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quotient group
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(Definition)
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Before defining quotient groups, some preliminary definitions must be introduced and a few propositions established.
Given a group $G$ and a subgroup $H$ of $G$ , the relation $\sim_L$ 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 right congruence modulo $H$ (observe that these two relations coincide if $G$ is abelian).
Proof. We will only give the proof for left congruence modulo $H$ , as the argument 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 reflexive. If $b\in G$ satisfies
$a\sim_L b$ , so that $b^{-1}a\in H$ , then by the closure 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 transitive, hence an equivalence relation. 
It follows from the preceding proposition 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{equation*} \begin{split} \bar{a}&=\set{b\in G\mid b\sim_L a}\\ &=\set{b\in G\mid a^{-1}b\in H}\\ &=\set{b\in G\mid
b=ah\text{ for some }h\in H}=\set{ah\mid h\in H}\text{.} \end{split} \end{equation*} Thus the equivalence class under $\sim_L$ containing $a$ is simply the left coset $aH$ of $H$ in $G$ . Similarly the equivalence class under $\sim_R$ containing $a$ is the right coset $Ha$ of $H$ in $G$ (when the binary operation of $G$ is written additively, our notation for left and right cosets becomes $a+H=\set{a+h\mid h\in H}$ and $H+a=\set{h+a\mid h\in H}$ ). 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 $\abs{G:H}$ , 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 $\abs{G:H}$ 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 product of the left cosets containing the elements $a$ and $b$ , respectively, to be the left coset containing the element $ab$ ; 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 operation of left coset multiplication 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 satisfaction of the above condition is the content of the following proposition:
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 multiplication 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 $h^\prime\in H$ . Multiplication 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 \begin{equation*} a^\prime b^\prime=ah_1bh_2=a(bb^{-1})h_1bh_2 =ab(b^{-1}h_1b)h_2\text{,} \end{equation*}and because $b^{-1}h_1b\in H$ 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 multiplication 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 satisfaction of each group axiom 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 additive group 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 $Q_8/\vbrack{-1}$ , where $\vbrack{-1}=\set{1,-1}$ is the cyclic subgroup of $Q_8$ generated by $-1$ , is of order $4$ , with elements $\vbrack{-1},i\vbrack{-1}=\set{i,-i},k\vbrack{-1}=\set{k,-k}$ , and $j\vbrack{-1}=\set{j,-j}$ . Since each non-identity element of $Q_8/\vbrack{-1}$ is of order $2$ , it is isomorphic to the Klein $4$ -group $V$ . Because each of $\vbrack{i}$
, $\vbrack{j}$ , and $\vbrack{k}$ 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 presentation $\vbrack{r,s\mid r^6=s^2=1,r^{-1}s=sr}$ ) is $\vbrack{r^3}=\set{1,r^3}$ ; the elements of the quotient $D_6/\vbrack{r^3}$ are $\vbrack{r^3}$ , $r\vbrack{r^3}=\set{r,r^4}$ , $r^2\vbrack{r^3}=\set{r^2,r^5}$ , $s\vbrack{r^3}=\set{s,sr^3}$ , $sr\vbrack{r^3}=\set{sr,sr^4}$ , and $sr^2\vbrack{r^3}=\set{sr^2,sr^5}$ ;
because \begin{equation*} sr^2\vbrack{r^3}r\vbrack{r^3}=sr^3\vbrack{r^3}=s\vbrack{r^3}\neq sr\vbrack{r^3}=r\vbrack{r^3}sr^2\vbrack{r^3}\text{,} \end{equation*}$D_6/\vbrack{r^3}$ is non-abelian, hence must be isomorphic to $S_3$ .
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Cross-references: non-Abelian, dihedral group, center, cyclic, isomorphic, order, generated by, cyclic subgroup, integers, multiplication, disjoint, conversely, implies, hypothesis, normal subgroup, well-defined, bijection, cardinality, right coset, left coset, equivalence classes, mutually disjoint, binary operation, closure, symmetric, inverses, identity, contains, proof, equivalence relation, right, abelian, relation, congruence, subgroup, group, definitions
There are 147 references to this entry.
This is version 31 of quotient group, born on 2001-12-21, modified 2008-08-04.
Object id is 1127, canonical name is QuotientGroup.
Accessed 24344 times total.
Classification:
| AMS MSC: | 20-00 (Group theory and generalizations :: General reference works ) |
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Pending Errata and Addenda
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