Show that if $|I| > \frac{3}{4}|G|$ then $G$ is abelian. I've self studying Nathan Jacobson's Basic Algebra and I came across this question:
Let $G$ be a finite group with $\alpha$ a automorphism of the group. Denote,
$$I =\{g \in G : \alpha(g) = g^{-1}\}$$
show that if  $|I| > \frac{3}{4}|G|$ then $G$ is abelian. Show that if $|I| = \frac{3}{4}|G|$ show that $G$ has an abelian subgroup of order 2.
So far what I've done is tried a proof by contradiction, saying suppose that $G$ is not abelian. Well then I computed for $g,h \in I$,
$$\alpha(gh) = \alpha(g) \alpha(h) = g^{-1}h^{-1} \neq (gh)^{-1}$$
since $G$ is assumed not to be abelian.
I need to find a contradiction from this, i'm just not sure how to do that.
Any help is appreciated, thanks.
 A: 1)Let G be the group and T is automorphism which send more than $3/4th$ elements goes to there inverse then G is abelian.
Claim:$T(x)=x^{-1} \forall x\in G$
Let S={$x\in G|T(x)=x^{-1}$}
|S|$\geq 3/4|G|$      ......Given
Case 1: If S is subgroup Then by Lagrange theorem |S| divides |G| 
As |S|$\geq 3/4|G|$ 
then  S=G  Hence We are Done
Case 2: If S is not a subgroup.
As $T(e)=e$ For any automorphism T Hence
$e \in S$ 
Consider $x\in S$ then $T(x)=x^{-1}$  $T(x^{-1})=x\in S$ Hence $x^{-1} \in S$ That is for every $x\in S$ $x^{-1} \in S$  As S is not subgroup as assumed Therefore It must not satisfy closure. 
Define $a^{-1}S$={$a^{-1}x|\forall x \in S$}
There exist some $ a \in S $ such that $a^{-1}S \neq S$ as Closure of elements of S has not been Guarrented.
$a^{-1}S \neq S$
 So there is some $t\notin S$ but $ t\in a^{-1}S$ 
$t=a^{-1}x $ for some$ x \in S$ Therefore $at=x$ i.e $at\in S$
 So $T(at)=t^{-1}a^{-1}$=$T(t)T(a)=$$T(ta)$ 
This is true for any $t\in a^{-1}S$ 
For all $t\in a^{-1}x $ $ta=at$
$t \in N(a)$ and for any $t \in a^{-1}S$ this is true.
 there for  $a^{-1}S \subset N(a)$
We can show bijection between S and $a^{-1}S$ so Cardinality of both are same
$|a^{-1}S|=|S| \geq 3/4||G|$ 
 And N(a) is subgroup
$N(a)=G$. i.e $a \in Z(G)$ 
That means every $a \in S ,a \in Z(G)$ This implies $S=Z(G)$ which is subgroup 
which is contradiction to our assumption Hence Done . 
S is Subgroup S=G 
2) Example : $S_3$ take map 13 to 12 and 12 to 13 , 23 fix. You get 4 element in I . And so done
