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.

First implies second

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 $b^{- 1} a$

Second implies first

Suppose $H$ is a nonempty subset closed under left quotient of elements. Then, pick an element $a$ from $H$ .

$a^{- 1} a$ is contained in $H$ , hence $e$ is in $H$
Now that $e$ is in $H$ , $a^{- 1} e$ is also in $H$ , so $a^{- 1}$ is in $H$
Suppose $a, b$ are in $H$ . Then, $a^{- 1}$ is also in $H$ . Hence, $(a^{- 1})^{- 1} b$ is in $H$ , which tells us that $a b$ is in $H$ .

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