Let $G$ be a compactly generated locally compact group, let $H$ be a subgroup (not necessarily closed) and let $U$ be a compact identity neighbourhood in $G$. Suppose that $G = HU$. Does there exist a finitely generated subgroup $K$ of $H$ such that $G = KU$?

If not, what happens if we assume $H$ is dense? What if $G$ is a connected Lie group?


The answer is 'yes'. I found a lemma in T. Tao's book 'Hilbert's Fifth Problem and Related Topics' (although probably the argument is much older) that is close to what I want:

Lemma 4.3.1: Let $G$ be a locally compact group. Then there exists an open subgroup $G'$ of $G$ that contains a cocompact finitely generated subgroup.

Actually, Tao's proof gives $G' \subseteq \langle F_1 \rangle U$ where $F_1$ is a finite subset of $H$ (Indeed, $F_1$ can be taken from any subset $A$ of $G$ such that $U^2 \subseteq AU$.) In particular, we can ensure $\langle F_1 \rangle U$ contains an almost connected open subgroup $V$ of $G$.

We can now pass to the totally disconnected quotient $G/G^\circ$ of $G$, where $V/G^\circ$ is a compact open subgroup. It's then a standard exercise in totally disconnected locally compact groups to obtain a finite subset $F_2$ of $H$ such that $VF_2V = F_2V = VF_2$ and such that $G = \langle F_2,V\rangle = \langle F_2 \rangle V$, so $G = \langle F_1 \cup F_2 \rangle U$. See for instance Lemma 2 of this paper: https://homepage.univie.ac.at/bernhard.kroen/KroenMoeller_lastversion.pdf; to take the elements from $H$, we just need to use the fact that $H$ is transitive on left cosets of $V$.


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