Quotient Groups

Recall the construction of $\mathbb{Z}/n\mathbb{Z}$: we put an equivalence relation on $\mathbb{Z}$, take equivalence classes, and then add classes by choosing representatives. This was our first quotient group. We now seek to generalize and understand this construction for arbitrary groups $G$. This will be another way to obtain a “smaller” group from $G$, and it can also be used to study the structure of $G$, like subgroups did.

Here is a perspective that helped me a lot: The study of quotient groups of $G$ is essentially equivalent to the study of homomorphisms of $G$ to another group. Let $\phi :G\to H$ be a homomorphism and recall that the fibers of $\phi$ are the sets of elements of $G$ projecting to single elements of $H$. You can picture this as in the following figure.

The group operation in $H$ provides a way to multiply two elements $a$ and $b$ in the image of $\phi$. This gives a natural multiplication of the fibers lying above those two points. This then makes the set of fibers into a group. This is a quotient group — $G$ has been partitioned into pieces and these pieces have the structure of a group.

Since the multiplication of fibers is defined from multiplication in $H$, it is easy to see an important fact: this quotient group is isomorphic to $\mathrm{im}\phi$. We now need to make all of this precise: how do we form a quotient group from a group $G$ without having a homomorphism to another group? The answer is that we need to find a way to partition $G$ into pieces that “behave like fibers”, and we will see that the answer is normal subgroups.

Cosets and Lagrange’s Theorem

Definition

Let $\sim$ be an equivalence relation on a group $G$. We say that $\sim$ is compatible with multiplication if, whenever $x\sim y$, we also have

$$xz\sim yz\text{and}zx\sim zy$$

for every $z\in G$.

Lemma

Let $G$ be a group and let $\sim$ be an equivalence relation on $G$. The rule

$$[x]\cdot [y]=[xy]$$

is well-defined on the set of equivalence classes $G/\sim$ if and only if $\sim$ is compatible with multiplication. In that case, $G/\sim$ is a group under this operation.

Proof

Suppose first that $\sim$ is compatible with multiplication. If $[x]=[x^{\prime }]$ and $[y]=[y^{\prime }]$, then $x\sim x^{\prime }$ and $y\sim y^{\prime }$. Compatibility gives

$$xy\sim x^{\prime }y\text{and}{x}^{\prime }y\sim x^{\prime }{y}^{\prime },$$

so by transitivity $xy\sim x^{\prime }{y}^{\prime }$. Hence $[xy]=[x^{\prime }{y}^{\prime }]$, and the product of classes is well-defined.

Conversely, suppose $[x]\cdot [y]=[xy]$ is well-defined. If $x\sim x^{\prime }$, then $[x]=[x^{\prime }]$, so for any $z\in G$,

$$[xz]=[x][z]=[x^{\prime }][z]=[x^{\prime }z].$$

Thus $xz\sim x^{\prime }z$. Similarly,

$$[zx]=[z][x]=[z][x^{\prime }]=[z{x}^{\prime }],$$

so $zx\sim z{x}^{\prime }$. Therefore $\sim$ is compatible with multiplication.

Finally, if the operation is well-defined, the group axioms are inherited from $G$:

$$[x]([y][z])=[x][yz]=[x(yz)]=[(xy)z]=[xy][z]=([x][y])[z],$$

the identity is $[e_G]$, and the inverse of $[x]$ is $[x^{-1}]$.

Definition

If $\sim$ is an equivalence relation on a group $G$ compatible with multiplication, then the group $G/\sim$ of equivalence classes is called the quotient group of $G$ by $\sim$.

Example

Let $G=\mathbb{Z}$ and fix an integer $n\ge 1$. Let ${\sim }_n$ be congruence modulo $n$. This equivalence relation is compatible with addition, and

$$\mathbb{Z}/{\sim }_n=\mathbb{Z}/n\mathbb{Z}.$$

Now we need a systematic way to produce equivalence relations on groups. Subgroups give the most important examples.

Definition

Let $H\le G$ be a subgroup. A left coset of $H$ in $G$ is a subset of $G$ of the form

$$gH=\{gh\mid h\in H\}$$

for some $g\in G$. A right coset of $H$ in $G$ is a subset of $G$ of the form

$$Hg=\{hg\mid h\in H\}$$

for some $g\in G$.

Example

Let $G=\mathbb{Z}$ and let $H=n\mathbb{Z}=\{nk\mid k\in \mathbb{Z}\}$. Since $\mathbb{Z}$ is abelian, the left and right cosets are the same:

$$a+n\mathbb{Z}=\{a+nk\mid k\in \mathbb{Z}\}.$$

These are exactly the congruence classes modulo $n$.

Coset criterion

Let $H\le G$ and let $x,y\in G$. Then

$$xH=yH⟺y^{-1}x\in H⟺x^{-1}y\in H.$$

Equivalently, $x$ and $y$ lie in the same left coset of $H$ exactly when $y^{-1}x\in H$. The same statement holds for right cosets.

Proof

If $xH=yH$, then $x\in yH$, so $x=yh$ for some $h\in H$ and hence $y^{-1}x=h\in H$. Conversely, if $y^{-1}x=h\in H$, then $x=yh$, and

$$xH=yhH=yH.$$

Since $H$ is closed under inverses, $y^{-1}x\in H$ if and only if $x^{-1}y=(y^{-1}x{)}^{-1}\in H$. The right coset statement is exactly the same.

Proposition

Let $H\le G$. The left cosets of $H$ in $G$ partition $G$, and the right cosets of $H$ in $G$ also partition $G$.

Proof

We prove the statement for left cosets. Every element $g\in G$ belongs to at least one left coset, namely $gH$, since $g=g{e}_G$ and $e_G\in H$.

Now suppose two left cosets $xH$ and $yH$ intersect. Choose $z\in xH\cap yH$. Then $z$ lies in the same left coset as $x$ and also in the same left coset as $y$. By Lemma Coset criterion, $xH=zH$ and $zH=yH$, so $xH=yH$. Thus any two left cosets are either equal or disjoint.

Proposition

Let $H\le G$. Every left coset and every right coset of $H$ has the same cardinality as $H$. More precisely, for every $g\in G$,

$$|gH|=|H|=|Hg|.$$

Proof

The map

$$H\to gH,h\mapsto gh$$

is surjective by definition of $gH$. It is injective because $g{h}_1=g{h}_2$ implies $h_1=h_2$ after multiplying on the left by $g^{-1}$. Thus $|gH|=|H|$.

Similarly, the map

$$H\to Hg,h\mapsto hg$$

is a bijection, so $|Hg|=|H|$.

Definition

The index of a subgroup $H$ in a group $G$, denoted $[G:H]$, is the number of left cosets of $H$ in $G$. Equivalently, it is the number of right cosets of $H$ in $G$.

Now, we prove a statement older than the modern definition of a group. Lagrange’s theorem says that the order of a subgroup must divide the order of the group.

Lagrange's theorem

Let $G$ be a finite group and let $H\le G$. Then

$$|G|=|H|[G:H].$$

In particular, $|H|$ divides $|G|$.

Proof

The left cosets of $H$ partition $G$, and each left coset has $|H|$ elements. If there are $[G:H]$ left cosets, then

$$|G|=|H|[G:H].$$

One reason Lagrange’s theorem is important is that it turns counting information about a group into arithmetic information about its elements.

Fermat's little theorem

Let $p$ be prime. The nonzero congruence classes modulo $p$ form a group under multiplication:

$$(\mathbb{Z}/p\mathbb{Z}{)}^{\times }=\{\overline{1},\overline{2},\dots ,\overline{p-1}\}.$$

This group has $p-1$ elements. If $a$ is an integer not divisible by $p$, then $\overline{a}\in (\mathbb{Z}/p\mathbb{Z}{)}^{\times }$. By Lagrange’s theorem, the order of the subgroup $⟨\overline{a}⟩$ divides $p-1$. Therefore

$${\overline{a}}^{p-1}=\overline{1},$$

or equivalently

$$a^{p-1}\equiv 1(\mathrm{mod}p).$$

This is Fermat’s little theorem. If $p\nmid a$, multiplying both sides by $a$ gives

$$a^p\equiv a(\mathrm{mod}p)$$

and if $p\mid a$, the same congruence is immediate.

The converse of Lagrange’s theorem is false: if $d$ divides $|G|$, there need not be a subgroup of $G$ of order $d$. The strongest general converse is given by the Sylow theorem, which we will not prove.

Sylow

If $p$ is prime and $p^a$ divides $|G|$, then $G$ has a subgroup of order $p^a$.

At this point, these cosets are looking a lot like the equivalence classes we need to form a quotient group, or the fibers of a homomorphism we introduced in the beginning. That is, what we’d like to do is use $x\sim y$ if $x$ and $y$ are in the same coset. But there is a critical problem: the left and right cosets need not be the same. This is exactly the obstruction to $\sim$ being compatible with multiplication.

Example

Let $G=D_n$ and let $H=⟨s⟩=\{e,s\}$, where $s$ is a reflection. The left cosets of $H$ are

$$\{e,s\},\{r,rs\},\{r^2,r^{2}s\},\dots ,\{r^{n-1},r^{n-1}s\}.$$

The right cosets are

$$\{e,s\},\{r,r^{-1}s\},\{r^2,r^{-2}s\},\dots ,\{r^{n-1},r^{-(n-1)}s\}.$$

These lists need not be the same. For example, $r$ lies in the left coset $\{r,rs\}$, while its right coset is $\{r,r^{-1}s\}$. Since $|D_n|=2n$ and $|H|=2$, we have $[D_n:H]=n$.

If we choose certain subgroups it works out:

Example

Again let $G=D_n$, but now let $K=⟨r⟩$ be the subgroup of rotations. The left cosets are

$$K\text{and}sK=\{s,sr,\dots ,s{r}^{n-1}\},$$

while the right cosets are

$$K\text{and}Ks=\{s,rs,\dots ,r^{n-1}s\}.$$

In this case $sK=Ks$, so the left and right cosets agree. Since $|K|=n$, we have $[D_n:K]=2$.

This is because $K$ is a normal subgroup of $D_n$, which we will explore in the next section.

Exercise

In $S_3$, let $H=⟨(12)⟩$. List the left cosets of $H$. Then show directly that the rule

$$(xH)(yH)=xyH$$

is not well-defined.

Normal subgroups

The examples at the end of the previous section show that left and right cosets do not always agree. This is exactly the obstruction to multiplying cosets in a sensible way. The subgroups for which this obstruction disappears are called normal subgroups.

Definition

We say that $N\le G$ is a normal subgroup of $G$, written $N⊴G$, if

$$gN{g}^{-1}=N$$

for every $g\in G$.

The element $gn{g}^{-1}$ is called the conjugate of $n$ by $g$. Thus a subgroup is normal if it is closed under conjugation by every element of the ambient group.

Remark

Just to be clear, you dont ask if a group is normal. Normality is a property of a subgroup inside a group, and it has very little to do with the actual structure of the subgroup.

Here are many examples of normal subgroups.

Example

Every group $G$ has two trivial normal subgroups:

$$\{e_G\}⊴G\text{and}G⊴G.$$

Example

Every subgroup of an abelian group is normal. Indeed, if $G$ is abelian and $H\le G$, then

$$gh{g}^{-1}=g{g}^{-1}h=h$$

for all $g\in G$ and $h\in H$. Hence $gH{g}^{-1}=H$ for every $g\in G$.

Example

More generally, for any group $G$, the center

$$Z(G)=\{z\in G\mid zg=gz\text{ for all }g\in G\}$$

is normal in $G$.

Exercise

Prove that the kernel of a group homomorphism is a normal subgroup.

The following proposition gives some useful criteria for checking normality.

Proposition

Let $N\le G$. The following conditions are equivalent:

1. $N⊴G$.
2. $gN{g}^{-1}\subseteq N$ for every $g\in G$.
3. $gN=Ng$ for every $g\in G$.

Proof

Clearly (1) implies (2). Conversely, if $gN{g}^{-1}\subseteq N$ for every $g\in G$, then applying this to $g^{-1}$ gives $g^{-1}Ng\subseteq N$. Conjugating by $g$ gives $N\subseteq gN{g}^{-1}$, so $gN{g}^{-1}=N$. Thus (1) and (2) are equivalent.

Also, $gN{g}^{-1}=N$ if and only if $gN=Ng$, by multiplying on the right by $g$. Thus (1) and (3) are equivalent.

Exercise

Prove that every subgroup of index $2$ is normal.

Exercise

Let $f:G\to H$ be a group homomorphism and let $K⊴H$. Prove that

$$f^{-1}(K)⊴G.$$

So how do we find normal subgroups? We’ve already show that kernels of homomorphism are normal, but in fact, these are the only examples.

Proposition

A subgroup $N\le G$ is normal if and only if it is the kernel of some homomorphism $f:G\to H$.

Proof

Deferred till later.

Remark

The relation “is a normal subgroup of” is not transitive. Let

$$V=\{e,(12)(34),(13)(24),(14)(23)\}\le S_4.$$

One can show that $V⊴S_4$. Since $V$ is abelian, the subgroup

$$A=\{e,(12)(34)\}$$

is normal in $V$. But $A$ is not normal in $S_4$, because

$$(13)(12)(34)(13{)}^{-1}=(14)(23)\notin A.$$

We will end with an example which is part of the motivation for Galois to create group theory.

Definition

A permutation in $S_n$ is even if it can be written as a product of an even number of transpositions. The alternating group $A_n$ is the set of all even permutations in $S_n$.

Remark

In Chapter 1 we proved that every permutation is a product of transpositions. What is not obvious is that the parity of the number of transpositions is well-defined: a permutation cannot be written both as a product of an even number of transpositions and as a product of an odd number of transpositions. Equivalently, there is a well-defined homomorphism

$$\mathrm{sgn}:S_n\to \{1,-1\},$$

which sends even permutations to $1$ and odd permutations to $-1$.

Proposition

For $n\ge 2$, the alternating group $A_n$ is normal in $S_n$.

Proof

By the previous remark, $A_n=\ker (\mathrm{sgn}).$ Since kernels of homomorphisms are normal, $A_n⊴S_n$.

Remark

At the time, formulas for solving polynomial equations were being sought (like the quadratic formula). Such formulas were known for polynomials of degree $2$, $3$, and $4$, but not for degree $5$ or higher. He wanted to show that a general quintic polynomial is not solvable by radicals. We will return to this story later and sketch how he proved this.

Quotient groups

We now arrive at the point of normal subgroups: they are exactly the subgroups for which we can multiply cosets in a well-defined way.

Theorem

Let $N\le G$. The operation

$$xN\cdot yN=xyN$$

makes the set of left cosets of $N$ in $G$ into a group if and only if $N⊴G$.

Proof

Suppose $N⊴G$. We first check that the operation is well-defined. If $xN=x^{\prime }N$ and $yN=y^{\prime }N$, then $x^{\prime }=x{n}_1$ and $y^{\prime }=y{n}_2$ for some $n_1,n_2\in N$. Since $N$ is normal, $y^{-1}{n}_{1}y\in N$, and hence

$$x^{\prime }{y}^{\prime }=x{n}_{1}y{n}_2=xy(y^{-1}{n}_{1}y)n_2\in xyN.$$

Therefore $x^{\prime }{y}^{\prime }N=xyN$. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity follows from associativity in $G$.

Conversely, suppose the formula gives a well-defined group operation on the set of left cosets. If $n\in N$ and $g\in G$, then $gN=gnN$, so well-definedness gives

$$N=(gN)(g^{-1}N)=(gnN)(g^{-1}N)=gn{g}^{-1}N.$$

Hence $gn{g}^{-1}\in N$. By Proposition normality criteria, $N⊴G$.

Definition

Let $N⊴G$. The quotient group $G/N$ is the group whose elements are the left cosets of $N$ in $G$, with multiplication

$$(xN)(yN)=(xy)N.$$

The identity element is $e_{G}N=N$ and the inverse of $gN$ is $(gN{)}^{-1}=g^{-1}N$.

This is the same quotient group as $G/{\sim }_N$, where ${\sim }_N$ is the equivalence relation defined by $x{\sim }_{N}y$ if and only if $x^{-1}y\in N$. The equivalence classes of ${\sim }_N$ are exactly the left cosets of $N$ in $G$.

Remark

Since $N$ is normal, the left and right cosets of $N$ agree. Thus we can also view $G/N$ as the set of right cosets of $N$ in $G$, with multiplication

$$(Nx)(Ny)=N(xy).$$

Remark

The order of the quotient group is the index of $N$ in $G$:

$$|G/N|=[G:N].$$

If $G$ is finite, then Lagrange’s theorem gives

$$|G/N|=\frac{|G|}{|N|}.$$

Example

Since $A_n=\ker (\mathrm{sgn})$, the quotient $S_n/A_n$ has two elements. The nontrivial coset is the set of odd permutations, and

$$S_n/A_n\cong \mathbb{Z}/2\mathbb{Z}.$$

Exercise

Let $N⊴G$ and let $g\in G$. Prove that the order of $gN\in G/N$ is the smallest positive integer $m$ such that $g^m\in N$, if such an $m$ exists. Use this to compute the orders of all elements of

$$(\mathbb{Z}/12\mathbb{Z})/⟨\overline{4}⟩.$$

Remember the motivation from the beginning of this chapter: we should think of the quotient group $G/N$ as the set of fibers of a homomorphism from $G$ to another group. The next lemma makes this precise.

Definition

Let $N⊴G$. The map

$$\pi :G\to G/N,\pi (g)=gN,$$

is called the canonical quotient map, the canonical surjection, or the canonical projection from $G$ to $G/N$.

Lemma

Let $N⊴G$. The canonical quotient map

$$\pi :G\to G/N$$

is a surjective group homomorphism with kernel $N$.

Proof

Surjectivity is immediate: every element of $G/N$ has the form $gN$ for some $g\in G$. The map $\pi$ is a homomorphism since

$$\pi (g_1{g}_2)=g_1{g}_{2}N=(g_{1}N)(g_{2}N)=\pi (g_1)\pi (g_2).$$

Using Lemma Coset criterion, its kernel is

$$\begin{array}{rl}\ker (\pi )& =\{g\in G\mid \pi (g)=N\}\\ & =\{g\in G\mid gN=N\}\\ & =\{g\in G\mid g\in N\}=N.\end{array}$$

Proposition normal groups are kernels

A subgroup $N\le G$ is normal if and only if it is the kernel of some homomorphism $f:G\to H$.

Proof

If $N$ is the kernel of a homomorphism, then the earlier exercise shows that $N$ is normal. Lemma quotient homomorphism gives the converse.

Example

The infinite dihedral group $D_{\infty }$ is the set

$$D_{\infty }=\{r^i,r^{i}s\mid i\in \mathbb{Z}\},$$

with multiplication determined by

$$\begin{array}{rlrl}{r}^i{r}^j& =r^{i+j},& r^i(r^{j}s)& =r^{i+j}s,\\ (r^{i}s)r^j& =r^{i-j}s,& (r^{i}s)(r^{j}s)& =r^{i-j}.\end{array}$$

Informally, this is the group with presentation

$$D_{\infty }=⟨r,s\mid s^2=e,sr{s}^{-1}=r^{-1}⟩.$$

For $n\ge 3$, the subgroup $⟨r^n⟩$ is normal in $D_{\infty }$, and

$$D_{\infty }/⟨r^n⟩\cong D_n.$$

The isomorphism is induced by sending

$$r⟨r^n⟩⟼r\text{and}s⟨r^n⟩⟼s.$$

Exercise

Let $m\mid n$ with $m\ge 3$. In $D_n=⟨r,s⟩$, prove that $⟨r^m⟩⊴D_n$, and prove that

$$D_n/⟨r^m⟩\cong D_m.$$

Remark

In the previous example, both $D_{\infty }$ and $⟨r^n⟩$ are infinite, but the quotient is finite. Thus a quotient of an infinite group by an infinite subgroup can be finite.

A quotient of an infinite group by an infinite subgroup can also be infinite. For example, in the additive group $\mathbb{Z}\times \mathbb{Z}$, the subgroup $\mathbb{Z}\times \{0\}$ is normal, and the quotient has one coset for each possible second coordinate. In contrast, every quotient of a finite group is finite.

Definition

Let $G$ be a group. For $x,y\in G$, the commutator of $x$ and $y$ is the element

$$[x,y]=xy{x}^{-1}{y}^{-1}.$$

The commutator subgroup, or derived subgroup, of $G$ is the subgroup generated by all commutators:

$$[G,G]=⟨[x,y]\mid x,y\in G⟩.$$

We have $[x,y]=e_G$ if and only if $xy=yx$. More generally, $[G,G]=\{e_G\}$ if and only if $G$ is abelian. Thus the commutator subgroup measures how far $G$ is from being abelian.

Exercise

Show that $[G,G]$ is a normal subgroup of $G$. Hint: first show that

$$g[x,y]g^{-1}=[gx{g}^{-1},gy{g}^{-1}].$$

Exercise

Let $G$ be a group. Define the abelianization of $G$ as $G_{\mathrm{ab}}=G/[G,G]$. Prove that the abelianization is the largest abelian quotient of $G$, in the following sense: if $N⊴G$ and $G/N$ is abelian, then $[G,G]\le N$.

Exercise

Compute the commutator subgroups $[S_3,S_3]$ and $[D_4,D_4]$. Then determine $S_3^{\mathrm{ab}}$ and $D_4^{\mathrm{ab}}$.

It is now time to prove the isomorphism theorems, which explain how homomorphisms and quotient groups fit together.

Isomorphism Theorems

Take any homomorphism $\phi :G\to H$. As we showed, the kernel $\ker \phi$ is a normal subgroup of $G$ (and in fact all normal subgroups arise as kernels of homomorphisms), so we may take the quotient group $G/\ker \phi$. Now, think back to the picture we have of quotient groups as fibers of a homomorphism. This picture makes it clear that the quotient group $G/\ker \phi$ is isomorphic to the image of $\phi$ in $H$. This is the first fundamental isomorphism theorem, which we now state and prove.

First Isomorphism Theorem

If $\phi :G\to H$ is a homomorphism of groups, then $\ker \phi ⊴G$ and $G/\ker \phi \cong \mathrm{im}\phi$.

Proof

We have already shown that $\ker \phi ⊴G$. Define a map

$$\psi :G/\ker \phi \to \mathrm{im}\phi ,\psi (g\ker \phi )=\phi (g).$$

This is well-defined: if $g\ker \phi =g^{\prime }\ker \phi$, then $g^{-1}{g}^{\prime }\in \ker \phi$, so $\phi (g^{-1}{g}^{\prime })=e_H$ and hence $\phi (g)=\phi (g^{\prime })$. The map $\psi$ is surjective by definition of the image, and it is injective because if $\psi (g\ker \phi )=e_H$, then $g\in \ker \phi$ and hence $g\ker \phi =\ker \phi$.

Example

Suppose $\phi :V\to W$ is a linear transformation of finite-dimensional vector spaces. Then the first isomorphism theorem implies that $|V:\ker (\phi )|=|\mathrm{im}\phi |$. This is the the rank-nullity theorem, which says that

$$\dim V=\dim \ker (\phi )+\dim \mathrm{im}(\phi ).$$

Example

The First Isomorphism Theorem gives a second proof of the classification of cyclic groups. Let $G=⟨x⟩$ and define

$$\varphi :\mathbb{Z}\to G,\varphi (m)=x^m.$$

This is a surjective homomorphism. If $x$ has infinite order, then $\ker \varphi =\{0\}$, so $\mathbb{Z}\cong G$. If $x$ has finite order $n$, then $\ker \varphi =n\mathbb{Z}$, so the First Isomorphism Theorem gives

$$\mathbb{Z}/n\mathbb{Z}=\mathbb{Z}/\ker \varphi \cong G.$$

Example

Let $F$ be a field and let $n\ge 1$. The determinant map

$$\det :{\mathrm{GL}}_n(F)\to F^{\times }$$

is a surjective group homomorphism. Its kernel is

$${\mathrm{SL}}_n(F)=\{A\in {\mathrm{GL}}_n(F)\mid \det (A)=1\}.$$

Therefore ${\mathrm{SL}}_n(F)⊴{\mathrm{GL}}_n(F)$, and the First Isomorphism Theorem gives

$${\mathrm{GL}}_n(F)/{\mathrm{SL}}_n(F)\cong F^{\times }.$$

Example

Let $G={\mathbb{R}}^{\times }$ and let $N=\{1,-1\}$. Since $G$ is abelian, $N⊴G$. The absolute value map

$$|\cdot |:{\mathbb{R}}^{\times }\to {\mathbb{R}}_{>0}^{\times }$$

is a surjective homomorphism with kernel $N$. Hence

$${\mathbb{R}}^{\times }/\{1,-1\}\cong {\mathbb{R}}_{>0}^{\times }.$$

To state the Second Isomorphism Theorem, we first need a small piece of notation.

Definition

Let $H$ and $K$ be subgroups of a group $G$. Define

$$HK=\{hk\mid h\in H,k\in K\}.$$

In general, $HK$ is only a subset of $G$, not necessarily a subgroup.

Remark

The subsets $H$ and $K$ are both contained in $HK$: for example, $h=h{e}_G\in HK$ for every $h\in H$, and $k=e_{G}k\in HK$ for every $k\in K$.

Exercise

Let $H$ and $K$ be subgroups of $G$.

1. Prove that $HK$ is a subgroup of $G$ if and only if $HK=KH$.
2. Prove that if at least one of $H$ or $K$ is normal in $G$, then

$$HK\le G\text{and}HK=KH=⟨H\cup K⟩.$$

Remark

The equality $HK=KH$ does not mean that every element of $H$ commutes with every element of $K$. It is an equality of subsets.

Second Isomorphism Theorem

Let $G$ be a group, let $H\le G$, and let $N⊴G$. Then

$$HN\le G,H\cap N⊴H,N⊴HN,$$

and there is an isomorphism $H/(H\cap N)\cong HN/N$.

Proof

Since $N$ is normal in $G$, Exercise products of subgroups gives $HN\le G$. Also, $H\cap N⊴H$ because $N⊴G$, and $N⊴HN$ because $HN\le G$.

Let

$$\phi :H\to HN/N,\phi (h)=hN.$$

This is the composition of the inclusion $H\hookrightarrow HN$ with the canonical projection $HN\twoheadrightarrow HN/N$, so it is a homomorphism. The map $\phi$ is surjective because every element of $HN/N$ has the form $hnN=hN$, with $h\in H$ and $n\in N$.Finally,

$$\ker \phi =\{h\in H\mid hN=N\}=H\cap N.$$

The result follows from the First Isomorphism Theorem.

Corollary

If $H$ and $N$ are finite subgroups of $G$ and $N⊴G$, then

$$|HN|=\frac{|H||N|}{|H\cap N|}.$$

Lattice Isomorphism Theorem

Let $G$ be a group, let $N⊴G$, and let $\pi :G\twoheadrightarrow G/N$ be the canonical projection. There is an order-preserving bijection

$$\left\{\begin{array}{c}\text{subgroups }H\le G\\ \text{such that }N\le H\end{array}\right\}⟷\left\{\begin{array}{c}\text{subgroups of }G/N\end{array}\right\}$$

given by

$$H⟼H/N\text{and}A⟼{\pi }^{-1}(A).$$

This correspondence has the following properties:

1. $H\le K$ if and only if $H/N\le K/N$.
2. $H⊴G$ if and only if $H/N⊴G/N$.
3. $[G:H]=[G/N:H/N]$.
4. If $N\le H,K\le G$, then

$$(H/N)\cap (K/N)=(H\cap K)/N$$

and

$$⟨H/N\cup K/N⟩=⟨H\cup K⟩/N.$$

Exercise

Prove the Lattice Isomorphism theorem.

Example

Let $G=\mathbb{Z}$ and let $N=12\mathbb{Z}$. Since $\mathbb{Z}$ is abelian, $12\mathbb{Z}⊴\mathbb{Z}$. The Lattice Isomorphism Theorem says that the subgroups of $\mathbb{Z}$ containing $12\mathbb{Z}$ correspond exactly to the subgroups of $\mathbb{Z}/12\mathbb{Z}$.

The subgroups of $\mathbb{Z}$ containing $12\mathbb{Z}$ are

$$\mathbb{Z},2\mathbb{Z},3\mathbb{Z},4\mathbb{Z},6\mathbb{Z},12\mathbb{Z}.$$

Under the correspondence $H\mapsto H/12\mathbb{Z}$, these become the subgroups of $\mathbb{Z}/12\mathbb{Z}$:

$$\mathbb{Z}/12\mathbb{Z},⟨\overline{2}⟩,⟨\overline{3}⟩,⟨\overline{4}⟩,⟨\overline{6}⟩,\{\overline{0}\}.$$

The two lattices have exactly the same shape:

For example, on the left

$$2\mathbb{Z}\cap 3\mathbb{Z}=6\mathbb{Z}.$$

On the right, the corresponding statement is

$$⟨\overline{2}⟩\cap ⟨\overline{3}⟩=⟨\overline{6}⟩.$$

The theorem says that every inclusion, intersection, and generated subgroup visible in the left lattice is preserved in the quotient lattice.

Of course, the actual subgroup structure of $\mathbb{Z}$ is much much larger than the one shown here — you should think that $G/N$ describes the structure of $G$ “above” the normal subgroup $N$.

The canonical projection $\pi :G\twoheadrightarrow G/N$ is the basic map out of $G$ that kills every element of $N$. The next theorem says that any homomorphism out of $G$ that kills $N$ must come from a unique homomorphism out of $G/N$.

Universal mapping property of quotient groups

Let $G$ be a group and let $N⊴G$. Let $f:G\to H$ be a group homomorphism such that $N\le \ker f$. Then there exists a unique group homomorphism

$$\overline{f}:G/N\to H$$

such that the diagram

commutes. Equivalently, $\overline{f}(gN)=f(g)$ for every $g\in G$.

Moreover,

$$\mathrm{im}(\overline{f})=\mathrm{im}(f)\text{and}\ker (\overline{f})=\ker (f)/N.$$

Proof

If such a map exists, then it must satisfy

$$\overline{f}(gN)=\overline{f}(\pi (g))=f(g),$$

so uniqueness is forced.

To prove existence, define

$$\overline{f}(gN)=f(g).$$

We must check that this is well-defined. Suppose $xN=yN$. Then $y^{-1}x\in N\le \ker f$, so

$$f(y{)}^{-1}f(x)=f(y^{-1}x)=e_H.$$

Hence $f(x)=f(y)$, so the value of $\overline{f}$ does not depend on the chosen representative.

The map $\overline{f}$ is a homomorphism because

$$\overline{f}((xN)(yN))=\overline{f}((xy)N)=f(xy)=f(x)f(y)=\overline{f}(xN)\overline{f}(yN).$$

By construction, $\overline{f}\circ \pi =f$.

The image statement follows from the formula $\overline{f}(gN)=f(g)$. Finally,

$$\begin{array}{rl}gN\in \ker (\overline{f})& ⟺\overline{f}(gN)=e_H\\ & ⟺f(g)=e_H\\ & ⟺g\in \ker f.\end{array}$$

Therefore $\ker (\overline{f})=\ker (f)/N$.

In short, to define a homomorphism out of $G/N$, it is enough to define a homomorphism out of $G$ whose kernel contains $N$.

Exercise

The braid group $B_n$ is generated by symbols ${\sigma }_1,\dots ,{\sigma }_{n-1}$ subject to the relations

$${\sigma }_i{\sigma }_j={\sigma }_j{\sigma }_i\text{if }|i-j|\ge 2$$

and

$${\sigma }_i{\sigma }_{i+1}{\sigma }_i={\sigma }_{i+1}{\sigma }_i{\sigma }_{i+1}.$$

The generator ${\sigma }_i$ represents the braid where strand $i$ crosses over strand $i+1$. Products of braids are formed by stacking one braid on top of another:

The second displayed relation is a three-strand braid move. It is not the relation ${\sigma }_i^2=e$: doing the same crossing twice is usually not the identity braid.

Show that forgetting the crossing information and recording only the final order of the endpoints defines a surjective homomorphism

$$B_n\to S_n\text{with}{\sigma }_i⟼(ii+1).$$

Describe the kernel geometrically; it is called the pure braid group $P_n$. Then prove that imposing the additional relations ${\sigma }_i^2=e$ on $B_n$ gives the symmetric group $S_n$. We often think of an element of $S_n$ as performing a shuffle on a set of points — $B_n$ remembers the “movie” of this shuffle.

Exercise

Let $G$ be a group and let $A$ be an abelian group. Show that every group homomorphism

$$f:G\to A$$

factors uniquely through the abelianization $G_{\mathrm{ab}}=G/[G,G]$.

Exercise

Prove that if $G/Z(G)$ is cyclic, then $G$ is abelian.

The third isomorphism theorem considers the question of taking quotient groups of quotient groups. In short, they cancel like fractions and we gain no new structural information from taking quotients of a quotient group.

Third Isomorphism Theorem

Let $G$ be a group and suppose

$$M\le N\le G,M⊴G,N⊴G.$$

Then $M⊴N$, $N/M⊴G/M$, and there is an isomorphism

$$(G/M)/(N/M)\cong G/N.$$

Proof

Since $M⊴G$, we have $M⊴N$. Consider the canonical projection

$$\pi :G\twoheadrightarrow G/N.$$

Since $M\le N=\ker \pi$, the universal mapping property gives a homomorphism

$$\overline{\pi }:G/M\to G/N,\overline{\pi }(gM)=gN.$$

This map is surjective, and its kernel is $\ker (\overline{\pi })=N/M$. Therefore $N/M⊴G/M$, and the First Isomorphism Theorem gives

$$(G/M)/(N/M)\cong G/N.$$

Composition Series and the Hölder Program

The isomorphism theorems tell us how to understand a group by passing to quotients. Composition series push this idea as far as it can reasonably go: break a group into simple quotient groups, in the same spirit that an integer is broken into prime factors.

Definition

A nontrivial group $G$ is called simple if its only normal subgroups are

$$\{e_G\}\text{and}G.$$

Simple groups are the groups which cannot be broken down further using normal subgroups and quotients. They are the “atoms” or “prime numbers” of group theory.

Example

If $p$ is prime, then $\mathbb{Z}/p\mathbb{Z}$ is simple. Indeed, its only subgroups are

$$\{\overline{0}\}\text{and}\mathbb{Z}/p\mathbb{Z},$$

and every subgroup of an abelian group is normal.

Proposition

The simple abelian groups are exactly the cyclic groups of prime order.

Proof

We have already seen that $\mathbb{Z}/p\mathbb{Z}$ is simple when $p$ is prime.

Conversely, let $G$ be a simple abelian group, and choose a nonidentity element $g\in G$. Since $G$ is abelian, every subgroup of $G$ is normal. The subgroup $⟨g⟩$ is nontrivial, so simplicity forces

$$G=⟨g⟩.$$

Thus $G$ is cyclic. If $G$ were infinite cyclic, then it would have a proper nontrivial subgroup, for example $⟨g^2⟩$. Hence $G$ is finite cyclic. If $|G|$ were composite, then $G$ would have a proper nontrivial subgroup. Therefore $|G|$ is prime.

If simple groups are the “primes”, we would wish for a “unique factorization theorem” for finite groups. We now describe this program, which was initiated by Hölder in 1889.

Definition

A composition series for a group $G$ is a finite chain of subgroups

$$\{e_G\}=G_0⊴G_1⊴\cdots ⊴G_n=G$$

such that each quotient $G_i/G_{i-1}$ is simple. The simple groups $G_i/G_{i-1}$ are called the composition factors of the series.

The notation $G_{i-1}⊴G_i$ is important. We do not require every $G_i$ to be normal in all of $G$; we only require it to be normal in the next group in the chain.

Example

If $G$ is simple, then

$$\{e_G\}⊴G$$

is a composition series. Its only composition factor is $G$ itself.

Example

In $\mathbb{Z}/12\mathbb{Z}$, one composition series is

$$\{\overline{0}\}⊴⟨\overline{6}⟩⊴⟨\overline{3}⟩⊴\mathbb{Z}/12\mathbb{Z}.$$

The composition factors are

$$\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/3\mathbb{Z}.$$

Another composition series is

$$\{\overline{0}\}⊴⟨\overline{4}⟩⊴⟨\overline{2}⟩⊴\mathbb{Z}/12\mathbb{Z},$$

whose composition factors occur in the order

$$\mathbb{Z}/3\mathbb{Z},\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z}.$$

Note: the order changed, but the list of factors did not.

Example

Note that in $D_8$, we have two composition series

$$1⊴⟨r^2⟩⊴⟨r⟩⊴D_8\text{and}1⊴⟨s⟩⊴⟨r^2,s⟩⊴D_8.$$

In each series there are three composition factors, each of which is isomorphic to the simple group $\mathbb{Z}/2\mathbb{Z}$.

Proposition

Every finite group has a composition series.

Proof

We argue by induction on $|G|$. If $G$ is simple, then $\{e_G\}⊴G$ is a composition series. If $G$ is not simple, choose a proper normal subgroup $N⊴G$ which is maximal among proper normal subgroups of $G$. Such an $N$ exists because $G$ is finite. By the Lattice Isomorphism Theorem, the quotient $G/N$ has no nontrivial proper normal subgroups, so $G/N$ is simple.

Since $|N|<|G|$, the induction hypothesis gives a composition series

$$\{e_G\}=G_0⊴G_1⊴\cdots ⊴G_k=N$$

for $N$. Appending $G$ gives

$$\{e_G\}=G_0⊴G_1⊴\cdots ⊴G_k=N⊴G,$$

and the new final quotient $G/N$ is simple. Therefore this is a composition series for $G$.

Jordan--Hölder theorem

Let $G$ be a finite group. Any two composition series for $G$ have the same length, and after reordering their factors, the two lists of composition factors are isomorphic term-by-term.

We will not prove the Jordan—Hölder theorem here. The proof is a refinement argument: one compares two normal chains by cutting each chain with the subgroups in the other chain, then uses the isomorphism theorems to identify the new quotient factors.

Remark

Composition factors behave like prime factors of an integer. A composition series is not unique, just as a factorization can be written in different orders. The Jordan—Hölder theorem says that the simple factors themselves are unique up to reordering.

Remark

The analogy with prime factorization is useful but imperfect. Knowing the composition factors of a group does not determine the group. For example, both

$$\mathbb{Z}/4\mathbb{Z}\text{and}\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$$

have two composition factors, both isomorphic to $\mathbb{Z}/2\mathbb{Z}$, but the groups are not isomorphic. Composition factors describe the simple pieces; they do not describe all the ways those pieces can be glued together.

After this, the Hölder program was to:

1. classify all finite simple groups, and

2. find all ways of putting them together to form other groups.

Step (1) was completely in about 1980, 100 years after the program was launched. The efforts of over 100 matehamticians covering about 5000-10000 pages of journal pages over 300-500 individual papers.

Theorem

There is a list consisting of 18 infinite families of simple groups and 26 simple groups not belonging to this family (the sporadic groups) such that every finite simple group is isomorphic to one of the groups in the list.

Definition

A finite group $G$ is called solvable if it has a composition series whose composition factors are all cyclic of prime order.

By the Jordan—Hölder theorem, if one composition series for a finite group has only cyclic prime-order factors, then every composition series does. Thus solvability is a property of the group, not of the chosen composition series.

Example

Every finite abelian group is solvable. Indeed, every composition factor of a finite abelian group is abelian and simple, hence cyclic of prime order.

Remark

The name “solvable” points back to Galois’s original goal: understanding when polynomial equations can be solved by radicals.

Very roughly, a polynomial has a group attached to it, called its Galois group. This group records the permutations of the roots that preserve all algebraic relations among those roots. For example, the roots of a polynomial may be labeled

$${\alpha }_1,\dots ,{\alpha }_n,$$

and the Galois group is a subgroup of $S_n$ describing which permutations of these roots are compatible with the algebraic structure of the equation.

Solving a polynomial by radicals means building its roots by repeatedly adjoining expressions such as square roots, cube roots, and more generally $m$th roots. On the group-theoretic side, these radical adjunctions contribute only cyclic prime-order composition factors. Thus Galois theory proves the following fundamental principle:

$$\text{solvable by radicals}⟹\text{solvable Galois group}.$$

This turns the existence of radical formulas into a group-theoretic question. The general quintic has Galois group $S_5$. In the following exercise, you will show that $S_5$ is not solvable, achieving Galois’s goal.

Exercise

A fact that we will not prove $A_5$ is simple — this is easy, but messy. Using this fact, prove that $S_5$ is not solvable.

Exercise

Let

$$V=\{e,(12)(34),(13)(24),(14)(23)\}.$$

Prove that

$$\{e\}⊴⟨(12)(34)⟩⊴V⊴A_4⊴S_4$$

is a composition series, and list its composition factors.

The classification of finitely generated abelian groups

For finite abelian groups, composition factors are especially simple: they are all cyclic groups of prime order. The Jordan—Hölder theorem then records the primes that appear, counted with multiplicity. For instance, every abelian group of order $12$ has composition factors

$$\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/3\mathbb{Z}$$

in some order.

But now we need to understand goal (2); how these can be glued together. Fortunately, we know how to do this in this case.

Theorem

Every finitely generated abelian group is isomorphic to a group of the form

$${\mathbb{Z}}^r\oplus \mathbb{Z}/n_1\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}/n_k\mathbb{Z},$$

where $r\ge 0$, each $n_i>1$, and

$$n_1\mid n_2\mid \cdots \mid n_k.$$

Moreover, the integer $r$ and the integers $n_1,\dots ,n_k$ are uniquely determined by the group.

Example

What are all the possible finite abelian groups of order $12$? The prime factorization is

$$12=2^2\cdot 3^1.$$

The $2$-part of the group can be either $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$, and the $3$-part of the group is $\mathbb{Z}/3\mathbb{Z}$. Therefore the two possible abelian groups of order $12$ are

$$\mathbb{Z}/12\mathbb{Z}\cong \mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}\text{and}\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}.$$