[Algebra] A nice group automorphism

[Paolo90]

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

[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...).

[Paolo90]

Ok, dissonance.

Do not worry, take your time and enjoy group theory

[Paolo90]

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 $G$?

[Valerio Capraro]

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

[Paolo90]

@ Valerio: right, of course.