TITLE: If $g\neq g^{-1}$ for all $g\in G\setminus \{e\}$, then the order of $G$ is odd
QUESTION [1 upvotes]: Given a group $G$ where $g\neq g^{-1}$ for all $g$ other than the identity, show the order of $G$ is odd.

What does it mean for the order of group $G$ to be odd?
Any help welcome.

REPLY [2 votes]: If $p$ is prime and divides the order of group $G$ then $G$ will contain an element with order $p$ (Cauchy). Note that $2$ is prime. So if the order of $G$ is even then $g^{2}=e$ (or equivalently $g=g^{-1}$) for some $g\in G\backslash\left\{ e\right\} $. This is not the case, so the order of $G$ must be odd.
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