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:

$H$ 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 $H$
$g \in H ⟹ g^{- 1} \in H$ : 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 $H$
$x, y \in H ⟹ x y \in H$ : 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 $x y$ is in $H$ .

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

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