# centralizer of a k-cycle

###### Theorem 1.

Let $\sigma$ be a $k$-cycle in $S_{n}$. Then the centralizer of $\sigma$ is

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

where $S_{n-k}$ 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 $\lvert C_{S_{n}}(\sigma)\rvert$ we need only count the number of conjugates of $\sigma$, i.e. the number of $k$-cycles.

In a $k$-cycle $(a_{1}~{}\ldots~{}a_{k})$, there are $n$ choices for $a_{1}$, $n-1$ choices for $a_{2}$, and so on. So there are $n(n-1)\cdots(n-k+1)$ 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

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

Finally,

 $n!=\lvert S_{n}\rvert=\frac{n(n-1)\cdots(n-k+1)}{k}\lvert C_{S_{n}}(\sigma)\rvert$

so that

 $\lvert C_{S_{n}}(\sigma)\rvert=\frac{k\cdot n!}{n(n-1)\cdots(n-k+1)}=k\cdot(n-% k)!$

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

Title centralizer of a k-cycle CentralizerOfAKcycle 2013-03-22 17:18:00 2013-03-22 17:18:00 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 20M30