Locally closed subgroups are closed
In a topology group \(G\), an open subgroup \(H\) is also closed. The proof of this statement is not hard: \(G=\bigcup_{g_i} g_iH\) is a disjoint union of open left cosets, where \(\{g_i\}\) is a complete representatives set of \(G/H\). Then \(\bigcup_{g_i \ne 1}g_iH\) is open and \(H=G-\bigcup_{g_i \ne 1}g_iH\) is the complement of an open set, and therefore \(H\) is closed. In this post, I will prove a slightly more general theorem:
We will see that an open subgroup \(H\) is locally closed, so it’s closed by the Theorem. Before the proof of the Theorem, let’s talk about locally closed sets first.