# $\sigma$-compactness of some locally compact Hausdorff topological groups

Is the topological group $$(\mathbf{Q}_p/\mathbf{Z}_p)^{\oplus k}$$, $$k\ge 1$$, a $$\sigma$$-compact topological group when endowed with its natural $$p$$-adic topology?

More generally, I'm looking for a criterion for locally compact Hausdorff topological groups to not contain a nested exhaustive sequence of compact subgroups (i.e. locally compact Hausdorff topological groups $$G$$ with a countable family of nested compact subsets $$K_n$$ such that $$\bigcup_{n\ge 0}K_n = G$$).

Example Real and complex Lie groups are $$\sigma$$-compact exactly when they are compact.

Example The same group $$(\mathbf{Q}_p/\mathbf{Z}_p)^{\oplus k}$$, endowed with the discrete topology. It is uncountable, so it's not $$\sigma$$-compact when endowed with the discrete topology, if I'm not missing anything.

• A second comment: I think $\sigma$-compact means a union of compact subspaces, which seems very different to an increasing sequence of compact subgroups. I think many non-compact Lie groups are $\sigma$-compact. Jun 15, 2022 at 9:56

I agree with @MatthewDaws ... for example the real Lie group $$(\mathbb R,+)$$ with the usual topology is locally compact and sigma-compact but not compact.

Take any locally compact group $$G$$. There is a neighborhood $$V$$ of $$e$$ with compact closure. The subgroup $$H$$ generated by $$V$$ is an open, locally compact, sigma-compact subgroup of $$G$$. The coset space $$G/H$$ is discrete.

The example. The group $$\mathbb Q_p$$ of $$p$$-adic numbers is locally compact. The subgroup $$\mathbb Z_p$$ of $$p$$-adic integers is an open, compact, subgroup. The quotient $$\mathbb Q_p/\mathbb Z_p$$ is countable and discrete.

Indeed, $$\mathbb Q_p = \bigcup_{n=0}^\infty p^n \mathbb Z_p .$$

For a fixed $$k \in \mathbb N$$, also $$(\mathbf{Q}_p/\mathbf{Z}_p)^{\oplus k}$$ is countable and discrete.

• Yes, I understand. You're of course correct. Thanks
– user480741
Jun 15, 2022 at 10:47