I would like to say that $\Bbb Q / \Bbb Z$ is not discrete (when $\Bbb Q$ has euclidean topology), since $\Bbb Z \subset \Bbb Q$ is not open. But OTOH we have $$\Bbb Q / \Bbb Z \cong \bigoplus_p \Bbb Q_p / \Bbb Z_p$$ which is a direct sum of discrete groups, so it should be a discrete group. Maybe the issue is that the above isomorphism is only as abstract groups, but not as topological groups.

Could anyone confirm/elaborate on this?

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    $\begingroup$ I think you're right, the above isomorphism is only an isomorphism of the underlying group structures. $\endgroup$ – user279515 Nov 29 '18 at 9:53
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    $\begingroup$ Note that $\Bbb Q$ with the Euclidean topology is isomorphic to $\Bbb Q$ with the discrete topology, if the isomorphism only requires to respect the group action and not the topology. $\endgroup$ – Asaf Karagila Nov 29 '18 at 9:54
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    $\begingroup$ The completion of the metric space $\mathbb{Q}/\mathbb{Z}, d(a,b) = \min_n |a-bn|$ is $\mathbb{R}/\mathbb{Z}$ while with the discrete metric $\tilde{d}(a,b) = \sup_p \inf_n |a-b n|_p$ it is complete and $\cong \bigoplus_p \Bbb Q_p / \Bbb Z_p$. $\endgroup$ – reuns Nov 29 '18 at 10:20
  • $\begingroup$ @reuns Frequently complete and non-complete-metrics on a set $X$ .induce the same topology. $\endgroup$ – Paul Frost Nov 29 '18 at 11:25
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    $\begingroup$ @PaulFrost For two group-invariant metrics it should be sufficient that the completion are not the same, I should have said that ? $\endgroup$ – reuns Nov 29 '18 at 11:52

A non-trival group $G$ can always be endowed with various distinct topologies making it a topological group. Two "universal choices" are the discrete topology and the indiscrete topology. See What is, exactly, a discrete group?.

However, you consider $G = \Bbb Q / \Bbb Z$ and emphasize that $\Bbb Q$ has the Euclidean topology. In that case the only reasonable topology on $G$ will be the quotient topology which is certainly not discrete.

The isomorphism $\Bbb Q / \Bbb Z \cong \bigoplus_p \Bbb Q_p / \Bbb Z_p$ is therefore only an algebraic isomorphism, not an isomorphism of topological groups.

Edited: I just considered which topology is given to an infinite sum $\bigoplus_{\alpha \in A} G_\alpha$ of abelian topological groups. There are various approaches, see for example

Higgins, P. J. "Coproducts of topological Abelian groups." Journal of Algebra 44.1 (1977): 152-159.


Chasco, M. J., and X. Dominguez. "Topologies on the direct sum of topological abelian groups." Topology and its Applications 133.3 (2003): 209-223.


In my opinion the conclusion is that an infinite sum of discrete abelian topological groups is not given the discrete topology. Whether one of the "reasonable" topologies on $\bigoplus_p \Bbb Q_p / \Bbb Z_p$ makes it isomorphic as a topological group to $\Bbb Q / \Bbb Z$ is not known to me.


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