# Connected implies no proper closed subgroup of finite index

## Contents

## Statement

In a connected topological group, there cannot be any proper closed subgroup of finite index (i.e., a closed subgroup that is also a subgroup of finite index).

## Related facts

## Facts used

## Proof

The proof is direct from Fact (1), and the observation that the existence of a *proper* subgroup (and hence a proper nonempty subset) that is both closed and open means that the group is not connected.