Proper subgroup of infinite group is coinfinite

Statement

Verbal statement

In an infinite group, the set-theoretic complement of any proper subgroup is infinite.

Statement with symbols

Let $G$ be an infinite group and $H$ be a proper subgroup of $G$ (i.e., a subgroup that is not the whole of $G$). Then, the set-theoretic complement of $H$ in $G$, namely, the set $G\setminus H$, is infinite.

Facts used

1. Left cosets partition a group
2. Left cosets are in bijection via left multiplication

Related facts

Generalizations to other algebraic structures

The result generalizes to quasigroups. Although the left cosets of a quasigroup need not be pairwise disjoint, they all have the same size; moreover, the left coset of any element outside the subquasigroup cannot intersect the subquasigroup itself.

Further information: Proper subquasigroup of infinite quasigroup is coinfinite

Finite version

The finite version of this result states that a proper subgroup cannot have size more than half that of the group.

Further information: subgroup of size more than half is whole group

Proof

Given: An infinite group $G$, a proper subgroup $H$.

To prove: $G\setminus H$ is infinite.

Proof: In case $H$ is finite, the $G\setminus H$ must be infinite for the union to be infinite.

Consider the case that $H$ is infinite. Then, since $H$ is proper, there exists $g\in G\setminus H$. Consider the left coset $gH$. By fact (1), $gH$ is disjoint from $H$, and by fact (2), $|gH|=|H|$. Thus:

$|H|=|gH|\leq |G\setminus H|$

Since $H$ is infinite, so is $G\setminus H$.