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.

Related facts

• Closed subgroup of finite index implies open
• Open subgroup implies closed
• Connected implies no proper open subgroup
• Compact implies every open subgroup has finite index

Proof

Example of 3-adic rationals

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

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

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

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

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 $g$.