# 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 $σ$.

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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 $\tau = \sigma^{i}$.
I'm afraid I still don't really get it. I think this is because I don't really understand my own question. Do I need to prove that: $τ=σ^i \implies στ=τσ$ ? What do you mean with $σ^τ$? –  90intuition Mar 19 '13 at 12:54
@90intuition, I have spelled out the conjugate. Yes, you have to prove $\sigma \tau = \tau \sigma$ if and only if $\tau = \sigma^{i}$ for some $i$. –  Andreas Caranti Mar 19 '13 at 12:57
Okay, so $⇐$ is trivial. And you prove ⇒ ? Aah I understand it, thanks ! You use the fact they commute to rewrite $σ$ right ? –  90intuition Mar 19 '13 at 13:05
And $τ(i)\overset{τ^{-1}}{↦} i\overset{σ}{↦}i+1\overset{τ}{↦}τ(i+1)$ –  90intuition Mar 19 '13 at 13:09
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 $\sigma$.