I am having trouble with Exercise 11, Section 1.10 of Basic Algebra 1 by Nathan Jacobson (pub. Freeman & Co. 1985). The statement to prove is:
Let $G$ be a finite group and $\phi$ an automorphism of $G$. Let
$ I = \{ g \in G : \phi g = g^{-1} \} $
If $|I| > {3\over4} |G|$ , $G$ is abelian.
If $|I| = {3\over4} |G|$ , $G$ has an abelian subgroup of index 2.
I'm trying to attack item 2 first, thinking there will be a way from 2 to 1, but I am not even at a point where that matters.
Facts I can see:
$\phi^2 = id_G$ , because the set of elements fixed by $\phi^2$ is a subgroup containing $I$.
Since $|G|$ is even (working on item 2!) , so must be the order of $K$ where
$ K = \{ k \in G : \phi k = k \} $
because we can partition $G$ into classes $\pi_k = \{ k, \phi k \}$ of size either 2 or 1, and $K$ is the union of all the singleton classes.
Hence, $K$ contains an element $i$ of order 2, so $ \phi i = \phi i^{-1} = i^{-1} $ , i.e.
$ 1 \neq i \in K \cap I $
That's already some nice information, but I still have no clue where to look for the abelian subgroup :-(