§9 Quotient Groups

§9.1 Quotient Groups

Let $H\subset G$ be a subgroup.

Question: When can the orbit set

G/H

be given a group structure?

Remark $Hg$ is the orbit of $g$ under the action of $H$ on $G$ . Thus

Hg=\mathcal{O}_{g}.

Therefore, the question is: is there a natural group operation on the set of cosets of $H$ ?

One candidate is

\begin{aligned} G/H\times G/H&\to G/H\\ (Hg_{1},Hg_{2})&\mapsto Hg_{1}g_{2}. \end{aligned}

Is this map well defined?

In general, no. But in a special case, it is!

Theorem 9.1 If $H\subset G$ is a normal subgroup, then the operation

\begin{aligned} G/H\times G/H&\to G/H\\ (Hg_{1},Hg_{2})&\mapsto Hg_{1}g_{2} \end{aligned}

is well defined and makes $G/H$ into a group.

Proof: To prove that this operation is well defined, we need to show that if there exist

g_{1},g_{1}^{\prime},g_{2},g_{2}^{\prime}\in G

such that

Hg_{1}=Hg_{1}^{\prime}

and

Hg_{2}=Hg_{2}^{\prime},

then

Hg_{1}g_{2} = Hg_{1}^{\prime}g_{2}^{\prime}.

Indeed,

Hg_{1}^{\prime}g_{2}^{\prime} = \{hg_{1}^{\prime}g_{2}^{\prime}\mid h\in H\},

where

\begin{aligned} Hg_{1}=Hg_{1}^{\prime} &\Rightarrow \mathcal{O}_{g_{1}} = \mathcal{O}_{g_{1}^{\prime}}\\ &\Rightarrow g_{1},g_{1}^{\prime} ~\text{belong to the same orbit}\\ &\Rightarrow g_{1}^{\prime} = h_{1}g_{1}, \quad \text{for some }h_{1}\in H. \end{aligned}

Similarly,

Hg_{2}=Hg_{2}^{\prime} \Rightarrow g_{2}^{\prime}=h_{2}g_{2}, \quad \text{for some }h_{2}\in H.

Therefore,

\begin{aligned} Hg_{1}^{\prime}g_{2}^{\prime} &= \{h\cdot h_{1}g_{1}\cdot h_{2}g_{2}\mid h\in H\}\\ &= \{h\cdot h_{1}g_{1}\cdot h_{2}g_{1}^{-1}g_{1}g_{2}\mid h\in H\}\\ &= \{h\cdot h_{1}h_{3}g_{1}g_{2}\mid h\in H\} \subset Hg_{1}g_{2}. \end{aligned}

Since $H$ is normal, we used

h_{3}:=g_{1}h_{2}g_{1}^{-1}\in H.

Since $Hg_{1}g_{2}$ and $Hg_{1}^{\prime}g_{2}^{\prime}$ are orbits/equivalence classes,

Hg_{1}g_{2} \supset Hg_{1}^{\prime}g_{2}^{\prime}.

Therefore,

Hg_{1}g_{2} = Hg_{1}^{\prime}g_{2}^{\prime}.

Thus the operation is well defined.

Why is this a group?

(1) Associativity:

\begin{aligned} (Hg_{1}\cdot Hg_{2})\cdot Hg_{3} &= Hg_{1}g_{2}\cdot Hg_{3}\\ &= H(g_{1}g_{2})g_{3}\\ &= Hg_{1}(g_{2}g_{3})\\ &= Hg_{1}\cdot Hg_{2}g_{3}\\ &= Hg_{1}\cdot(Hg_{2}\cdot Hg_{3}). \end{aligned}

(2) Identity:

H1_{G}\cdot Hg = Hg = Hg\cdot H1_{G}.

(3) Inverses:

\begin{aligned} Hg\cdot Hg^{-1} &= Hgg^{-1}\\ &= H1_{G}\\ &= Hg^{-1}g\\ &= Hg^{-1}\cdot Hg. \end{aligned}

~\tag*{$\square$}

Let us look at some examples.

Example 9.1

H=\{\mathrm{id}_{G}\}\subset G.

$H$ is normal because for every $g\in G$ ,

\begin{aligned} gHg^{-1} &= \{ghg^{-1}\mid h\in H\}\\ &= \{g\mathrm{id}_{G}g^{-1}\}\\ &= \{\mathrm{id}_{G}\}\\ &= H. \end{aligned}

But $G/H$ is not a new or particularly different group, since there is an isomorphism

\begin{aligned} G/H&\to G\\ Hg&\mapsto g. \end{aligned}

Example 9.2 Let

H=n\mathbb{Z} = \{\text{multiples of }n\} = \{\ldots,-2n,-n,0,n,2n,\ldots\}.

Then

Ha = \{a^{\prime}\in\mathbb{Z} \mid a^{\prime}=kn+a, ~\text{for some }k\in\mathbb{Z}\} = \mathcal{O}_{a}.

Proposition 9.2 There is a bijection

\mathbb{Z}/n\mathbb{Z} \to \{0,1,\ldots,n-1\}

for every $n\geqslant1$ .

Proof: Given

\mathcal{O}_{a} \in \mathbb{Z}/n\mathbb{Z},

let $r_{a}$ be the unique number such that

a=kn+r_{a}, \quad k\in\mathbb{Z}, \quad r_{a}\in\{0,1,\ldots,n-1\}.

That is, $r_a$ is the remainder obtained when dividing $a$ by $n$ —a concept familiar from elementary school.

Define the bijection by

\begin{aligned} \mathbb{Z}/n\mathbb{Z} &\to \{0,1,\ldots,n-1\}\\ \mathcal{O}_{a} &\mapsto r_{a}. \end{aligned}

Well defined?

\mathcal{O}_{a} = \mathcal{O}_{a^{\prime}},

then

a^{\prime} = a+k^{\prime}n \tag{by the definition of an orbit}.

Therefore,

\begin{aligned} a^{\prime} &= k^{\prime}n+kn+r_{a}\\ &= (k^{\prime}+k)n+r_{a}. \end{aligned}

Hence

r_{a^{\prime}} = r_{a},

where $r_{a^{\prime}}$ is the unique element of ${0,1,\ldots,n-1}$ appearing in

a^{\prime}=kn+r_{a^{\prime}}.

Thus the map is well defined.

Injective?

Given $\mathcal{O}*{a}$ and $\mathcal{O}*{b}$ , suppose

r_{a}=r_{b}.

Then

\begin{aligned} a&=kn+r_{a}\\ b&=ln+r_{b}\\ &=ln+r_{a}. \end{aligned}

Therefore,

a-b=(k-l)n,

a\in\mathcal{O}_{b},

and hence

\mathcal{O}_{a} = \mathcal{O}_{b}.

Surjective?

\begin{aligned} \mathcal{O}_{0}&\mapsto0\\ \mathcal{O}_{1}&\mapsto1\\ \mathcal{O}_{2}&\mapsto2\\ &~\,\vdots\\ \mathcal{O}_{n-1}&\mapsto n-1. \end{aligned}

~\tag*{$\square$}

By this proposition,

\mathbb{Z}/n\mathbb{Z}

is a group of order

n = |\{0,1,\ldots,n-1\}|.

What is the structure of this group?

The group operation is: add the numbers and then take the remainder modulo $n$ .

Definition 9.1 Let $a,b\in\mathbb{Z}$ . We write

a\equiv b\pmod n

a=b\pmod n

a-b=kn \quad \text{for some }k\in\mathbb{Z}.

Equivalently,

a\equiv b\pmod n \Longleftrightarrow \mathcal{O}_{a} = \mathcal{O}_{b}.

When we write

a\pmod n,

we mean the equivalence class

\mathcal{O}_{a} = Ha \in G/H.

Remark It may seem inconvenient to keep track of large equivalence classes $\mathcal{O}*{a}$ , $\mathcal{O}*{b}$ , and so on. Instead, we may simply regard $\mathcal{O}_{a}$ as a number: namely, the remainder $r$ obtained when $a$ is divided by $n$ :

a=kn+r.

This is justified by the bijection

\mathbb{Z}/n\mathbb{Z} \cong \{0,\ldots,n-1\}.

Thus, whenever you see “ $a\pmod n$ ”, you may simply think of the number $r$ .

Similarly, the group operation is just “clock arithmetic”:

(a,b)\mapsto a+b\pmod n.

You can think of this as the remainder of $(a+b)$ upon division by $n$ .

§9.2 Subgroups Descend to Quotient Groups

Let $G$ be an arbitrary group and let

H\triangleleft G.

Proposition 9.3 There is a bijection between the set of subgroups of $G$ containing $H$ and the set of subgroups of $G/H$ .

Proof: Let

p:G\to G/H

be the group homomorphism sending $g$ to $[g]$ .

Given a subgroup $K\subset G$ , consider the composition of group homomorphisms

K\xhookrightarrow{}G\to G/H.

Since the image of any group homomorphism is a subgroup, this shows that $p(K)$ is a subgroup of $G/H$ .

Thus we obtain a map

\{\text{subgroups of }G\} \to \{\text{subgroups of }G/H\}

given by

K\mapsto p(K).

We prove that this map is surjective. Given

K^{\prime}\subset G/H,

consider the preimage

p^{-1}(K^{\prime})\subset G.

This is a subgroup of $G$ , because if

p(x),p(y)\in K^{\prime},

then

p(xy) = p(x)p(y) \in K^{\prime},

since $K^{\prime}$ is closed under multiplication.

Now it remains only to prove that for every $K\subset G$ ,

p^{-1}(p(K)) = K.

Clearly,

K\subset p^{-1}(p(K)).

To prove the other inclusion, let

x\in p^{-1}(p(K)).

By the definition of $p(K)$ , there exists some $y\in K$ such that

p(x)=p(y).

Then

p(xy^{-1}) = 1_{G/H},

xy^{-1}\in H.

Since $K$ contains $H$ ,

xy^{-1}\in K,

and therefore

x\in K.

~\tag*{$\square$}

Proposition 9.4 There is a bijection between the set of normal subgroups of $G$ containing $H$ and the set of normal subgroups of $G/H$ .

Proof: We first prove that if $K$ is normal, then $p(K)$ is normal. This gives a map

\{\text{normal subgroups of }G\} \to \{\text{normal subgroups of }G/H\}.

Indeed, if

[k]\in p(K),

then

[g][k][g]^{-1} = [gkg^{-1}] = [k^{\prime}]

for some $k^{\prime}\in K$ , because $K$ is normal in $G$ .

Therefore,

p(K)\subset G/H

is a normal subgroup.

Notice that here we used the fact that

G\to G/H

is surjective; otherwise, we would not know that every element of $G/H$ lies in the image $p(G)$ .

Surjectivity: We prove that if $p(K)$ is normal, then

K=p^{-1}(p(K))

is normal.

k\in K, \qquad g\in G,

then

[gkg^{-1}] = [g][k][g^{-1}] = [k^{\prime}]

for some

[k^{\prime}]\in p(K),

i.e. for some $k^{\prime}\in K$ .

Therefore,

gkg^{-1} \in p^{-1}(p(K)) = K.

We already know that this assignment is injective.

~\tag*{$\square$}

§9.3 Commutative Diagrams

Definition 9.2 Suppose we have groups $K,G,H,X$ and maps between these groups:

\psi\circ\phi = \beta\circ\alpha: K\to X,

we say that the diagram commutes, or that it is commutative.

Remark The symbol $1$ will denote the trivial group. This group is uniquely determined up to isomorphism.

Now consider the diagram

What does it mean for this diagram to commute?

It means precisely that

\psi\circ\phi

is the map sending every element to the identity.

Why?

Because the maps along the left and bottom send every element of $K$ to

1_X\in X

assuming they are group homomorphisms. Thus $\psi\circ\phi$ is a constant map.

Equivalently,

\mathrm{im}(\phi) \subset \ker(\psi).

This is very close to the notion of exactness that we discussed earlier.

Example 9.3

K=G=X=1.

Example 9.4

K=SL_n(\mathbb{R}), \qquad G=GL_n(\mathbb{R}), \qquad X=\mathbb{R}^{\times},

where $\phi$ is the inclusion map and

\psi=\det.

Example 9.5 Let

K\subset G

be a normal subgroup, so $\phi$ is the inclusion map.

Let

X=G/K.

Since $K$ is normal in $G$ , this is a group. Let $\psi$ be the canonical quotient projection.

§9.4 The Universal Property of Quotient Groups

Theorem 9.5 Given a commutative diagram

there exists a unique homomorphism

\tilde{\psi}:G/X\to X

such that the diagram

commutes.

In other words, if

\psi:G\to X

is such that

K\subset\ker(\psi)

for some normal subgroup $K\subset G$ , then there exists a unique map

\tilde{\psi}:G/K\to X

such that

\psi = \tilde{\psi}\circ\pi,

where

\pi:G\to G/K

is the canonical quotient map.

This is called the universal property of the quotient group.

Proof: We want to define the map

\tilde{\psi}:G/K\to X.

Define it by taking a coset

Kg\mapsto\psi(g),

or equivalently by choosing any preimage in $G$ of a coset in $G/K$ under the canonical map $\pi$ .

We need to prove that this is well defined.

Suppose

Kg=Kg^{\prime},

or equivalently

g=k\cdot g^{\prime}.

Do we have

\psi(g)=\psi(g^{\prime})?

Indeed,

\psi(g) = \psi(k\cdot g^{\prime}) = \psi(k)\cdot\psi(g^{\prime}) = \psi(g^{\prime}),

so everything works.

We now need to verify that $\tilde{\psi}$ is unique and that

\psi = \tilde{\psi}\circ\pi.

But

\psi(g) = \tilde{\psi}(Kg) = \tilde{\psi}(\pi(g)).

Thus it remains only to prove uniqueness.

Suppose there is another such map

\gamma:G/K\to X.

Then

\psi(g) = \gamma(\pi(g)) = \gamma(Kg).

Hence $\gamma$ and $\tilde{\psi}$ must agree on every coset $K$ , and we are done.

~\tag*{$\square$}

Remark The uniqueness follows from the requirement in the theorem that the diagram commute.

§9.5 Generalization of Quotient Groups

The idea of a universal property allows us to generalize the notion of a quotient group.

Let

G\overset{\chi}{\twoheadrightarrow}H

be any surjective homomorphism. We now use a double arrow to indicate surjectivity.

Then $H$ is a “quotient group” of $G$ , in the sense that a diagram of the form

has a similar universal property.

To answer the question, what should $K$ be? We simply take it to be the kernel of $\chi$ .

Thus:

Theorem 9.6 Given

G\overset{\chi}{\twoheadrightarrow}H,

let

K=\ker(\chi).

Then the diagram

where

i:K\to G

is the inclusion map, has the same universal property:

given any group homomorphism

\psi:G\to X

whose kernel contains $K$ , there exists a unique map

\tilde{\psi}:H\to X

such that

\psi = \tilde{\psi}\circ\chi.

Proof: How do we define

\tilde{\psi}:H\to X?

Since $\chi$ is surjective, for every $h\in H$ there exists some $g\in G$ such that

h=\chi(g),

although $g$ need not be unique.

Now define

\tilde{\psi}:=\psi(g).

We need to check that this is a well-defined homomorphism.

Suppose

\chi(g)=\chi(g^{\prime}).

Then

\chi(g)=\chi(g^{\prime})

implies

\chi(g\cdot(g^{\prime})^{-1}) = 1_H.

Therefore,

g\cdot(g^{\prime})^{-1} \in K,

and hence

g=g^{\prime}\cdot k

for some $k\in K$ .

The remainder of the argument is exactly the same as in the proof of the previous theorem.

~\tag*{$\square$}

Corollary 9.7 Suppose

\chi:G\to H

is surjective and

K=\ker(\chi).

Then

H\cong G/K.

Remark This is the so-called First Isomorphism Theorem. Let us prove it using universal properties. We will see it again in the next chapter.

Proof: Theorem 9.3 gives a map from

G/K

H,

and Theorem 9.4 gives a map from

H

G/K.

Composing these maps gives a map

G/K\to H\to G/K.

This composite satisfies the required commutativity constraints.

The composite is given by the two maps obtained from the two theorems, but the identity map

G/K\to G/K

also satisfies the same constraints.

Since these maps are uniquely determined, the composite must be the identity.

Similarly, in the other direction, we obtain a composite

H\to G/K\to H,

and the same argument shows that this composite is the identity on $H$ .

Thus we have maps

G/K\to H

and

H\to G/K

whose composites in both directions are identity maps.

Therefore there is a bijection between $G/K$ and $H$ . Since the maps are group homomorphisms, they are in fact isomorphic as groups.

~\tag*{$\square$}

Remark The notion of a universal property allows us to characterize certain objects uniquely, up to isomorphism, just as we did in the proof of the corollary, because a universal property is a statement of existence and uniqueness.

If you are thinking categorically, two objects are “isomorphic in a category” precisely when they satisfy the same universal property.

In our example, two groups—the quotient group and any group onto which $G$ maps surjectively—satisfy the same universal property in the “category of groups”, so they are isomorphic.

This is a very important idea in category theory and is related to a consequence of the Yoneda lemma.

Proposition 9.8 Let $K,G$ be arbitrary groups and let

\phi:K\to G

be any group homomorphism.

Then there exists another group homomorphism

\psi:G\to H

such that we have a universal property represented by the diagram

where

\epsilon:K\to1

is the trivial homomorphism sending every element of $K$ to the identity of the trivial group $1$ , and the bottom arrow

1\to H

is the unique homomorphism from the trivial group to $H$ .

Proof:

Construct $H$ and $\psi$ .

Define

H = G/\phi(K),

where $\phi(K)$ is the image of $K$ under $\phi$ , and hence is a subgroup of $G$ .

The quotient

G/\phi(K)

consists of all cosets

g\phi(K), \qquad g\in G.

Define

\psi:G\to H

to be the natural projection

\psi(g) = g\phi(K).

Thus $\psi$ sends each element $g$ to its coset in $H$ , and $\psi$ is a surjective homomorphism.

Verify commutativity of the diagram.

We need to show

\psi\circ\phi = \epsilon.

For any $k\in K$ , compute

\psi(\phi(k)) = \phi(k)\phi(K) = \phi(K).

Since

\phi(k)\in\phi(K),

the coset

\phi(k)\phi(K)

is equal to $\phi(K)$ , which is the identity element of $H$ .

Thus $\psi(\phi(k))$ sends every element of $K$ to the identity in $H$ .

This agrees with the composition through $\epsilon$ :

(\text{the unique map from }1\text{ to }H) \circ \epsilon(k) = 1_H.

Therefore, the diagram commutes.

This construction satisfies the following universal property:

For any group $L$ and any homomorphism

f:G\to L

such that

f\circ\phi = \epsilon,

there exists a unique homomorphism

\overline{f}:H\to L

such that

f = \overline{f}\circ\psi.

Indeed:

(1) Since

f\circ\phi = \epsilon,

for every $k\in K$ ,

f(\phi(k)) = 1_L.

This means

\phi(k)\in\ker(f)

for all $k\in K$ .

Hence

\phi(K) \subset \ker(f).

(2) Define

\overline{f}:H\to L

\overline{f}(g\phi(K)) = f(g).

This is well defined because if

g\phi(K) = g^{\prime}\phi(K),

then

g^{-1}g^{\prime} \in \phi(K).

Therefore,

f(g^{-1}g^{\prime}) = 1_L,

which implies

f(g) = f(g^{\prime}).

Hence $\overline{f}$ is a well-defined homomorphism.

(3) Uniqueness:

Suppose there is another homomorphism

\overline{f}^{\prime}

such that

f = \overline{f}^{\prime}\circ\psi.

Then for every $g\in G$ ,

\overline{f}(g\phi(K)) = f(g) = \overline{f}^{\prime}(g\phi(K)).

Therefore,

\overline{f} = \overline{f}^{\prime}.

~\tag*{$\square$}

Contents

§9 Quotient Groups

§9.1 Quotient Groups

§9.2 Subgroups Descend to Quotient Groups

§9.3 Commutative Diagrams

§9.4 The Universal Property of Quotient Groups

§9.5 Generalization of Quotient Groups