Suppose $G$ is a compact metrizable abelian group, is it true that $G$ has no finite index subgroups iff $G$ is connected?
Any reference or help is appreciated.
Thanks in advance!
Here are my thought on this problem.
Since $G$ has a finite index subgroup iff we have a group homomorphism forom $G$ to a finite group, I think maybe(?) it is automically continuous, so $G$ is not connected.
If $G$ is not connected, the motivating example is $G=\prod_{n>0}\mathbb{Z}/2\mathbb{Z}$, clearly it has finite index subgroup, but how to prove the general case?
The question is a bit sloppy since you don't specify whether finite index subgroups are meant to be closed. In general this makes a big difference. Anyway here it doesn't:
Proposition. Let $G$ be a compact group. Equivalent statements:
To check this I need the following lemma making use of Pontryagin duality: if $G$ is a compact abelian group with $pG=0$ and is nontrivial then $G$ is not connected.
Proof of the lemma: Pontryagin duality (or a weak form of it) implies that if $G\neq 0$, then there is a nonzero continuous homomorphism from $G$ to the circle. Since its image has exponent $p$, it must be the cyclic group of order $p$. Hence $G$ is not connected.
Proof of the proposition. 1 implies 2 obviously. Suppose 2. then $G$ admits some cyclic group of order $p$ as a quotient. Hence $pG\neq G$. Since $pG$ is closed (by compactness) it follows that $G/pG$ is nontrivial, and by the lemma it's not connected. Hence $G$ is not connected either, showing 3. Suppose 3. then replacing $G$ by the quotient of $G$ by its zero component, we can assume $G\neq 0$ and $G$ totally disconnected. Then $G$ is profinite, and 1. follows.