# Compact abelian topological group G. Then $G$ is connected iff its dual $\hat G$ is torsion-free.

Here I was trying to prove the following statement.

Let $$G$$ be a compact abelian topological group. Then $$G$$ is connected iff its dual $$\hat G$$ is torsion-free.

The proof in the link above goes as follows.

Suppose $$\phi:G\to S^1$$ is an element of $$\hat G$$ and $$n\geq1$$ are such that $$\phi^n$$ is the unit element in $$\hat G$$, that is, $$\phi(g)^n=1$$ for all $$g\in G$$. Then $$\phi$$ takes values in the subgroup of $$S^1$$ of elements of order divisible by $$n$$. This subgroup is finite, so it is discrete, so if $$G$$ is connected, then $$\phi$$ must be constant (because it is continuous!). We thus see that $$G$$ connected implies $$\hat G$$ is torsion-free. Conversely, suppose $$G$$ is not connected, and let $$G_0$$ be the connected component of $$1_G$$. Then $$G/G_0$$ is a finite abelian group and non-trivial. In particular, the dual group $$(G/G_0)^\wedge$$ is also finite and non-trivial (it is non-canonically isomorphic to $$G/G_0$$, in fact) so it has torsion. To see that $$\hat G$$ has torsion, we need only observe that there is an injective morphism of groups $$(G/G_0)^\wedge\to\hat G$$.

However I cannot understand why there exists $$n$$ such that $$\phi^n(g)=1$$? Can anyone explain?

• It's not like it exists for some reason, it is just supposed to exist. You're proving that $\hat G$ is torsion-free, so you suppose there is a torsion element $\phi \in \hat G$ and work from there. Jun 14, 2021 at 5:05
• @lisyarus, okay then what is the contradiction? Jun 14, 2021 at 5:08
• Probably, it would imply $\phi$ is the identity map and $n=1$? but we are choosing $\phi$ randomly? Jun 14, 2021 at 5:09

If $$K$$ is a finite non-zero abelian group then every non-zero element in it is torsion, so if you embed it in some group $$\hat G$$, the image under the embedding of any non-zero element would have the same order so is torsion.
Note however that the proof is incorrect because $$G/G_0$$ is not generally a finite discrete group. It is for Lie groups. As the link shows you have to find a subgroup $$H$$ of $$G$$, possibly bigger than $$G_0$$ such that $$G/H$$ is a non-zero finite abelian group.