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# quotient group

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).

###### 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 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

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

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 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

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

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 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 $Q_{8}/\langle-1\rangle$, 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 presentation $\langle r,s\mid r^{6}=s^{2}=1,r^{{-1}}s=sr\rangle$) 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 $S_{3}$.

## Mathematics Subject Classification

20-00*no label found*

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