Property of the order of an element
Let
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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 $g$.
Let
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 $g$.