Sufficiency of subgroup criterion

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This article gives a proof/explanation of the equivalence of multiple definitions for the term subgroup
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Statement

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

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