The centralizer of a cycle is the group generated by the elements of $S_n$ that are disjoint from the cycle and the group generated by the cycle.

Question

I am having a hard time trying to prove that the centralizer of a cycle is the group generated by the elements of $S_n$ that are disjoint from the cycle and the group generated by the cycle.

It is clear to me that these elements are in the centralizer of the cycle, but why any element in the centralizer belongs to this group?

Thanks!

Answer 1

Hint 1: Let $\sigma$ be a cycle in $S_n$, where $S_n$ permutes the set $\{x_1,\dots, x_n\}$. We have that the centraliser $C_{S_n}(\sigma)=\{\rho\in S_n\mid \rho\sigma=\sigma\rho\}$. Observe that, by definition, any power $\sigma^k$ of $\sigma$ is an element of $C_{S_n}(\sigma)$, since $\sigma^k\sigma=\sigma^{k+1}=\sigma\sigma^k$. So let's consider $\rho\in C_{S_n}(\sigma)\setminus\langle\sigma\rangle$, where $\langle\sigma\rangle$ is the subgroup generated by $\sigma$. We can write $\sigma=(s_1\dots s_m)$ for some $1\le m\le n$. In order for $\rho$ to commute with $\sigma$, it must send each $s_i$ to the same $x_j$ (dependent on $i$), regardless of which side of $\sigma$ the $\rho$ is on.

Hint 2: For any $\rho\in C_{S_n}(\sigma)$, we have $$\rho\sigma\rho^{-1}=\sigma$$ and there's a convenient lemma for conjugating permutations . . .