Let $G$ be a group. If there exists a positive integer $m$ such that $a^m=1$ for all $a\in G$, then is $G$ finite?

Question

Let $G$ be a group. If there exists a positive integer $m$ such that $a^m=1$ for all $a\in G$, then is $G$ finite? That is, all elements should have order that divides the fixed integer $m$. (The set of orders of elements of $G$ is a finite set.)

I have no idea how to prove or disprove this statement. Is there any counterexample for this?

Answer 1

What about $(\mathbb{Z}_2[x],+)$? Every element has order $2$.