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I edited the question to remove the first part as it is already answered here.

Let $G$ be a finite group and $f$ an automorphism of $G$ and $A = \{a\in G: f(a) = a^{-1}\}$.

Prove that if $|A| = 3/4 |G|$ then $G$ has an abelian subgroup of index $2$.

Here's something (very) related.

Hints or solutions much appreciated.

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  • $\begingroup$ Sorry, that's the second time today - but this one really is an exact duplicate. $\endgroup$ – Derek Holt Jun 9 '17 at 13:24
  • $\begingroup$ The question in your title is not the same as the one in the body, could you clarify? $\endgroup$ – Arnaud D. Jun 9 '17 at 13:24
  • $\begingroup$ @DerekHolt why are these duplicates so damn hard to find?! Thanks anyway! $\endgroup$ – Cauchy Jun 9 '17 at 13:25
  • $\begingroup$ @DerekHolt still, though, I'd like to prove the second part. $\endgroup$ – Cauchy Jun 9 '17 at 13:30
  • $\begingroup$ @DerekHolt never mind, it's an easy one. $\endgroup$ – Cauchy Jun 9 '17 at 13:32
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Hint If $a,b,ab \in A$ then show that $ab=ba$.

Now, pick some $a \in A$. Use the above hint to show that $|C(a)|>\frac{1}{2}|G|$ (there are less than a quarter bad choices for $b$ and less than a quarter bad choices for $ab$). This shows that $A \subset Z(G)$.

Since $|Z(G)| >1/2 |G|$ the center is $G$.

For the second part, try to show that if $a \in A$ then $$C(a) \cap A \geq \frac{1}{2}|G|$$

Show that $C(a) \cap A$ is an Abelian subgroup of $G$. Since $A \neq G$, this subgroup cannot be everything.

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  • $\begingroup$ I couldn't show that $C(a) \cap A$ is abelian. Can I get some help with this? $\endgroup$ – Cauchy Jun 11 '17 at 3:35
  • $\begingroup$ @Cauchy try to use the fact that if $b, c, bc \in A$ then $bc=cb$. $\endgroup$ – N. S. Jun 11 '17 at 3:59

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