Let $G$ be a compact abelian group. A compact abelian group is said to have dimension $n$ if $\dim\_\mathbb{Q} \mathbb{Q}\otimes \hat G = n$. Equivalently one can show that this holds if $G$ is isomorphic to $(\mathbb{R}^n\times \Delta)/\Gamma$ where $\Delta$ is a zero dimensional compact abelian group and $\Gamma$ is a countable discrete subgroup.


The locally connected component of $G$ is the image of $\mathbb{R}^n\times\{0\}$ under this quotient and it is dense in the connected component of $G$. If the locally connected component and the connected component of $G$ coincide then they become a torus in which case $G$ can be written as $G\_0\times G/G\_0$.


But, clearly this does not have to be the situation.


My question is what about the case where $\Delta$ is a direct product of finite $p$-groups (not always the same prime, say $\Delta = \prod\_{p\in \mathbb{Prime}} \mathbb{Z}/p\mathbb{Z}$)? Note that in this case any closed subgroup $H\leq \Delta$ satisfies that $\Delta = H\times \Delta/H$ so this might lead to the desired result.



Finite dimensional compact abelian group that is not a product of connected and a totally disconnected
Here's, for any positive integer $k$, a second countable, compact group, of dimension $k$, whose zero connected component is not a topological direct factor. (Edit: below I construct such groups with some additional requirement.)


Let $I$ be any infinite set of primes (all primes if you like). Consider $G=\prod\_{p\in I}\mathbf{Z}/p\mathbf{Z}$. This is an abelian group, whose torsion subgroup is $\bigoplus\_{p\in I}\mathbf{Z}/p\mathbf{Z}$, and the quotient is a torsion-free divisible abelian group, hence isomorphic to some $\mathbf{Q}$-vector space of uncountable dimension. Choose any integer $k>0$, and pick a subgroup of the latter quotient, isomorphic to $\mathbf{Q}^k$, and let $H$ be its inverse image in $G$. (With some little effort, one can construct explicitly such $H$.)


So the torsion subgroup in $H$ is $\bigoplus\_{p\in I}\mathbf{Z}/p\mathbf{Z}$, and the quotient by the torsion is isomorphic to $\mathbf{Q}^k$. This is not splittable as direct product of torsion and torsion-free, since $H$ is residually finite and $\mathbf{Q}$ is not.


Let $K$ be the Pontryagin dual of the discrete group $H$, let $S$ be the Pontryagin dual of $\mathbf{Q}$ (this is a connected, 1-dimensional, torsion-free compact group). Then $K$ admits $S^k$ as closed subgroup (equal to its 0 connected component), and the quotient is topologically isomorphic to the Pontryagin dual of $\bigoplus\_{p\in I}\mathbf{Z}/p\mathbf{Z}$, namely $G$ itself (viewed now as topological group, profinite). This is not part of a splitting as topological direct product, because it was not the case at the dual level. 




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Edit (1) Concerning your question with additional requirement:


You want a compact abelian group such that (1) it has a closed profinite subgroup, isomorphic to a product of finite groups, such that quotient is a (finite-dimensional) torus, and (2) then $K$ is not direct product of its zero component $K^\circ$ with any closed subgroup.


By Pontryagin duality, this is equivalent to finding a (discrete) abelian group $A$ with the properties: (1') it has a subgroup that is free of finite rank, such that the quotient is a direct sum of finite groups. (2') The torsion subgroup in $A$ is not a direct factor.


The groups $H$ above do not answer this, because the quotient of such $H$ by any nonzero finitely generated subgroup contains a Prüfer group at some prime.




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Edit (2). Here is now an example for the question with additional requirement. 



> 
> Claim: there exists a (discrete) countable abelian group satisfying (1') and (2'). 
> 
> 
> 


First, consider the subgroup $M$ of $\mathbf{Q}$ generated by all $1/p$ for $p$ ranging over all primes (note that $1/4\notin M$). Then $M/\mathbf{Z}$ is isomorphic to $\bigoplus\_p\mathbf{Z}/p\mathbf{Z}$. We consider a copy of $M$ in $\prod\_p \mathbf{Z}/p\mathbf{Z} / \bigoplus\_p \mathbf{Z}/p\mathbf{Z}$ and consider its inverse image $N$ in $\prod\_p \mathbf{Z}/p\mathbf{Z}$. So, if $T$ is the torsion subgroup in $N$, we have $T=\bigoplus\_p \mathbf{Z}/p\mathbf{Z}$ and $N/T\simeq M$.


I claim that this is not split. Indeed, let $x$ be a nontorsion element in $P=\prod\_p \mathbf{Z}/p\mathbf{Z}$. Nontorsion means that its support $I$ is infinite. Then $x\in pP$ if and only if $p\notin I$. This shows that $\bigcap\_p pP=0$. Hence $P$ has no subgroup isomorphic to $M$ (since $\bigcap\_p pM=\mathbf{Z}$). 


We have proved (2'). For (1'), lift the copy of $\mathbf{Z}$ in $N$. The quotient $N/C$ by this cyclic subgroup $C$ lies in an extension with kernel $T$, and quotient $M/\mathbf{Z}$, which is also isomorphic to $T$. It follows that $N/C\simeq\bigoplus\_p F\_p$, where $F\_p$ is an abelian group of order $p^2$ for each $p$.


