# Algebra, Problems For Fun (9)

• May 22nd 2009, 05:27 AM
NonCommAlg
Algebra, Problems For Fun (9)
Let $G$ be a group and suppose there exists $n \geq 1$ such that the map $g \mapsto g^n$ is an automorphism of $G.$ Prove that $g^{n-1} \in Z(G)$ for all $g \in G.$
• May 23rd 2009, 04:28 AM
Swlabr
Note that $\phi: G \rightarrow G, g \mapsto g^n$ is a homomorphism and so $(gh)^n = g^{n}h^n$. Also, as it is onto every $h' \in G$ can be written in this form, $h' = h^n$ for some $h \in G$.

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

Now, note that $g^{n}h^{n} = (gh)^n = g(hg)^{n-1}h$
$\Rightarrow g^{n-1}h^{n-1} = (hg)^{n-1} = \underbrace{(hg)(hg)...(hg)}_{n-1} = (hg)^{n}g^{-1}h^{-1} = h^{n}g^{n}g^{-1}h^{-1}$.

Thus, $g^{n-1}h^n = (hg)^{n-1}h = h^{n}g^{n}g^{-1}h^{-1}h = h^{n}g^{n-1}$ 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 AM
NonCommAlg
Quote:

Originally Posted by Swlabr
Note that $\phi: G \rightarrow G, g \mapsto g^n$ is a homomorphism and so $(gh)^n = g^{n}h^n$. Also, as it is onto every $h' \in G$ can be written in this form, $h' = h^n$ for some $h \in G$.

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

Now, note that $g^{n}h^{n} = (gh)^n = g(hg)^{n-1}h$
$\Rightarrow g^{n-1}h^{n-1} = (hg)^{n-1} = \underbrace{(hg)(hg)...(hg)}_{n-1} = (hg)^{n}g^{-1}h^{-1} = h^{n}g^{n}g^{-1}h^{-1}$.

Thus, $g^{n-1}h^n = (hg)^{n-1}h = h^{n}g^{n}g^{-1}h^{-1}h = h^{n}g^{n-1}$ 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!

it's correct. go eat your lunch before you get in trouble! haha