I've been working on a group theory problem but hit a roadblock, and I'd appreciate some guidance.
The problem revolves around a finite group $G$ and a non-trivial automorphism $f$ in $G$. Specifically, for every element $x$ in $G$, $f(x)$ is either $x$ or $x^{-1}$.
I'm attempting to demonstrate that $G$ is a solvable group, indicating it can be systematically reduced to its identity element. I've made some initial attempts, identifying fixed elements under $f$ and those transforming into their inverses, but I'm struggling to tie everything together to prove the group's solvability.
Any hints, strategies, or insights on how to proceed further would be immensely helpful! Thank you in advance for your assistance!
Let $H = \{ x \in G \mid f(x)=x \}$. Then $H$ is a subgroup of $G$ and, since $f$ is nontrivial, there exists $g \in G \setminus H$.
Then for all $h \in H$, $gh \not\in H$, and so $f(gh) = g^{-1}h= (gh)^{-1} = h^{-1}g^{-1}$ and hence $g^{-1}hg=h^{-1}$. So $H$ is an abelian normal subgroup of $G$, and since the induced action of $f$ on $G/H$ inverts every element, $G/H$ is abelian. So $G$ is metabelian and hence solvable.