# 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 $$G_0\times G/G_0$$.

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.

• In your 3rd paragraph, the solenoid $G=R\times Z_p/Z$ is connected, because the image of $R\times 0$ is dense.
– YCor
Nov 4 '18 at 12:19
• Also note that your definition of dimension is rather a characterization of the topological dimension in this setting.
– YCor
Nov 4 '18 at 13:14
• @Ycor right it is dense, my bad. I deleted this section. Nov 4 '18 at 13:21

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.

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.

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 $$p$$.

• When you say "isomorphic to $\mathbb{Q}^k$ or some vector space over $\mathbb{Q}$" you mean isomorphic as abstract groups am I right? Because the direct sum is not a closed subgroup. I'm also confused about the last paragraph, the dual group of $G$ is the direct sum, which is discrete, then how come it admits a connected group as a closed subgroup? Nov 4 '18 at 13:20
• I fixed a typo ($K$ is dual of $H$, not of $G$). Except in the first and last paragraph, i.e. when I construct $H$, I consider no topology on the groups.
– YCor
Nov 4 '18 at 13:27
• Right I see now. You find a compact abelian group $K$ such that $K/K_0$ is a direct product, $\prod_{i\in I}\mathbb{Z}/p_i\mathbb{Z}$ and $K \not =K_0\times \prod_{i\in I}\mathbb{Z}/p_i\mathbb{Z}$. But this doesn't answer my question, is it? To answer my question I think you need to find a group $K$ and a direct product of finite groups $\Delta$ such that $K/\Delta$ is a torus but $K\not = K_0 \times K/K_0$. Nov 4 '18 at 13:40
• Your question was somewhat unclear: I answered the question suggested by the title, for which you provided a wrong example in your post.
– YCor
Nov 4 '18 at 13:45
• Oh I see, in this case I will accept this answer. But will leave the comments just to ensure nobody gets confused. I will try to use this example to construct what I need. Thanks Nov 4 '18 at 13:46