Show if $\pi,\sigma$ are any permutation s.t $(\pi\sigma)^2=\pi^2\sigma^2$, then $\pi\sigma=\sigma\pi$...

Question

As the title says, the problem is:

Show if $\pi,\sigma$ are any permutation s.t $(\pi\sigma)^2=\pi^2\sigma^2$, then $\pi\sigma=\sigma\pi$.

There is a theorem that states

If $\pi,\sigma$ are any permutations s.t $\pi\sigma=\sigma\pi$, then $(\pi\sigma)^r=\pi^r\sigma^r$

So, it appears for $r=2$, we're proving the converse of the above theorem. To be frank, and I hate to give you guys a "no work" problem, I'm not even sure how to begin this. It seems so basic, but my head can't make a sense of any clear path. Can anyone lead me down the right path? It's just been hours of no progress with this...

Answer 1

This is true for any group. The condition you're working with is $gghh=ghgh$. See anything that can be cancelled on the left and/or on the right side of this equation?

Answer 2

$$(\pi\sigma)^2 = \pi\sigma\pi\sigma =\pi^2\sigma^2 \Rightarrow \pi^{-1}(\pi\sigma\pi\sigma)\sigma^{-1} = \pi^{-1}(\pi^2\sigma^2)\sigma^{-1} \Rightarrow\\ (\pi^{-1}\pi)\sigma\pi(\sigma\sigma^{-1}) = (\pi^{-1}\pi)\pi\sigma(\sigma\sigma^{-1})\Rightarrow \boxed{\sigma\pi = \pi\sigma}$$