Sufficiency of subgroup criterion: Difference between revisions

Latest revision as of 19:43, 31 March 2022

This article gives the statement, and possibly proof, of a basic fact in group theory.
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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 $H$ , $ab^{-1}$ 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.

@@ Line 1: / Line 1: @@
 {{basic fact}}
-{{definition equivalence|[[subgroup]]}}
+{{definition equivalence|subgroup}}
+<section begin=main/>
 ==Statement==
 For a [[subset of a group|subset]] <math>H</math> of a group <math>G</math>, the following are equivalent:
-* <math>H</math> is a subgroup, viz <math>H</math> is closed under the binary operation of multiplication, the inverse map, and contains the identity element
+# <math>H</math> is a subgroup, viz <math>H</math> is closed under the binary operation of multiplication, the inverse map, and contains the identity element
-* <math>H</math> is a nonempty set closed under [[left quotient of elements]] (that is, for any <math>a, b</math> in <math>H</math>, <math>b^{-1}a</math> is also in <math>H</math>)
+# <math>H</math> is a nonempty set closed under [[left quotient of elements]] (that is, for any <math>a, b</math> in <math>H</math>, <math>b^{-1}a</math> is also in <math>H</math>)
-* <math>H</math> is a nonempty set closed under [[right quotient of elements]] (that is, for any <math>a, b</math> in <math>H</math>, <math>ab^{-1}</math> is also in <math>H</math>)
+# <math>H</math> is a nonempty set closed under [[right quotient of elements]] (that is, for any <math>a, b</math> in <math>H</math>, <math>ab^{-1}</math> is also in <math>H</math>)
 ==Proof==
@@ Line 15: / Line 15: @@
 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===
+===(1) implies (2)===
 Clearly, if <math>H</math> is a subgroup:
@@ Line 22: / Line 22: @@
 * Whenever <math>a, b</math> are in <math>H</math> so is <math>b^{-1}</math> and hence <math>b^{-1}a</math>
-===Second implies first===
+===(2) implies (1)===
-Suppose <math>H</math> is a nonempty subset closed under left quotient of elements. Then, pick an element <math>a</math> from <math>H</math>.
+Suppose <math>H</math> is a nonempty subset closed under left quotient of elements. Then, pick an element <math>u</math> from <math>H</math>. ('''VIDEO WARNING''': In the embeddded video, the letter <math>a</math> is used in place of <math>u</math>, which is a little unwise, but the spirit of reasoning is the same).
-* <math>a^{-1}a</math> is contained in <math>H</math>, hence <math>e</math> is in <math>H</math>
+* <math>e</math> is in <math>H</math>: Set <math>a = b = u</math> to get <math>u^{-1}u</math> is contained in <math>H</math>, hence <math>e</math> is in <math>H</math>
-* Now that <math>e</math> is in <math>H</math>, <math>a^{-1}e</math> is also in <math>H</math>, so <math>a^{-1}</math> is in <math>H</math>
+* <math>g \in H \implies g^{-1} \in H</math>: Now that <math>e</math> is in <math>H</math>, set <math>b = g, a =e </math> to get <math>b^{-1}a = g^{-1}e</math> is also in <math>H</math>, so <math>g^{-1}</math> is in <math>H</math>
-* Suppose <math>a, b</math> are in <math>H</math>. Then, <math>a^{-1}</math> is also in <math>H</math>. Hence, <math>(a^{-1})^{-1}b</math> is in <math>H</math>, which tells us that <math>ab</math> is in <math>H</math>.
+* <math>x,y \in H \implies xy \in H</math>: Set <math>a = y, b= x^{-1}</math>. The previous step tells us both are in <math>H</math>. So <math>b^{-1}a = (x^{-1})^{-1}y</math> is in <math>H</math>, which tells us that <math>xy</math> is in <math>H</math>.
 Thus, <math>H</math> satisfies all the three conditions to be a subgroup.
+<section end=main/>