Subgroup of finite index need not be closed in T0 topological group

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Statement

It is possible to have a T0 topological group and a subgroup of finite index that is not a closed subgroup of the whole group. In fact, we could choose a subgroup of finite index that is a dense subgroup in the whole group.

Related facts

Proof

Example of 3-adic rationals

Suppose G is the group defined as the following subgroup of the additive group of rational numbers:

G := \{ \frac{a}{3^n} \mid a,n \in \mathbb{Z}, n \ge 0 \}

Let H be the subgroup of G comprising those elements that can be expressed with even numerators:

H := \{ \frac{2a}{3^n} \mid a,n \in \mathbb{Z}, n \ge 0 \}

Then:

  1. G is a T0 topological group: In fact, it is a subgroup of the rationals, hence a metrizable group.
  2. H is a subgroup of index two in G: For every g \in G, either g \in H or g + 1 \in H. Thus, H has a coset space of size two, hence has index two.
  3. H is dense in G. In particular, it is not closed in G: For any point goutside H, that point plus elements of the form 1/3^n are all in H, and these sums approach g.