If $G$ is a LCA, does always exist a discrete subgroup $H$ of $G$ such that $G/H$ is compact? Following an idea about proving Fourier inversion formula for arbitrary LCA groups from the corresponding result for compact groups, I stumbled upon this problem: is it true that, given an arbitrary topological locally compact abelian group $G$, there always exists a discrete subgroup of $H$ such that $G/H$ is compact?
My knowledge about examples of LCA groups is so little that basically I know only cylinders, their duals and finite groups, so I couldn't work out a counter-example (if such a counter-example exists)... On the other hand, I've basically have no clue about how to prove such a result (if such a result is true).
 A: The answer is no. In fact, this shows that the following group is an LCA with no infinite discrete subgroups whatsoever!
We will construct a counter example due to I. Kaplansky from "Walter Rudin, Fourier analysis on groups, Interscience Tracts in Pure and Applied Mathematics" Example 2.4.7:
Set $G:= \{ x\in \{0,1,2,3\}^{\mathbb{N}} \mid \: \  x_n\in \{1,3\}\ \text{finite amount of times} \} $.
The group operation is coordinate-wise addition.
And the topology comes from the restricted product structure where $G_i=\{0,1,2,3\}$ and $K_i = \{0,2\}$ is a compact open subset. 
Look at $K= \{x\in G \mid \ 2x=0\}$ (i.e all sequences with no 1's or 3's).
Observe that if $H$ is infinite then $H \cap K$ is infinite ($x\in H  $ implies $2x\in K$ ) hence is not discrete. If it were discrete then $H \cap K$ would be closed and from compactness of $K$ it would be compact, hence finite.
We get that $G$ has no infinite discrete subgroup, but since $G$ is not compact if $H$ is finite $G/H$ will not be compact.
Moreover, in the same chapter Rudin shows that for $G$ compactly generated your assertion is true.
