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Here's yet another exercise that stumped me:

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

Any hints/solutions will be appreciated.

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2 Answers 2

up vote 4 down vote accepted

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.

Conersely, 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 $(G/G_0)^\wedge\to\hat G$.

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You use compactness to show that $G/G_0$ is finite, right? –  Chera Sep 22 '11 at 7:19
    
Indeed. It is discrete and compact. –  Mariano Suárez-Alvarez Sep 22 '11 at 7:19
    
Thank you very much! –  Chera Sep 22 '11 at 7:20
    
Why should $G_0$ be open? A linearly topologized group is totally disconnected; if it is compact and infinite, the connected component of $1$ is not open. Example: the $p$-adic integers. –  egreg Dec 10 '14 at 17:07

Mariano's answer is not quite right. For instance if $G=\mathbf{Z}_p$ then $G$ is compact and disconnected, but $G_0$ is trivial so $G/G_0=\mathbf{Z}_p$ is not finite. Instead one must argue as follows.

Let $U$ be a nontrivial closed and open neighbourhood of $1$. Since $G$ is compact we know that $U$ is compact, so by continuity of multiplication there is an open neighbourhood $V$ of $1$ such that $V\cdot U \subset U$. Let $H$ be the group generated by $V$. Then $H$ is open and $H\cdot U\subset U$, so in particular $H\subset U$. So $H$ is a proper open subgroup, so $G/H$ is finite and nontrivial. Now any character of $G/H$ induces a torsion character of $G$.

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