Sufficiency of subgroup criterion

Statement
For a subset $$H$$ of a group $$G$$, the following are equivalent:


 * 1) $$H$$ is a subgroup, viz $$H$$ is closed under the binary operation of multiplication, the inverse map, and contains the identity element
 * 2) $$H$$ is a nonempty set closed under left quotient of elements (that is, for any $$a, b$$ in $$H$$, $$b^{-1}a$$ is also in $$H$$)
 * 3) $$H$$ is a nonempty set closed under right quotient of elements (that is, for any $$a, b$$ in $$H$$, $$ab^{-1}$$ is also in $$H$$)

Proof
We shall here prove the equivalence of the first two conditions. Equivalence of the first and third conditions follows by analogous reasoning.

(1) implies (2)
Clearly, if $$H$$ is a subgroup:


 * $$H$$ is nonempty since $$H$$ contains the identity element
 * Whenever $$a, b$$ are in $$H$$ so is $$b^{-1}$$ and hence $$b^{-1}a$$

(2) implies (1)
Suppose $$H$$ is a nonempty subset closed under left quotient of elements. Then, pick an element $$u$$ from $$H$$. (VIDEO WARNING: In the embeddded video, the letter $$a$$ is used in place of $$u$$, which is a little unwise, but the spirit of reasoning is the same).


 * $$e$$ is in $$H$$: Set $$a = b = u$$ to get $$u^{-1}u$$ is contained in $$H$$, hence $$e$$ is in $$H$$
 * $$g \in H \implies g^{-1} \in H$$: Now that $$e$$ is in $$H$$, set $$b = g, a =e $$ to get $$b^{-1}a = g^{-1}e$$ is also in $$H$$, so $$g^{-1}$$ is in $$H$$
 * $$x,y \in H \implies xy \in H$$: Set $$a = y, b= x^{-1}$$. The previous step tells us both are in $$H$$. So $$b^{-1}a = (x^{-1})^{-1}y$$ is in $$H$$, which tells us that $$xy$$ is in $$H$$.

Thus, $$H$$ satisfies all the three conditions to be a subgroup.