[Algebra] A nice group automorphism
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 $Tx=x^{-1}$, $forall x in G$.
Prove that $G$ is abelian and $Tx=x^{-1}$, $forall x in G$.
Risposte
@ Valerio: right, of course.

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: $\frac{6}{8}=\frac{3}{4}$!!
Ah.. I was forgotten: $\frac{6}{8}=\frac{3}{4}$!!
Question. What if $Tx=x^{-1}$ for exactly $3/4$ of the elements of $G$?

Ok, dissonance.
Do not worry, take your time and enjoy group theory
Do not worry, take your time and enjoy group theory

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...).