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I know this question is already answered but My argument is different I wanted t check that is I am giving right arrgument?
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
Any Help will be appreciated

Curious student
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    Your proof looks correct mathematically but your presentation could use some work. – Asvin Aug 07 '18 at 07:11
  • It is very hard to read. Just try to use the point at the many right points. The idea of proof is not transparent. "This is true for any $t$..." is unclear. What is "This"? Why not state the corresponding property first. The inclusion $a^{-1}S\subset N(a)$ is unclear, and it is very unfortunate to have this written without point and new sentence. How should people digest $$ a^{-1}S \subset N(a)\ |a^{-1}S|=|S| \geq 3/4||G\ ?$$ – dan_fulea Aug 07 '18 at 09:14
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    @dan_fulea Sorry Sir For My Mistake .I had edited question so now it become quite readable.Please Suggest me my mistake .Thanks for helping me. – Curious student Aug 07 '18 at 11:12
  • It is still useful, I think, if you compare your proof with the proofs given at the duplicates, where several people put real effort to explain it. – Dietrich Burde Aug 08 '18 at 10:23

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