Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $(G/G_0)^\wedge\to\hat G$.

share|cite|improve this answer
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 $G$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.