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Let $A$ and $H$ be closed subgroups of a $\sigma$-compact locally compact group $G$. Assume further that $A$ is abelian. Is the group $AH$ locally compact subgroup in the subspace topology?

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closed as off-topic by Ricardo Andrade, Jack Huizenga, Andrey Rekalo, Stefan Kohl, Noah S Dec 1 '13 at 16:20

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  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Jack Huizenga, Andrey Rekalo, Stefan Kohl, Noah S
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$G=\mathbb R$, $A=\mathbb Z$, $H=h\mathbb Z$ for some $h\in\mathbb R\smallsetminus\mathbb Q$. –  Emil Jeřábek Apr 2 '13 at 17:10
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Also, AH need not be a subgroup. –  Misha Apr 2 '13 at 18:57
    
Voting to close, since it has been answered in the comments. The alternate solution would be to copy these answers into the answer box so that the software realizes this question has been answered. –  David White Apr 3 '13 at 12:34
    
OK, I’ve made this an answer. –  Emil Jeřábek Apr 3 '13 at 13:53
    
This question is a reaction to my answer to your Eisenstein question? I have edited my answer this morning with some reference for GL(2). –  plusepsilon.de Apr 3 '13 at 14:00
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1 Answer 1

The is not true in general. For example, if $G=\mathbb R$, $A=\mathbb Z$, and $H=h\mathbb Z$ for some $h\in\mathbb R\smallsetminus\mathbb Q$, then $AH$ (that is, $\mathbb Z+h\mathbb Z$) is a countable dense subgroup of $\mathbb R$, and as such it is not locally compact.

Furthermore, as pointed out by Misha, the product $AH$ need not even be a subgroup: for instance, take $G$ to be the discrete free group on two generators $a,h$, and $A$ and $H$ the cyclic subgroups generated by $a$ and $h$, respectively. Instead of $A$ being abelian, you should require that $AH=HA$; a sufficient condition is that one of the subgroups is normal (or more generally, that one of the subgroups is included in the normalizer of the other).

If $AH=HA$ and both subgroups are compact, then $AH$ is a compact subgroup of $G$, being a continuous image of a compact space.

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Actually, I stated my original question wrong. What I had in mind at the beginning is that $A$ is a subgroup of the center of $G$. But in either case, you answered my question. Thanks. –  Windi Apr 4 '13 at 21:54
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