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.

  • $\begingroup$ I have a strong suspicion that if $x\mapsto x^n$ is an automorphism, then applying that map twice gives the identity map. $\endgroup$ – Arthur Mar 17 at 9:43
  • $\begingroup$ @Arthur how does that help ? $\endgroup$ – Eklavya Mar 17 at 9:48
  • $\begingroup$ Because that means that some elements are sent to their inverses, and some elements are sent to themselves. The subgroup of elements sent to their inverse is in the center and the subgroup of elements sent to themselves means $x^{n-1}=e$. Every other element is the product of two such elements. Again, I haven't actually checked any of these claims (otherwise I would post it as an answer), but it feels right. $\endgroup$ – Arthur Mar 17 at 9:54

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}


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