Let $G$ be a group. If $g\in G$ has finite order $m$ and $H$ is a normal subgroup of $G$, then show that the order of the element $Hg\in G/H$ is finite and divides $m$.
2026-03-31 07:57:18.1774943838
Divisibility of the an Element of a Subgroup
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Hint:
If $H$ is a normal subgroup, for any $g\in G$, and any $k\in \mathbf Z$, $\;(Hg)^k=Hg^k$.