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 parity of $\rho$.)

Let $\sigma =c_1\dots c_k$, and $\tau =c_1^{\prime }\dots c_k^{\prime }$ of the same cycle type: $c_i=(a_{i1}\dots a_{i{r}_i})$, $c_i^{\prime }=(b_{i1}\dots b_{i{r}_i})$.

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 $c_1^{\prime }=(b_{11}\dots b_{1r_1})$ as $(b_{12}{b}_{13}\dots b_{1r_1}{b}_{11})$, which is the same cycle expressed slightly differently.

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

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

(It is the case that $c_1^{\prime -1}\tau c_1^{\prime }=\tau$, because $c_1^{\prime }$ commutes with all of $\tau$ except with the first cycle $c_1^{\prime }$, so the expression is $c_1^{\prime -1}{c}_1^{\prime }{c}_1^{\prime }{c}_2^{\prime }\dots c_k^{\prime }$, where we have pulled the final $c_1^{\prime }$ through to the beginning of the expression $\tau =c_1^{\prime }{c}_2^{\prime }\dots c_k^{\prime }$.)

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 ${\rho }^{\prime }=\rho \left(b_1{d}_1\right)\left(b_2{d}_2\right)\dots \left(b_r{d}_r\right)$. Since $r$ is odd and $\rho$ is an odd permutation, ${\rho }^{\prime }$ is an even permutation.

Now conjugating by ${\rho }^{\prime }$ is the same as first conjugating by $\left(b_1{d}_1\right)\left(b_2{d}_2\right)\dots \left(b_r{d}_r\right)$ and then by $\rho$.

But conjugating by $\left(b_1{d}_1\right)\left(b_2{d}_2\right)\dots \left(b_r{d}_r\right)$ 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 ${\rho }^{\prime }$ yields $\sigma$, and ${\rho }^{\prime }$ is 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 $\left(123\right)$ to the cycle $\left(231\right)$), 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.

Example

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