I know that in $S_n$, two cycles are conjugate if and only if they have the same cycle structure. This isn't true of $A_n$ though because apparently $(123)$ and $(213)$ aren't conjugate in $A_3$.

My question is firstly, how can we prove that $(123)$ and $(213)$ aren't conjugate in $A_3$, and secondly is there some way of generalising that for arbitrary $n$?

By that I mean for any $n$, can we find two permutations of the same cycle type that aren't conjugate in $A_n$?

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    $\begingroup$ Well, $A_3$ is abelian, so the conjugacy classes are the singleton sets. $\endgroup$ – Travis Willse Apr 15 '16 at 14:33
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    $\begingroup$ The size of the conjugacy class of an element $x$ in a group $G$ is equal to the index of the centralizer, $|[x]|=|G|/|C_G(x)|$. We know what happens with $G=S_n$. When we look at $A_n$ instead, the numerator $|G|$ is halved. The question is what happens to the denominator, i.e. to the centralizer. Clearly $C_{A_n}(x)=C_{S_n}(x)\cap A_n$. So if $C_G(x)$ consists of only even permutations, then the centralizers are equal, and conjugacy classes are halved. If $x$ is centralized by any odd permutation, then the centralizer is halved, too. And in that case the conjugacy class remains the same. $\endgroup$ – Jyrki Lahtonen Apr 15 '16 at 15:11

It turns out that the conjugacy class of an even permutation $g \in S_n$, $n > 1$, decomposes into smaller conjugacy classes in $A_n$ (always into two classes) iff the cycle decomposition of $g$ is into odd cycles of distinct length.

So, if $n > 2$ is odd, then the $n$-cycles split into two conjugacy classes. If $n > 2$ is even, then the $(n - 1)$-cycles split into two classes.

It follows from the above characterization that for $2 < n < 8$ only one cycle type splits, but for $n = 8$ two do (those corresponding to cycle types $(1, 7)$ and $(3, 5)$).

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  • $\begingroup$ Thanks for fixing the bugged link, Jyrki. $\endgroup$ – Travis Willse Apr 15 '16 at 15:13

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