Recall that in the Symmetric group $~$S_n$~$, 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 $~$A_n$~$, the conjugacy classes nearly all remain the same; the exception is those classes in $~$S_n$~$ which have cycle type all odd and all distinct. Those classes split into exactly two classes in $~$A_n$~$.

# 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 $~$\tau \in A_n$~$ with the same cycle type as $~$\sigma \in A_n$~$ conjugate (in $~$A_n)$~$ to $~$\sigma$~$? If so, then the conjugacy class in $~$A_n$~$ of $~$\sigma$~$ is just the same as that of $~$\sigma$~$ in $~$S_n$~$; if not, then the conjugacy class in $~$S_n$~$ must break into two pieces in $~$A_n$~$, namely $~$\{ \rho \sigma \rho^{-1} : \rho \ \text{even} \}$~$ and $~$\{ \rho \sigma \rho^{-1} : \rho \ \text{odd} \}$~$. (Certainly these are conjugacy classes: they only contain even permutations so they are subsets of $~$A_n$~$, while everything in either class is conjugate in $~$A_n$~$ because the definition only depends on the [even_odd_parity parity] of $~$\rho$~$.)

Let $~$\sigma = c_1 \dots c_k$~$, and $~$\tau = c_1' \dots c_k'$~$ of the same cycle type: $~$c_i = (a_{i1} \dots a_{i r_i})$~$, $~$c_i' = (b_{i1} \dots b_{i r_i})$~$.

Define $~$\rho$~$ to be the permutation which takes $~$a_{ij}$~$ to $~$b_{ij}$~$, but otherwise does not move any other letter. Then $~$\rho \sigma \rho^{-1} = \tau$~$, so if $~$\rho$~$ lies in $~$A_n$~$ then we are done: $~$\sigma$~$ and $~$\tau$~$ are conjugate in $~$A_n$~$.

That is, we may assume without loss of generality that $~$\rho$~$ is odd.

### If any of the cycles is even in length

Suppose without loss of generality that $~$r_1$~$, the length of the first cycle, is even. Then we can rewrite $~$c_1' = (b_{11} \dots b_{1r_1})$~$ as $~$(b_{12} b_{13} \dots b_{1 r_1} b_{11})$~$, which is the same cycle expressed slightly differently.

Now $~$c_1'$~$ is even in length, so it is an odd permutation (being a product of an odd number of transpositions, $~$(b_{1 r_1} b_{11}) (b_{1 (r_1-1)} b_{11}) \dots (b_{13} b_{11})(b_{12} b_{11})$~$). Hence $~$\rho c_1'$~$ is an even permutation.

But conjugating $~$\tau$~$ by $~$\rho c_1'$~$ yields $~$\sigma$~$: $$~$\sigma = \rho \tau \rho^{-1} = \rho c_1' (c_1'^{-1} \tau c_1') c_1'^{-1} \rho^{-1}$~$$ which is the result of conjugating $~$c_1'^{-1} \tau c_1' = \tau$~$ by $~$\rho c_1'$~$.

(It is the case that $~$c_1'^{-1} \tau c_1' = \tau$~$, because $~$c_1'$~$ commutes with all of $~$\tau$~$ except with the first cycle $~$c_1'$~$, so the expression is $~$c_1'^{-1} c_1' c_1' c_2' \dots c_k'$~$, where we have pulled the final $~$c_1'$~$ through to the beginning of the expression $~$\tau = c_1' c_2' \dots c_k'$~$.)

%%hidden(Example): Suppose $~$\sigma = (12)(3456), \tau = (23)(1467)$~$.

We have that $~$\tau$~$ is taken to $~$\sigma$~$ by conjugating with $~$\rho = (67)(56)(31)(23)(12)$~$, which is an odd permutation so isn't in $~$A_n$~$. But we can rewrite $~$\tau = (32)(1467)$~$, and the new permutation we'll conjugate with is $~$\rho c_1' = (67)(56)(31)(23)(12)(32)$~$, where we have appended $~$c_1' = (23) = (32)$~$. It is the case that $~$\rho$~$ is an even permutation and hence is in $~$A_n$~$, because it is the result of multiplying the odd permutation $~$\rho$~$ by the odd permutation $~$c_1'$~$.

Now the conjugation is the composition of two conjugations: first by $~$(32)$~$, to yield $~$(32)\tau(32)^{-1} = (23)(1467)$~$ (which is $~$\tau$~$ still!), and then by $~$\rho$~$. But we constructed $~$\rho$~$ so as to take $~$\tau$~$ to $~$\sigma$~$ 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 $~$r_1 = r_2$~$ (and label them both $~$r$~$), so the first two cycles are of the same length: say $~$\sigma$~$'s version is $~$(a_1 a_2 \dots a_r)(c_1 c_2 \dots c_r)$~$, and $~$\tau$~$'s version is $~$(b_1 b_2 \dots b_r)(d_1 d_2 \dots d_r)$~$.

Then define $~$\rho' = \rho (b_1 d_1)(b_2 d_2) \dots (b_r d_r)$~$. Since $~$r$~$ is odd and $~$\rho$~$ is an odd permutation, $~$\rho'$~$ is an even permutation.

Now conjugating by $~$\rho'$~$ is the same as first conjugating by $~$(b_1 d_1)(b_2 d_2) \dots (b_r d_r)$~$ and then by $~$\rho$~$.

But conjugating by $~$(b_1 d_1)(b_2 d_2) \dots (b_r d_r)$~$ takes $~$\tau$~$ to $~$\tau$~$, because it has the effect of replacing all the $~$b_i$~$ by $~$d_i$~$ and all the $~$d_i$~$ by $~$b_i$~$, so it simply swaps the two cycles round.

Hence the conjugation of $~$\tau$~$ by $~$\rho'$~$ yields $~$\sigma$~$, and $~$\rho'$~$ is in $~$A_n$~$.

%%hidden(Example): Suppose $~$\sigma = (123)(456), \tau = (154)(237)$~$.

Then conjugation of $~$\tau$~$ by $~$\rho = (67)(35)(42)(34)(25)$~$, an odd permutation, yields $~$\sigma$~$.

Now, if we first perform the conjugation $~$(12)(53)(47)$~$, we take $~$\tau$~$ to itself, and then performing $~$\rho$~$ yields $~$\sigma$~$. The combination of $~$\rho$~$ and $~$(12)(53)(47)$~$ is an even permutation, so it does line in $~$A_n$~$. %%

## 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 $~$\sigma$~$ and $~$\tau$~$ have the same cycle type, and they are conjugate in $~$S_n$~$ by an odd permutation (so they are not obviously conjugate in $~$A_n$~$); we need to show that indeed they are *not* conjugate in $~$A_n$~$.

Indeed, the only ways to rewrite $~$\tau$~$ into $~$\sigma$~$ (that is, by conjugation) involve taking each individual cycle and conjugating it into the corresponding cycle in $~$\sigma$~$. There is no choice about which $~$\tau$~$-cycle we take to which $~$\sigma$~$-cycle, because all the cycle lengths are distinct. But the permutation $~$\rho$~$ which takes the $~$\tau$~$-cycles to the $~$\sigma$~$-cycles is odd, so is not in $~$A_n$~$.

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 $~$(123)$~$ to the cycle $~$(231)$~$), 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 $~$\rho$~$ from being odd to being even.

Therefore $~$\tau$~$ and $~$\sigma$~$ genuinely are not conjugate in $~$A_n$~$, so the conjugacy class splits.

%%hidden(Example): Suppose $~$\sigma = (12345)(678), \tau = (12345)(687)$~$ in $~$A_8$~$.

Then conjugation of $~$\tau$~$ by $~$\rho = (87)$~$, an odd permutation, yields $~$\sigma$~$.
Can we do this with an *even* permutation instead?

Conjugating $~$\tau$~$ by anything at all must keep the cycle type the same, so the thing we conjugate by must take $~$(12345)$~$ to $~$(12345)$~$ and $~$(687)$~$ to $~$(678)$~$.

The only ways of $~$(12345)$~$ to $~$(12345)$~$ are by conjugating by some power of $~$(12345)$~$ itself; that is even. The only ways of taking $~$(687)$~$ to $~$(678)$~$ are by conjugating by $~$(87)$~$, or by $~$(87)$~$ and then some power of $~$(678)$~$; all of these are odd.

Therefore the only possible ways of taking $~$\tau$~$ to $~$\sigma$~$ involve conjugating by an odd permutation $~$(678)^m(87)$~$, possibly alongside some powers of an even permutation $~$(12345)$~$; therefore to get from $~$\tau$~$ to $~$\sigma$~$ requires an odd permutation, so they are in fact not conjugate in $~$A_8$~$. %%

# Example

In $~$A_7$~$, the cycle types are $~$(7)$~$, $~$(5, 1, 1)$~$, $~$(4,2,1)$~$, $~$(3,2,2)$~$, $~$(3,3,1)$~$, $~$(3,1,1,1,1)$~$, $~$(1,1,1,1,1,1,1)$~$, and $~$(2,2,1,1,1)$~$. The only class which splits is the $~$7$~$-cycles, of cycle type $~$(7)$~$; it splits into a pair of half-sized classes with representatives $~$(1234567)$~$ and $~$(12)(1234567)(12)^{-1} = (2134567)$~$.