Let $\tau \in S_{n} $. Show that there exists $\sigma \in S_{n} $ such that $\sigma \circ \tau =\tau ^{-1}\circ \sigma$

Question

Let $\tau \in S_{n} $. Show that there exists $\sigma \in S_{n} $ such that $\sigma \circ \tau =\tau ^{-1}\circ \sigma$.

Here $S_{n}$ is the symmetric group.

Answer 1

Hint: Prove the result for when $\tau$ is a cycle, and make sure $\sigma$ doesn't permute any elements apart from those cycled by $\tau$. Once you've done that, take an arbitrary $\tau$, write it in disjoint cycle form, and apply the result to each of those cycles in turn.

Answer 2

Since any permutation is the product of two involutions, write $\tau=\rho\sigma$ where $\rho^2=\sigma^2=I;$ then $$\tau^{-1}\sigma=(\rho\sigma)^{-1}\sigma=(\sigma\rho)\sigma=\sigma(\rho\sigma)=\sigma\tau.$$

Answer 3

As a complement to Arthur's answer, suppose that $\tau$ is a cycle on $n$ elements. You can view $\tau$ as the map $x\mapsto x+1$ on $\frac{\mathbb Z}{n\mathbb Z}$. Then, consider $\sigma(x)=2-x$ which gives a permutation of $\frac{\mathbb Z}{n\mathbb Z}$ also.

You have

$$ \sigma(\tau(x))=2-(x+1)=1-x $$

and

$$ \tau^{-1}(\sigma(x))=(2-x)-1=1-x $$

So $\sigma\tau=\tau^{-1}\sigma$ as wished.