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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.

Added: concerning your question with additional requirement:

The property "be a direct sum of finite abelian groups" (or product, at a dual level) is a bit unstable. I see from the examples you have in mind that you mainly think of a direct sum $\prod_pF_p$ with $F_p$ finite $p$-group; the latter smaller class is better-behaved as it passes to quotients (i.e., at the compact side, the corresponding class passes to closed subgroups).

Edit (2). Here is now an example for the question with additional requirement. By the above, we want a discrete 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 $N/T\simeq M$.

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.)

Edit (1) Concerning your question with additional requirement:

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 $N/T\simeq M$.

Added: concerning your additional question:

The property "be a direct sum of finite abelian groups" (or product, at a dual level) is a bit unstable. I see from the examples you have in mind that you mainly think of a direct sum $\prod_pF_p$ with $F_p$ finite $p$-group; the latter smaller class is better-behaved as it passes to quotients (i.e., at the compact side, the corresponding class passes to closed subgroups).

Added: concerning your question with additional requirement:

The property "be a direct sum of finite abelian groups" (or product, at a dual level) is a bit unstable. I see from the examples you have in mind that you mainly think of a direct sum $\prod_pF_p$ with $F_p$ finite $p$-group; the latter smaller class is better-behaved as it passes to quotients (i.e., at the compact side, the corresponding class passes to closed subgroups).

Edit (2). Here is now an example for the question with additional requirement. By the above, we want a discrete 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 $p$.

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.

Let $I$ be any infinite set of primes (all primes if you like). Consider $G=\prod_{i\in I}\mathbf{Z}/p_i\mathbf{Z}$. This is an abelian group, whose torsion subgroup is $\bigoplus_{i\in I}\mathbf{Z}/p_i\mathbf{Z}$, 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 $\bigoplus_{i\in I}\mathbf{Z}/p_i\mathbf{Z}$, 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 $\bigoplus_{i\in I}\mathbf{Z}/p_i\mathbf{Z}$, 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.

Added: concerning your additional question:

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.

The property "be a direct sum of finite abelian groups" (or product, at a dual level) is a bit unstable. I see from the examples you have in mind that you mainly think of a direct sum $\prod_pF_p$ with $F_p$ finite $p$-group; the latter smaller class is better-behaved as it passes to quotients (i.e., at the compact side, the corresponding class passes to closed subgroups).

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.

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.

Added: concerning your additional question:

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.

The property "be a direct sum of finite abelian groups" (or product, at a dual level) is a bit unstable. I see from the examples you have in mind that you mainly think of a direct sum $\prod_pF_p$ with $F_p$ finite $p$-group; the latter smaller class is better-behaved as it passes to quotients (i.e., at the compact side, the corresponding class passes to closed subgroups).