Let $\displaystyle G$ be a group and suppose there exists $\displaystyle n \geq 1$ such that the map $\displaystyle g \mapsto g^n$ is an automorphism of $\displaystyle G.$ Prove that $\displaystyle g^{n-1} \in Z(G)$ for all $\displaystyle g \in G.$

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- May 22nd 2009, 05:27 AMNonCommAlgAlgebra, Problems For Fun (9)
Let $\displaystyle G$ be a group and suppose there exists $\displaystyle n \geq 1$ such that the map $\displaystyle g \mapsto g^n$ is an automorphism of $\displaystyle G.$ Prove that $\displaystyle g^{n-1} \in Z(G)$ for all $\displaystyle g \in G.$

- May 23rd 2009, 04:28 AMSwlabr
Note that $\displaystyle \phi: G \rightarrow G, g \mapsto g^n$ is a homomorphism and so $\displaystyle (gh)^n = g^{n}h^n$. Also, as it is onto every $\displaystyle h' \in G$ can be written in this form, $\displaystyle h' = h^n$ for some $\displaystyle h \in G$.

Thus, it is sufficient to prove that $\displaystyle g^{n-1}h^n = h^{n}g^{n-1}$.

Now, note that $\displaystyle g^{n}h^{n} = (gh)^n = g(hg)^{n-1}h $

$\displaystyle \Rightarrow g^{n-1}h^{n-1} = (hg)^{n-1} = \underbrace{(hg)(hg)...(hg)}_{n-1} = (hg)^{n}g^{-1}h^{-1} = h^{n}g^{n}g^{-1}h^{-1}$.

Thus, $\displaystyle g^{n-1}h^n = (hg)^{n-1}h = h^{n}g^{n}g^{-1}h^{-1}h = h^{n}g^{n-1}$ as required.

I hope that's correct - haven't had a chance to check it properly as my wife is badgering me to go through and eat lunch! - May 23rd 2009, 04:38 AMNonCommAlg