# If $|\lbrace g \in G: \pi (g)=g^{-1} \rbrace|>\frac{3|G|}{4}$, then $G$ is an abelian group.

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.

• My first thought is that whenever $x\in S$, $T(x)$ is also in $S$. I don't know if that helps you. Sep 8 '13 at 7:10
• Second vague thought: Prove somehow that $S$ is a subgroup of $G$, and that $S$ is too big to be a proper subgroup. Sep 8 '13 at 21:34

Fix any $s\in S$. Define three sets: $T=\{\,t\mid t,st\in S\,\}$, $T_1=\{\,t\mid t\not\in S\,\}$ and $T_2=\{\,t\mid st\not\in S\,\}$. Then, clearly, $T=G\setminus(T_1\cup T_2)$. Hence $$|T|=|G|-|T_1|-|T_2|+|T_1\cap T_2|>|G|-\frac{|G|}{4}-\frac{|G|}{4}=\frac{|G|}2$$ Moreover, if $t\in T$ then $$st=((st)^{-1})^{-1}=\pi(t^{-1}s^{-1})=\pi(t^{-1})\pi(s^{-1})=ts$$ Hence $T\subseteq C_G(s)$. Hence $|C_G(s)|>|G|/2$ and so $C_G(s)=G$ for any $s\in S$, which implies that $S\subseteq Z(G)$. So $|Z(G)|>\frac34|G|>\frac12|G|$ thus $Z(G)=G$
• I think you mean "fix any $s$" in line 1, and $-|G|/4$ in line 2.And why $\pi(t^{-1})=t$? Notice that $\pi$ is not assumed to be an involution here. Mar 14 '13 at 16:20
• Notice that we can modify the proof by $$st=\pi^{-1}(\pi(st))=\pi^{-1}((st)^{-1})=\pi^{-1}(t^{-1}s^{-1})=\pi^{-1}(t^{-1})\pi^{-1}(s^{-1})=ts.$$ And other parts are the same. Mar 14 '13 at 16:31
• Since $\pi$ is an isomorphism, if $t\in S$ then $\pi(t^{-1})=\pi(t)^{-1}=(t^{-1})^{-1}=t$ hence $t^{-1}\in S$. I left it out on purpose - the OP shoud fill in at least some detailes. Mar 14 '13 at 18:26
• I see how your arguments work now. BTW, in line 2, it should be $$|G|-\frac{|G|}{4}-\frac{|G|}{4}=\frac{|G|}{2}.$$ In any case, thanks for the answer. Mar 15 '13 at 1:23
Hint. This answer is exactly the proof for $\pi=\text{id}_{\text{Aut}(G)}$. Can you change the wording to extend it to an arbitrary automorphism?
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 $$S$$.