# If $f$ is an automorphism and $|\{a: f(a) = a^{-1}\}| = 3/4 |G|$ then $G$ has an abelian subgroup of index $2$

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.

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

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