Group of even order contains an element of order $2$ (Explanation)

Question

I am working on the problem:

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

which already has solution (in the link below), please explain the solution in that link

Group of even order contains an element of order 2

Answer 1

OK, the most popular answer defines a relation $x \sim y$ iff $x=y$ or $x=y^{-1}$. You can check that this is indeed an equivalence relation and so we have the set of classes $C_i$, $i \in I$, which form a disjoint partition of $G$. Every class has size $1$ or $2$ (size $1$ iff $C_i = \{x\}$ with $x = x^{-1}$ or equivalently $x^2 = e$. If $x$ is not $e$ and does not have order $2$, it is in a class of size $2$, namely $\{x, x^{-1}\}$. There is at least one class of size $1$, namely $\{e\}$.

If $a$ is the number of classes of size $2$, it has $b:=|I|-a $ classes of size $1$, and $|G| = 2a + b$ and it follows that $b=|G|-2a$ is even, so $b \ge 2$ and so there is at least 1 class besides $\{e\}$ that has size $1$, and this is an element of order $2$.