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Let $H$ be subgroup of a topological group $G$. Suppose $H$ is the (internal) direct product of two of its subgroups $K_1$ and $K_2$. Does it follow that $\bar{H}$ is the direct product of $\bar{K_1}$ and $\bar{K_2}$?

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Well, depends on which direct product you mean, but if you mean only group product, you can find a very simple example.

Let $G=(\mathbb{R},+)$. Let $K_1=\mathbb{Z}$ and $K_2=\alpha\mathbb{Z}$ where $\alpha$ is irrational. Then $H = \{u+v\alpha: u,v\in\mathbb{Z}\} \cong K_1\oplus K_2$. But $H$ is dense in $\mathbb{R}$, so $\bar{H}=\mathbb{R}$, but $K_1$ and $K_2$ are closed in $\mathbb{R}$.

In this case $H$ is not homeomorphic to $K_1\times K_2$ as a topological space. I'm not sure what you can say if you require that the group isomorphism between $H$ and $K_1\oplus K_2$ is also a homeomorphism.

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Yes this is what I meant. Thanks for the example. –  Mustafa Gokhan Benli Sep 30 '11 at 15:58
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I think some topological abelian groups have dense free abelian subgroups, but are not themselves free. A direct factor of a free abelian group is free, and a direct product of two free abelian groups is free, so this would give a counterexample.

A specific example should be the direct product (with product topology, I think, but I could be wrong) of copies of the discrete infinite cyclic group. The result is not a free abelian group, but the direct sum of those copies of the discrete infinite cyclic group is dense (in the appropriate topology).

I think lots of "algebraically complete" abelian groups should have similar properties, but I have never really considered the topology, just the algebraic versions that sometimes use topological words. The p-adic integers should be another example, but I was even less sure I had the topology right there.

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