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Suppose $n \ge 3$ and $g,h \in S_{n}$ are such that the only non-fixed point of both $g$ and $h$ is $k$. I am trying to prove that the commutator of $g$ and $h$ is a 3-cycle with the only non-fixed points are $k, g(k)$, and $h(k)$. I think you would only need to look at four cases, computing each of these points under the commutator and one other point that is not any of these. So first, I used the convention that for any $\sigma, \tau \in S_{n}$, $\sigma\tau$ is the composition $\sigma \circ \tau$.

Now, assume $g(k) = i$ for some $i \neq k$ and $h(k) = j$ for some $j \neq k$. The first intuition is to apply the commutator to $k$: $ghg^{-1}h^{-1}(k)$.

I am not sure on how to proceed and deal with the inverses, but I think I am on the right track.

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In order to evaluate this, you would need to know $h^{-1}(k)$ first. Should one just let this be another integer in $\{1,2,...,n\}$ or can this be explicitly found with what I have assumed? – afedder Nov 2 '12 at 5:32

$h^{-1}(k)$ is not $k$, so it is different from $k$ and it is not a fixed point of $h$. This means that it is a fixed point of $g$ and thus of $g^{-1}$.

Therefore, $ghg^{-1}h^{-1}(k)=ghh^{-1}(k)= g(k)$.

A similar calculation with argument $h(k)$, $g(k)$ and finally $m$ different than the three numbers will give you the desired result.

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