How to show the elements ${x^p}$ and $y$ are commute based on relation ${x^y}={x^{1+p}}$

Question

This is a follow up to this question.

Let $G = \langle x,y,z\mid{x^{{p^2}}} = {y^p} = {z^p} = 1,{x^y} = {x^{1+p}},[x,z] = [y,z] = 1\rangle$.

How to show ${x^p}$ and $y$ are commute based on relation ${x^y}={x^{1+p}}$?

I have tried to solve it but i'm stuck. Please help me.

if ${x^y}=x.{x^p}$, then ${y^{-1}}xy=x.{x^p}$. from this i got $xy=yx.{x^p}$.

we have $x^y = x^{1+p}$

let $(x^p)^y = x^{1+p}$, $(x^y)^p = ({x^{1+p})^p}= x^p.x^{p^2}$

let $x^p$ be the element in $G$,

$(x^p)y = y(x^p).x^{p^2}=y(x^p)$ since $x^{p^2}=1$.

is it true?

Answer 1

The solution is correct, although I'd say it is a little more lengthy than necessary. I'd summarize it as $$ y^{-1}x^p y = (y^{-1}x y)^p = (x^{1 + p})^p = x^p x^{p^2} = x^p. $$ Multiplying this equation by $y$ from the left, we get $x^p y = y x^p$.