Is it true that every subgroup of finite non-zero Haar measure of an abelian locally compact group should be open and compact? This is obviously true for the case of discrete abelian groups. Thanks.
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2$\begingroup$ I think that it's classical that if $X$ is a measurable subset of nonzero measure then $X-X$ has non-empty interior? If so it's clear that any measurable subgroup of positive measure has to be open (and hence compact open if it has finite measure). $\endgroup$– YCorCommented Jan 31, 2016 at 13:54
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$\begingroup$ By $X-X$ I mean the set of $x-y$ when $x,y\in X$, which of course is the same as $X+X$ if $X$ is symmetric. $\endgroup$– YCorCommented Jan 31, 2016 at 14:35
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Let $G$ be an LCA group, $\mu$ be a Haar measure on $G$ and $H$ be a (closed) subgroup of $G$ with $0<\mu(H)<\infty$. First, by restricting $\mu$ to $H$, we obtain a finite Haar measure on $H$. This means that $H$ is compact. Further, by projecting $\mu$ to $G/H$ via the quotient morphism $G\to G/H$, we obtain a Haar measure $\nu$ on $G/H$, which is positive on singletons. This means that $G/H$ is discrete, hence $H$ is open in $G$.
answered Sep 9, 2016 at 18:00
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$\begingroup$ How do you show that $\nu$ is a Haar measure? namely, that it takes finite values on finite subsets? $\endgroup$– YCorCommented Sep 10, 2016 at 6:47
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$\begingroup$ Formally, $\nu$ is defined by $\nu(A)=\mu(p^{-1}(A))$ for $A\subseteq G/H$ Borel, where $p\colon G\to G/H$ is the quotient morphism. Then $\nu(\text{singleton in $G/H$})=\mu(\text{$H-$coset in $G$})=\mu(H)$, which is positive and finite by the assumptions. $\endgroup$ Commented Sep 10, 2016 at 7:02
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$\begingroup$ oops sorry I did a typo, I meant "how do you show that $\nu$ takes finite values on compact subsets"? $\endgroup$– YCorCommented Sep 10, 2016 at 7:14
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$\begingroup$ This follows from compactness of $H$ (which is established in the beginning). If $K\subseteq G/H$ is compact then $p^{-1}(K)$ is also compact. Indeed, first, there is a compact set $C\subseteq G$ with $p(C)=K$. Now $p^{-1}(K)=CH$ is a product of compact sets and hence is compact. Thus, $\nu(K)=\mu(CH)$ is finite. $\endgroup$ Commented Sep 10, 2016 at 9:06
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$\begingroup$ Ah ok sure. This is properness of the quotient morphism when the kernel is compact. $\endgroup$– YCorCommented Sep 10, 2016 at 9:16