Compactly generated locally compact groups

Let $G$ be a locally compact Hausdorff topological group (Topological group: multiplication and inversion are continuous, Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not). It is compactly generated if some compact subset $K\subseteq G$ (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right) generates $G$ as an abstract group, that is, $⟨K⟩=G$ (The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups). Equivalently, there is a compact subset $Q\subseteq G$ with $e\in Q$ and $Q=Q^{-1}$ such that $G={\bigcup }_{n\ge 1}{Q}^n,Q^n=\{q_1\cdots q_n:q_1,\dots ,q_n\in Q\}\mathrm{.}$

For the equivalence, given a compact generating set $K$, take $Q=K\cup K^{-1}\cup \{e\}$. Inversion is a homeomorphism, so $K^{-1}$ is compact; a finite union of compact subsets is compact by the open-cover definition. The set $Q$ is symmetric and contains $e$. Its positive finite products form a subgroup: they contain $e$, are closed under concatenation, and are closed under inverses by symmetry. This subgroup contains $K$ and is contained in every subgroup containing $K$, so it is exactly $⟨K⟩$; the copies of $e$ pad shorter words to any larger exponent. Conversely, if a symmetric identity-containing $Q$ has $G={\bigcup }_{n\ge 1}{Q}^n$, then every element of $G$ belongs to the subgroup generated by $Q$.

Every compact group is compactly generated, by taking $K=G$. More generally, every connected locally compact Hausdorff group is compactly generated, so in particular this holds for the connected $\sigma$-compact case. Indeed, choose a compact neighbourhood $K$ of the identity (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open) and let $H=⟨K⟩$. Since $K$ contains an open neighbourhood $U$ of the identity, $H$ contains $U$ and is open: for each $h\in H$, the open set $hU$ lies in $H$. Every coset of an open subgroup is open, so the complement of $H$ is open as well. Thus $H$ is clopen, and connectedness forces $H=G$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets). No choice principle is used.