Is there a LCA group $G$ such that $G/\mathbb{T}\cong\mathbb{R}$?

Question

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 $\mathbb{T}\cong \mathbb{R}/\mathbb{Z}$.

Such that the quotient $G/H$ induced with the quotient topology is isomorphic to $\mathbb{R}$, but $G\not = \mathbb{R}\times\mathbb{T}$.

Answer 1

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 $B\cong\mathbb{R}$ such that $A/B\cong\mathbb{Z}$ with the discrete topology, but $A\not\cong \mathbb{R}\times\mathbb{Z}$. Since $A/B$ is discrete, $A$ is topologically just the disjoint union of the cosets of $B$. Picking any element $a\in A$ 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 $B\times\mathbb{Z}\to A$ by mapping $(b,n)$ to $b+na$.

(Note that in fact this argument does not require that $B\cong\mathbb{R}$, 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 $\mathbb{T}$, and we can conclude that $G\cong H\times G/H$.)

Answer 2

If $f\colon\mathbb{T}\to G$ is an embedding of LCA groups, then Pontryagin duality gives an epimorphism $f^*\colon G^*\to\mathbb{T}^*=\mathbb{Z}$. Since the codomain is discrete and the image of $f^*$ is dense, $f^*$ is surjective. Therefore there exists $g\colon\mathbb{Z}\to G^*$ such that $f^*\circ g$ is the identity.

Apply Pontryagin duality again to get $h\colon G\to\mathbb{T}$ such that $h\circ f$ is the identity. Hence $G$ is topologically isomorphic to $\mathbb{T}\times G/\mathbb{T}$, by abstract nonsense.