# Cycle types of elements of centralizers of transitive permutation groups

I think the following is true, but I'm not sure how to prove it. Any help would be greatly appreciated.

Suppose $$G$$ is a transitive subgroup of $$S_n$$ and $$C$$ is the centralizer of $$G$$ in $$S_n$$. If $$\sigma\in C$$ has order $$k$$, then $$\sigma$$ is a product of $$n/k$$ disjoint $$k$$ cycles.

• It suffices (easy check) to show that if $\sigma \in C$ fixes any point, then $\sigma$ is trivial. Suppose $\sigma(x) = x$. Since $G$ is transitive, for any other point $y$, there exists a $g \in G$ such that $g(x) = y$. But now $$\sigma(y) = \sigma(g(x)) = \sigma g(x) = g \sigma(x) = g(x) = y.$$ Thus $\sigma$ fixes every point and so $\sigma = e$. – user670344 May 10 at 0:25

This is true and well known. The property that $$C$$ has in your conclusion is usually just called "semiregular".