$x \mapsto x^n $ is a automorphism of group $G$, show that for all $x$ in $G$, $x^{n-1} \in Z(G)$.

Question

If $x \mapsto x^n $ is a automorphism of group $G$, show that for all $x$ in $G$, $x^{n-1} \in Z(G)$.

This mean $G=\{x^n:x \in G\}$ and $x^n=e$ if and only if $x= e$.

Now let $y\in G$. I Want to show that $$yx^{n-1}=x^{n-1}y.$$ We know that $y$ is $n^{\text {th}}$ power of some element of $G$, but how to proceed from here?

Any hint is appreciated. Thanks in Advance.

Answer 1

Note that $(yx)^n=y^nx^n$(by definition of homomorphism). So $$\underbrace{(yx)(yx)\cdots (yx)}_{\text {$n-1$ times}}\;y=y^n x^{n-1}.$$ Now

\begin{align} y^n x^{n-1}&=x ^{-1}x\;\underbrace{(yx)(yx)\cdots (yx)}_{\text {$n-1$ times}}\;y\\ &=x ^{-1}\underbrace{(xy)(xy)\cdots (xy)}_{\text {$n$ times}}\\ &=x^{-1}(xy)^n\\ &=x^{-1}x^ny^n\, (\text{again by homomorphism definition})\\ &=x^{n-1}y^n. \end{align}

Answer 2

As $x \mapsto x^n $ is an automorphism, by the homomorphism's condition, we have that $(xy)^n=x^ny^n$ and:

$$\begin{align} y^nx^{n-1}y^{-n}&=y^nx^{n-1}xx^{-1}y^{-n}\\ &=(yx)^nx^{-1}y^{-1}y^{-n+1}\\ &=(yx)^{n-1}y^{-n+1}\\ &=y^{n-1}x^{n-1}y^{-n+1}\ \end{align}$$

And now:

$$y^nx^{n-1}y^{-n}=y^{n-1}x^{n-1}y^{-n+1} \implies yx^{n-1}y^{-1}=x^{n-1}; x^{n-1} \in Z(G)$$