Every locally compact topological group is normal space
Let $U$ be a neighbourhood of the unity element $e$ of the locally compact topological group $G$. $U$ can be taken open, relatively compact and symmetric, since $U=V \cap V^{-1}$ for some open neighbouhood $V$ of $e$. Let $H=\cup_{n\in N} U^{n} = \cup_{n\in N} \bar{U}^{n}$ be an open, therefore closed subgroup of $G$. [: It is easy to see $\cup_{n\in N} U^{n}$ is an open subgroup. Moreover, $\cup_{n\in N} U^{n}= \overline{\cup_{n\in N} U^{n}}\supset \cup_{n\in N} \overline{U^{n}}\supset \cup_{n\in N} U^{n}$]. It is a countable union of the compact sets $\bar{U}^{n}$, so it is a σ-compact space, therefore Lindelof, therefore normal, since Lindelof and regularity gives normality. And topological groups are always regular: Let $U$ be an open neighbourhood of $e$ and $V$ a symmetric open neighbourhood of $e$ such that $V^{2}\subset U$. Let $x\in \bar{V}$. Then $xV\cap V \neq \emptyset$, so there exist $v_{1},v_{2} \in V$ with $xv_{1}=v_{2}$, which gives $x=v_{2}(v_{1})^{-1}\in VV^{-