Proper subgroup of infinite group is coinfinite

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) uses::Left cosets partition a group
 * 2) uses::Left cosets are in bijection via left multiplication

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.

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

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

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