# quotient group

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  $aH$ 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 $a+H=\{a+h\mid h\in H\}$ 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 $ab$; 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 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