[Algebra] A nice group automorphism

Paolo902
Problem. Let $G$ be a finite group and let $T \in "Aut"(G)$ (the group of the automorphisms of $G$). Suppose that $T$ maps more than 3/4 of the elements of $G$ to their inverse, i.e. $Tx=x^{-1}$ for more than 3/4 of the elements of $G$.

Prove that $G$ is abelian and $Tx=x^{-1}$, $forall x in G$.

Risposte
Paolo902
@ Valerio: right, of course. :wink:

Principe2
If I am not wrong, the result does not hold anymore, by virtue of the following counter-example. Let $Q$ be the groups of quaternions and extend to an automorphism of $Q$ the mapping $T(i)=-i$ and $T(j)=-j$. This automorphism will map $1,-1,i,-i,j,-j$ to their own inverses and will fix $k$ and $-k$.

Ah.. I was forgotten: $\frac{6}{8}=\frac{3}{4}$!!

Paolo902


Question. What if $Tx=x^{-1}$ for exactly $3/4$ of the elements of $G$? :wink:

Paolo902
Ok, dissonance.

Do not worry, take your time and enjoy group theory :wink:

dissonance
Dear Paolo, please do not reveal the solution to this exercise! I'm willing to test myself against it in an attempt to revive my long-time dormant group theory skills. (But I'll need some time as I'm busy...).

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