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

## 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.

## Proof

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 $g$outside $H$, that point plus elements of the form $1/3^n$ are all in $H$, and these sums approach $g$.