centralizer of a k-cycle

Theorem 1.

Let $\sigma$ be a $k$-cycle in $S_n$. Then the centralizer of $\sigma$ is

$$C_{S_n}(\sigma )=\{{\sigma }^i\tau \mid 0\le i\le k-1,\tau \in S_{n-k}\}$$

where $S_{n\mathrm{-}k}$ is the subgroup of $S_n$ consisting of those permutations that fix all elements appearing in $\sigma$.

Proof.

This is fundamentally a counting argument. It is clear that $\sigma$ commutes with each element in the set given, since $\sigma$ commutes with powers of itself and also commutes with disjoint permutations. The size of the given set is $k\cdot (n-k)!$. However, the number of conjugates of $\sigma$ is the index of $C_{S_n}(\sigma )$ in $S_n$ by the orbit-stabilizer theorem, so to determine $|C_{S_n}(\sigma )|$ we need only count the number of conjugates of $\sigma$, i.e. the number of $k$-cycles.

In a $k$-cycle $(a_1\mathrm{\dots }{a}_k)$, there are $n$ choices for $a_1$, $n-1$ choices for $a_2$, and so on. So there are $n(n-1)\mathrm{\cdots }(n-k+1)$ choices for the elements of the cycle. But this counts each cycle $k$ times, depending on which element appears as $a_1$. So the number of $k$-cycles is

$$\frac{n(n-1)\mathrm{\cdots }(n-k+1)}{k}$$

Finally,

$$n!=|S_n|=\frac{n(n-1)\mathrm{\cdots }(n-k+1)}{k}|C_{S_n}(\sigma )|$$

so that

$$|C_{S_n}(\sigma )|=\frac{k\cdot n!}{n(n-1)\mathrm{\cdots }(n-k+1)}=k\cdot (n-k)!$$

and we see that the elements in the list above must exhaust $C_{S_n}(\sigma )$.∎