I was reading this post here . Why is it sufficient to say that $\ker(\pi)$ contains an open subgroup to conclude that $\ker(\pi)$ is an open subgroup?
2026-04-05 20:37:41.1775421461
Proof Question on Open Kernel
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Let $H$ be an open subgroup contained in $G=\ker(\pi)$. Then we can find an open neighbourhood $U\subseteq H$ containing the identity. Then $$G=\bigcup_{g\in G}Ug,$$ and is open.