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Finite dimensional compact abelian group that is not a product of connected and a totally disconnected

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 $G_0\times G/G_0$.

Clearly this does not have to be the situation. For instance consider $G=\mathbb{R}\times \mathbb{Z}_p/\mathbb{Z}$ where $\mathbb{Z}_p$ are the $p$-adic integers and $\mathbb{Z}$ lies in $\mathbb{R}$ in the obvious way and in $\mathbb{Z}_p$ diagonally with the first coordinate being zero. Then $G$ is not connected (because the image of $\mathbb{R}\times\{0\}$ is not dense) but clearly can't split to a direct product of $G_0$ and $G/G_0$ because $\mathbb{Z}_p$ has no finite subgroups.

My question is what about the case where $\Delta$ is a direct product of finite groups, say $\Delta = \prod_{p\in \mathbb{Prime}} \mathbb{Z}/p\mathbb{Z}$? (This time any closed subgroup $H\leq \Delta$ satisfies that $\Delta = H\times \Delta/H$ so this might lead to the desired result)