Tour:Sufficiency of subgroup criterion

This article adapts material from the main article: sufficiency of subgroup criterion

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WHAT YOU NEED TO DO:

• Recall the various definitions of subgroup, including the subgroup criterion
• Try proving that the subgroup criterion is necessary and sufficient
• Read below the proof of sufficiency
• Compare with the result we just saw for finite groups (any nonempty multiplicatively closed subset is a subgroup)
  

PONDER:

• What axioms of group structure play a role in this proof?
• In what ways are things different for finite groups, and why?

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.

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