Is it true that every subgroup of finite nonzero 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.

2$\begingroup$ I think that it's classical that if $X$ is a measurable subset of nonzero measure then $XX$ has nonempty 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). $\endgroup$ – YCor Jan 31 '16 at 13:54

$\begingroup$ By $XX$ I mean the set of $xy$ when $x,y\in X$, which of course is the same as $X+X$ if $X$ is symmetric. $\endgroup$ – YCor Jan 31 '16 at 14:35
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 $G$.

$\begingroup$ How do you show that $\nu$ is a Haar measure? namely, that it takes finite values on finite subsets? $\endgroup$ – YCor Sep 10 '16 at 6:47

$\begingroup$ 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. $\endgroup$ – William of Baskerville Sep 10 '16 at 7:02

$\begingroup$ oops sorry I did a typo, I meant "how do you show that $\nu$ takes finite values on compact subsets"? $\endgroup$ – YCor Sep 10 '16 at 7:14

$\begingroup$ 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. $\endgroup$ – William of Baskerville Sep 10 '16 at 9:06

$\begingroup$ Ah ok sure. This is properness of the quotient morphism when the kernel is compact. $\endgroup$ – YCor Sep 10 '16 at 9:16