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?

  • $\begingroup$ What does the idea of index mean here if $G$ is possibly infinite? $\endgroup$ – Cameron Williams Oct 23 '14 at 2:05
  • $\begingroup$ @CameronWilliams, I mean $K\leq G$ is of finite index if the left coset $G/H$ is finite. Equivalently, if the Haar measure of $K$ is nonzero. Maybe there is better name for this? $\endgroup$ – ougao Oct 23 '14 at 2:15

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:

  1. $G$ has a proper closed finite index subgroup
  2. $G$ has a proper finite index subgroup
  3. $G$ is not connected

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.

  • $\begingroup$ thanks a lot for your answer! The subgroup I am interested in is closed, coming from fixed point of a group element in some group acting on this $G$. My knowledge on topological groups, especially lie groups is almost zero..., I apologize for the inaccuracy. Any books on this stuff to recommend? Thanks again! $\endgroup$ – ougao Oct 23 '14 at 19:06
  • 1
    $\begingroup$ Hewitt and Ross, Abstract harmonic analysis I, should contain the necessary information. $\endgroup$ – YCor Oct 23 '14 at 20:57

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