conjugacy in ${A}_{n}$
Recall that conjugacy classes^{} in the symmetric group^{} ${S}_{n}$ are determined solely by cycle type. In the alternating group^{} ${A}_{n}$, however, this is not always true. A single conjugacy class in ${S}_{n}$ that is contained in ${A}_{n}$ may split into two distinct classes when considered as a subset of ${A}_{n}$. For example, in ${S}_{3}$, $(123)$ and $(132)$ are conjugate, since
$$(23)(123)(23)=(132)$$ 
but these two are not conjugate in ${A}_{3}$ (note that $(23)\notin {A}_{3}$).
Note in particular that the fact that conjugacy in ${S}_{n}$ is determined by cycle type means that if $\sigma \in {A}_{n}$ then all of its conjugates in ${S}_{n}$ also lie in ${A}_{n}$.
The following theorem fully characterizes the behavior of conjugacy classes in ${A}_{n}$:
Theorem 1.
A conjugacy class in ${S}_{n}$ splits into two distinct conjugacy classes under the action of ${A}_{n}$ if and only if its cycle type consists of distinct odd integers. Otherwise, it remains a single conjugacy class in ${A}_{n}$.
Thus, for example, in ${S}_{7}$, the elements of the conjugacy class of $(12345)$ are all conjugate in ${A}_{7}$, while the elements of the conjugacy class of $(123)(456)$ split into two distinct conjugacy classes in ${A}_{7}$ since there are two cycles of length $3$. Similarly, any conjugacy class containing an evenlength cycle, such as $(1234)(56)$, splits in ${A}_{7}$.
We will prove the above theorem by proving the following statements:

•
A conjugacy class in ${S}_{n}$ consisting solely of even permutations^{} (i.e. that is contained in ${A}_{n}$) either is a single conjugacy class or is the disjoint union^{} of two equalsized conjugacy classes when considered under the action of ${A}_{n}$.

•
If $\sigma \in {A}_{n}$, then the elements of the conjugacy class of $\sigma $ in ${S}_{n}$ (which is just all elements of the same cycle type as $\sigma $) are conjugate in ${A}_{n}$ if and only if $\sigma $ commutes with some odd permutation.

•
$\sigma \in {S}_{n}$ does not commute with an odd permutation if and only if the cycle type of $\sigma $ consists of distinct odd integers.
Throughout, we will denote by ${\mathcal{C}}_{S}(\sigma )$ the conjugacy class of $\sigma $ under the action of ${S}_{n}$.
To prove the first statement, note that conjugacy is a transitive action. By the theorem that orbits of a normal subgroup^{} are equal in size when the full group acts transitively, we see that if $\sigma \in {A}_{n}$, then ${\mathcal{C}}_{S}(\sigma )$ splits into ${S}_{n}:{A}_{n}{C}_{{S}_{n}}(\sigma )$ classes under the action of ${A}_{n}$ (recall that ${C}_{G}(x)$, the centralizer^{} of $x$, is simply the stabilizer^{} of $x$ under the conjugation action of $G$ on itself). But since ${S}_{n}:{A}_{n}$ is either $1$ or $2$, we see that the conjugacy class of $\sigma $ either remains a single class in ${A}_{n}$ or splits into two classes.
Note also that the elements of ${\mathcal{C}}_{S}(\sigma )$ are all conjugate in ${A}_{n}$ if and only if ${A}_{n}{C}_{{S}_{n}}(\sigma )={S}_{n}$, which happens if and only if ${C}_{{S}_{n}}(\sigma )\u2288{A}_{n}$, which in turn is the case if and only if some odd permutation is in the centralizer of $\sigma $, which means precisely that $\sigma $ commutes with some odd permutation. This proves the second statement.
To prove the third statement, suppose first that $\sigma $ does not commute with an odd permutation. Clearly $\sigma $ commutes with any cycle in its own cycle decomposition, so if $\sigma $ contains a cycle of even length, that is an odd permutation with which $\sigma $ commutes. So $\sigma $ must consist solely of [disjoint] cycles of odd length. If two of these cycles have the same length, say $({a}_{1}{a}_{2}\mathrm{\dots}{a}_{2k+1})$ and $({b}_{1}{b}_{2}\mathrm{\dots}{b}_{2k+1})$, then
$$\begin{array}{c}(({a}_{1}{b}_{1})\mathrm{\dots}({a}_{2k+1}{b}_{2k+1}))({a}_{1}{a}_{2}\mathrm{\dots}{a}_{2k+1})({b}_{1}{b}_{2}\mathrm{\dots}{b}_{2k+1}){(({a}_{1}{b}_{1})\mathrm{\dots}({a}_{2k+1}{b}_{2k+1}))}^{1}=\hfill \\ \hfill ({a}_{1}{a}_{2}\mathrm{\dots}{a}_{2k+1})({b}_{1}{b}_{2}\mathrm{\dots}{b}_{2k+1})\end{array}$$ 
so the product^{} of $({a}_{1}{a}_{2}\mathrm{\dots}{a}_{2k+1})$ and $({b}_{1}{b}_{2}\mathrm{\dots}{b}_{2k+1})$, and thus $\sigma $, commutes with the product of $2k+1$ transpositions^{}, which is an odd permutation. Thus all the cycles in the cycle decomposition of $\sigma $ must have different [odd] lengths.
To prove the converse^{}, we show that if the cycles in the cycle decomposition all have distinct lengths, then $\sigma $ commutes precisely with the group generated by its cycles. It follows then that if all the distinct lengths are odd, then $\sigma $ commutes only with these permutations^{}, which are all even. Choose $\sigma $ with distinct cycle lengths in its cycle decomposition, and suppose that $\sigma $ commutes with some element $\tau \in {S}_{n}$. Conjugation^{} preserves cycle length, so since $\tau $ commutes with $\sigma $ and $\sigma $ has all its cycles of distinct lengths, each cycle in $\tau $ must commute with each cycle in $\sigma $ individually.
Now, choose a nontrivial cycle ${\tau}_{1}$ of $\tau $, and choose $j\in \tau $ such that $\sigma $ moves $j$ (we can do this, since $\sigma $ can have at most one cycle of length $1$ and the cycle length of $\tau $ is greater than $1$). Let ${\sigma}_{1}$ be the cycle of $\sigma $ containing $j$. Then ${\tau}_{1}$ commutes with ${\sigma}_{1}$ since $\tau $ commutes with $\sigma $, so ${\tau}_{1}$ is in the centralizer of ${\sigma}_{1}$, and it is not disjoint from ${\sigma}_{1}$. But the centralizer of a $k$cycle $\rho $ consists of products of powers of $\rho $ and cycles disjoint from $\rho $. Thus ${\tau}_{1}$ is a power of ${\sigma}_{1}$. So each cycle in $\tau $ is a power of a cycle in $\sigma $, and we are done.
Title  conjugacy in ${A}_{n}$ 

Canonical name  ConjugacyInAn 
Date of creation  20130322 17:18:04 
Last modified on  20130322 17:18:04 
Owner  rm50 (10146) 
Last modified by  rm50 (10146) 
Numerical id  6 
Author  rm50 (10146) 
Entry type  Theorem 
Classification  msc 20M30 