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

This article gives a proof/explanation of the equivalence of multiple definitions for the term subgroup
View a complete list of pages giving proofs of equivalence of definitions

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:

Definitions used

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 $H$.

Proof

(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.