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 onlyif $b^{-1}a\in H$ is called left congruence modulo $H$; similarly the relation defined by $a{\sim }_{R}b$ ifand only if $a{b}^{-1}\in H$ 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 equivalenceclasses 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 $\overline{a}$ the equivalence class containing $a$ under ${\sim }_L$, we see that

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 $|G:H|$, isthe 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 bijectionbetween 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 $\mathrm{(}a\mathit{}H\mathrm{,}b\mathit{}H\mathrm{)}\mathrm{\mapsto }a\mathit{}b\mathit{}H$ gives a well-defined binary operation on $G\mathrm{/}H$ if and only if $H$ is a normal subgroupof $G$.

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 $h{g}^{-1}H=e{g}^{-1}H=g^{-1}H$; this implies that $h{g}^{-1}\in g^{-1}H$, hence that $h{g}^{-1}=g^{-1}{h}^{\prime }$ for some $h^{\prime }\in H$. on the left by $g$ gives $gh{g}^{-1}=h^{\prime }\in H$, and because $g$ and $h$ were chosen arbitrarily, we may conclude that $gH{g}^{-1}\subseteq H$ for all $g\in G$, from which it follows that $H⊴G$. 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 $a^{\prime }=a{h}_1$ and $b^{\prime }=b{h}_1$; now, we have

$$a^{\prime }{b}^{\prime }=a{h}_{1}b{h}_2=a(b{b}^{-1})h_{1}b{h}_2=ab(b^{-1}{h}_{1}b)h_2\text{,}$$

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 cosetsare either disjoint or equal, we may conclude that $a^{\prime }{b}^{\prime }H=abH$, so that multiplicationof 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-definedbinary operation of left coset multiplication (the of each group follows from that of $G$), and is called a quotient or factor group (more specificallythe 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 thedistinct left cosets of the group are $n\mathbb{Z},1+n\mathbb{Z},\mathrm{\dots },(n-1)+n\mathbb{Z}$.

Example.

Although the group $Q_8$ is not abelian, each of its subgroups its normal, so any will suffice for the formationof quotient groups; the quotient $Q_8/⟨-1⟩$, where $⟨-1⟩=\{1,-1\}$ is the cyclic subgroup of $Q_8$ generated by $-1$, is of order $4$, with elements $⟨-1⟩,i⟨-1⟩=\{i,-i\},k⟨-1⟩=\{k,-k\}$ , and $j⟨-1⟩=\{j,-j\}$. Since each non-identity element of $Q_8/⟨-1⟩$ is of order $2$, it is isomorphic to the Klein $4$-group $V$. Because each of $⟨i⟩$, $⟨j⟩$, and $⟨k⟩$ 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 http://planetmath.org/node/2182presentation $⟨r,s\mid r^6=s^2=1,r^{-1}s=sr⟩$) is $⟨r^3⟩=\{1,r^3\}$; the elements of the quotient $D_6/⟨r^3⟩$ are $⟨r^3⟩$, $r⟨r^3⟩=\{r,r^4\}$, $r^2⟨r^3⟩=\{r^2,r^5\}$, $s⟨r^3⟩=\{s,s{r}^3\}$, $sr⟨r^3⟩=\{sr,s{r}^4\}$, and $s{r}^2⟨r^3⟩=\{s{r}^2,s{r}^5\}$; because

$$s{r}^2⟨r^3⟩r⟨r^3⟩=s{r}^3⟨r^3⟩=s⟨r^3⟩\ne sr⟨r^3⟩=r⟨r^3⟩s{r}^2⟨r^3⟩\text{,}$$

$D_6/⟨r^3⟩$ is non-abelian, hence must be isomorphic to $S_3$.