$G$ is a non-abelian finite group.$f$ is an automorphism of $G$ such that $f^2 = I_G$.Show $f(x) = x$ for some $x \neq e$. I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.  
There is the following problem in this book:  

Let $G$ be a non-abelian finite group.
  Let $f$ be an automorphism of $G$ such that $f^2 = I_G$.
  Show that there exists $x \in G$ such that $f(x) = x$ and $x \neq e$.  

My attempt is here:  
(1)
If $x' = f(x)$, then $x = f(f(x)) = f(x')$.
So, we can write as follows:
$x \stackrel{\mathrm{f}}{\longleftrightarrow} x'$.
This problem says there exists $y \neq e$ such that
$y \stackrel{\mathrm{f}}{\longleftrightarrow} y$.  
(2)
$H := \{x \in G | f(x) = x\}$ is a subgroup of $G$.
So, we need to show $H \neq \{e\}$.  
(3)
Let $G = \{x_1, x_2, \cdots, x_n\}$.
Let $f(x_i) = x_i'$ for $i \in \{1, 2, \cdots, n\}$.  
Then,  
$x_1 \stackrel{\mathrm{f}}{\longleftrightarrow} x_1'$.
$x_2 \stackrel{\mathrm{f}}{\longleftrightarrow} x_2'$.
$\cdots$
$x_n \stackrel{\mathrm{f}}{\longleftrightarrow} x_n'$.  
And  
$G = \{x_1, x_2, \cdots, x_n\} = \{x_1', x_2', \cdots, x_n'\}$.  
So,  
$x_1 x_2 \cdots x_n  \stackrel{\mathrm{f}}{\longleftrightarrow}  x_1' x_2' \cdots x_n' = x_1 x_2 \cdots x_n$.  
So,  
$x_1 x_2 \cdots x_n \in H$.  
But I cannot show that $x_1 x_2 \cdots x_n \neq e$.  
(4)
I don't use the assumption that $G$ is non-abelian yet.  
 A: Suppose $x_1 x_2 \neq x_2 x_1$. Then $x_2x_1...x_n\stackrel{\mathrm{f}}{\longleftrightarrow} x'_2x'_1...x'_n=x_2x_1...x_n$. Since $x_1x_2...x_n \neq x_2x_1...x_n$, we must have $H \neq \{e\}$.
A: Use the line of reasoning in  Show that a finite group with certain automorphism is abelian by Arturo Magidin [AM] and user9413 [U]: If the only fixed point of $f$ is $e$ then $G$ is abelian.
However, I found the first step to show that showing Claim 1 below [AM] [U], to be nontrivial so I worked out the proof for myself.

Claim 1: If $f$ as specified has only $\{e\}$ as the fixed point, then every $g \in G$ can be written as $x^{-1}f(x)$ for some $x \in G$.  

So here is my proof of Claim 1: Suppose $x$ and $y$ in $G$ satisfy $x^{-1}f(x) = y^{-1}f(y)$ [AM]. Then applying $f$ to both sides yields:
$f(x^{-1}f(x)) = f(x^{-1})f(f(x)) = f(x^{-1})x = f(y^{-1})y$.
Then this implies $f(y)f(x^{-1}) = yx^{-1}$, which implies $f(yx^{-1}) = yx^{-1}$, which would imply that $yx^{-1}$ is a fixed point of $f$, which, by the assumption that $e$ is the only fixed point, implies $yx^{-1} = e$, which implies $y=x$. Thus $h: x \mapsto x^{-1}f(x)$ is a one-to-one of $G$ onto itself, and [as $G$ is finite], implies that the image of $h$ is $G$ itself, implying Claim 1. 
A: Thank you very much, Izralbu, Tsemo Aristide, Mike.  
I write this answer for my memo.  
If $x' = f(x)$, then $x = f(f(x)) = f(x')$.
So, we can write as follows:  
$x \stackrel{\mathrm{f}}{\longleftrightarrow} x'$.  
This problem says there exists $y \neq e$ such that  
$y \stackrel{\mathrm{f}}{\longleftrightarrow} y$.  
(1)
Assume that there doesn't exist $y \neq e$ such that  
$y \stackrel{\mathrm{f}}{\longleftrightarrow} y$.  
(2-1)
We show that $G \ni x \longmapsto x^{-1} x' \in G$ is injective.  
Assume that $x_1^{-1} x_1' = x_2^{-1} x_2'$.
Then, $x_1 x_2^{-1} = x_1' x_2'^{-1}$.  
(2-2)
$x_1 \stackrel{\mathrm{f}}{\longleftrightarrow} x_1'$.
$x_2^{-1} \stackrel{\mathrm{f}}{\longleftrightarrow} x_2'^{-1}$.  
So, $x_1 x_2^{-1} \stackrel{\mathrm{f}}{\longleftrightarrow} x_1' x_2'^{-1} = x_1 x_2^{-1}$ by (2-1).  
(2-3)
By assumption (1), $x_1 x_2^{-1} = e$.
So, $x_1 = x_2$.
So, $G \ni x \longmapsto x^{-1} x' \in G$ is injective. 
(3)
By assumption, $G$ is finite.
So, $G \ni x \longmapsto x^{-1} x' \in G$ is surjective.
So, $G \ni x \longmapsto x^{-1} x' \in G$ is bijective.  
(4)
$\{x_1, x_2, \cdots, x_n\} = \{x_1^{-1} x_1', x_2^{-1} x_2', \cdots, x_n^{-1} x_n'\}$ by (3).  
(5)
Let $g$ be an arbitrary element of $G$.
Then, there exists $h \in G$ such that $g = h^{-1} h'$.
Then, $g = h^{-1} h' \stackrel{\mathrm{f}}{\longleftrightarrow} h'^{-1} h = (h^{-1} h')^{-1} = g^{-1}$.  
(6)
Let $x, y$ be arbitrary elements of $G$.
$x \stackrel{\mathrm{f}}{\longleftrightarrow} x^{-1}$ by (5).
$y \stackrel{\mathrm{f}}{\longleftrightarrow} y^{-1}$ by (5).
So, $x y \stackrel{\mathrm{f}}{\longleftrightarrow} x^{-1} y^{-1}$.
And, $x y \stackrel{\mathrm{f}}{\longleftrightarrow} (x y)^{-1}$ by (5).
$\therefore x^{-1} y^{-1} = (x y)^{-1}$.
$\therefore x^{-1} y^{-1} = y^{-1} x^{-1}$.
$\therefore x y = y x$.
$\therefore G$ is abelian.  
This contradicts the assumption that $G$ is non-abelian.  
