11.Quotient group

Overview

The left and right cosets of a normal subgroup are the same, and the set of all cosets forms a partition of the group. Using the same equivalence relation to form the quotient group, it is denoted by $G/H$. A group can be defined on $G/H$, called the quotient group.

$H$ is a normal subgroup of $G$. Let the set $G/H=\{aH\mid a\in G\}(\text{or}G/H=\{Ha\mid a\in G\}),$ which is the set formed by all cosets of $H$ in $G$.

Define an operation "$\cdot$" on $G/H$: $\mathrm{\forall }aH,bH\in G/H,aH\cdot bH=(ab)H.$

Then $G/H$​ forms a group under this operation.

Definition

Let $H$ be a normal subgroup of $G$. The set of all cosets of $H$ in $G$ is a group, denoted as $G/H$, called the quotient group of $G$ by $H$, where the operation is defined as:

$$\mathrm{\forall }aH,bH\in G/H,aH\cdot bH=(ab)H.$$

In the group $<\mathbb{Z},+>$, $H_m=<m>=\{...,-2m,-m,0,m,2m,...\}$ is a normal subgroup, then the quotient group $\mathbb{Z}/H_m=\{k+<m>\}$ is ${\mathbb{Z}}_m$, where $x+H_m=\{nm+x\mid n\in \mathbb{Z}\}$.

For simplicity, it can be written as $\{0,1,2,\dots ,m-1\}$, which is the integer modulo $m$ addition group $<{\mathbb{Z}}_m,+>$​.

For example, when $m=6,H_6=\{...,-12,-6,0,6,12,...\}$​,

$\mathbb{Z}/H_6=\{a{H}_6|a\in \mathbb{Z}\}$

$a=1,1H_6=1+H_6=\{...,-11,-5,1,7,13,...\}$​,

Everything in this set divided by 6 has a residual 1.

$a=2,2H_6=2+H_6=\{...,-10,-4,2,9,14,...\}$​

Everything in this set divided by 6 has a residual 2.

So they are integer modulo 6 addition group $<{\mathbb{Z}}_6,+>$.

In general, $G/H$ is also called the coset group of $G$ modulo $H$. Thus $G/H$ can also be represented as $G/H=\{\bar{a}\mid a\in G\}$, where $\bar{a}=aH$.

The order of the quotient group $G/H$ is the index of $H$ in $G$. When $G$ is finite, $|G/H|=|G|/|H|$​​​.

Natural mapping

Let $H$ be a normal subgroup of $G$, $G/H$ is the quotient group of $G$ by $H$. Define a mapping:

$$g:G\to G/H,\mathrm{\forall }a\in G,g(a)=aH,$$

then $g$ is a surjective homomorphism from $G$ to the quotient group $G/H$. This $g$ is called the natural projection.

Proof:

Clearly, $g$ is surjective, so we only need to prove that $g$ is a homomorphism. For $\mathrm{\forall }a,b\in G$, we have

$$g(ab)=(ab)H=aH\cdot bH=g(a)\cdot g(b),$$

so $g$ is a homomorphism.

This Theory shows that the group $G$ and its arbitrary quotient group $G/H$ are isomorphic.

Fundamental theorem of homomorphism

Let $G,G_1$ be groups, $e,e^{\prime }$ be the identities of $G$ and $G_1$, respectively. If $f:G\to G_1$ is a surjective homomorphism, and the kernel $K=\text{Ker}f$ is a normal subgroup of $G$, then the quotient group $<G/K,\cdot >$ is isomorphic to $G_1$.

Proof: Construct a mapping $g:G/K\to G_1$, $\mathrm{\forall }aK\in G/K,g(aK)=f(a)$​.

Why?

If there's an element in $G$, there must be a coset $aK$ in $G/K$, and a $f(a)$ in $G_1$.

Since we need to find a mapping from $G/K$ to $G_1$,...

Below, we show that $g$ is a homomorphism.

Because quotient groups are all cosets, different representative elements may represent different cosets, so it is necessary to prove that different representative elements have no effect on the final result. So we need to prove: if $aK=bK$ then $f(a)=f(b)$.

(1) First prove that $g$ is a mapping from $G/K$ to $G_1$: $\mathrm{\forall }aK,bK\in G/K,\text{if}aK=bK,\text{then}{a}^{-1}b\in K,\text{so}f(a^{-1}b)=e^{\prime },$ which implies $f(a{)}^{-1}f(b)=(f(a){)}^{-1}f(b)=e^{\prime },$ thus $f(a)=f(b)$.

Therefore, $g(aK)=g(bK)$, so $g$ is a mapping from $G/K$ to $G_1$.

(2) Next, prove that $g$ is bijective. For $\mathrm{\forall }aK,bK\in G/K$, if $g(aK)=g(bK)$, then $f(a)=f(b)$, thus $f(a)(f(b){)}^{-1}=e^{\prime }$, i.e.,

$$f(a)(f(b){)}^{-1}=f(a)f(b^{-1})=f(a{b}^{-1})=e^{\prime },$$

which means $a{b}^{-1}\in K$, so $aK=bK$, making $g$ injective.

For $\mathrm{\forall }c\in G_1$, since $f$ is surjective, there exists $a\in G$, such that $f(a)=c$, then take $aK\in G/K$, and $g(aK)=f(a)=c$, so $g$ is surjective, and thus $g$ is bijective.

(3) Finally, prove that $g$ is a homomorphism. For $\mathrm{\forall }aK,bK\in G/K$, we have

$$g(aK\cdot bK)=g((ab)K)=f(ab)=f(a)f(b)=g(aK)g(bK),$$

so $g$ is a homomorphism. In summary, $g$ is an isomorphism, that is, the quotient group $G/K$ is isomorphic to the group $G_1$.