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Let $G$ be a topological group and $H$ and $N$ subgroups. Suppose $H$ is contained in the normalizer of $N$, then by using arguments of the second isomorphism theorem we can show that there is a canonical continuous isomorphism $$\phi:H/H\cap N\rightarrow HN/N$$ Are there cases in which this fails to be an actual homeomorphism?

Josué Tonelli-Cueto
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1 Answers1

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Presumably $H$ and $HN$ take the subspace topology from $G$ and project it onto the quotient groups?

Would the following be a counterexample? Let $G=\mathbb{R}$ be the additive group of reals, $H=\mathbb{Z}\cdot\sqrt2$ and $N=\mathbb{Q}$. Then $H\cap N$ is trivial, so $H/(H\cap N)$ inherits the discrete topology from $H$. On the other hand $N$ is dense in $G$, so $HN/N$ has only trivial closed sets, i.e. it has the trivial topology.

Jyrki Lahtonen
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  • Yes, I suppose subspace topology on subgroups and quotient topology in quotients. I think the example answers completely my question. Thank you very much. – Josué Tonelli-Cueto Jul 20 '13 at 21:06
  • I guess assuming that $N$ is closed would remove some pathological topologies. For example, both groups will then be $T_1$, hence Hausdorff. Don't know if that is sufficient (or necessary) for this to be a homeomorphism? – Jyrki Lahtonen Jul 20 '13 at 21:07
  • What I know is that a sufficient condition is $H$ to be open. However, I don't know any other condition. – Josué Tonelli-Cueto Jul 20 '13 at 21:13
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    Assuming that $N$ is closed doesn't help: Let $G = \mathbb{R}$, $H = \alpha \mathbb{Z}$ with $\alpha$ irrational and $N = \mathbb{Z}$. Then $H \cap N$ is trivial, $H/(H\cap N) \cong \mathbb{Z}$ with the discrete topology and $HN/N$ is a dense subgroup of $G/N = \mathbb{R}/\mathbb{Z}$. – Martin Jul 20 '13 at 22:05
  • I have open a new question to treat that matter. http://math.stackexchange.com/questions/448667/strengthening-the-second-isomorphism-theorem-for-topological-groups – Josué Tonelli-Cueto Jul 21 '13 at 09:40