For any compact abelian group $K$.$$K\cong H_0\times \mathrm{U}(1)^k,$$where $H_0$ is a finite group.

closed as off topic by Henry Cohn, Emil Jeřábek, Stefan Geschke, Felipe Voloch, Todd Trimble♦ Oct 9 '12 at 15:53
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Any topological group of the form $K\cong H_0\times U(1)^k$ (with $H_0$ finite and $k$ a positive integer) is a closed subgroup of $U(1)^h$ (for some positive integer $h\geq k$). Furthermore, all the closed subgroups of $U(1)^h$ are of this form. The proof is an easy application of the PontryaginVan Kampen duality. In fact, such groups are the duals of the finitely generated Abelian groups (which are quotients of $\mathbb Z^h$). It is well known that such groups are of the form $F\times \mathbb Z^k$ (with $F$ a finite group). To find a general form for a compact abelian group is as difficult as giving a structure theorem for discrete abelian group (which is known to be quite a difficult, and fairly open, problem in general, even if there are nice results for countable torsion groups). EDIT: just to answer also to the comment of Stefan Geschke. Finitely generated groups in $\mathrm{Mod}(\mathbb Z)$ can be characterized as the Noetherian objects of the category. So I guess that (by duality) the objects of the form $K\cong H_0\times U(1)^k$ should be the Artinian objects in the category of compact abelian topological ($T_2$) groups. (N.B.= here by Artinian I mean the categorytheoretical notion) 

