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Let $G$ be a finite group and let $\phi: G \to G$ be an automorphism of $G$ such that $\phi(g) \ne g$ for all non-identity elements of G.

i) Show that each element $h$ of G can be written in the form $h=g^{-1}\phi(g)$ for some $g\in G$.

ii)If $\phi^2:G\to G$ is the identity map, show that $\phi(a)=a^{-1}$ for all $a \in G$

Just want some hints.

Nicky Hekster
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user121819
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  • For (i), I am probably wrong, but it would seem to me that it is only true if $h$ is identity – Lemon Jan 20 '14 at 07:33

1 Answers1

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Hint for (i): show that the map $f: G \rightarrow G$ defined by $f(g)=g^{-1}\phi(g)$ is injective. Draw the conclusion that $f$ has to be surjective.

Hint for (ii): from (i) one has $\phi(g)=gh$ for some $h \in G$. Hence $\phi((\phi(g))=g=\phi(gh)=\phi(g)\phi(h)=gh\phi(h)$, so $1=h\phi(h)$.

Nicky Hekster
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