About conjugacy in $A_n$ I know that $\sigma$, $\tau$ $\in$ $S_n$ are conjugate if and only if they have the same cycle structure. Is there any explicit way that we can determine whether two elements in $A_n$ are conjugates?  
 A: For two elements $x,y \in A_n$ to be conjugate within $A_n$ they have to have the same cycle type. Suppose $axa^{-1}=y$ for some $a\in S_n$.
Suppose $x$ contains a cycle $c$ which moves an even number of points (elements of the underlying set of size $n$), and is therefore an odd permutation. We have $cxc^{-1}=x$ so that $acxc^{-1}a^{-1}=axa^{-1}=y$. One of $ac$ and $a$ must be even, so $x$ and $y$ are conjugate within $A_n$.
Suppose $x$ contains two cycles $d,e$ of the same odd length (including the case where two or more elements of the underlying set are fixed). For example, $(1 2 3)$ and $(456)$. In the example it will be found that the odd permutation $f=(14)(25)(36)$ of order $2$ exchanges the two cycles, and fixes every other element of the underlying set. Then $fxf^{-1}=x$ and we can proceed as before. This clearly works in every such case.
The remaining case has $x$ a product of cycles of distinct odd lengths. Write $y$ underneath $x$ with the same cycle order. If the permutation which takes the elements of $x$ to the corresponding elements of $y$ is even, it will conjugate one to the other within $A_n$. If it is odd, then there will be no even permutation available, and the conjugacy class will split in two.
Take $A_5$ for an example. The only elements with an even length cycle are of form $(12)(34)$ and these are all conjugate within the group. The only other elements which move at least $4$ points are the $5-$cycles - and these fit the criterion for splitting, which indeed they do. The $5-$cycles also split in $A_6$, but not in $A_7$ where they fix two points.
It is thus easy to classify the cycle types where the conjugacy classes split (or stay the same). But you have to compute explicitly to see whether two elements in a split conjugacy class are conjugate or not.
