Proof involving permutations

Question

Let $\delta = \begin{pmatrix} 1 & 2 & 3 & 4 & 5\\ 2 & 3 & 4 & 5 & 1 \end{pmatrix} \in S_5$ be a permutation. If $\tau \in S_5$ has the property that $\tau \delta^2 = \delta^2 \tau$, prove that $\tau \delta = \delta \tau$.

I've tried to multiply $\tau$ and $\delta$, but got nowhere.

Thank you in advance!

Answer 1

Write as cycles: $\;\delta=(12345)\;$ . The claim follows at once from the fact that this kind of permutations ( which are cycles of full length in $\;S_n\;$ and $\;n\;$ is odd ) commute only with their powers, which are also $\;5\,-$ cycles, but most probably you haven't yet covered this, so let us give some hints assuming you've already seen what happens with cycles that are conjugated:

Suppose

$$\;\delta\sigma=\sigma\delta\iff \sigma\delta\sigma^{-1}\stackrel{(*)}=\delta$$

But we know (hopefully) that

$$\sigma (12345)\sigma^{-1}=(\sigma(1)\;\sigma(2)\;\sigma(3)\;\sigma(4)\;\sigma(5))\stackrel{(*)}=(12345)\implies\;\sigma$$

must be a $\;5\,-$ cycle again and, in fact, one that is a cyclic repetition of $\;(12345)\;$ , and there are only five like this:

$$\delta=(12345),\,\delta^2=(13524),\,\delta^3=(14253),\,\delta^4=(15432),\,\delta^5=(1) =\text{ the identity.}$$

Answer 2

$\tau\delta^2=\delta^2\tau$,
$\tau\delta^2\tau^{-1}=\delta^2$,
$(\tau\delta^2\tau^{-1})^3=(\delta^2)^3$,
$\tau\delta^6\tau^{-1}=\delta^6$, but $\delta^5=1$, so $\delta^6=\delta$, so
$\tau\delta\tau^{-1}=\delta$,
$\tau\delta=\delta\tau$.