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.

  • 2
    $\begingroup$ 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. $\endgroup$ Jun 15, 2022 at 9:56

1 Answer 1


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.

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

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