Recall that in the Symmetric group , the notion of "Conjugacy class" coincides with that of "has the same cycle type" (proof). It turns out to be the case that when we descend to the Alternating group , the conjugacy classes nearly all remain the same; the exception is those classes in which have cycle type all odd and all distinct. Those classes split into exactly two classes in .
Proof
Conjugate implies same cycle type
This direction of the proof is identical to the same direction in the proof of the corresponding result on symmetric groups.
Converse: splitting condition
Recall before we start that an even-length cycle can only be written as the product of an odd number of transpositions, and vice versa.
The question is: is every with the same cycle type as conjugate (in to ? If so, then the conjugacy class in of is just the same as that of in ; if not, then the conjugacy class in must break into two pieces in , namely and . (Certainly these are conjugacy classes: they only contain even permutations so they are subsets of , while everything in either class is conjugate in because the definition only depends on the [even_odd_parity parity] of .)
Let , and of the same cycle type: , .
Define to be the permutation which takes to , but otherwise does not move any other letter. Then , so if lies in then we are done: and are conjugate in .
That is, we may assume without loss of generality that is odd.
If any of the cycles is even in length
Suppose without loss of generality that , the length of the first cycle, is even. Then we can rewrite as , which is the same cycle expressed slightly differently.
Now is even in length, so it is an odd permutation (being a product of an odd number of transpositions, ). Hence is an even permutation.
But conjugating by yields : which is the result of conjugating by .
(It is the case that , because commutes with all of except with the first cycle , so the expression is , where we have pulled the final through to the beginning of the expression .)
%%hidden(Example): Suppose .
We have that is taken to by conjugating with , which is an odd permutation so isn't in . But we can rewrite , and the new permutation we'll conjugate with is , where we have appended . It is the case that is an even permutation and hence is in , because it is the result of multiplying the odd permutation by the odd permutation .
Now the conjugation is the composition of two conjugations: first by , to yield (which is still!), and then by . But we constructed so as to take to on conjugation, so this works just as we needed. %%
If all the cycles are of odd length, but some length is repeated
Without loss of generality, suppose (and label them both ), so the first two cycles are of the same length: say 's version is , and 's version is .
Then define . Since is odd and is an odd permutation, is an even permutation.
Now conjugating by is the same as first conjugating by and then by .
But conjugating by takes to , because it has the effect of replacing all the by and all the by , so it simply swaps the two cycles round.
Hence the conjugation of by yields , and is in .
%%hidden(Example): Suppose .
Then conjugation of by , an odd permutation, yields .
Now, if we first perform the conjugation , we take to itself, and then performing yields . The combination of and is an even permutation, so it does line in . %%
If all the cycles are of odd length, and they are all distinct
In this case, we are required to check that the conjugacy class does split. Remember, we started out by supposing and have the same cycle type, and they are conjugate in by an odd permutation (so they are not obviously conjugate in ); we need to show that indeed they are not conjugate in .
Indeed, the only ways to rewrite into (that is, by conjugation) involve taking each individual cycle and conjugating it into the corresponding cycle in . There is no choice about which -cycle we take to which -cycle, because all the cycle lengths are distinct. But the permutation which takes the -cycles to the -cycles is odd, so is not in .
Moreover, since each cycle is odd, we can't get past the problem by just cycling round the cycle (for instance, by taking the cycle to the cycle ), because that involves conjugating by the cycle itself: an even permutation, since the cycle length is odd. Composing with the even permutation can't take from being odd to being even.
Therefore and genuinely are not conjugate in , so the conjugacy class splits.
%%hidden(Example): Suppose in .
Then conjugation of by , an odd permutation, yields . Can we do this with an even permutation instead?
Conjugating by anything at all must keep the cycle type the same, so the thing we conjugate by must take to and to .
The only ways of to are by conjugating by some power of itself; that is even. The only ways of taking to are by conjugating by , or by and then some power of ; all of these are odd.
Therefore the only possible ways of taking to involve conjugating by an odd permutation , possibly alongside some powers of an even permutation ; therefore to get from to requires an odd permutation, so they are in fact not conjugate in . %%
Example
In , the cycle types are , , , , , , , and . The only class which splits is the -cycles, of cycle type ; it splits into a pair of half-sized classes with representatives and .