quotient group

Before defining quotient groups, some preliminary definitions must be introduced and a few established.

Given a group $G$ and a subgroup $H$ of $G$ , the http://planetmath.org/node/122relation $\sim_{L}$ on $G$ defined by $a\sim_{L}b$ if and only if $b^{-1}a\in H$ is called left congruence modulo $H$ ; similarly the relation defined by $a\sim_{R}b$ if and only if $ab^{-1}\in H$ is called congruence modulo $H$ (observe that these two relations coincide if $G$ is abelian).

Proposition.

Left (resp. right) congruence modulo $H$ is an equivalence relation on $G$ .

Proof.

We will only give the proof for left congruence modulo $H$ , as the for right congruence modulo $H$ is analogous. Given $a\in G$ , because $H$ is a subgroup, $H$ contains the identity $e$ of $G$ , so that $a^{-1}a=e\in H$ ; thus $a\sim_{L}a$ , so $\sim_{L}$ is http://planetmath.org/node/1644reflexive. If $b\in G$ satisfies $a\sim_{L}b$ , so that $b^{-1}a\in H$ , then by the of $H$ under the formation of inverses, $a^{-1}b=(b^{-1}a)^{-1}\in H$ , and $b\sim_{L}a$ ; thus $\sim_{L}$ is symmetric. Finally, if $c\in G$ , $a\sim_{L}b$ , and $b\sim_{L}c$ , then we have $b^{-1}a,c^{-1}b\in H$ , and the closure of $H$ under the binary operation of $G$ gives $c^{-1}a=(c^{-1}b)(b^{-1}a)\in H$ , so that $a\sim_{L}c$ , from which it follows that $\sim_{L}$ is http://planetmath.org/node/1669transitive, hence an equivalence relation. ∎

It follows from the preceding that $G$ is partitioned into mutually disjoint, non-empty equivalence classes by left (resp. right) congruence modulo $H$ , where $a,b\in G$ are in the same equivalence class if and only if $a\sim_{L}b$ (resp. $a\sim_{R}b$ ); focusing on left congruence modulo $H$ , if we denote by $\bar{a}$ the equivalence class containing $a$ under $\sim_{L}$ , we see that

\begin{split}\displaystyle\bar{a}&\displaystyle=\{b\in G\mid b\sim_{L}a\}\\ &\displaystyle=\{b\in G\mid a^{-1}b\in H\}\\ &\displaystyle=\{b\in G\mid b=ah\text{ for some }h\in H\}=\{ah\mid h\in H\}% \text{.}\end{split}

Thus the equivalence class under $\sim_{L}$ containing $a$ is simply the left coset $a H$ of $H$ in $G$ . Similarly the equivalence class under $\sim_{R}$ containing $a$ is the right coset $H a$ of $H$ in $G$ (when the binary operation of $G$ is written additively, our notation for left and right cosets becomes $a+H=\{a+h\mid h\in H\}$ and $H+a=\{h+a\mid h\in H\}$ ). Observe that the equivalence class under either $\sim_{L}$ or $\sim_{R}$ containing $e$ is $eH=H$ . The index of $H$ in $G$ , denoted by $|G:H|$ , is the cardinality of the set $G/H$ (read “ $G$ modulo $H$ ” or just “ $G$ mod $H$ ”) of left cosets of $H$ in $G$ (in fact, one may demonstrate the existence of a bijection between the set of left cosets of $H$ in $G$ and the set of right cosets of $H$ in $G$ , so that we may well take $|G:H|$ to be the cardinality of the set of right cosets of $H$ in $G$ ).

We now attempt to impose a group on $G/H$ by taking the of the left cosets containing the elements $a$ and $b$ , respectively, to be the left coset containing the element $a b$ ; however, because this definition requires a choice of left coset representatives, there is no guarantee that it will yield a well-defined binary operation on $G/H$ . For the of left coset to be well-defined, we must be sure that if $a^{\prime}H=aH$ and $b^{\prime}H=bH$ , i.e., if $a^{\prime}\in aH$ and $b^{\prime}\in bH$ , then $a^{\prime}b^{\prime}H=abH$ , i.e., that $a^{\prime}b^{\prime}\in abH$ . Precisely what must be required of the subgroup $H$ to ensure the of the above condition is the content of the following :

Proposition.

The rule $(aH,bH)\mapsto abH$ gives a well-defined binary operation on $G/H$ if and only if $H$ is a normal subgroup of $G$ .

Proof.

Suppose first that of left cosets is well-defined by the given rule, i.e, that given $a^{\prime}\in aH$ and $b^{\prime}\in bH$ , we have $a^{\prime}b^{\prime}H=abH$ , and let $g\in G$ and $h\in H$ . Putting $a=1$ , $a^{\prime}=h$ , and $b=b^{\prime}=g^{-1}$ , our hypothesis gives $hg^{-1}H=eg^{-1}H=g^{-1}H$ ; this implies that $hg^{-1}\in g^{-1}H$ , hence that $hg^{-1}=g^{-1}h^{\prime}$ for some $h^{\prime}\in H$ . on the left by $g$ gives $ghg^{-1}=h^{\prime}\in H$ , and because $g$ and $h$ were chosen arbitrarily, we may conclude that $gHg^{-1}\subseteq H$ for all $g\in G$ , from which it follows that $H\unlhd G$ . Conversely, suppose $H$ is normal in $G$ and let $a^{\prime}\in aH$ and $b^{\prime}\in bH$ . There exist $h_{1},h_{2}\in H$ such that $a^{\prime}=ah_{1}$ and $b^{\prime}=bh_{1}$ ; now, we have

a^{\prime}b^{\prime}=ah_{1}bh_{2}=a(bb^{-1})h_{1}bh_{2}=ab(b^{-1}h_{1}b)h_{2}% \text{,}

and because $b^{-1}h_{1}b\in H$ by assumption, we see that $a^{\prime}b^{\prime}=abh$ , where $h=(b^{-1}hb)h_{2}\in H$ by the closure of $H$ under in $G$ . Thus $a^{\prime}b^{\prime}\in abH$ , and because left cosets are either disjoint or equal, we may conclude that $a^{\prime}b^{\prime}H=abH$ , so that multiplication of left cosets is indeed a well-defined binary operation on $G/H$ . ∎

The set $G/H$ , where $H$ is a normal subgroup of $G$ , is readily seen to form a group under the well-defined binary operation of left coset multiplication (the of each group follows from that of $G$ ), and is called a quotient or factor group (more specifically the quotient of $G$ by $H$ ). We conclude with several examples of specific quotient groups.

Example.

A standard example of a quotient group is $\mathbb{Z}/n\mathbb{Z}$ , the quotient of the of integers by the cyclic subgroup generated by $n\in\mathbb{Z}^{+}$ ; the order of $\mathbb{Z}/n\mathbb{Z}$ is $n$ , and the distinct left cosets of the group are $n\mathbb{Z},1+n\mathbb{Z},\ldots,(n-1)+n\mathbb{Z}$ .

Example.

Although the group $Q_{8}$ is not abelian, each of its subgroups its normal, so any will suffice for the formation of quotient groups; the quotient $Q_{8}/\langle-1\rangle$ , where $\langle-1\rangle=\{1,-1\}$ is the cyclic subgroup of $Q_{8}$ generated by $-1$ , is of order $4$ , with elements $\langle-1\rangle,i\langle-1\rangle=\{i,-i\},k\langle-1\rangle=\{k,-k\}$ , and $j\langle-1\rangle=\{j,-j\}$ . Since each non-identity element of $Q_{8}/\langle-1\rangle$ is of order $2$ , it is isomorphic to the Klein $4$ -group $V$ . Because each of $\langle i\rangle$ , $\langle j\rangle$ , and $\langle k\rangle$ has order $4$ , the quotient of $Q_{8}$ by any of these subgroups is necessarily cyclic of order $2$ .

Example.

The center of the dihedral group $D_{6}$ of order $12$ (with http://planetmath.org/node/2182presentation $\langle r,s\mid r^{6}=s^{2}=1,r^{-1}s=sr\rangle$ ) is $\langle r^{3}\rangle=\{1,r^{3}\}$ ; the elements of the quotient $D_{6}/\langle r^{3}\rangle$ are $\langle r^{3}\rangle$ , $r\langle r^{3}\rangle=\{r,r^{4}\}$ , $r^{2}\langle r^{3}\rangle=\{r^{2},r^{5}\}$ , $s\langle r^{3}\rangle=\{s,sr^{3}\}$ , $sr\langle r^{3}\rangle=\{sr,sr^{4}\}$ , and $sr^{2}\langle r^{3}\rangle=\{sr^{2},sr^{5}\}$ ; because

sr^{2}\langle r^{3}\rangle r\langle r^{3}\rangle=sr^{3}\langle r^{3}\rangle=s% \langle r^{3}\rangle\neq sr\langle r^{3}\rangle=r\langle r^{3}\rangle sr^{2}% \langle r^{3}\rangle\text{,}

$D_{6}/\langle r^{3}\rangle$ is non-abelian, hence must be isomorphic to $S_{3}$ .

Title	quotient group
Canonical name	QuotientGroup
Date of creation	2013-03-22 12:04:06
Last modified on	2013-03-22 12:04:06
Owner	azdbacks4234 (14155)
Last modified by	azdbacks4234 (14155)
Numerical id	35
Author	azdbacks4234 (14155)
Entry type	Definition
Classification	msc 20-00
Synonym	factor group
Synonym	quotient
Related topic	Group
Related topic	NormalSubgroup
Related topic	Subgroup
Related topic	EquivalenceRelation
Related topic	Coset
Related topic	NaturalProjection
Defines	left congruence modulo a subgroup
Defines	right congruence modulo a subgroup
Defines	index