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?
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.