Subset containing identity whose left translates partition the group is a subgroup

Statement
Suppose $$G$$ is a group and $$H$$ is a nonempty subset of $$G$$ containing the identity element of $$G$$. Then, the following are equivalent:


 * 1) $$H$$ is a subgroup of $$G$$
 * 2) For any $$x,y \in H$$, either $$xH = yH$$ or $$xH \cap yH$$ is empty
 * 3) For any $$g \in G$$, either $$H = gH$$ or $$H \cap gH$$ is empty

Related facts
A reformulation, or immediate corollary, of this is the statement that:

Left congruence on a group equals left coset space relation from a subgroup

Subgroup
We shall use the definition of subgroup in terms of the subgroup criterion. A nonempty subset $$H$$ of a group $$G$$ is termed a subgroup if for any $$a,b \in H$$, the element $$a^{-1}b$$ is also in $$H$$.

(1) implies (2) implies (3)
This is a direct consequence of the fact that for any subgroup, the left cosets partition the group.

(3) implies (1)
Given: A group $$G$$ and nonempty subset $$H$$ such that for every $$g \in G$$ either $$H = gH$$ or $$H \cap gH$$ is empty.

To prove: For any $$a,b \in H$$, the element $$a^{-1}b$$ is also in $$H$$

Proof: Let $$a,b \in H$$. Consider the set $$aH$$. Since $$H$$ contains the identity element, $$a \in aH$$, so $$aH \cap H$$ is nonempty. Thus, by assumption, $$aH = H$$, so there exists $$h \in H$$ such that $$ah = b$$, so $$h = a^{-1}b$$. Thus, $$a^{-1}b$$ is in $$H$$, completing the proof.