11.4.1 Conjugacy Classes in
By Theorem 3.4.6, two permutations are conjugate in if and only if they have the same cycle type; the key computation is So the conjugacy classes of correspond to the partitions of , and the class equation of is obtained by counting the permutations of each cycle type.
Example 11.4.1 (Conjugacy classes in and ). For , the partitions , and give the identity, the 3 transpositions and the 2 three-cycles: .
For , the five partitions give the identity (1 element), the transpositions (), the products of two disjoint transpositions (3), the 3-cycles () and the 4-cycles (). The class equation of is .
Example 11.4.2 (Normal subgroups of ). A normal subgroup of is a union of conjugacy classes containing , and its order divides 24. Adding 1 to sums of distinct class sizes among , the only divisors of 24 we obtain besides 1 and 24 are and . Both unions are subgroups. The first is , the Klein four-group: each non-identity element has order 2, and the product of two distinct ones is the third (for instance ), so . The second consists of the identity, the elements and the 3-cycles, which are exactly the even permutations; it is . So the normal subgroups of are , , and .
11.4.2 Conjugacy Classes in
Recall the sign homomorphism (Definition 3.4.10) and the alternating group (Definition 3.4.15), a normal subgroup of index 2 in for . Transpositions are odd, and a -cycle is a product of transpositions, , so it has sign . Hence if and only if its cycle decomposition has an even number of cycles of even length.
A conjugacy class of contained in is a union of conjugacy classes of , but it need not be a single class: conjugating by elements of only, we may reach fewer elements.
Proposition 11.4.3 (Splitting of conjugacy classes in ). Let be a conjugacy class of , . Then either or . In the second case, let . Then exactly one of the following holds.
- Some odd permutation commutes with , that is, ; then is a single conjugacy class of .
- ; then splits into two conjugacy classes of , each with elements.
Proof. Since is normal, it contains the -conjugacy class of each of its elements; so if meets , then . Suppose , and fix an odd permutation . Then , so where is the -conjugacy class of and is the -conjugacy class of . Two conjugacy classes of are equal or disjoint, and if and only if , that is, for some with . The elements are exactly the odd permutations, so if and only if some odd permutation commutes with . In case (b), conjugation by maps into (it sends to , and ), and conjugation by maps into in the same way. These maps are inverse to each other, so . ∎
Proposition 11.4.4 (When a class splits). Let , and count the fixed points of as cycles of length 1. The -conjugacy class of splits into two conjugacy classes of if and only if the cycle lengths of are all odd and pairwise distinct. (In particular, if it splits, has at most one fixed point.)
Proof. Suppose first that has a cycle of even length. Then commutes with (it commutes with itself and with the other cycles of , which are disjoint from it), and is odd. Suppose next that has two cycles and of the same odd length ; for these are two fixed points and . Then is a product of disjoint transpositions, hence odd, and by the key computation : conjugation by exchanges the two cycles and fixes the other cycles of , whose entries does not move. In both cases , and the class does not split by Proposition 11.4.3.
Conversely, suppose the cycle lengths of are odd and pairwise distinct, let be the supports of the cycles of (including the 1-cycles), and let commute with . By the key computation, the cycles of are obtained by applying to the entries of the cycles of ; by the uniqueness of the cycle decomposition (Theorem 3.4.1), maps each onto the support of a cycle of of the same length, which is itself because the lengths are distinct. Write . Then , so for each ; if , it follows that for all . So agrees with on for some , and . Each is a cycle of odd length, hence even, so is even. Thus , and the class splits by Proposition 11.4.3. ∎
Example 11.4.5 (Conjugacy classes of and ). In , the three elements form one class (they have cycles of even length), while the eight 3-cycles, with cycle lengths , split into two classes of 4; for instance and are not conjugate in . The class equation of is . The Klein four-group is normal in , hence in ; so is not simple.
In , the 15 elements form one class (for instance, the odd permutation commutes with ), and so do the 20 three-cycles, with cycle lengths (for instance, commutes with ). The 24 five-cycles split into two classes of 12. The class equation of is
Corollary 11.4.6 (3-cycles are conjugate in ). For , the 3-cycles form a single conjugacy class of .
Proof. The 3-cycles form a conjugacy class of , contained in . A 3-cycle has fixed points, that is, two cycles of length 1, so by Proposition 11.4.4 this class does not split. ∎
For the corollary fails, as the example shows; this is where the hypothesis of the next theorem enters.
11.4.3 Simplicity of
Recall that a group is simple if its only normal subgroups are and the group itself.
Theorem 11.4.7 (Simplicity of ). The alternating group is simple.
Proof. A normal subgroup of is a union of conjugacy classes of containing , and divides 60. By Example 11.4.5, , where is a sum of distinct terms of the list . The possible values of are , and of these only 1 and 60 divide 60. So or . ∎
For larger we argue with 3-cycles.
Lemma 11.4.8 ( is generated by 3-cycles). For , every element of is a product of 3-cycles.
Proof. By induction on . For , and the claim is clear. Let , assume the claim for , and let . If , then is an even permutation of , so it is a product of 3-cycles by induction. Otherwise, let and let be any element distinct from and . The 3-cycle sends to , so satisfies , and is a product of 3-cycles by the first case. Hence so is . ∎
Theorem 11.4.9 (Simplicity of ). For every , the alternating group is simple.
Proof. Let be a normal subgroup of with . If contains a 3-cycle, it contains all 3-cycles, since they form one conjugacy class of (Corollary 11.4.6); then by Lemma 11.4.8. So it suffices to find a 3-cycle in . Throughout, we use that for and , the element lies in , because . Products are composed from right to left.
Choose , . If has order and is a prime dividing , then has order ; so we may assume that has prime order . Its disjoint cycles commute, so ; the move disjoint sets of points, so each . A -cycle has order , so every cycle of of length greater than 1 has length : is a product of disjoint -cycles.
Case 1: . Write , where is the product of the other -cycles of and fixes . Let , so . Applying , , , in turn, Any other point is fixed: neither nor lies in , so and do not move them. Hence
Case 2: . If is a 3-cycle we are done. Otherwise , with fixing . With , the same kind of computation gives , an element of of order 5, and Case 1 applies to it.
Case 3: , and is a product of two transpositions, . Since , there is a point . With , we get , of order 5, and Case 1 applies.
Case 4: , and is a product of more than two transpositions. Since is even, it has at least four, and with fixing . With , we get , an element of of order 3, and Case 2 applies. ∎
Corollary 11.4.10 (Normal subgroups of ). For , the only normal subgroups of are , and .
Proof. Let . Then is normal in , hence in , so is or by Theorem 11.4.9. If , then or , since has index 2. If , then contains at most one odd permutation (if are odd, then , so ), and . Suppose with . Since is normal, for every . But has a cycle of length at least 2, and as there is a point ; for , , a contradiction. So . ∎