Let $H,G,K$ be three topological groups, we say that $G$ is an extension of $K$ by $H$ if the following short sequence 
$$0\rightarrow H\rightarrow G\rightarrow K\rightarrow 0$$
is exact. (If $H$ is a subgroup of $G$ this is equivalent to $K\cong G/H$)


Now, assume that $H,G,K$ are compact abelian topological groups, therefore, by the structure theorem for compact abelian Lie groups, we have that $K$ is a Lie group if and only if it is isomorphic to $\mathbb{T}^n\times C\_k$ where $\mathbb{T}$ is the circle group ($S^1$ or $\mathbb{R}/\mathbb{Z}$), $C\_k$ is a finite abelian group and $n\in\mathbb{N}$. 


It is well known that every extension of a Lie group by a Lie group is again a Lie group. In particular, every extension of a Lie group by a finite group is a Lie group.


My question is: whether every extension of a compact abelian Lie group by a pro-finite group is a subgroup of $\mathbb{T}^n\times D$ where $D$ is a profinite group for some $n\in\mathbb{N}$.


I am most interested in the case where $H$ is a direct product of finite groups.


Any proof / reference is appreciated.


Edit: I also assume that the groups are Hausdorff. Also every morphism is both topological and algebraic.



Profinite extension of a Lie group
Here is a counterexample.


Take $H=\prod\_{n\geq 1}\mathbb{Z}/n\mathbb{Z}$ (or any other profinite group into which $\mathbb{Z}$ injects). Let $\tilde{G}=H\times\mathbb{R}$, and let $G=\tilde{G}/\mathbb{Z}$, where $\mathbb{Z}$ sits inside $\tilde{G}$ diagonally. We get a short exact sequence
$$
1\to H\to G\to\mathbb{R}/\mathbb{Z}\to 1.
$$
Then $G$ is compact, and the connected component of the identity in $G$ is isomorphic to $\mathbb{R}$ as an abstract group (though not as a topological group).


Now suppose $Q$ is a compact subgroup of $\mathbb{T}^n\times D$. The connected component of the identity in $Q$ is a closed subgroup of $\mathbb{T}^n\times \{1\_D\}$. An infinite closed subgroup of $\mathbb{T}^n$ must have a non-trivial element of finite order, so cannot be isomorphic to $\mathbb{R}$ as a group. So $Q$ cannot be isomorphic to $G$.


