## How to prove this equation [closed]

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

-
I'm guessing that this question won't last long in its current form, but let me point out that you at very least need to add some hypotheses. If the scope of $K$ is "compact topological abelian group" then your statement is false. – Ramsey Oct 9 at 14:34
This is clearly not true, consider for example $\mathbb Z_2^{\mathbb N}$. I am trying to think of some context where this question makes sense. Is $k$ finite? – Stefan Geschke Oct 9 at 15:05
Presumably he means compact Hausdorff abelian Lie groups, where $k$ is finite. – Todd Trimble Oct 9 at 15:49
Try looking here for instance: hopf.math.purdue.edu/Dwyer-Wilkerson/lie/… – Todd Trimble Oct 9 at 15:51
Oh sorry, two comments ago, I should have said "he or she". – Todd Trimble Oct 9 at 15:52

## closed as off topic by Henry Cohn, Emil Jeřábek, Stefan Geschke, Felipe Voloch, Todd TrimbleOct 9 at 15:53

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).
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)