Converse of "every finite group has a finite exponent"?

Question

A group $G$ is finite iff all its elements have the same finite order?

One direction (when $G$ is finite) is trivial following from Lagrange's theorem. But if a each element of $G$ has the same finite order (says $n<\infty$), i.e. $\forall g\in G$, $g^n=e_G$. Will that implies $G$ is finite? I try to find some counter examples like infinite group consists of only nilpotent elements (not nilpotent group), but it doesn't work. So the statement is true?

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Edit: The statement should be "A group $G$ is finite iff the order of any elements divides a fixed finite number n?" So one direction should be: Given $n<\infty$, $\forall g\in G$, if $g^{m}=e_G\neq g^{m-1}$, then $m\mid n\implies |G|<\infty$

Answer 1

Try $$\mathbb Z /2\mathbb Z \times \mathbb Z /2\mathbb Z \times\mathbb Z /2\mathbb Z \times\dots.$$ Every element has order 2.

Answer 2

No, take for example the infinite group $G=\prod_{n=1}^\infty{\mathbb Z}_2$ where every non-identity element has order 2.