Let $G$ be a finite group. Let $T$ be an element of $\operatorname{Aut}(G)$ such that $T^2=I$, and $xT=x\Leftrightarrow x=e$. Then, $G$ is abelian.

Question

I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
The following problem is Problem 11 on p.70 in Herstein's book.

Let $G$ be a finite group, $T$ an automorphism of $G$ with the property that $xT=x$ if and only if $x=e$. Suppose further that $T^2=I$. Prove that $G$ must be abelian.

I cannot solve this problem.

How to use the assumption $T^2=I$ to solve this problem?

My observation:
If $a\neq e$, then $b:=aT\neq a$.
And $bT=(aT)T=aT^2=aI=a$.
So, we can write $G=\{e,a_1,a_1T,a_2,a_2T,\dots,a_n,a_nT\}$.
So, $\#G=2n+1$.

Answer 1

My solution of this problem is the following:

Let $\phi$ be a mapping $\phi: G\ni x\mapsto x^{-1}(xT)\in G$.
If $\phi(x)=\phi(y)$, then $x^{-1}(xT)=y^{-1}(yT)$.
Then, $(xT)(yT)^{-1}=xy^{-1}$.
Then, $(xT)(y^{-1}T)=xy^{-1}$.
Then, $(xy^{-1})T=xy^{-1}$.
By the assumption of Problem 10, $xy^{-1}=e$.
So, $x=y$.
So, $\phi$ is injective.
Since $G$ is a finite set, $\phi$ is also surjective.
So, for every $g\in G$, there exists $x\in G$ such that $g=x^{-1}(xT)$.

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I will write Karthik Kannan's answer here:

Let $g$ be an arbitrary element of $G$.
Then, by Problem 10, there exists $x\in G$ such that $g=x^{-1}(xT)$.
Then, $gT=(x^{-1}(xT))T=(x^{-1}T)(xT^2)=(xT)^{-1}(xI)=(xT)^{-1}x=(x^{-1}(xT))^{-1}=g^{-1}$.
So, $gT=g^{-1}$ holds for every $g\in G$.
Let $a,b$ be arbitrary two elements of $G$.
Then, $(ab)^{-1}=(ab)T=(aT)(bT)=a^{-1}b^{-1}=(ba)^{-1}$.
So, $ab=ba$.