$A_n$ and the conjugacy classes I am asked to prove that if n is odd, then the set of n-cycles in $A_{n}$ consists of two conjugacy classes of the same size.
I have been starting at my notes regarding the conjugacy classes for $A_{5}$, but I still have no idea how to begin this proof. Can someone help me or at least give me a push in the right direction?
 A: There is an easy way to see this. Use the conjugacy-class-centralizer theorem. If $g\in A_n$ then $|g^G|=|G|/|C_G(g)|$. So understanding how the class $g^{S_n}$ splits into $A_n$-classes revolves around whether $C_{S_n}(g)=C_{A_n}(g)$ or $|C_{S_n}(g):C_{A_n}(g)|=2$, i.e, $g$ commutes with an odd permutation.
We see that the $A_n$- and $S_n$-classes of $g$ coincide if and only if $g$ commutes with an odd permutation, and the $S_n$-class splits into two $A_n$-classes if and only if $g$ does not commute with an odd permutation.
To find the two classes, a representative for the other class is the conjugate of $g$ by any odd permutation. (To see this, if $x$ is odd and $g^x=g^y$ for some even $y$, then $yx^{-1}$, an odd permutation, centralizes $g$, a contradiction.)
A: We fix an odd $n\ge 3$. (For $n=1$...)
Let $H$ be the alternate group $A_n$, a subgroup of the symmetric group $G:=S_n$ (of permutaions of the symbols $1,2,\dots,n$), its index is two. The group $G$ has $n!$ elements, the group $H$ has $n!/2$ elements. Two cycles $c,d$ of order $n$ are conjugated in $G$, simply write them as
$$
\begin{aligned}
c &= (c_1, c_2,\dots, c_n) ,\\
d &= (d_1, d_2,\dots, d_n)
\end{aligned}
$$
and consider the permutation $s$ which maps $c_1\to d_1$, $c_2\to d_2$, ... and so on. (It is well defined, since any symbol is exactly once in the cyclic writings above.) Then we have for instance:
$$
scs^{-1}(d_1)=sc(c_1)=s(c_2)=d_2\ ,
$$
and similarly for the other indices, so $scs^{-1}=d$.
If we restrict to $H$, then we can use only even permutations to conjugate.
We can moreover "norm" the representations of $c,d$ above, asking for the condition $c_1=d_1=1$. Now assume some $s\in H$ can be used to conjugate $s$ to $d$. We do not have necessarily $s(1)=1$, but we can arrange so by inserting on the one or other side of $s$ a power of $c$ or $d$. Now $s$ is determined by the above rule of mapping $c_k$ to $d_k$. If this map is even, we can conjugate, if not we cannot.

Here are some experiments in the case $H=A_7$, subgroup of $G=S_7$. The programming language used is sage:
sage: H = AlternatingGroup(7)
sage: H.order()
2520
sage: factorial(7)/2
2520
sage: classes = H.conjugacy_classes()
sage: c = classes[0]
sage: c
Conjugacy class of () in Alternating group of order 7!/2 as a permutation group
sage: c.representative()
()
sage: classes7 = [ c for c in classes if c.representative().order() == 7 ]
sage: classes7
[Conjugacy class of (1,2,3,4,5,6,7) in Alternating group of order 7!/2 as a permutation group,
 Conjugacy class of (1,2,3,4,5,7,6) in Alternating group of order 7!/2 as a permutation group]

The chosen representatives (in sage) "differ by a transposition", if we map the symbols mot-a-mot in each other, as they come. And this transposition is odd, outside the $A_7$.
