Let $x$ be $(1\dots n)\in S_n$ show that there exists $y\in x^{S_n}$ such that $x^{S_n} = x^{A_n}\sqcup y^{A_n}$.

Question

Denote $x^G$ the conjugacy class,

$$x^G := \left\{ gxg^{-1}:g\in G\right\}.$$

Let $x$ be $(1\dots n)\in S_n$ show that there exists $y\in x^{S_n}$ such that $x^{S_n} = x^{A_n} \sqcup y^{A_n}$.

Basically I want to take $y\in x^{S_n} \setminus x^{A_n}$, and to show that $\left|y^{A_{n}}\right|=\left|x^{A_{n}}\right|$ and because $y^{A_{n}}\cap x^{A_{n}}=\varnothing$ we're done. My issue is with showing that $\left|y^{A_{n}}\right|=\left|x^{A_{n}}\right|$, still have no idea how to prove this one.

Answer 1

Hints.

If $n$ is even and $x$ is a cycle of length $n$, then $x^{A_n}=x^{S_n}$. Indeed, if $\sigma\in S_n$, then $x^\sigma=x^{x\sigma}$.

If $n$ is odd, then for any cycle $x\in S_n$ of length $n$ and any odd permutation $\tau$ we have $$ x^{S_n}=x^{\tau A_n\cup A_n}=y^{A_n}\cup x^{A_n} $$ Here $y=x^\tau$. Well, the fact that $y^{A_n}\cap x^{A_n}=\varnothing$ you already proved.