If G is a group of even order, prove it has an element $a \neq e$ satisfying $a^2 = e$. [duplicate]

If $G$ is a group of even order, prove it has an element $a \neq e$ satisfying $a^2 = e$.

My proof:

Let $|G| = 2n$. Since $G$ is finite, there exists, $a \in G$ such that $a^p = e$ and by Lagrange's Theorem, p divides 2n. By Euclid's lemma, since p does not divide 2, p divides n. Let $n = pk$. Hence, $(a^n)^2 = (a^{pk})^2 = ((a^p)^k)^2 = (e^k)^2 = e$. Therefore, $a^n$ is an element that satisfy the condition.

Is my solution OK?

For this problem, I am just wondering how I can solve this problem without using Lagrange's Theorem, as this problem is an exercise before the Lagrange's Theorem was taught.

Solution 1:

The following is perhaps one of most simple proofs:

Pair up if possible each element of $\;G\;$ with its inverse, and observe that

$$g^2\neq e\iff g\neq g^{-1}\iff \;\text{there exists the pair}\;\;(g, g^{-1})$$

Now, there is one element that has no pairing: the unit $\;e\;$ (since indeed $\;e=e^{-1}\iff e^2=e$), so since the number of elements of $\;G\;$ is even there must be at least one element more, say $\;e\neq a\in G\;$ , without a pairing, and thus $\;a=a^{-1}\iff a^2=e\;$

Solution 2:

This is not a strict proof, but you may find it helpful when you want proof without Lagrange's theorem:

We have that for every $g\in G$ there a unique $g^{-1} \in G$ such that $gg^{-1}=e$. If you suppose that there is no $a \in G$ such that $a^2=e$, so that $a=a^{-1}$ (i.e. there is no self-inverse element), then for every $x\neq e$ in $G$ we can assign unique $y\in G$ such that $xy=e$. So the set of pairs of elements that are inverses to each other form a partition for $G$.

But then $|G|$ is odd since $e$ is only self-inverse element.