If $\phi\in{\rm Aut}(G)$ for finite $G$ s.t. $\phi^2=id_G$ and $\phi(g)=g$ only if $g=1$, then $\forall g\in G\exists h\in G$ s.t. $g=h^{-1}\phi(h)$.

Question

Let $\phi:G\rightarrow G$ be an automorphism of a finite group $G$, with $\phi^2=id_G$, and $\phi(g)=g$ only if $g=1$. Prove that for each $g\in G$, there exists $h \in G$ such that $g=h^{-1}\phi (h)$.

I believe that this isn't a very difficult question. But I can't seem to see how to do it. In particular, I do not see the significance of "$\phi(g)=g$ only if $g=1$"

Any hints will be appreciated.

Answer 1

What you need to show is that $h \mapsto h^{-1}\phi(h)$ maps $G$ onto $G$. Since $G$ is finite, it will suffice to show that this map is one-to-one. Note that you can argue this theoretically, and do not need to give an explicit formula for the $h$ that gets mapped to each $g$.

The significance of the requirement that $\phi(g) = g$ if an only if $g = 1$ is that, if $\phi(h)=$ then $h^{-1}\phi(h) = h^{-1}h = 1$. This will immediately disqualify the map from being one-to-one if there is some $h \neq 1$ with this property.