# 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:

**Theorem.**Let \(G\) be a topological group. If \(H\) is a locally closed subgroup in \(G\), then \(H\) is closed.

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.

**Remark.** Let \(X\) be a topological space and \(S\) be a subset of \(X\). If \(S\) is open, then for any \(s \in S\), there exists an open neighborhood \(U\) of \(s\) in \(X\) such that \(U \subset S\). But this implies \(S \cap U=U\) is closed in \(U\), so \(S\) is locally closed. This shows any open set is locally closed. Hence, if \(H\) is an open subgroup in a topological group \(G\), then \(H\) is forced to be closed by the **Theorem**.

**Proposition 1.**Let \(X\) be a topological space and \(S\) be a subset of \(X\). Let \(\overline{S}\) be the closure of \(S\) in \(X\). Then the following are equivalent.

(1) \(S\) is locally closed.

(2) \(S\) is open in \(\overline{S}\).

(3) \(S\) is an intersection of an open set and a closed set.

*Proof.* (1)\(\Longrightarrow\)(2): Suppose \(S\) is locally closed. For any \(s \in S\), since \(S\) is locally closed, there exists a neighborhood \(U\) of \(s\) in \(X\) such that \(U \cap S\) is closed in \(U\). Let \(V\) be the open set containing \(s\) in \(U\), then \(V \cap S\) is also closed in \(V\). Since \(V\) is open, \(V \cap \overline{S}\) is the closure of \(V \cap S\) in \(V\), so \(V \cap \overline{S} = V \cap S \subset S\). Hence, we find an open neighborhood \(V \cap \overline{S}\) of \(s\) in \(\overline{S}\) that is contained in \(S\), so \(S\) is open in \(\overline{S}\).

(2)\(\Longrightarrow\)(3): Suppose \(S\) is open in \(\overline{S}\). Then there exists an open set \(U\) in \(X\) such that \(S = U \cap \overline{S}\). Therefore, \(S\) is an intersection of an open set and a closed set.

(3)\(\Longrightarrow\)(1): Suppose \(S=U \cap C\), where \(U\) is open and \(C\) is closed. Then for any \(s \in S\), take \(U\) as the neighborhood of \(s\) in \(X\), then \(U \cap S=U \cap U \cap C=U \cap C\) is closed in \(U\), because \(C\) is closed. Hence, \(S\) is locally closed. \(\square\)

**Proposition 2.**Let \(X\) be a Hausdorff topological space. If \(S\) is a locally compact subspace in \(X\), then \(S\) is locally closed.

*Proof.* For any \(s \in S\), since \(X\) is Hausdorff and \(S\) is locally compact, there exists an open set \(U\) containing \(s\) in \(X\) such that \(\overline{U} \cap S\) is compact in \(S\), where \(\overline{U}\) is the closure of \(U\) in \(X\). In particular, it is compact in \(X\). Therefore, \(\overline{U} \cap S\) is closed in \(X\). Then \(U \cap S = U \cap (\overline{U} \cap S)\) is closed in \(U\). Hence, \(S\) is locally closed.
\(\square\)

We need some lemmas before going into the proof of the **Theorem**.

*Pf.* Clearly, \(\overline{H}\) is closed. It remains to prove \(\overline{H}\) is a subgroup, i.e., for \(a, b \in \overline{H}\), we need to show \(a^{-1} \in \overline{H}\) and \(ab \in \overline{H}\).

Let \(a \in \overline{H}\). Since \(g \mapsto g^{-1}\) is homeomorphism of \(G\) to itself, for any open neighborhood \(U\) of \(a^{-1}\) in \(G\), \(U^{-1}=\{u^{-1}: u \in U\}\) is an open neighborhood of \(a\). Then \((U^{-1}-\{a\}) \cap H \ne \varnothing\), since \(a \in \overline{H}\). Therefore, \((U-\{a^{-1}\}) \cap H \ne \varnothing\), which means \(a^{-1} \in \overline{H}\).

Let \(a, b \in \overline{H}\). Since \(m: G \times G \to G\) defined by \(m(xy)=xy\) is continuous, for any open neighborhood \(U\) of \(ab\) in \(G\), \(m^{-1}(U)\) is open in \(G \times G\). Since \(ab \in U\), \((a, b) \in m^{-1}(U)\), there there exists \(V\) and \(W\), open neighborhood of \(a\) and \(b\) respectively, such that \(V \times W \subset m^{-1}(U)\). Since \(a, b \in \overline{H}\), \((V-\{a\}) \cap H \ne \varnothing\) and \((W-\{b\}) \cap H \ne \varnothing\), so \(({m(V \times W)-ab}) \cap H \ne \varnothing\). Therefore, \((U-\{ab\}) \cap H \ne \varnothing\). This shows \(ab \in \overline{H}\). \(\vartriangleleft\)

*Pf.* Since \(U\) and \(V\) are open, so is \(U \cap V\). For any \(x \in X\), for any open neighborhood \(W\) of \(x\), since \(U\) is dense, \(U \cap W\) is a nonempty open set. Hence, \(U \cap V \cap W\) is nonempty, because \(V\) is dense. Therefore, \(U \cap V\) is open dense in \(X\).
\(\vartriangleleft\)

*Proof of the Theorem.* Let \(\overline{H}\) be the closure of \(H\) in \(G\). \(\overline{H}\) is a subgroup by

**Lemma 1**. Because \(H\) is locally closed, we can apply the

**Proposition 1**above to conclude that \(H\) is open dense in \(\overline{H}\). Suppose \(H \ne \overline{H}\), then \(\overline{H}= \bigcup_{g_i} g_i H\), a disjoint union of left cosets of \(H\) in \(\overline{H}\). Since \(H \ne \overline{H}\), there exists at least two disjoint left cosets. But \(H\) is open dense in \(\overline{H}\), all its left cosets are also open dense in \(\overline{H}\), so by

**Lemma 2**, any two of them have nonempty intersection, which contradicts the fact that they are disjoint. Hence, \(H=\overline{H}\), so \(H\) is closed. \(\square\)

**Corollary.**Let \(G\) be a topological group and \(H\) be a subgroup of \(G\). Then if either one of the following conditions holds, then \(H\) is closed:

(1) \(H\) is an open subgroup;

(2) \(G\) is a \(T_1\) space and \(H\) is discrete;

(3) \(G\) is a Hausdorff space and \(H\) is locally compact.

*Proof.* (1) See **Remark** above.

(2) If \(G\) is \(T_1\), then sets containing single point are closed, so discrete sets are locally closed. Hence, \(H\) is locally closed, and therefore closed.

(3) This follows immediately from **Proposition 2** and the **Theorem**.
\(\square\)