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?
closed as off-topic by Ricardo Andrade, Jack Huizenga, Andrey Rekalo, Stefan Kohl, Noah S Dec 1 '13 at 16:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
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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.