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

From Groupprops

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

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

Then:

- is a T0 topological group: In fact, it is a subgroup of the rationals, hence a metrizable group.
- is a subgroup of index two in : For every , either or . Thus, has a coset space of size two, hence has index two.
- is dense in . In particular, it is not closed in : For any point outside , that point plus elements of the form are all in , and these sums approach .