Vtyjkyuk

I'm looking for an example to the following situation:

A locally compact abelian (LCA) group $G$ (I assume that the groups are Hausdorff) .

A (closed) subgroup $H$ of $G$ which is isomorphic (as topological group) to the circle $mathbbTcong mathbbR/mathbbZ$.

Such that the quotient $G/H$ induced with the quotient topology is isomorphic to $mathbbR$, but $Gnot = mathbbRtimesmathbbT$.

This is not possible. By taking Pontryagin duals of everything, this is equivalent to asking for a locally compact abelian group $A$ with a closed subgroup $BcongmathbbR$ such that $A/BcongmathbbZ$ with the discrete topology, but $Anotcong mathbbRtimesmathbbZ$. Since $A/B$ is discrete, $A$ is topologically just the disjoint union of the cosets of $B$. Picking any element $ain A$ whose image in $A/B$ is a generator, then every element of $A$ is in the same connected component as a unique multiple of $a$ so we get a topological isomorphism $BtimesmathbbZto A$ by mapping $(b,n)$ to $b+na$.

(Note that in fact this argument does not require that $BcongmathbbR$, and requires only that $A/B$ is a free abelian group with the discrete topology. So dually, all that is required of your $H$ is that it is topologically isomorphic to a product of copies of $mathbbT$, and we can conclude that $Gcong Htimes G/H$.)

If $fcolonmathbbTto G$ is an embedding of LCA groups, then Pontryagin duality gives an epimorphism $f^*colon G^*tomathbbT^*=mathbbZ$. Since the codomain is discrete and the image of $f^*$ is dense, $f^*$ is surjective. Therefore there exists $gcolonmathbbZto G^*$ such that $f^*circ g$ is the identity.

Apply Pontryagin duality again to get $hcolon GtomathbbT$ such that $hcirc f$ is the identity. Hence $G$ is topologically isomorphic to $mathbbTtimes G/mathbbT$, by abstract nonsense.