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:

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.

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 $ab$ is in $H$.

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