# Proof that group of even order contains element of order 2

I tried to prove the following claim:

If $G$ is a group of even order then $G$ contains an element of order $2$.

Please could someone check my proof and tell me if it is correct?

My proof:

Let $|G|=2n$ for some $n$ and let $\{k_1, \dots, k_{2n}\}$ be the set of all orders of elements of $G$.

If there exists any $g\in G$ such that $k_i$ is even, say, $k_i = 2s$, then $g^{s}$ has order $2$.

If all $k_i$ are odd pick any $k_i$ and the corresponding element $g$. Then since $k_i$ divides $2n$ we have $2n = sk_i = 2tk_i$ for some $t$. Then $g^{tk_i}$ has order $2$.

This concludes the proof.

• This doesn't work. By hypothesis, $g$ has order $k_i$, so $g^{tk_i} = 1$. Note that because you assume that all of the $k_i$ are odd the only way this branch of the proof can end is via a contradiction. Nov 8, 2015 at 8:37
• Here's a cute strategy that doesn't obviously generalize: look at the map $g \mapsto g^{-1}$. An element of order $2$ is a nontrivial fixed point of this map. Can you show that there must be such a nontrivial fixed point? Nov 8, 2015 at 8:39
• @QiaochuYuan Regarding your second comment: Just to be sure, this suggestion is what the answer by Anurag A is doing, right? Nov 9, 2015 at 1:12
• Yes, that's correct. Nov 9, 2015 at 2:10

Consider the set $G-\{e\}$. Its cardinality must be odd because $|G|$ is even.
If no element has order $2$ in $G$ then for every $g \in G-\{e\}$ there exists its inverse $g^{-1} \in G$ with $g \neq g^{-1}$. But then we will have an even number of elements in $G-\{e\}$. A contradiction.