Let $G$ be a group. Let $N\unlhd G$ with $G/N$ being cyclic. Let $H\leq G$. Show that $HN\unlhd G$.
I cannot figure out how to use the fact that $G/N$ is cyclic in this proof. Any help is much appreciated.
2026-04-05 18:29:16.1775413756
Let $G$ be a group. Let $N\unlhd G$ with $G/N$ being cyclic. Let $H\leq G$. Show that $HN\unlhd G$.
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As $G/N$ is abelian, $HN/N$ is a normal subgroup of $G/N$. By the correspondence between subgroups of $G/N$ and subgroups of $G$ containing $N$ (which preserves normality), this implies that $HN$ is normal in $G$