For any compact abelian group $K$.$$K\cong H\_0\times \mathrm{U}(1)^k,$$where $H\_0$ is a finite group.



How to prove this equation
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 Pontryagin-Van 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 category-theoretical notion)


