Group with an automorphism of order 2 (Jacobson BA1) 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 :-(
 A: This is essentially the same argument as Tim's for part 1), but slightly less abstractly stated.
For $g,h\in I$ we have $\phi(gh)=\phi(g)\phi(h)=g^{-1}h^{-1}$. For fixed $g$, there are less than $|G|/4$ elements $h$ for which $gh$ isn't in $I$, and thus more than $|G|/2$ for which it is. But then $\phi(gh)=(gh)^{-1}=h^{-1}g^{-1}$, so $g^{-1}$ and $h^{-1}$ commute. Since $g^{-1}$ commutes with more than half of $G$, it commutes with all of $G$, and since $g$ was arbitrary, $G$ is abelian.
A: For part 2) Let $x\in I, y\in I \cap x^{-1} I.$ Then $y\in C_G (x),$ as can be seen by applying $\sigma $ to $xy.$ Now $|x^{-1} I| = |I| = \frac{3}{4} |G|,$ so applying inclusion-exclusion gives $|C_G (x) | \ge |I \cap x^{-1}S| = \frac{3}{2} |G| - |I \cup x^{-1} I| \ge \frac{|G|}{2}.$ Hence $[G: C_G(x) ] \le 2$ for all $x\in G.$ Now if $C_G(x)$ is properly contained in $G$ for some $x\in I$ we are done. Otherwise, we have $I \le Z(G)$; by the size constraint on $I,$ this forces that $G$ be abelian. From here it is easy to show that $I $ is necessarily a subgroup of $G,$ contradicting Lagrange's Theorem.
The techniques used for part 1) are quite similar, except that towards the end you need to shift your argument towards the center of the group.
