Let $σ∈S_n$ be a $n$ cycle. Prove that an element $τ∈S_n$ only commutes with $σ$ if $τ$ is a power of $σ$.

Let $σ∈S_n$ be a $n$ cycle. Prove that an element $τ∈S_n$ only commutes with $σ$ if $τ$ is a power of $σ$.

Suppose WLOG $$ \sigma = (1 2 3 \dots n). $$ Then $$ (1 2 3 \dots n) = \tau (1 2 3 \dots n) \tau^{-1} = (\tau(1) \tau(2) \dots \tau(n)). $$ So if $\tau(1) = i+1$ (for some $0 \le i < n$), then $\tau(2) = i+2$, etc , $\tau(j) = j+i$ (Kasper correctly notes - thanks! - in a comment that once we get to $n$, we should fold over to $1$, essentially arguing modulo $n$), so $\tau = \sigma^{i}$.

The conjugacy class of $\sigma$ is the set of all $n$-cycles, of which there are $(n-1)!$. Thus the index of the centralizer of $\sigma$ is $(n-1)!$, so the order of the centralizer is $n!/(n-1)!=n$. But there are $n$ powers of $\sigma$, so they alone commute with $\sigma$.