, so t(D) equals D or -D and nothing else. Define sgn(t) to be the resulting sign, so t(D) = sgn(t) D. Since acting by uv is acting by u and then by v, we get sgn(uv) = sgn(u) sgn(v): the sign is a homomorphism into {1, -1}.

Now compute the sign of a transposition (a b) with a < b. A factor (x_i - x_j) changes sign exactly when the swap reverses the relative order of i and j. That happens for the pair {a, b} itself, and for each c strictly between a and b it happens for both {a, c} and {c, b}. Pairs not involving a or b are untouched, and a pair involving one of them and a symbol outside the interval keeps its order. So the number of sign changes is 2(b - a - 1) + 1, an odd number, and sgn((a b)) = -1.

If a permutation is a product of k transpositions then its sign is (-1)^k. Since the sign depends only on the permutation, (-1)^k is the same for every such factorisation, so all factorisations of a given permutation have k of the same parity. This completes the proof.

Call a permutation even if sgn = 1 and odd if sgn = -1. From the unpacking formula, a k-cycle has sign (-1)^(k-1): three-cycles are even, transpositions and four-cycles are odd. The core of it: parity is the only invariant of a transposition factorisation, and it is the reason the alternating group exists.

Conjugacy classes and generators

The previous module showed that conjugation in S_n relabels the symbols inside cycles, so two permutations are conjugate exactly when they have the same cycle type. That makes the conjugacy classes of S_n correspond precisely to the partitions of n. Here is S_5 in full, whose seven partitions give seven classes.

The class sizes total 1 + 10 + 15 + 20 + 20 + 30 + 24 = 120 = 5 factorial, as they must, and the even classes total 1 + 15 + 20 + 24 = 60, which is the order of the alternating group A_5. That last column will do a great deal of work in the next lesson.

Generating sets are usefully small. All transpositions generate S_n, since every permutation factors into them. The adjacent transpositions (1 2), (2 3), ..., (n-1, n) already suffice, because any transposition (a b) can be built by walking a along to b and back: for instance (1 3) = (2 3)(1 2)(2 3). Even two elements are enough: (1 2) together with the long cycle (1 2 3 ... n) generate S_n, because conjugating (1 2) repeatedly by the long cycle produces every adjacent transposition in turn.

Common misconceptions

• "Cycle notation is ambiguous, so cycle type is not well defined." The notation has freedom in the order of the cycles and the starting symbol, but the multiset of cycle lengths is determined by the permutation, because each cycle is an orbit of the cyclic group the permutation generates.
• "An even permutation is one that moves an even number of symbols." A three-cycle moves three symbols and is even; a transposition moves two and is odd. Parity counts transposition factors, not moved symbols, and a k-cycle has sign (-1)^(k-1).
• "All cycles commute." Only disjoint ones. (1 2)(1 3) and (1 3)(1 2) are different three-cycles, and forgetting this is the most common computational error in this subject.
• "The order of a permutation is the number of symbols it moves." It is the least common multiple of its cycle lengths. In S_7 the permutation (1 2 3 4)(5 6 7) moves all seven symbols and has order 12, not 7.

Try it: shuffles and orders

Exercise. A permutation of eight objects has cycle type 4 + 2 + 2. Find its order, its sign, and the size of its conjugacy class in S_8.

Solution. The order is lcm(4, 2, 2) = 4. The sign is the product of the cycle signs: a four-cycle contributes (-1)^3 = -1 and each two-cycle contributes -1, so the total is (-1)(-1)(-1) = -1 and the permutation is odd. For the class size, count permutations of that type: choose and arrange the four-cycle in 8 * 7 * 6 * 5 / 4 = 420 ways, then split the remaining four symbols into two 2-cycles in 3 ways, giving 420 * 3 = 1260. As a check, 1260 divides 8 factorial = 40320, with quotient 32, which is the order of the centraliser.

Change one input. Change the type to 4 + 4. Now the order is still 4, but the sign is (-1)(-1) = +1, so the permutation is even. Two permutations of the same order, opposite parities: order does not determine sign, and sign does not determine order.

What to remember

Every permutation splits uniquely into disjoint cycles, which commute, so its order is the least common multiple of the cycle lengths. Every permutation is a product of transpositions in many ways, but always with a fixed parity, proved by watching what a permutation does to the product of all differences. Conjugacy classes of S_n are the cycle types, so they correspond to partitions of n, and small generating sets exist: the adjacent transpositions, or even a single transposition together with a long cycle.

Looking ahead

The sign map is a homomorphism from S_n onto a group of order 2, so by the first isomorphism theorem its kernel, the alternating group A_n, has index 2 and order n factorial over 2. The next lesson shows that for n at least 5 this group is simple, meaning it has no normal subgroup other than the two obvious ones. That fact, established for A_5 by a hands-on argument with the class table above, is what eventually blocks a radical formula for the quintic.

Sources

1. Weisstein, E. W. (n.d.). Symmetric group. Wolfram MathWorld. mathworld.wolfram.com
2. Judson, T. W. (n.d.). Chapter 5: Permutation groups. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
3. Wikipedia contributors. (n.d.). Parity of a permutation. Wikipedia. en.wikipedia.org
4. Wikipedia contributors. (n.d.). Faro shuffle. Wikipedia. en.wikipedia.org
5. Artin, M. (2011). Algebra (2nd ed.), Chapter 1 and Chapter 7. Pearson.

Hold a regular icosahedron, the twenty-faced solid, and count its rotations. Through each of the six pairs of opposite vertices runs an axis with four nontrivial turns, giving 24. Through each of the ten pairs of opposite faces runs an axis with two nontrivial turns, giving 20. Through each of the fifteen pairs of opposite edges runs an axis with one half turn, giving 15. Adding the identity: 1 + 24 + 20 + 15 = 60.

That group of 60 rotations is A_5, the alternating group on five symbols, and it is the smallest non-abelian group with no proper non-trivial normal subgroup. Everything you can do to an icosahedron is either nothing or everything, in the precise sense that there is no way to collapse part of its symmetry consistently. This lesson proves that, and the proof is a piece of arithmetic on a table of five numbers.

The alternating group

The sign map from S_n to {1, -1} is a surjective homomorphism for n at least 2, since transpositions map to -1 and the identity to 1. Its kernel is the alternating group A_n, the set of even permutations. Being a kernel, it is normal; by the first isomorphism theorem S_n / A_n has order 2, so |A_n| = n! / 2. For n = 5 that is 60, matching the icosahedron.

Proposition. For n at least 3, the three-cycles generate A_n.

Proof. An even permutation is a product of an even number of transpositions, so group the factors in consecutive pairs and handle each pair. A pair (a b)(a b) is the identity. A pair sharing one symbol gives (a b)(a c) = (a c b), a three-cycle: applying (a c) first sends a to c, then (a b) fixes c, so a goes to c; similarly c goes to a then to b, and b goes to a. A pair with no shared symbol gives (a b)(c d) = (a c b)(a c d), which you can verify by tracking each of the four symbols through the right side. So every pair is a product of at most two three-cycles, and the whole permutation is a product of three-cycles. This completes the proof.

That proposition converts questions about A_n into questions about three-cycles, and it is the reason the simplicity argument works at all: if a normal subgroup contains one three-cycle and all three-cycles are conjugate, it contains all of them and therefore everything.

The conjugacy classes of A5

From the class table of S_5 in the previous lesson, the even classes are the identity (size 1), the double transpositions (size 15), the three-cycles (size 20), and the five-cycles (size 24), and these total 60. But a class of S_5 can break into smaller classes of A_5, because A_5 has fewer elements available to conjugate with, and one class here does break.

The test is a centraliser count. If x lies in A_n, its S_n-class has size [S_n : C_{S_n}(x)] and its A_n-class has size [A_n : C_{A_n}(x)]. When the centraliser in S_n contains an odd permutation, the two indexes agree and the class does not split; when the centraliser lies entirely inside A_n, the A_n-index is half as large and the class splits into two of equal size.

Apply this. A five-cycle such as t = (1 2 3 4 5) has S_5-class of size 24, so |C_{S_5}(t)| = 120/24 = 5. But t itself generates a subgroup of order 5 inside that centraliser, so the centraliser is exactly <t>, all of whose elements are even. So the class splits: the 24 five-cycles fall into two A_5-classes of 12. By contrast, the transposition (4 5) commutes with (1 2 3), so the centraliser of a three-cycle contains an odd permutation and the class of 20 stays whole; similarly (1 2) commutes with (1 2)(3 4), so the class of 15 stays whole.

Total: 1 + 15 + 20 + 12 + 12 = 60. On the icosahedron these are, in order, the identity, the fifteen edge half turns, the twenty face rotations, and the twenty-four vertex rotations split by angle into twelve by 72 degrees and twelve by 144 degrees.

A5 is simple

Theorem. A_5 has no normal subgroup other than {e} and A_5.

Proof. Let N be normal in A_5. Since N is closed under conjugation by every element of A_5, it is a union of whole conjugacy classes, and it contains the class {e}. So |N| is 1 plus a sum of some of the numbers 15, 20, 12 and 12. By Lagrange, |N| must also divide 60.

Enumerate the possibilities. Taking no further class gives 1, which divides 60. Adding one class gives 16, 21, or 13. Adding two gives 36, 28, 28, 33, 33, or 25. Adding three gives 48, 48, 45, or 40. Adding all four gives 60. Of these values only 1 and 60 divide 60: 16, 21, 13, 36, 28, 33, 25, 48, 45 and 40 all fail. So |N| is 1 or 60, and N is trivial or everything. This completes the proof.

Read what made the argument work. Normal subgroups must be unions of classes, class sizes must divide the order, and the particular numbers 1, 12, 12, 15, 20 admit no sub-sum containing 1 that also divides 60 except the two extremes. Change any one of those class sizes and the argument collapses, which is exactly what happens one step down.

Why A4 is not simple

A_4 has order 12, with classes of sizes 1 (identity), 3 (double transpositions), and two classes of 4 (the eight three-cycles, split for exactly the centraliser reason above). Now 1 + 3 = 4, which does divide 12, and the corresponding set really is a subgroup:

V = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}.

It is closed, since the product of any two distinct double transpositions is the third, and it is a union of conjugacy classes, so it is normal. It is the Klein four-group, and it is why A_4 is not simple and why the general quartic can be solved by radicals. The chain {e} inside V inside A_4 inside S_4 has abelian quotients of orders 4, 3 and 2 at the successive steps, and Module 6 will show that such a chain is precisely what a radical formula needs.

Beyond five symbols

A_n is simple for every n at least 5. The proof has two halves, and the first is short enough to give here.

Lemma. For n at least 5, all three-cycles are conjugate in A_n.

Proof. Any two three-cycles are conjugate in S_n, since they have the same cycle type: there is some g in S_n with g (a b c) g^(-1) = (d e f). If g is even we are done. If g is odd, pick two symbols u and v not among a, b, c, which is possible because n is at least 5, and replace g by g' = g (u v). Then g' is even, and since (u v) commutes with (a b c), conjugating by g' gives the same result. This completes the proof.

The second half is the statement that any non-trivial normal subgroup of A_n contains a three-cycle. It is proved by taking a non-identity element of the normal subgroup, forming a suitable commutator with a well-chosen three-cycle, and checking that the result is closer to a three-cycle than what you started with, in a case analysis over cycle types. It is a genuine argument rather than a slogan, and it is carried out in full in Dummit and Foote's Section 4.6; a text course should say honestly that we are pointing at it rather than performing it. Granting it, the two halves combine: a non-trivial normal subgroup contains a three-cycle, hence all of them by the lemma, hence all of A_n by the generation proposition.

What simplicity is for

Simple groups are the pieces from which every finite group is assembled. Given G, choose a maximal proper normal subgroup G_1, so that G/G_1 is simple; repeat inside G_1. The resulting chain is a composition series, and the Jordan-Hölder theorem says the multiset of simple quotients does not depend on the choices made. For S_5 the series is {e} inside A_5 inside S_5, with quotients A_5 and Z_2. Because A_5 is simple and not abelian, no refinement can produce abelian quotients throughout, and that single fact is what Module 6 will convert into the impossibility of a quintic formula.

Why this matters: the classification of finite simple groups, completed in the 1980s and running to thousands of journal pages, lists all of these building blocks. The alternating groups A_n for n at least 5 form one of its infinite families, and you have just proved the first member belongs there.

Common misconceptions

• "A_n is simple for all n." Not for n = 4, where the Klein four-group is normal, and trivially not for n below 4. Simplicity begins at 5, and that boundary is the same boundary that separates solvable quartics from unsolvable quintics.
• "Conjugacy classes of S_n stay whole inside A_n." Some split. The 24 five-cycles become two A_5-classes of 12, because their centraliser in S_5 contains no odd permutation.
• "Simple means small or structureless." A_5 has 60 elements, five conjugacy classes and a rich subgroup lattice including copies of Z_5, S_3 and A_4. What it lacks is normal subgroups, not structure.
• "A_5 is the smallest simple group." The cyclic groups of prime order are simple and much smaller. A_5 is the smallest non-abelian one, and the qualification carries the whole content.

Try it: normality by classes

Exercise. Verify directly that V = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} is a normal subgroup of A_4.

Solution. Closure: the product of two distinct double transpositions is the third, for instance (1 2)(3 4) composed with (1 3)(2 4) sends 1 to 4, 4 to 1, 2 to 3 and 3 to 2, which is (1 4)(2 3). Each element is its own inverse, and the identity is present, so V is a subgroup of order 4. Normality: conjugation preserves cycle type, so a conjugate of a double transposition is a double transposition, and V contains every double transposition in S_4. Hence gVg^(-1) = V for every g, and indeed V is normal in the larger group S_4 as well.

Change one input. Try the same idea in A_5: is the set of the identity plus the fifteen double transpositions a subgroup? Its size is 16, which does not divide 60, so Lagrange says no. That single arithmetical obstruction is the whole difference between four symbols and five.

Looking back

A_n is the kernel of the sign map, of index 2 and order n!/2, and it is generated by three-cycles. In A_5 the classes have sizes 1, 15, 20, 12 and 12, the last two coming from the splitting of the five-cycles, and no sub-sum containing 1 divides 60 except 1 and 60 themselves, so A_5 is simple. A_4 fails because 1 + 3 = 4 does divide 12 and the corresponding set is the Klein four-group.

Looking ahead

The group half of this course is finished. Module 5 restarts from a second set of axioms, this time for a structure with two operations: rings, where you can add, subtract and multiply but not always divide. Ideals will play the role normal subgroups played, quotient rings the role of quotient groups, and the first isomorphism theorem will reappear almost word for word. The payoff is a theory of polynomial factorisation strong enough to support the field theory that finishes the course.

Sources

1. Weisstein, E. W. (n.d.). Alternating group. Wolfram MathWorld. mathworld.wolfram.com
2. Wikipedia contributors. (n.d.). Alternating group. Wikipedia. en.wikipedia.org
3. Wikipedia contributors. (n.d.). Icosahedral symmetry. Wikipedia. en.wikipedia.org
4. Wikipedia contributors. (n.d.). Composition series. Wikipedia. en.wikipedia.org
5. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Section 4.6. John Wiley and Sons.

On a twelve-hour clock, multiply 3 by 4. You get 12, which reads as 0. Neither factor was zero, and yet the product is. Nothing like that can happen with ordinary integers, and the difference is not a curiosity: it is the single property that separates the arithmetic you can trust from the arithmetic you cannot.

Groups have one operation. Almost every number system you have ever used has two, addition and multiplication, tied together by distributivity. This module axiomatises that situation, and the first job is to find out which familiar facts survive the axioms and which, like the impossibility of a product of nonzero things being zero, have to be assumed separately.

The ring axioms

A ring is a set R with two operations, written as addition and multiplication, such that:

• R with addition is an abelian group, with identity 0 and each element a having a negative -a;
• multiplication is associative, (ab)c = a(bc), and has an identity 1 with 1a = a1 = a;
• multiplication distributes over addition on both sides: a(b + c) = ab + ac and (b + c)a = ba + ca.

Multiplication is not assumed commutative; when it is, the ring is called commutative. Some books omit the multiplicative identity from the axioms, and then 2Z, the even integers, counts as a ring. We include it, so 2Z will be an example of something that just misses. Note also what is not assumed: no division, no cancellation, and no guarantee that a product of nonzero elements is nonzero.

That last row deserves a look, because it shows the clock phenomenon in a setting with nothing modular about it. Let f(x) = 0 for x at most 1/2 and f(x) = x - 1/2 beyond, and let g(x) = 1/2 - x for x at most 1/2 and 0 beyond. Both are continuous and neither is the zero function, yet at every point at least one factor vanishes, so fg is identically zero.

What the axioms force

Three familiar facts are consequences, not assumptions, and each is proved from distributivity plus the additive group structure.

Proposition. In any ring, 0a = a0 = 0, and (-a)b = -(ab), and (-a)(-b) = ab.

Proof. From 0 = 0 + 0 and distributivity, 0a = (0 + 0)a = 0a + 0a; adding -(0a) to both sides gives 0 = 0a, and the other side is symmetric. Next, (-a)b + ab = ((-a) + a)b = 0b = 0, so (-a)b is the additive inverse of ab. Finally (-a)(-b) = -(a(-b)) = -(-(ab)) = ab, using the previous fact twice. This completes the proof.

The rule that a negative times a negative is a positive is therefore not a convention adopted for convenience. It is forced by distributivity, and it holds in every ring, including rings whose elements are matrices or functions and where the words positive and negative have no meaning.

Units and zero divisors

An element u of R is a unit if it has a two-sided multiplicative inverse in R. The units form a group under multiplication, written R^* or U(R): closure holds because (uv)^(-1) = v^(-1)u^(-1), and the other axioms are immediate. In Z the units are 1 and -1. In Z_n the units are exactly the residues coprime to n, recovering the group U(n) of Module 1. In M_2(R) the units are the invertible matrices, which is GL(2,R).

A nonzero element a is a zero divisor if there is a nonzero b with ab = 0 or ba = 0. On the clock, 3 and 4 are zero divisors; among the continuous functions, the f and g above are. A commutative ring with 1 not equal to 0 and no zero divisors is an integral domain.

Proposition. Cancellation holds in an integral domain: if a is nonzero and ab = ac, then b = c.

Proof. From ab = ac we get a(b - c) = 0. Since a is nonzero and there are no zero divisors, b - c = 0. This completes the proof.

Worked example. In Z_12 the equation 3x = 3y does not force x = y: take x = 1 and y = 5, since 3 * 5 = 15 = 3. In Z_7, which is a domain because 7 is prime, cancellation is safe. This is exactly why solving linear congruences requires care about the modulus.

Proposition. Z_n is an integral domain if and only if n is prime.

Proof. If n = ab with 1 < a, b < n, then a and b are nonzero residues whose product is 0, so there are zero divisors. If n = p is prime and ab is congruent to 0 modulo p, then p divides ab, so by Euclid's lemma p divides a or b, meaning one of the residues is zero. This completes the proof.

Fields, and a theorem about finiteness

A field is a commutative ring in which 1 is not 0 and every nonzero element is a unit. Q, R, C and Z_p for prime p are fields; Z is not, because 2 has no reciprocal.

Proposition. Every field is an integral domain.

Proof. Suppose ab = 0 with a nonzero. Multiply on the left by a^(-1): the left side becomes (a^(-1)a)b = b and the right becomes a^(-1) 0 = 0. So b = 0, and there are no zero divisors. This completes the proof.

The converse fails: Z is a domain and not a field. But it fails only because Z is infinite.

Theorem. Every finite integral domain is a field.

Proof. Let D be a finite integral domain and let a be nonzero. Consider the map from D to D sending x to ax. It is injective, because ax = ay forces x = y by cancellation. An injective map from a finite set to itself is surjective, so some x satisfies ax = 1. Since D is commutative, x is a two-sided inverse of a. Hence every nonzero element is a unit. This completes the proof.

Worked example. Z_5 is a finite domain, so it must be a field, and indeed 2 * 3 = 6 = 1 and 4 * 4 = 16 = 1, giving reciprocals for every nonzero residue. The theorem also foreshadows Module 6: the finite fields are exactly the objects a finite domain can be, and there is one of each prime power order.

Characteristic

Write n . 1 for the sum of n copies of 1. The characteristic of a ring is the least positive n with n . 1 = 0, or 0 if no such n exists. So Z, Q and R have characteristic 0, while Z_n has characteristic n.

Proposition. The characteristic of an integral domain is 0 or a prime.

Proof. Suppose the characteristic is n with n = ab and 1 < a, b < n. Then 0 = n . 1 = (a . 1)(b . 1), using distributivity to expand the product of the two sums. In a domain one factor must vanish, so a . 1 = 0 or b . 1 = 0, contradicting the minimality of n. Hence n has no such factorisation and is prime. This completes the proof.

Key idea: in a field of characteristic p, adding any element to itself p times gives zero, and the map sending x to x^p turns out to be a ring homomorphism. That map, the Frobenius endomorphism, is why finite fields behave so uniformly.

Homomorphisms and subrings

A ring homomorphism from R to S is a map f with f(a + b) = f(a) + f(b), f(ab) = f(a)f(b), and f(1) = 1. The last condition is a genuine requirement, not a consequence: the map from Z to Z sending everything to 0 respects both operations and is not counted as a ring homomorphism here.

Two examples will carry a lot of weight later. Reduction modulo n is a ring homomorphism from Z onto Z_n. And for a fixed element c of a commutative ring R, evaluation at c is a ring homomorphism from R[x] to R sending a polynomial p(x) to p(c); that it respects multiplication is exactly the statement that you may substitute before or after multiplying and get the same answer. In Module 6 the kernel of an evaluation map will be the source of minimal polynomials.

A subring is a subset containing 1 that is closed under addition, negation and multiplication. Z is a subring of Q; Z[i] is a subring of C; the even integers are closed under everything except containing 1, which is why they fail our definition.

Common misconceptions

• "If ab = 0 then a = 0 or b = 0." True in a domain, false in a general ring. On the clock 3 * 4 = 0; among 2 by 2 matrices, the matrix with a single 1 in the top right multiplied by itself is zero.
• "A unit and a zero divisor are just opposites." In a finite commutative ring every nonzero element is one or the other, but in Z the element 3 is neither a unit nor a zero divisor. The dichotomy needs finiteness.
• "Every ring has division." Rings guarantee addition, subtraction and multiplication only. Fields are the rings where division by nonzero elements is available, and most rings are not fields.
• "Characteristic 0 means the ring is infinite in a boring way." Characteristic 0 means the copy of Z generated by 1 never wraps around, so the ring contains a copy of the integers. It says nothing about the rest of the ring, which may be enormous, as R shows.

Try it: units in a ring of integers with i

Exercise. Find all units of the Gaussian integers Z[i] = {a + bi : a, b integers}.

Solution. Define the norm N(a + bi) = a^2 + b^2, a non-negative integer. A direct computation with the multiplication rule shows N(zw) = N(z)N(w). If z is a unit with inverse w, then N(z)N(w) = N(1) = 1, and since both norms are non-negative integers, N(z) = 1. So a^2 + b^2 = 1 with a and b integers, which forces (a, b) to be one of (1, 0), (-1, 0), (0, 1), (0, -1). The units are therefore 1, -1, i, -i, a cyclic group of order 4.

Change one input. Consider instead Z[sqrt(2)] = {a + b sqrt(2)} with norm a^2 - 2b^2, which can be negative. Units now require |a^2 - 2b^2| = 1, and 1 + sqrt(2) qualifies since 1 - 2 = -1. Its powers are all distinct, so this ring has infinitely many units. Same construction, one sign changed in the norm, and a finite unit group becomes infinite.

Recap

A ring has an abelian group under addition, an associative multiplication with identity, and distributivity linking them. From those alone come 0a = 0 and the sign rules. Units form a group; zero divisors are the obstruction to cancellation; an integral domain is a commutative ring with neither 1 = 0 nor zero divisors, and cancellation is available there. Every field is a domain, every finite domain is a field, and the characteristic of a domain is 0 or prime.

Looking ahead

Normal subgroups were the kernels of group homomorphisms and the subgroups you could quotient by. The next lesson identifies the ring analogue, the ideal, proves that ideals are exactly the kernels of ring homomorphisms, builds quotient rings, and characterises when the quotient is a domain (prime ideals) and when it is a field (maximal ideals). That last characterisation is the engine of the entire field theory to come.

Sources

1. Weisstein, E. W. (n.d.). Ring. Wolfram MathWorld. mathworld.wolfram.com
2. Weisstein, E. W. (n.d.). Integral domain. Wolfram MathWorld. mathworld.wolfram.com
3. Judson, T. W. (n.d.). Chapter 16: Rings. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
4. Wikipedia contributors. (n.d.). Integral domain. Wikipedia. en.wikipedia.org
5. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Chapter 7. John Wiley and Sons.

Take a multiple of 6 and multiply it by any integer you like: 42 times 137 is still a multiple of 6. Nothing similar happens with, say, the odd numbers, or with the perfect squares. The multiples of 6 do not merely stay closed among themselves; they absorb whatever they are multiplied by. That absorption, not closure, is the property this lesson is built on.

In group theory the subgroups you could quotient by were the normal ones. In ring theory they are the ideals, and the reason is the same in outline and different in detail: coset multiplication has to survive a change of representative, and absorption is exactly what makes it survive.

Ideals

Let R be a commutative ring. A subset I is an ideal if it is a subgroup under addition and satisfies ra in I for every r in R and every a in I. Note that an ideal need not contain 1; in fact if it does, then r = r1 lies in I for every r and I is all of R. So a proper ideal is exactly a proper absorbing additive subgroup.

Three families of examples, each of which will be used later.

• Multiples. For any a in R, the set (a) = {ra : r in R} is an ideal, called the principal ideal generated by a. In Z this is aZ.
• Vanishing sets. In R[x], the polynomials with p(3) = 0 form an ideal: sums of such polynomials still vanish at 3, and multiplying by anything preserves the root.
• Kernels. If f is a ring homomorphism, the set of elements mapping to 0 is an ideal, since f(ra) = f(r)f(a) = f(r) 0 = 0.

Proposition. Every ideal of Z is principal.

Proof. Let I be an ideal. If I = {0} then I = (0). Otherwise I contains a nonzero element and hence, being closed under negation, a positive one; let d be the least positive member. Certainly (d) is contained in I by absorption. Conversely take a in I and divide: a = qd + r with 0 <= r < d. Then r = a - qd is a difference of two members of I, so it lies in I, and minimality of d forces r = 0. Hence a lies in (d). This completes the proof.

That is the same argument that showed every subgroup of a cyclic group is cyclic, and the same division algorithm drives it. A ring in which every ideal is principal and which is also a domain is called a principal ideal domain; Z is one, and so is F[x] for a field F, as the next lesson proves.

At the other extreme: a field F has exactly two ideals, {0} and F. For if an ideal contains a nonzero a, absorption gives a^(-1)a = 1 in the ideal, and then everything. That tiny observation will do real work below.

Quotient rings

Theorem. Let I be an ideal of a commutative ring R. The set R/I of additive cosets a + I, with operations

(a + I) + (b + I) = (a + b) + I and (a + I)(b + I) = ab + I,

is a commutative ring with zero 0 + I and identity 1 + I.

Proof. Addition is well defined because I is an additive subgroup of an abelian group, so R/I is already a quotient group. For multiplication, suppose a + I = a' + I and b + I = b' + I, so a' = a + i and b' = b + j with i, j in I. Then

a'b' = (a + i)(b + j) = ab + aj + ib + ij.

Each of aj, ib and ij lies in I by absorption, and I is closed under addition, so a'b' - ab lies in I and a'b' + I = ab + I. The ring axioms are then inherited from R one line at a time. This completes the proof.

Look at what absorption did. It was needed three times, once for each cross term, and no weaker condition would have covered aj, where a is an arbitrary ring element and j comes from I. A mere subring would leave that term outside.

Worked example. Z/6Z has six elements, the classes of 0 through 5, and multiplication of classes is multiplication of representatives reduced modulo 6. It is not a domain: the class of 2 times the class of 3 is the class of 0.

Worked example. Consider R[x]/(x^2 + 1). Every polynomial can be divided by x^2 + 1 to leave a remainder of degree at most 1, so every class has a unique representative a + bx. Multiplying two such and reducing uses x^2 = -1 in the quotient, giving (a + bx)(c + dx) = (ac - bd) + (ad + bc)x. That is the multiplication rule for complex numbers, and the quotient is a copy of C with x playing the role of i. The complex numbers are not a mysterious extension of the reals; they are a quotient ring.

Ideals are exactly the kernels

Theorem (first isomorphism theorem for rings). Let f be a ring homomorphism from R to S. Then ker(f) is an ideal of R, the image is a subring of S, and R/ker(f) is isomorphic to the image by the map sending a + ker(f) to f(a). Conversely every ideal is the kernel of some homomorphism.

Proof. That the kernel is an ideal was noted above, and that the image is a subring is immediate from the homomorphism properties. Write K = ker(f) and define F(a + K) = f(a). It is well defined and additive by the group version of the theorem. It is multiplicative: F((a+K)(b+K)) = F(ab + K) = f(ab) = f(a)f(b). It sends 1 + K to f(1) = 1. It is injective, since F(a + K) = 0 means a lies in K, and surjective onto the image by construction. For the converse, given an ideal I, the map from R onto R/I sending a to a + I is a ring homomorphism with kernel exactly I. This completes the proof.

Apply it twice. Evaluation at a real number c maps R[x] onto R, and its kernel is the set of polynomials vanishing at c, which is the principal ideal (x - c) by the factor theorem. So R[x]/(x - c) is isomorphic to R. Evaluation at i maps R[x] onto C, since a + bx hits a + bi; its kernel contains x^2 + 1, and if p(i) = 0 then dividing p by x^2 + 1 leaves a remainder ax + b with ai + b = 0, forcing a = b = 0. So the kernel is exactly (x^2 + 1) and R[x]/(x^2 + 1) is isomorphic to C, confirming the hand computation above.

Prime and maximal ideals

Two conditions on an ideal correspond exactly to two conditions on the quotient, and this correspondence is the reason the whole apparatus exists.

A proper ideal P is prime if ab in P implies a in P or b in P. A proper ideal M is maximal if no ideal sits strictly between M and R.

Theorem. For a commutative ring R with identity and a proper ideal P, the quotient R/P is an integral domain if and only if P is prime.

Proof. Suppose P is prime and (a + P)(b + P) = 0 + P. That says ab lies in P, hence a or b does, hence a + P or b + P is zero in the quotient. Since P is proper, 1 + P is not 0 + P. So R/P is a domain. Conversely if R/P is a domain and ab lies in P, then the product of the classes is zero, so one class is zero and the corresponding element lies in P. This completes the proof.

Theorem. For a commutative ring R with identity and a proper ideal M, the quotient R/M is a field if and only if M is maximal.

Proof. Suppose M is maximal and take a outside M, so that a + M is a nonzero element of the quotient. The set M + (a) = {m + ra} is an ideal containing M and containing a, so it strictly contains M and therefore equals R. In particular 1 = m + ra for some m in M and r in R. Reading that modulo M gives (r + M)(a + M) = 1 + M, so a + M is a unit. Hence every nonzero class is invertible and R/M is a field.

Conversely suppose R/M is a field, and let J be an ideal with M strictly inside J. Pick a in J but not in M. Then a + M is nonzero, so it has an inverse r + M, meaning ra - 1 lies in M, hence in J. Since ra lies in J by absorption, so does 1, and therefore J = R. So no ideal sits strictly between M and R. This completes the proof.

Corollary. Every maximal ideal is prime, because every field is an integral domain.

The converse fails, and Z shows how. The ideal (0) is prime, since Z is a domain, but it is not maximal because (0) sits inside (2) sits inside Z. For a nonzero ideal (n) of Z, the quotient is Z_n, which is a field exactly when n is prime and a domain under exactly the same condition. So in Z, apart from the zero ideal, prime and maximal coincide.

Why this matters: to build a field, find a maximal ideal and quotient by it. In Module 6 we will build a field containing a root of a chosen polynomial by taking F[x] and quotienting by the ideal generated by that polynomial, and the whole construction rests on the theorem just proved.

Common misconceptions

• "An ideal is just a subring." A subring must contain 1; an ideal usually does not, and must absorb multiplication by the whole ring. Z is a subring of Q but not an ideal of it, since (1/2)(3) leaves Z.
• "Prime ideals are the ideals generated by primes." That is true in Z, and it is where the name comes from, but the definition is about products landing inside the ideal. In Z[x] the ideal (x) is prime because the quotient is Z, a domain, and no prime number is involved.
• "Maximal means largest." It means no ideal strictly between it and the whole ring. A ring can have many maximal ideals, none of which contains the others: in Z every (p) is maximal.
• "The complex numbers must be constructed by declaring a square root of -1." They can be, but they also arise as a quotient ring R[x]/(x^2 + 1) with no declaration at all, which is why nothing about them is logically suspect.

Try it: an ideal that is prime and not maximal

Exercise. In Z[x], show that (x) is prime but not maximal.

Solution. Evaluation at 0 is a ring homomorphism from Z[x] onto Z, sending a polynomial to its constant term. Its kernel is the set of polynomials with zero constant term, which is exactly (x). By the first isomorphism theorem Z[x]/(x) is isomorphic to Z. Since Z is a domain but not a field, the ideal (x) is prime but not maximal.

Change one input. Replace the coefficients by a field: in Q[x], the quotient Q[x]/(x) is Q, a field, so (x) is maximal there. The same-looking ideal in two rings, prime in one and maximal in the other, and the difference is entirely in whether the coefficient ring was already a field. To make (x) maximal inside Z[x] you must enlarge it, for instance to (2, x), whose quotient is Z_2.

What to carry forward

An ideal is an additive subgroup that absorbs multiplication by the whole ring, and absorption is precisely what makes coset multiplication well defined. Ideals are exactly the kernels of ring homomorphisms, and the first isomorphism theorem transfers verbatim from groups. Every ideal of Z is principal, by the division algorithm. The quotient R/P is a domain exactly when P is prime, and R/M is a field exactly when M is maximal, so maximal ideals are the manufacturing plant for fields.

Looking ahead

Everything now depends on knowing which polynomials are irreducible, since those are the ones generating maximal ideals in F[x]. The next lesson develops the division algorithm for polynomials, proves that F[x] is a principal ideal domain with unique factorisation, and supplies the practical tests, the rational root theorem, Gauss's lemma and Eisenstein's criterion, that decide irreducibility for the polynomials you will actually meet.

Sources

1. Weisstein, E. W. (n.d.). Ideal. Wolfram MathWorld. mathworld.wolfram.com
2. Weisstein, E. W. (n.d.). Maximal ideal. Wolfram MathWorld. mathworld.wolfram.com
3. Wikipedia contributors. (n.d.). Ideal (ring theory). Wikipedia. en.wikipedia.org
4. Wikipedia contributors. (n.d.). Quotient ring. Wikipedia. en.wikipedia.org
5. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Section 7.4. John Wiley and Sons.

The polynomial x^4 + 1 is never zero for a real x, since x^4 is never negative. It is nevertheless not irreducible over the reals:

x^4 + 1 = (x^2 + sqrt(2) x + 1)(x^2 - sqrt(2) x + 1),

as you can check by multiplying out, since the middle terms cancel and (x^2 + 1)^2 - 2x^2 = x^4 + 1. So "has no roots" and "cannot be factored" are different statements, and confusing them is the single most common error in this part of the subject. Sorting out which polynomials really are unfactorable is this lesson's business, and the payoff is a machine for constructing fields.

Dividing polynomials

Throughout, F is a field and F[x] is the ring of polynomials with coefficients in F. Since F is a domain, degrees add: deg(fg) = deg(f) + deg(g) for nonzero f and g, because the product of the leading coefficients cannot vanish. In particular F[x] is itself an integral domain, and its units are exactly the nonzero constants.

Theorem (division algorithm). Let f and g be in F[x] with g nonzero. There exist unique q and r in F[x] with f = qg + r and either r = 0 or deg(r) < deg(g).

Proof. Existence, by induction on deg(f). If f = 0 or deg(f) < deg(g), take q = 0 and r = f. Otherwise write f = a x^n + ... and g = b x^m + ... with n at least m, and note b is invertible because F is a field. The polynomial f_1 = f - (a b^(-1)) x^(n-m) g has its degree-n term cancelled, so deg(f_1) < n, and by induction f_1 = q_1 g + r with r of the required kind. Then f = (q_1 + a b^(-1) x^(n-m)) g + r. For uniqueness, suppose q g + r = q' g + r'. Then (q - q')g = r' - r. If q and q' differ, the left side has degree at least deg(g), while the right side has degree less than deg(g) or is zero, which is impossible. So q = q' and then r = r'. This completes the proof.

Note where the field hypothesis entered: inverting the leading coefficient b. Over Z the algorithm fails, since you cannot divide x by 2x and stay in Z[x].

Corollary (factor theorem). For c in F, f(c) = 0 if and only if x - c divides f.

Proof. Divide f by x - c, getting f = q(x - c) + r with r constant. Evaluating at c gives f(c) = r. So f(c) = 0 exactly when r = 0, that is, exactly when x - c divides f. This completes the proof.

Corollary. A nonzero polynomial of degree n over a field has at most n roots in that field.

Proof. Induct on n. Degree 0 means a nonzero constant, with no roots. If f has degree n and a root c, write f = (x - c) q with deg(q) = n - 1. Any root d other than c gives 0 = (d - c) q(d) with d - c nonzero, so q(d) = 0 because F is a domain. By induction q has at most n - 1 roots, so f has at most n. This completes the proof.

The domain hypothesis is doing real work. In Z_8, which is not a domain, the polynomial x^2 - 1 has four roots: 1, 3, 5 and 7 all square to 1 modulo 8.

F[x] is a principal ideal domain

Theorem. Every ideal of F[x] is principal.

Proof. Let I be a nonzero ideal and let g be a nonzero element of least degree in I. Certainly (g) is inside I. For the reverse, take f in I and divide: f = qg + r. Then r = f - qg lies in I, and if r were nonzero its degree would be less than deg(g), contradicting minimality. So r = 0 and f lies in (g). This completes the proof.

This is the third time this argument has appeared: subgroups of cyclic groups, ideals of Z, ideals of F[x]. All three run on a division algorithm and the minimality of a chosen element, and recognising the pattern is worth more than memorising the three statements.

A polynomial p of degree at least 1 is irreducible over F if it cannot be written as a product of two polynomials in F[x] each of degree at least 1. Irreducibility depends on the field: x^2 + 1 is irreducible over R and factors as (x + i)(x - i) over C, and x^2 - 2 is irreducible over Q and factors over R.

Theorem. The ideal (p) is maximal in F[x] if and only if p is irreducible. Consequently F[x]/(p) is a field exactly when p is irreducible.

Proof. Suppose p is irreducible and (p) is contained in an ideal J. Since every ideal is principal, J = (g) for some g, and g divides p. Irreducibility leaves two cases: g is a nonzero constant, so J = F[x]; or g is a constant multiple of p, so J = (p). Hence (p) is maximal. Conversely if p = gh with both factors of positive degree, then (p) sits strictly inside (g), which sits strictly inside F[x], so (p) is not maximal. The final claim is the theorem from the previous lesson. This completes the proof.

Unique factorisation follows from the same machinery. Irreducible elements of F[x] generate maximal, hence prime, ideals, so an irreducible dividing a product must divide one factor; that is Euclid's lemma for polynomials. Induction on degree then gives every nonconstant polynomial as a product of irreducibles, unique up to order and up to constant factors.

Deciding irreducibility in practice

For low degrees there is a clean criterion. A polynomial of degree 2 or 3 over a field is irreducible if and only if it has no root in that field, because any factorisation must produce a linear factor. From degree 4 the criterion fails, and x^4 + 1 in the opening paragraph is the standard witness: no real roots, yet it splits into two real quadratics.

Over Q three tools cover most cases.

Rational root theorem. If f = a_n x^n + ... + a_0 has integer coefficients and p/q in lowest terms is a root, then p divides a_0 and q divides a_n. The proof is one line of clearing denominators: substituting and multiplying by q^n gives a_n p^n = -q(a_(n-1)p^(n-1) + ... + a_0 q^(n-1)), so q divides a_n p^n and hence a_n, since p and q are coprime; the symmetric rearrangement handles p.

Gauss's lemma. A polynomial with integer coefficients that factors over Q into two polynomials of positive degree also factors over Z into two polynomials of the same degrees. This is what lets you test irreducibility over the rationals using only integer arithmetic.

Eisenstein's criterion. Let f = a_n x^n + ... + a_0 have integer coefficients and suppose a prime p divides every a_i for i less than n, does not divide a_n, and p^2 does not divide a_0. Then f is irreducible over Q.

Proof of Eisenstein. By Gauss's lemma it is enough to rule out a factorisation f = gh over Z with both factors of positive degree. Reduce the equation modulo p. Every coefficient of f except the leading one dies, so the reduction of f is a_n x^n with a_n nonzero modulo p. In Z_p[x], which has unique factorisation, the only ways to factor a_n x^n are as constants times powers of x, so the reductions of g and h are constants times x^r and x^s with r + s = n and both r and s positive. Positive exponents mean the constant terms of both g and h reduce to zero, that is, both are divisible by p. But then their product a_0 is divisible by p^2, contradicting the hypothesis. This completes the proof.

Worked example 1. Is x^3 - 3x - 1 irreducible over Q? By the rational root theorem any rational root is an integer dividing 1, so it is 1 or -1. Substituting gives 1 - 3 - 1 = -3 and -1 + 3 - 1 = 1, neither zero. A cubic with no rational root is irreducible over Q, so yes.

Worked example 2. Is x^4 + 1 irreducible over Q? Eisenstein does not apply directly, since no prime divides 0 and 1 appropriately. Substitute x by x + 1, which cannot change irreducibility because it is an invertible change of variable:

(x + 1)^4 + 1 = x^4 + 4x^3 + 6x^2 + 4x + 2.

Now take p = 2: it divides 4, 6, 4 and 2, does not divide the leading 1, and 4 does not divide the constant 2. So the shifted polynomial is irreducible over Q, and therefore so is x^4 + 1. The same polynomial is reducible over R, as the opening showed, and reducible over every Z_p, which is a well-known curiosity.

Building a field with four elements

Take F = Z_2 and p(x) = x^2 + x + 1. Its only possible roots in Z_2 are 0 and 1, and p(0) = 1 and p(1) = 1 + 1 + 1 = 1, so it has no root and, being quadratic, is irreducible. Therefore K = Z_2[x]/(x^2 + x + 1) is a field.

How big is it? Every class has a unique representative of degree at most 1, namely 0, 1, a, or a + 1, where a denotes the class of x. So K has four elements. The multiplication rule comes from a^2 = a + 1, which is what x^2 + x + 1 = 0 says in characteristic 2.

Read the middle entry: a * a = a + 1. Read the corner: (a+1)(a+1) = a^2 + 2a + 1 = a^2 + 1 = (a + 1) + 1 = a, since 2a = 0. Every nonzero element has an inverse, visible as a 1 in its row. This is the field with four elements, and it is not Z_4, which is not a field at all. Bottom line: to build a field containing a root of an irreducible polynomial, quotient the polynomial ring by that polynomial and the root is the class of x.

Common misconceptions

• "No roots means irreducible." True only in degrees 2 and 3. x^4 + 1 has no real roots and factors over R; (x^2 + 1)^2 has no real roots and is obviously reducible.
• "Irreducibility is a property of the polynomial." It is a property of the polynomial together with the field. x^2 - 2 is irreducible over Q, reducible over R; x^2 + 1 is irreducible over R, reducible over C.
• "The field with four elements is Z_4." Z_4 has a zero divisor, since 2 * 2 = 0, so it is not a field. The field of order 4 is the quotient constructed above, and every finite field has prime power order.
• "Eisenstein fails, so the polynomial is reducible." Eisenstein is a sufficient condition only. When it fails directly, try a substitution such as x to x + 1, which preserves irreducibility and sometimes makes the criterion apply, as it did for x^4 + 1.

Try it: an irreducible cubic and the field it creates

Exercise. Show that x^3 - 2 is irreducible over Q, and describe the field Q[x]/(x^3 - 2).

Solution. Eisenstein at p = 2 applies at once: 2 divides the coefficients 0, 0 and -2, does not divide the leading 1, and 4 does not divide -2. So x^3 - 2 is irreducible over Q and the quotient is a field. Dividing any polynomial by x^3 - 2 leaves a remainder of degree at most 2, so every element of the quotient is uniquely a + b t + c t^2 with rational a, b, c, where t is the class of x and satisfies t^3 = 2. Concretely this field is a copy of Q with a cube root of 2 adjoined.

Change one input. Try x^3 - 8. Eisenstein at 2 now fails, because 4 does divide -8. And that is not an accident of the test: x^3 - 8 = (x - 2)(x^2 + 2x + 4), so the polynomial really is reducible. Two polynomials differing only in a constant, one generating a field of degree 3 over Q and the other splitting off a rational root.

Putting it together

Division with remainder works in F[x] because a field lets you invert a leading coefficient, and from it come the factor theorem, the bound of n roots for degree n, and the fact that every ideal is principal. Irreducible polynomials generate maximal ideals, so quotienting by one produces a field, and unique factorisation into irreducibles follows. Over Q, the rational root theorem handles small cases, Gauss's lemma reduces the question to integer coefficients, and Eisenstein's criterion settles many polynomials at a glance, sometimes only after a change of variable.

Looking ahead

The quotient Q[x]/(x^3 - 2) is a field containing Q, and as a vector space over Q it has dimension 3, with basis 1, t, t^2. That number 3 is called the degree of the extension, and the final module is organised around it: how degrees multiply in towers, what happens when you adjoin all the roots of a polynomial at once, and how the symmetries of the resulting field encode whether the polynomial can be solved by radicals.

Sources

1. Weisstein, E. W. (n.d.). Polynomial ring. Wolfram MathWorld. mathworld.wolfram.com
2. Weisstein, E. W. (n.d.). Eisenstein's irreducibility criterion. Wolfram MathWorld. mathworld.wolfram.com
3. Judson, T. W. (n.d.). Chapter 17: Polynomials. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
4. Wikipedia contributors. (n.d.). Eisenstein's criterion. Wikipedia. en.wikipedia.org
5. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Chapter 9. John Wiley and Sons.

Doubling a cube means constructing, from a segment of length 1, a segment of length equal to the cube root of 2. Greek geometers wanted it done with straightedge and compass alone, and for two thousand years nobody managed it or proved it impossible. The obstruction turned out to have nothing to do with geometry. It is a fact about a single number: the smallest polynomial with rational coefficients killing the cube root of 2 has degree 3, and 3 is not a power of 2.

To say that precisely we need a way to measure how much bigger one field is than another. The measurement is startlingly simple, and it is linear algebra.

An extension is a vector space

If F is a subfield of K, we call K an extension of F. Then K is a vector space over F: its elements can be added, and they can be scaled by elements of F, and the vector space axioms are just some of the field axioms read selectively. The degree [K : F] is the dimension of that vector space, finite or infinite.

The notation F(a) means the smallest subfield of the big field containing F and the element a. For Q(sqrt(2)) that is the set of numbers p + q sqrt(2) with rational p and q, and it is a field because 1/(p + q sqrt(2)) = (p - q sqrt(2))/(p^2 - 2q^2), with the denominator nonzero since sqrt(2) is irrational.

The tower law

Theorem. If F is inside K is inside L, with [K : F] and [L : K] both finite, then [L : F] = [L : K][K : F].

Proof. Let u_1, ..., u_m be a basis of K over F and v_1, ..., v_n a basis of L over K. We claim the mn products u_i v_j form a basis of L over F.

Spanning. Take x in L. Since the v_j span L over K, write x = c_1 v_1 + ... + c_n v_n with each c_j in K. Since the u_i span K over F, write c_j = d_(1j) u_1 + ... + d_(mj) u_m with each d_(ij) in F. Substituting gives x as an F-combination of the products u_i v_j.

Independence. Suppose a combination of the products with coefficients d_(ij) in F is zero. Group by j: the sum over j of (sum over i of d_(ij) u_i) v_j is zero, and each bracket lies in K. Independence of the v_j over K forces every bracket to vanish, and then independence of the u_i over F forces every d_(ij) to vanish.

So the products form a basis, and counting them gives mn. This completes the proof.

Notice the shape: this is Lagrange's theorem for fields. Degrees multiply along a tower exactly as indexes multiply along a chain of subgroups, and the consequence is the same. A degree in the middle of a tower must divide the degree at the top.

Worked example. Consider Q inside Q(sqrt(2)) inside Q(sqrt(2), sqrt(3)). The first step has degree 2. The second has degree 2 as well, since sqrt(3) is not in Q(sqrt(2)): if sqrt(3) = p + q sqrt(2) then squaring gives 3 = p^2 + 2q^2 + 2pq sqrt(2), forcing pq = 0 and then either sqrt(3) rational or sqrt(3/2) rational, both false. So the tower law gives [Q(sqrt(2), sqrt(3)) : Q] = 4, with basis 1, sqrt(2), sqrt(3), sqrt(6).

Minimal polynomials

An element a of an extension is algebraic over F if some nonzero polynomial in F[x] has a as a root, and transcendental otherwise. sqrt(2) and i are algebraic over Q; e and pi are transcendental, proved by Hermite in 1873 and Lindemann in 1882.

Theorem. Let a be algebraic over F. There is a unique monic irreducible polynomial m in F[x] with m(a) = 0, and it divides every polynomial in F[x] killing a. Moreover F(a) is isomorphic to F[x]/(m), and [F(a) : F] = deg(m), with basis 1, a, a^2, ..., a^(n-1) where n = deg(m).

Proof. Consider the evaluation homomorphism from F[x] sending p to p(a). Its kernel is an ideal, nonzero because a is algebraic, and every ideal of F[x] is principal, so the kernel is (m) for a unique monic m. That m divides every polynomial killing a is just membership in the kernel. The image of the map is F[a], the set of polynomial expressions in a, and by the first isomorphism theorem F[x]/(m) is isomorphic to F[a], which is a subring of a field and hence a domain. So (m) is prime, and in F[x] that makes m irreducible and (m) maximal, so the quotient is a field. A field containing F and a and contained in every such field is exactly F(a), so F[a] = F(a). Finally, remainders modulo m have degree less than n, and they are unique, so 1, a, ..., a^(n-1) is a basis. This completes the proof.

Worked example. Find the minimal polynomial of a = sqrt(2) + sqrt(3) over Q. Square: a^2 = 5 + 2 sqrt(6), so a^2 - 5 = 2 sqrt(6). Square again: (a^2 - 5)^2 = 24, that is, a^4 - 10a^2 + 1 = 0. So a is a root of x^4 - 10x^2 + 1, and [Q(a) : Q] divides 4. It cannot be 1, since a is irrational, and it cannot be 2, because a computation shows a^3 = 11 sqrt(2) + 9 sqrt(3), so sqrt(2) = (a^3 - 9a)/2 lies in Q(a), and then so does sqrt(3), giving Q(a) = Q(sqrt(2), sqrt(3)) of degree 4. Hence the quartic is the minimal polynomial and is irreducible over Q.

Proposition. Every finite extension is algebraic.

Proof. Let [K : F] = n and take a in K. The n + 1 elements 1, a, a^2, ..., a^n cannot be linearly independent over F in an n-dimensional space, so some nontrivial F-combination vanishes. That combination is a nonzero polynomial in F[x] with a as a root. This completes the proof.

The three classical impossibilities

A point is constructible from a given unit segment if it can be reached by a finite sequence of straightedge and compass steps. Each step intersects lines with lines, lines with circles, or circles with circles, and solving those systems requires at worst a square root of quantities already available. So every constructible number lies at the top of a tower

Q = F_0 inside F_1 inside ... inside F_k, with [F_(i+1) : F_i] = 2,

and by the tower law [F_k : Q] = 2^k. If a is constructible then Q(a) sits inside F_k, so [Q(a) : Q] divides 2^k and is therefore itself a power of 2.

Doubling the cube. The required length is cbrt(2), whose minimal polynomial over Q is x^3 - 2, irreducible by Eisenstein at 2. So [Q(cbrt(2)) : Q] = 3, which is not a power of 2. The construction is impossible.

Trisecting an angle. Take the specific angle of 60 degrees, which is itself constructible. Trisecting it means constructing cos(20 degrees). The triple angle identity cos(3t) = 4cos^3(t) - 3cos(t) with 3t = 60 degrees gives 1/2 = 4c^3 - 3c, that is, 8c^3 - 6c - 1 = 0. By the rational root theorem the only candidates are plus or minus 1, 1/2, 1/4, 1/8, and substituting each shows none is a root, so the cubic is irreducible over Q and [Q(c) : Q] = 3. Not a power of 2, so 60 degrees cannot be trisected. Note the shape of the claim: some angles can be trisected, for instance 90 degrees, and the theorem says only that no method works for every angle.

Squaring the circle. A square with the area of a unit circle has side sqrt(pi). Lindemann proved in 1882 that pi is transcendental, so it satisfies no polynomial over Q at all, and neither does sqrt(pi). A transcendental number has infinite degree over Q, so it cannot be constructible. Why this matters: three problems that resisted geometric attack for two millennia are settled by asking one arithmetic question about a degree.

Common misconceptions

• "Degree counts the number of new elements." It counts the dimension as a vector space. Q(sqrt(2)) has degree 2 over Q and contains infinitely many numbers not in Q.
• "Every extension is algebraic." R over Q is not: pi satisfies no rational polynomial. What is true is the converse direction, that every finite extension is algebraic.
• "Any number whose degree is a power of 2 is constructible." The degree condition is necessary, not sufficient. There are quartic numbers that are not constructible, because the tower of quadratic steps must actually exist, not merely be permitted by the arithmetic.
• "No angle can be trisected." Many can, including 90 degrees, whose third is 30 degrees and clearly constructible. What fails is a general method, and 60 degrees is the standard witness.

Try it: a degree six extension

Exercise. Compute [Q(sqrt(2), cbrt(5)) : Q].

Solution. Build the tower Q inside Q(sqrt(2)) inside Q(sqrt(2), cbrt(5)). The first step has degree 2, since x^2 - 2 is irreducible over Q. For the second step, cbrt(5) is a root of x^3 - 5, irreducible over Q by Eisenstein at 5, so its degree over Q is 3. Now use divisibility: [Q(sqrt(2), cbrt(5)) : Q] is a multiple of both 2 and 3, hence of 6, and by the tower law it is at most 2 * 3 = 6. So it is exactly 6, and the second step has degree 3.

Change one input. Replace cbrt(5) by sqrt(8). Now sqrt(8) = 2 sqrt(2) already lies in Q(sqrt(2)), the second step has degree 1, and the whole extension has degree 2. Adjoining an element sometimes adds nothing, and the degree is what tells you.

What to remember

An extension is a vector space over the smaller field, and its dimension is the degree. Degrees multiply along towers, which makes middle degrees divide top degrees, exactly as Lagrange's theorem does for groups. An algebraic element has a unique monic irreducible minimal polynomial, F(a) is F[x] modulo it, and the degree of the extension is the degree of that polynomial. Constructible numbers have degree a power of 2, which kills doubling the cube, trisecting 60 degrees, and squaring the circle.

Looking ahead

Adjoining one root of a polynomial is not always enough. Q(cbrt(2)) contains one cube root of 2 but not the two complex ones, so x^3 - 2 does not factor completely there. The next lesson builds the smallest field in which a given polynomial factors into linear pieces, its splitting field, and then studies the group of symmetries of that field. The correspondence between subgroups of that group and intermediate fields is the theorem the whole course has been heading toward.

Sources

1. Weisstein, E. W. (n.d.). Extension field. Wolfram MathWorld. mathworld.wolfram.com
2. Wikipedia contributors. (n.d.). Degree of a field extension. Wikipedia. en.wikipedia.org
3. Wikipedia contributors. (n.d.). Doubling the cube. Wikipedia. en.wikipedia.org
4. Wikipedia contributors. (n.d.). Angle trisection. Wikipedia. en.wikipedia.org
5. Judson, T. W. (n.d.). Chapter 21: Fields. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org

The field Q(cbrt(2)) contains a root of x^3 - 2. It does not contain the other two. Those are w * cbrt(2) and w^2 * cbrt(2) where w = (-1 + i sqrt(3))/2 is a primitive cube root of 1, and both are non-real, while Q(cbrt(2)) sits entirely inside the real numbers. So adjoining one root did not finish the job.

To factor x^3 - 2 completely you need Q(cbrt(2), w), which by the tower law has degree 3 * 2 = 6 over Q, since w satisfies the irreducible quadratic x^2 + x + 1 and is not real. That field is the natural home of the polynomial, and this lesson is about what such fields know that the polynomial does not.

Splitting fields

A polynomial f in F[x] splits in an extension K if it factors there into linear pieces. A splitting field of f over F is an extension in which f splits and which is generated over F by the roots, so nothing unnecessary has been added.

Theorem. Every nonconstant f in F[x] has a splitting field, of degree at most n factorial over F, where n = deg(f).

Proof. Induct on n. If n = 1 then f already splits over F. Otherwise pick an irreducible factor p of f and form F_1 = F[x]/(p). By the previous lesson this is a field, it contains a copy of F (the constants), and the class of x is a root of p, hence of f, with [F_1 : F] = deg(p), at most n. Over F_1 the factor theorem lets us write f = (x - a) g with deg(g) = n - 1. By induction g has a splitting field K over F_1 of degree at most (n-1) factorial, and in K the polynomial f splits. Discard anything not generated by the roots. The tower law bounds the degree by n times (n-1) factorial. This completes the proof.

This construction is due to Kronecker, and it is worth pausing on how little it assumes: no complex numbers, no analysis, no appeal to the fundamental theorem of algebra. Roots are manufactured out of the polynomial ring itself. One can also show that any two splitting fields of the same polynomial over the same base are isomorphic by an isomorphism fixing the base, so speaking of "the" splitting field is legitimate.

The symmetries of a field

An automorphism of K fixing F is a ring isomorphism from K to itself that leaves every element of F alone. These form a group under composition, written Gal(K/F) and called the Galois group of the extension.

Two constraints make these groups computable. First, an automorphism is determined by where it sends the generators of K over F, because everything else is built from them by field operations, which the map respects. Second, if a has minimal polynomial m over F, then 0 = s(m(a)) = m(s(a)) for any automorphism s fixing F, since the coefficients of m are fixed. So an automorphism permutes the roots of each minimal polynomial. The point: a Galois group is a group of permutations of roots, subject to whatever algebraic relations those roots satisfy.

Worked example. Gal(Q(sqrt(2))/Q). An automorphism must send sqrt(2) to a root of x^2 - 2, so to sqrt(2) or -sqrt(2). Both choices give genuine automorphisms, and the group has order 2, isomorphic to Z_2. Note that |Gal| = 2 = [Q(sqrt(2)) : Q].

Worked example. Gal(Q(sqrt(2), sqrt(3))/Q). Here sqrt(2) may go to plus or minus sqrt(2) and independently sqrt(3) to plus or minus sqrt(3), giving four automorphisms. Write s for the one negating sqrt(2) only and t for the one negating sqrt(3) only. Each has order 2 and they commute, so the group is the Klein four-group, of order 4, again matching the degree.

A cautionary example. Gal(Q(cbrt(2))/Q) is trivial. An automorphism must send cbrt(2) to a root of x^3 - 2 lying inside Q(cbrt(2)), and the only such root is cbrt(2) itself, since the others are not real. So the group has order 1 while the degree is 3. Here the group has lost track of the extension.

An extension is called Galois when |Gal(K/F)| = [K : F]. For fields of characteristic 0, which is where we work, this happens exactly when K is the splitting field of some polynomial over F. The failed example above is exactly a case where K is not a splitting field: Q(cbrt(2)) holds one root of x^3 - 2 and misses two.

Fixed fields and the correspondence

Given a subgroup H of Gal(K/F), its fixed field is

K^H = {x in K : s(x) = x for every s in H}.

It is a field, since the set of elements fixed by a collection of ring isomorphisms is closed under all four operations, and it contains F by definition of the Galois group.

Fundamental theorem of Galois theory. Let K over F be a finite Galois extension with group G = Gal(K/F). Then:

• The maps H to K^H and L to Gal(K/L) are mutually inverse bijections between the subgroups of G and the fields between F and K.
• Both maps reverse inclusion: a bigger subgroup fixes a smaller field.
• Degrees match indexes: [K : K^H] = |H| and [K^H : F] = [G : H].
• K^H is itself a Galois extension of F exactly when H is a normal subgroup of G, and in that case Gal(K^H / F) is isomorphic to G/H.

The full proof runs several pages, resting on Artin's lemma that [K : K^H] = |H| for any finite group H of automorphisms, and it is carried out completely in Dummit and Foote's Chapter 14 and in Judson's Chapter 23. Rather than gesture at it, we will verify every clause on two examples where each object can be written down.

First verification: two square roots

Take K = Q(sqrt(2), sqrt(3)) over F = Q, with G the Klein four-group {1, s, t, st} as above. The subgroups of G are the trivial one, three of order 2, and G itself: five in all. So the theorem predicts exactly five intermediate fields, with degrees over Q equal to the indexes 4, 2, 2, 2, 1.

Check the third row from the bottom. The automorphism st negates both square roots, so it sends sqrt(6) = sqrt(2) sqrt(3) to (-sqrt(2))(-sqrt(3)) = sqrt(6), fixing it. Inclusion reversal is visible: the smallest subgroup fixes the largest field. Since G is abelian, every subgroup is normal, and correspondingly every intermediate field is a Galois extension of Q, which you can confirm directly: each is the splitting field of a quadratic.

Second verification: the cubic

Take K = Q(cbrt(2), w), the splitting field of x^3 - 2, with [K : Q] = 6. Every automorphism permutes the three roots r_1 = cbrt(2), r_2 = w cbrt(2), r_3 = w^2 cbrt(2), and an automorphism is determined by that permutation because the roots generate K. Since the group has order 6 and embeds in S_3, which has order 6, the Galois group is all of S_3.

Every degree in the fourth column is the index of the corresponding subgroup, as promised. The last column is where the theorem earns its keep. The subgroup of three-cycles is normal, and correspondingly Q(w) is a Galois extension of Q: it is the splitting field of x^2 + x + 1, and the quotient S_3 / A_3, of order 2, is its Galois group. The order-2 subgroups are not normal, and correspondingly Q(cbrt(2)) is not Galois over Q, which is precisely the failure we noticed at the start: it contains one root of x^3 - 2 and misses two.

Remember: normal subgroup on the group side means splitting field on the field side, and that dictionary is what converts a question about equations into a question about groups.

Common misconceptions

• "The Galois group is the group of all permutations of the roots." It is the subgroup of permutations that respect every algebraic relation the roots satisfy over F. For x^4 - 10x^2 + 1, whose roots are the four numbers plus or minus sqrt(2) plus or minus sqrt(3), the group has order 4, not 24.
• "Adjoining a root gives the splitting field." Only when the polynomial has degree 2, or when the extension happens to be normal. Q(cbrt(2)) is the standard counterexample.
• "Every finite extension is Galois." Galois means the automorphism group is as large as the degree. Q(cbrt(2)) over Q has degree 3 and a trivial group, so it is not Galois.
• "The correspondence preserves inclusion." It reverses it. Larger subgroups impose more conditions and therefore fix smaller fields.

Try it: a quartic splitting field

Exercise. Find the splitting field of x^4 - 2 over Q, its degree, and the order of its Galois group.

Solution. The four roots are 2^(1/4), -2^(1/4), i * 2^(1/4) and -i * 2^(1/4), so the splitting field is Q(2^(1/4), i). Build a tower: [Q(2^(1/4)) : Q] = 4, since x^4 - 2 is irreducible by Eisenstein at 2; then adjoining i has degree 2, because Q(2^(1/4)) is inside the real numbers and so does not already contain i. The tower law gives degree 8, and since the extension is a splitting field it is Galois, so the group has order 8. It turns out to be D4, the symmetry group of a square, acting on the four roots as they sit at the corners of a square in the complex plane.

Change one input. Replace x^4 - 2 by x^4 - 1. Its roots are the four fourth roots of unity, and they already lie in Q(i), so the splitting field has degree 2 and the Galois group has order 2. Same degree of polynomial, a factor of four difference in the size of the group.

Where this leaves us

Every polynomial has a splitting field, built by quotienting a polynomial ring by an irreducible factor and iterating, and it is unique up to isomorphism. The automorphisms of a splitting field fixing the base form the Galois group, which acts faithfully on the roots. When the extension is Galois, the fundamental theorem sets up an inclusion-reversing bijection between subgroups and intermediate fields, matching orders to degrees and normal subgroups to Galois subextensions. Two examples, the biquadratic and the cubic, exhibit every clause explicitly.

Looking ahead

Solving an equation by radicals means building its roots by a sequence of root extractions, and each extraction adds a layer to a tower of field extensions. Translated through the correspondence, such a tower becomes a chain of subgroups with abelian quotients: a solvable group. The last lesson makes that precise, shows the Galois group of the general quintic is S_5, and uses the simplicity of A_5 from Module 4 to prove that S_5 is not solvable, and hence that no radical formula can exist.

Sources

1. Weisstein, E. W. (n.d.). Splitting field. Wolfram MathWorld. mathworld.wolfram.com
2. Wikipedia contributors. (n.d.). Fundamental theorem of Galois theory. Wikipedia. en.wikipedia.org
3. Judson, T. W. (n.d.). Chapter 23: Galois theory. Abstract Algebra: Theory and Applications, Mathematics LibreTexts. math.libretexts.org
4. Wikipedia contributors. (n.d.). Splitting field. Wikipedia. en.wikipedia.org
5. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Chapter 14. John Wiley and Sons.

Gerolamo Cardano's Ars Magna appeared in 1545. It contains a formula for the roots of any cubic, learned from Tartaglia and originally found by Scipione del Ferro, and a method for the quartic due to Cardano's student Lodovico Ferrari. Both express the roots using only the coefficients, the four arithmetic operations, and extraction of roots.

Nothing comparable was ever added for degree 5. Not because nobody tried: Euler, Lagrange, Bezout and Vandermonde all worked on it during the eighteenth century, and Lagrange's 1771 memoir, the one that opened Module 1, was an attempt to find the pattern behind the successes and extend it. The pattern turned out to be a group, and the group at degree 5 is the wrong shape. This lesson explains exactly what that means.

What a formula would have to be

Say that a polynomial f over a field F of characteristic 0 is solvable by radicals if its roots lie in some field reached from F by a chain

F = F_0 inside F_1 inside ... inside F_r, where each F_(i+1) = F_i(a_i) with some power a_i^(n_i) lying in F_i.

Each step adjoins one root of one element already available. That is precisely what a formula built from the coefficients using arithmetic and radicals does: it takes a square root here, a cube root there, and combines. The cubic formula, for instance, adjoins a square root and then a cube root, so the corresponding tower has two steps.

Solvable groups

A group G is solvable if there is a chain

G = G_0 containing G_1 containing ... containing G_r = {e},

with each G_(i+1) normal in G_i and each quotient G_i / G_(i+1) abelian. The name is not a coincidence, and the theorem tying it to equations is stated below.

Every abelian group is solvable, with the one-step chain G containing {e}. Every group of prime power order is solvable, because its centre is nontrivial by the class equation and one can induct. And the small symmetric groups are solvable:

• S_3 containing A_3 containing {e}, with quotients of orders 2 and 3, both cyclic and so abelian.
• S_4 containing A_4 containing V containing {e}, where V is the Klein four-group of double transpositions from Module 4. The quotients have orders 2, 3 and 4, and the last one is V itself, which is abelian.

Those two chains are the group-theoretic shadow of the Cardano and Ferrari formulas. The three steps of the S_4 chain correspond to the resolvent cubic and the two square roots that Ferrari's method extracts.

S5 is not solvable

The proof rests on one computation with three-cycles, and it is worth doing by hand once.

Lemma. Let n be at least 5, and let H and K be subgroups of S_n with K normal in H and H/K abelian. If H contains every three-cycle, so does K.

Proof. Since H/K is abelian, every commutator x y x^(-1) y^(-1) with x, y in H lies in K: the classes commute, so xyK = yxK, which says (yx)^(-1)(xy) is in K.

Now compute one such commutator. In S_5, take x = (1 2 3) and y = (3 4 5), and evaluate x y x^(-1) y^(-1) by applying the factors right to left. Symbol 1: y^(-1) = (3 5 4) fixes it, x^(-1) = (1 3 2) sends it to 3, y sends 3 to 4, and x fixes 4, so 1 goes to 4. Symbol 4: to 3, then to 2, fixed, then to 3, so 4 goes to 3. Symbol 3: to 5, fixed, to 3, then to 1, so 3 goes to 1. Symbols 2 and 5 return to themselves. The commutator is (1 4 3), a three-cycle.

Given any three-cycle (r s t) and any two further symbols u and v, relabel the computation by matching 1, 4, 3, 2, 5 to r, s, t, u, v. Then x = (r u t) and y = (t s v) are three-cycles, both in H by hypothesis, and their commutator is exactly (r s t). Since commutators lie in K, every three-cycle lies in K. This completes the proof.

Theorem. S_n is not solvable for n at least 5.

Proof. Suppose S_n = G_0 containing G_1 containing ... containing G_r = {e} is a solvable chain. The group G_0 = S_n contains every three-cycle. By the lemma applied to the pair G_0 and G_1, so does G_1. Applying the lemma repeatedly along the chain, every G_i contains every three-cycle, including G_r = {e}. But the identity group contains no three-cycle, and three-cycles exist because n is at least 3. This contradiction shows no such chain exists. This completes the proof.

Compare with S_4, where the same lemma would fail at the last step: V contains no three-cycle, and the lemma does not apply because n = 4 leaves no two spare symbols for the relabelling. Why this matters: the entire difference between the quartic and the quintic is the availability of two extra symbols.

Galois's criterion

Theorem (Galois). Let F have characteristic 0 and let f be in F[x]. Then f is solvable by radicals if and only if the Galois group of the splitting field of f over F is a solvable group.

The correspondence of the previous lesson explains the shape of this. Adjoining an n-th root, once the relevant roots of unity are present, produces a Galois extension with cyclic group. Stacking such steps produces a tower whose corresponding chain of subgroups has abelian quotients, which is exactly the definition of solvable. The converse direction runs the argument backwards, building a radical tower out of a solvable chain, and its technical heart is the theory of cyclic extensions. Judson's Chapter 23 and Dummit and Foote's Chapter 14 both carry the proof in full.

Combining the criterion with the previous theorem gives the classical result. If the Galois group of a quintic is S_5, the quintic is not solvable by radicals. And the general quintic, with independent transcendental coefficients, has Galois group S_5. So no formula in radicals can express the roots of a general quintic in terms of its coefficients.

A specific quintic with no radical solution

The general statement is about a generic polynomial. Here is a concrete one with integer coefficients.

Claim. f(x) = x^5 - 6x + 3 has Galois group S_5 over Q, so its roots cannot be written using radicals.

Argument. First, f is irreducible over Q by Eisenstein at the prime 3: it divides the coefficients 0, 0, 0 and -6 and the constant 3, misses the leading 1, and 9 does not divide 3.

Second, count real roots. The derivative f'(x) = 5x^4 - 6 vanishes at x around plus or minus 1.047. At the negative critical point f is about 8.0, and at the positive one it is about -2.0, so the graph rises to a positive local maximum, falls to a negative local minimum, and rises again. With the two ends going to minus and plus infinity, that gives exactly three real roots and hence exactly two non-real ones, a complex conjugate pair.

Third, identify the group G, viewed as a subgroup of S_5 permuting the five roots. Irreducibility makes the action transitive, so 5 divides |G| by orbit-stabiliser, and Cauchy's theorem supplies an element of order 5, which in S_5 must be a five-cycle. Complex conjugation restricts to an automorphism of the splitting field: it fixes the three real roots and swaps the conjugate pair, so it acts as a transposition. A subgroup of S_5 containing a five-cycle and a transposition is all of S_5, because conjugating the transposition by powers of the five-cycle produces a set of transpositions that generate. Hence G = S_5, which is not solvable, and by Galois's criterion f is not solvable by radicals.

Note what is and is not being claimed. The roots exist; three of them are real numbers you can locate to any accuracy you like, and Newton's method will produce them in a few iterations. What does not exist is a finite expression for them built from 5, 6, 3 and root extractions.

The people who settled it

Paolo Ruffini published a book in 1799 arguing that the general equation of degree greater than four cannot be solved algebraically. His argument had a gap and was largely ignored. Niels Henrik Abel proved the impossibility in 1824, publishing at his own expense as a six-page pamphlet compressed to save printing costs. Both results say that no formula exists; neither explains which particular equations do have radical solutions.

Évariste Galois supplied the explanation between 1829 and 1832, in memoirs that were repeatedly lost or rejected by the Paris Academy, and he was dead at 20 after a duel in May 1832. What he added was the group: a criterion that decides, for any given polynomial, whether radicals suffice. The reason his name is on the whole subject rather than Abel's is that his answer is a mechanism, not a verdict.

Common misconceptions

• "Quintic equations cannot be solved." Every quintic with complex coefficients has five complex roots, by the fundamental theorem of algebra, and numerical methods find them quickly. What fails is a general formula in radicals.
• "No quintic can be solved by radicals." Many can. x^5 - 2 has roots the fifth roots of 2 times fifth roots of unity, all expressible in radicals, and its Galois group has order 20, which is solvable. The criterion is about the group, not the degree.
• "Abel and Galois proved the same thing." Abel proved the general quintic is not solvable by radicals. Galois gave a necessary and sufficient condition covering every polynomial, which also explains why degrees up to 4 succeed.
• "Radicals are the only closed forms." The quintic can be solved using more general functions: Hermite in 1858 solved it with elliptic modular functions, and the Bring radical gives a normal form. The theorem restricts the toolkit to arithmetic and root extraction, and nothing more.

Try it: a solvable quintic

Exercise. Explain why x^5 - 2 is solvable by radicals, and identify the order of its Galois group over Q.

Solution. The roots are z, zc, zc^2, zc^3, zc^4, where z is the real fifth root of 2 and c is a primitive fifth root of unity. The splitting field is Q(z, c). Now [Q(z) : Q] = 5 by Eisenstein at 2, and [Q(c) : Q] = 4, since c satisfies the irreducible x^4 + x^3 + x^2 + x + 1. Both degrees divide the total, so the degree is a multiple of 20, and it is at most 5 * 4 = 20, so it is exactly 20 and the Galois group has order 20. That group has a normal cyclic subgroup of order 5 with cyclic quotient of order 4, so it is solvable, and the criterion confirms what the explicit radical expressions already showed.

Change one input. Compare with x^5 - 6x + 3, whose group is S_5 of order 120. Two quintics, both irreducible over Q, and the only difference that matters is whether the group of the splitting field admits a chain with abelian quotients.

Looking back

Solvability by radicals means the roots lie at the top of a tower of root extractions. Solvability of a group means a chain of subgroups with abelian quotients. Galois's theorem says the two conditions match. S_3 and S_4 are solvable, which is why Cardano and Ferrari succeeded; S_n for n at least 5 is not, because a commutator of two three-cycles is a three-cycle whenever two spare symbols exist, so no chain can ever shed them all. A specific quintic, x^5 - 6x + 3, has group S_5, and so has roots that no radical formula can name.

Where the course has taken you

Seventeen lessons ago the subject was a triangle of card on a table and four axioms. Those axioms produced Lagrange's theorem, quotients, the isomorphism theorems, actions, the class equation and Sylow. A second set of axioms produced rings, ideals and the factorisation theory of polynomials. Field degrees closed three problems from antiquity, and the correspondence between subgroups and subfields converted a question about formulas into a question about the shape of a group. That is what abstraction bought: one argument about three-cycles, made once, settles a question that resisted direct attack for two hundred and eighty-seven years after Cardano.

Sources

1. Weisstein, E. W. (n.d.). Abel's impossibility theorem. Wolfram MathWorld. mathworld.wolfram.com
2. Wikipedia contributors. (n.d.). Abel-Ruffini theorem. Wikipedia. en.wikipedia.org
3. O'Connor, J. J., & Robertson, E. F. (2000). Niels Henrik Abel. MacTutor History of Mathematics Archive. mathshistory.st-andrews.ac.uk
4. O'Connor, J. J., & Robertson, E. F. (1996). Evariste Galois. MacTutor History of Mathematics Archive. mathshistory.st-andrews.ac.uk
5. Wikipedia contributors. (n.d.). Solvable group. Wikipedia. en.wikipedia.org
6. Dummit, D. S., & Foote, R. M. (2004). Abstract algebra (3rd ed.), Sections 14.7 and 3.4. John Wiley and Sons.