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automorphism of order 2 fixing just identity. Prove that G is abelian.


Well-known member
MHB Math Scholar
Mar 10, 2012
Let $G$ be a finite group, $T$ an automorphism of $G$ with the property that $T(x)=x$ if and only if $x=e$. Suppose further that $T^2=I$, that is, $T(T(x))=x$ for all $x\in G$. Show that $G$ is abelian.

I approached this problem using the permutation representation afforded by $T$ on $G$. Its easy to deduce that the cycle representation of the permutation of $G$ caused by $T$ has $(n-1)/2$ disjoint transpositions, where $n=|G|$. We know, from this, that $n$ is odd but so what? I am not able to exploit the homomorphism property of $T$.


Well-known member
MHB Math Scholar
Feb 15, 2012
hint: (i don't want to spoil your fun because this is a beautiful theorem)

define $f:G \to G$ by:

$f(g) = g^{-1}T(g)$

show $f$ is injective (and thus bijective).

now...what is $T\circ f(g)$?