Condition for two elements of An to be conjugated

Question

I want to find how the conjugation classes partition the A5, I know I have 24 5-cycles, but since 24 does not divide 60, then those 24 5-cycles are certainly not in the same class of conjugation, a teacher told me that two elements In Sn are conjugated, if and only if it has the same cyclic structure, I imagined that this definition would be equivalent to the number of disjoint cycles in which the permutation can be decomposed, but that would put all 5-cycles in the same class. Could anyone give a more precise definition and say why these 24 5-cycles divide into two classes of 12 elements each?

Answer 1

Let $\sigma \in A_n$, and let $\sigma^{S_n}$ and $\sigma^{A_n}$ denote the conjugacy classes of $\sigma$ in $S_n$ and $A_n$ respectively. Then, one has

If $\sigma$ commutes with an odd permutation, then $\sigma^{A_n} = \sigma^{S_n}$. If $\sigma$ does not commute with any odd permutation, then $$ \sigma^{S_n} = \sigma^{A_n}\sqcup ((12)\sigma(12))^{A_n} $$ and in particular, $|\sigma^{A_n}| = |\sigma^{S_n}|/2$ in this case.

Now, consider the 5-cycle $\sigma = (12345) \in A_5$, and let $C_{S_n}(\sigma)$ denote the centralizer of $\sigma$ in $S_n$. Since $|\sigma^{S_n}| = 24$, we have $$ |C_{S_n}(\sigma)| = 5 $$ Hence the only elements that commute with $\sigma$ as powers of $\sigma$ itself. In particular, $\sigma$ does not commute with an odd permutation, so $$ |\sigma^{A_n}| = 12 $$ by the above result.

Answer 2

Here is a more general answer which might be helpful to you. This concerns any index 2 subgroup of a finite group G.

Let $H$ be a index two subgroup of $G$. Since H is necessarily normal it is a union of conjugacy classes. Let $H$ and $Hs$ be two right cosets of $H$ in $G$. Let $x \in H$. Denote $C_{G}(x)$ and $C_{H}(x)$ as the conjugacy class of $x$ in $G$ and $H$ respectively.

Claim: $C_{G}(x)=C_{H}(x) \cup C_{H}(sxs^{-1})$

proof: $C_{G}(x)=\{\ gxg^{-1}|g \in G \}\ $

But since $G=H \cup Hs$ we can write $C_{G}(x)=\{\ hxh^{-1}|h \in H \}\ \cup \{\ hsxs^{-1}h^{-1}|h \in H \}\ $

which clearly implies the claim.

Now, if $sxs^{-1}$ is H-conjugate to $x$, then there exists $h \in H $ s.t $sxs^{-1}=hxh^{-1}$, which implies that $h^{-1}s$ is in the centraliser of $x$. So, $sxs^{-1}$ is H-conjugate to $x$ iff $Z_{G}(x) \cap Hs \neq \phi$, where $Z_{G}(x)$ denotes the centraliser of $x$ in $G$. So, the conjugacy class $C_{G}(x)$ will either remain intact inside H if $Z_{G}(x) \cap Hs \neq \phi$, or else if $Z_{G}(x) \cap Hs =\phi$, then it will split into two conjugacy classes having one representative as $x$ and the other representative as $sxs^{-1}$, where $s$ is an arbitrarily chosen non-trivial right coset representative of $H$ in $G$.

Now How can you use this discussion in your question. Clearly $A_{n}$ is index-2 subgroup of $S_{n}$. Suppose $\sigma \in S_{n}$ is a permutation which also belongs to $A_{n}$. You want to find out what happens to its conjugacy class inside $A_{n}$.

There is a great theorem which goes as follows:

If $\sigma \in S_{n}$. Then the cycle type of $\sigma$ consists of distinct odd integers in its cycle type iff $\sigma$ doesn't commute with any odd permutation.

This above theorem along with the discussions above shows that if $\sigma \in A_{n}$ , then conjugacy class of $\sigma$ breaks into two classes of equal size iff the cycle type of $\sigma$ consists of distinct odd integers.

In your case $(1 2 3 4 5) \in A_{n}$ has cycle type (5) which obviously has cycle type comprising of distinct odd integers. So, the conjugacy class breaks into two equal classes in $A_{n}$. You can find the other representative easily by choosing any odd permutation s, say (1 2). Then $(1 2)(1 2 3 4 5)(1 2)^{-1}=(2 1 3 4 5).$

But for $A_{5}$, we don't have to go through the theorem that I have stated. You can just see the above discussion and realise that $(1 2 3 4 5)$ doesn't commute with any odd permutation, hence giving your result.