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.
Best Answer
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.