Property of the order of an element

#group_theory

Let $G$ be a group, $m \in N$ , and $g \in G$ be an element such that $g^{m} = e$ . Prove that $o (g) ∣ m$ , where $o (g)$ is the order of $g$ .

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Let $G$ be a group, $m\in {\bf N}$, and $g\in G$ be an element such that $g^m=e$. Prove that $\operatorname{o}(g)\mid m$, where $\operatorname{o}(g)$ is the order of $g$.