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How can I prove directly that a locally compact topological group G is normal? I have done this by showing that every locally compact topological group is strongly Paracompact. But I could not prove it directly.

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What do you mean by "normal group"? – Davide Giraudo Jan 14 '12 at 19:37
@Davide I believe he means the topology is normal. – David Mitra Jan 14 '12 at 19:58
yes, I mean normal topological space. – Abcd J Jan 15 '12 at 6:20
Yes, as every locally compact topological group is a disjoint sum of $\sigma$-compact groups, and each of these is regular Lindelöf, hence normal, and a sum of normal spaces is normal. – Henno Brandsma Jan 15 '12 at 10:16
I think hausdorffness is a necessary condition for group $G$. – M.Sina Jan 13 '13 at 10:11


Start with a precompact symmetric open neighbourhood $U$ of the identity and form the subgroup $H = \cup_{n \in \mathbb N} U^n$. Show that $H$ is clopen and $\sigma$-compact and conclude that it's Lindelöf.

Now take an open covering $\mathcal V$ of $G$. Use that $H$ is Lindelöf to find a $\sigma$-locally finite open refinement $\mathcal W$ of $\mathcal V$.

Finishing up the proof of the result, you should know a theorem that says that preregular paracompact spaces are normal.

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Hausdorfness is not required for the group. Let $H$ be a clopen $\sigma$-compact subgroup of $G$. $H$ exists and is normal. Thus $G$ is normal. This requires the knowledge that given $G$ a topological group and $H \subseteq G$ an open subgroup, $G$ is normal iff $H$ is normal.

Proof of my last statement is fairly easy as well. $(\Rightarrow)$, $H$ is closed, done. $(\Leftarrow)$ Each coset of $H$ is homeomorphic to $H$ and clopen. Done

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