Construction of a given neighbourhood in a locally compact group

Question

Let $G$ be a locally compact group. Why is it possible to select a compact neighbourhood $U$ of $e \in G$ such that $U=U^{-1}$ and $gU^2 \subset V$?

This is a construction quickly stated by Helgason, but I don't understand how it is done.

Answer 1

It seems the following.

I assume that you mean $G$ is a locally compact (Hausdorff) topological group (that is, both the multiplication $\cdot:G\times G\to G$, $(x,y)\to xy$ and the inversion $\cdot:G\to G$, $x\to x^{-1}$ on the group $G$ are continuous).*

Now let $V\subset G$ be an arbitrary open set and $g\in V$ be an arbitrary element. Then $g\cdot e\cdot e =g\in V$. Since the space $G$ is locally compact (Hausdorff), the unit $e$ has a closed compact neighborhood $U_0$. The continuity of the multiplication on the group $G$ implies that there is a closed neighborhoods $U_1$ of the unit such that $g\cdot U_1\cdot U_1\subset V$. The continuity of the inversion at the unit of the group $G$ implies that there is a closed neighborhoods $U_2$ of the unit such that $U_2^{-1}\subset U_1\cap U_0$. It rests to put $U=U_2\cap U_2^{-1}$.

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* These conditions are conditional. :-) First of all, each $T_0$ topological group is Tychonoff. Next, by Ellis Theorem, if $G$ is a locally compact regular group endowed with a topology, making shifts $l_a:G\to G$, $x\to ax$ and $r_a:G\to G$, $x\to xa$ continuous for each element $a\in G$, then $G$ is a topological group. Moreover, if the group $G$ is locally compact and the multiplication on $G$ is continuous, then we don’t even need the regularity, because in this case, by my theorem, :-) the inversion on the group $G$ is continuous too.