Group of even order contains an element of order 2

Hint. Define an equivalence relation on $G$ by $x\sim y$ if and only if $x=y$ or $x=y^{-1}$. Then remember that an equivalence relation partitions a set.

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.