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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.

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

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    $\begingroup$ 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. $\endgroup$ – awllower Mar 14 '13 at 16:20
  • $\begingroup$ 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. $\endgroup$ – awllower Mar 14 '13 at 16:31
  • $\begingroup$ 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. $\endgroup$ – Dennis Gulko Mar 14 '13 at 18:26
  • $\begingroup$ 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. $\endgroup$ – awllower Mar 15 '13 at 1:23
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    $\begingroup$ When one is asked to show that a group is abelian, the most natural thing is to show that the centre is the whole group. $\endgroup$ – Dennis Gulko Mar 26 '13 at 7:10
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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?

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Let $t \in S$ and $x \in S \cap tS$. There exists $s \in S$ such that $x = ts$. Note that we have $s^{-1}t^{-1} = x^{-1} = T(x) = T(ts) = T(t)T(s) = t^{-1}s^{-1}$. 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 $T$ because they agree on a generating set, namely $S$.

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