TITLE: Group of even order contains an element of order 2
QUESTION [41 upvotes]: I am working on the following problem from group theory:

If $G$ is a group of order $2n$, show that the number of elements of $G$ of order $2$ is odd.

That is, for some integer $k$, there are $2k+1$ elements $a$ such that $a \in G,\;\; a*a = e$, where $e$ is the identity element of $G$.

REPLY [5 votes]: If $G$ is a group of order $2n$, show that the number of elements of $G$ of order $2$ is odd.  

Define an equivalence relation on $G$ by $x\sim y$ if and only if $x=y$ or $x=y^{-1}$.
Let $C_a := \{x \in G | a \sim x\}$.
$\#C_a = 1$ if and only if the inverse of $a$ is $a$.
$\#C_a = 2$ if and only if the inverse of $a$ is not $a$.
Note that the inverse of $a$ is unique.  
Let $X := \{C_a | \#C_a = 1\} = \{C_{a_1}, \cdots, C_{a_l}\}$.
Let $Y := \{C_b | \#C_b = 2\} = \{C_{b_1}, \cdots, C_{b_m}\}$.
$C_e = \{e\}$.
So, $l \geq 1$.  
Then, $\#G = l + 2 m$.
Since $\#G$ is even, $l$ must be even.  
$\#C_a = 1$
$\Leftrightarrow$
the inverse of $a$ is $a$
$\Leftrightarrow$
$a^2 = e$,
and  
$e^1 = e$. 
So, the number of elements of $G$ of order $2$ is $l-1$.
And $l-1$ is odd.