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Another question on Mathoverflow (here: Complete topological groups in which all subgroups are closed) asks if there exists a complete, nondiscrete topological group $G$ such that all subgroups of $G$ are closed.

Consider a finite group $G'.$ Consider the group $G=G'^\mathbb Z$ with the Tychonoff distance $$d(c,c')=\begin{cases}0,\mbox{ if }c=c'\\ 2^{-\min\{|j|:j\in \mathbb Z, c(j)\neq c'(j)\}}\mbox{ otherwise}\end{cases}.$$

(the topology induced by this (ultra)metric is the prodiscrete topology). $G$ with this topology is a complete, non-discrete topological group. The fact that the underlying set with this topology is complete and non-discrete can be found in the monographs by Tullio Ceccherini-Silberstein and Michel Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics. Springer, 2010 (MR2683112, Zbl 1218.37004). The fact that it is a topological group is almost trivial if you consider the above distance.

Is it true that every subgroup of this group is closed?

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    $\begingroup$ As per Moishe's comment on the linked answer, $\bigoplus_\Bbb Z G'$ is a dense, proper subgroup. More generally your group is uncountable and Polish, so for every $2<\alpha<\omega_1$ it will have a Borel subgroup which is $\mathbf{\Pi}^0_\alpha$ but not of lower complexity, by Farah, Ilijas, and Sławomir Solecki. "Borel subgroups of Polish groups." Advances in Mathematics 199.2 (2006): 499-541. $\endgroup$
    – Alessandro Codenotti
    Commented Nov 5 at 10:28
  • $\begingroup$ When you write the power $F^X$, it means all functions $X\to F$, not only finitely supported ones. One notation for the restricted power (elements of the power with finite support) is $F^{(X)}$. $\endgroup$
    – YCor
    Commented Nov 5 at 12:12
  • $\begingroup$ However, this countable group ($F^{(X)}$, $F$ nontrivial finite, $X$ infinite countable, with induced topology from the product topology on $F^X$) has a non-closed abelian subgroup. First if $F$ is abelian, the homomorphism $F^{(X)}\to F$, $f\mapsto \sum_x f(x)$, has a dense kernel. In general, let $C$ be a nontrivial cyclic subgroup of $F$, and then $C^{(X)}\subset F^{(X)}$ has a non-closed subgroup. $\endgroup$
    – YCor
    Commented Nov 5 at 12:17
  • $\begingroup$ And in any case, $F^{(X)}$ is not complete then (it's a dense proper subgroup of the compact group $F^X$). $\endgroup$
    – YCor
    Commented Nov 5 at 12:18
  • $\begingroup$ I meant all the functions, not only the finitely supported ones. However, now I see the point. If any of you post the answer, I’ll accept it. $\endgroup$
    – Nick Belane
    Commented Nov 5 at 12:45

1 Answer 1

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The metric induces the product topology, so the group $G$ is compact. The direct sum of $\mathbb{Z}$ many copies of $G'$ is a countable dense subgroup, but not the whole group, so it is not closed.

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