# Subgroup of finite index need not be closed in algebraic group

From Groupprops

## Statement

It is possible to have an algebraic group (in fact, a one-dimensonal algebraic group) and a subgroup of finite index that is not a closed subgroup of the group.

## Related facts

## Proof

### One-dimensional examples

- Consider the field of real numbers. Take the multiplicative group. This has a subgroup of index two given by the positive reals under multiplication. However, since the topology is a cofinite topology, the only proper closed subsets are the finite subsets, so this subgroup is not closed. In fact, it is a dense subgroup.