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

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