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It is not true for $G/H\times H\simeq G$ to hold for a subgroup $H\leq G$ when we talk about group isomorphisms.

However, what about topological groups and a homeomorphism in the above? (without being a group morphism)

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No. Take $G=\mathbb{R}$ and $H = \mathbb{Z}$. If you consider the quotient $G/H$ as the group version correctly topologized, then $\mathbb{R}/\mathbb{Z} \times \mathbb{Z}$ is not homeo to $\mathbb{R}$. For one, the former isn't even connected (it's a stack of circles).

I'm also sure there are simpler examples than this one.

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  • $\begingroup$ It is a simple enough example, I think. It doesn't work for finite groups though, as with the discrete topology any set bijection would work. I wonder if there's any conditions under which it would be true? $\endgroup$
    – Feelix
    Apr 23, 2018 at 13:33
  • $\begingroup$ If $H$ is nontrivial and discrete, you're going to have problems with the number of connected components. $\endgroup$
    – Randall
    Apr 23, 2018 at 13:35
  • $\begingroup$ Still, some silly condition could be G/H, H, R have one connected component. Probably false but it's something to think about. $\endgroup$
    – Feelix
    Apr 23, 2018 at 13:36

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