TITLE: If $|\lbrace g \in G: \pi (g)=g^{-1} \rbrace|>\frac{3|G|}{4}$, then $G$ is an abelian group.
QUESTION [19 upvotes]: Assume that $\pi$ is an automorphism of a finite group $G$. Let $S$ denote the set $\lbrace g \in G: \pi (g)=g^{-1} \rbrace$. Show that if $|S|>\frac{3|G|}{4}$, then $G$ is an abelian group. 

Anyone has any idea on how to solve this ? I have no idea how to start.

REPLY [6 votes]: Let $t \in S$ and $x \in S \cap tS$. There exists $s \in S$ such that $x = ts$. Note that we have $s^{-1}t^{-1} = x^{-1} = \pi(x) = \pi(ts) = \pi(t)\pi(s) = t^{-1}s^{-1}$. Therefore $t$ and $s$ commute. This implies $t$ and $x$ commute.
We have shown that $C_G(t) \supset S \cap tS$. But $S \cap tS$ generates $G$ since its size exceeds $|G|/2$. So, $t \in Z(G)$. This proves $S \subset Z(G)$. Now, since $S$ generates $G$, we have $Z(G) = G$, i.e., $G$ is abelian. This implies $x \mapsto x^{-1}$ is an automorphism of $G$ and it must be equal to $\pi$ because they agree on a generating set, namely $S$.