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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 $m\mid n\implies |G|<\infty$

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    $\begingroup$ No, there are infinite groups with all non-identity of the same order. $\endgroup$ – Tobias Kildetoft Mar 15 '18 at 11:00
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    $\begingroup$ Note that $g^n=e_G$ does not mean that all elements have order $n$, but only that all elements have order a divisor of $n$. $\endgroup$ – Arnaud D. Mar 15 '18 at 11:02
  • $\begingroup$ Recall that in the definition of order, the $n$ such that $g^{n}=e$ is taken to be minimal. So, in $C_{4} = \langle g \rangle$, the element $g^{2}$ has order $2$, not order $4$. $\endgroup$ – preferred_anon Mar 15 '18 at 11:04
  • $\begingroup$ What is a nilpotent element in a group? $\endgroup$ – lhf Mar 15 '18 at 11:24
  • $\begingroup$ The word you're looking for is "exponent". Your question is whether a group is finite if and only if it has finite exponent. The answer is no (as some people have already pointed out.) $\endgroup$ – verret Mar 15 '18 at 17:42
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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.

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

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