[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 $Tx=x^{-1}$, $forall x in G$.

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...).

Ok, dissonance.

Do not worry, take your time and enjoy group theory

Testo nascosto, fai click qui per vederlo

Proof. Let $Y:={g \in G: Tg=g^{-1}}$. Then $|Y|>\frac{3}{4}|G|$ by hypothesis. Choose $a \in Y$ and set $aY=:Z$. It's easy to see that $|Z|=|Y|$. Now we want to compute $|Y cap Z|$. Since $|G setminus Y|=|G setminus Z| \le 1/4|G|$, simply using the De Morgan laws, we get $|Y cap Z| > \frac{|G|}{2}$. Now pick $s in Y cap Z$ and $y \in Y$: $s$ must be of the form $ay$ and since $s \in Y$, $y^{-1}a^{-1}=T(ay)=T(a)T(y)=a^{-1}y^{-1}$, i.e. $y \in C(a)$ (with $C(cdot)$ we've indicated the centralizer of $cdot$).

Now remember that $C(cdot)$ is a subgroup, and in particular we have just shown that $forall a in Y$, $|C(a)|>|G|/2$; by Lagrange's theorem, $C(a)=G$. This means that every element of $Y$ commutes with all the element of the group: in other words, $Y subseteq Z(G)$. But $3/4|G| <|Y| \le |Z(G)|$ which gives $Z(G)=G$ so $G$ is abelian.

In this case, we can easily prove that $Y$ is indeed a subgroup and, since it has got more than $3/4$ of the elements of $G$, it must be $G$ itself, i.e. $Tx=x^{-1}, forall x in G$.

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

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: $\frac{6}{8}=\frac{3}{4}$!!

@ Valerio: right, of course.