FREE GROUPS

We now turn tothe

coproduct

^{in the} categoryof groups. First^aremark. Let G ⁼

n

^Gibe a direct

product

of groups.

We observe that each G_j admits an

injective homomorphism

^{into the}

product,

^{on the}

j-th component, namely

^the map

Aj:

^{Gj -.}

n

^{Gi such that}

i

forxin G

j^, the i-th component of

Aj(X)

^{is the unit element ofGi if i =F}

^j,

^and

is

equal

^{to x} itself if i ⁼

j.

^This

embedding

will be called the canonical one.

But westill don't havea

coproduct

^{of the}

family,

because the factorscommute with each other. To

get

^a

coproduct

^{one has to} work somewhat harder.

Let G be agroupand Sa subset ofG. We recall that G is

generated by

^S

if_everyelement ofGcanbewrittenas afinite

product

of elements ofSandtheir inverses

(the empty product being always

^{taken as the} unit element of

G).

ElementsofSarethencalled

generators.

^Ifthere existsafinitesetof

generators

forGwecall G

finitely generated.

^{IfS isaset and} qJ: S ^-. G isa map,^wesay that_qJ

generates

^Gifits

image generates

^G.

Let Sbe aset,

and/:

^{S-.F a}map into agroup. Let g: S^-. G be another map.

Iff(S) (or

^{as wealso}

say,f) generates F,

^{then it is}obvious that there exists at most one

homomorphism t/J

^{ofF intoG} which makes the

following diagram

commutative:

S ^{f )}F

\)

G

We now consider the

category

^{e whose}

objects

^are the maps of S into groups.

Iff:

^{S-.G and}

f'

^{:S-. G' are two}

objects

^{in this}

category,

^wedefine

a

morphism fromfto f'

^tobea

homomorphism

qJ^:G ^-. G^'such that_qJ⁰

f

⁼

f',

i.e. the

diagram

is commutative:

G

s j

^'"

G'

By

^{a free}group determined

by S,

^{we shall mean a} universal element in this

category.

Proposition

^{12.1. Let S be a set.} Then there exists a

free

group

(F, f)

determined

by

^S.

Furthermore, f

^is

injective,

^{and F is}

generated by

^the

image off.

Proof (I

^owethis

proof

^{to J.}

Tits.)

^We

begin

^withalemma.

Lemma 12.2. There exists asetIand^a

family of

groups

{G;hEI

^such

^that, if

g: S G is a map

of

^{S into a} group

G,

and ₉ generates

G,

^{then G is}

isomorphic

^{to some}

G;.

Proof

^{This is a}

simple

^{exercise in}

cardinalities,

^{which we} carry ^out. IfS is

finite,

^{thenGis finiteor}denumerable. If S is

infinite,

thenthe

cardinality

^ofG

is ^< the

cardinality

^ofSbecauseGconsists of finite

products

^ofelements of

g(S).

Let Tbeasetwhich is infinite denumerable ifSis

finite,

^{and hasthesamecardin-}

ality

^{asS if S}is infinite. Foreach

non-empty

^subsetHof

T,

^{let r}H be thesetof group^{structures on H. For}each _yEr_H, let

Hy

^{be theset}

^{H, together}

^withthe

group^structurey. Then the

family {H y}

^{for yEr}

Hand

^H

ranging

^oversubsets

ofTis the desired

family.

Wereturn to the

proof

^ofthe

proposition.

^{Foreach iEI welet Mi be the}

set of

mappings

^{ofS into Gi. For each} map ({J^EM_i, we let G_i

,qJ be the set- theoretic

product

^of

G;

^{and the set with one element}

{qJ},

^{so that}

G;,

qJ is the

"same" group^as

G;

^indexed

by

qJ. We let

Fo= n n G;,qJ

ielqJeMi

be the Cartesian

product

^ofthegroups

Gi,qJ.

^Wedefineamap

10

^:S-+F0

by sending

^{Son thefactor G}i,qJ

by

^{means of}qJitself. Wecontend that

given

^a

map g: S G of S into agroup

G,

there exists a

homomorphism t/!*:

^F0 G

making

^{the usual}

diagram

commutative:

Fo

j

^*

G

That

is, t/!.

⁰

fo

⁼ g. Toprove

this,

^wemay ^assumethat 9

generates G, simply by restricting

^{our attention to the}

subgroup

^ofG

generated by

^the

image

^ofg.

By

^the

lemma,

there exists an

isomorphism

^{A.:G -+}

G;

^{for some}

^i,

^{and A.0g} isan element

t/J

^of

Mi.

^Welet ni,'"be the

projection

^{on the}

(i, t/J) ^factor,

^andwe

let

t/J.

^{= A.}

-1

0n;,",. Then the _map

t/1.

^{makes the}

following diagram

^com-

mutative.

gI

_G

) ^1:;.

A )

G;,,,,

We letF be the

subgroup

^of

Fo generated by

^the

image of/

^{o, and welet}

I

simply

^be

equal

^to

10'

^{viewedas a}map of Sinto F. Welet_g. be the restriction of

t/J.

^{toF. Inthisway,we seeatoncethat}the map g.is the

unique

^one

making

our

diagram commutative,

and thus that

(F,f)

^{is the}

required

free group.

Furthermore,

it is clear that

f

^is

injective.

For each set S we select one free _group determined

by S,

and denote it

by (F(s),ls)

^or

briefly by F(S).

^{It is}

generated by

^the

image

^of

fs.

^One may view Sas contained in

F(S),

^{and the} elements ofS are called free

generators

of

F(S).

^Ifg: S G isamap, ^wedenote

by

g.:

F(S)

^Gthe

homomorphism realizing

^the

universality

^ofourfreegroup

F(S).

If A:S S' is a map ofone set into

another,

^{we let}

F(A)

^:

F(S) F(S')

^be

the map

(fs'

⁰

A)..

S ^{Is )}

F(S)

Al IA.

^=F(A)

S' ⁾ F'

(

^S'

)

Is'

Thenwemay

regard

^{Fas a}functor from the

category

^{ofsets tothe}

category

^of groups

(the

^functorial

properties

^are

trivially verified,

^{and will be left to the}

reader).

If

^{A. is}

surjective,

^then

F(A.)

^isalso

surjective.

We

again

^{leave the}

proof

^{tothe reader.}

Iftwo sets

S,

^{S' have thesame}

cardinality,

^then

they

^are

isomorphic

^{in the}

category

^ofsets

(an isomorphism being

^{in this case a}

bijection !),

^{and hence}

F(S)

^is

isomorphic

^to

F(S').

^{IfS has n}

elements,

^{we call}

F(S)

^the free group

on n

generators.

LetGbeagroup,and letSbethesamesetasG

(i.e.

^{Gviewedas a}set, without group

structure).

^{We have the}

identity

map g^:S

G,

^{and hencea}

surjective homomorphism

g.^:

F(S)

^G

which will be called canonical. Thus every group is a factor _group ofa free group.

Onecanalsoconstructgroups

by

^whatiscalled

generators

and relations. Let S be a set, and

F(S)

^{the free} group. We assume that

f:

^S

F(S)

^{is an in-} clusion. Let R be aset of elements of

F(S).

Each element ofRcanbe written

as a finite

product

n

UXv

v=1

where eachXvisan element ofSoraninverse ofanelement ofS. Let Nbe the smallestnormal

subgroup

^of

F(S) containing R,

i.e. the intersectionof all normal

subgroups

^of

F(S) containing

^{R. Then}

F(S)/N

^{will becalled the} group deter- mined

by

^the

generators

^{S and the}relations R.

Example.

^{One shows}

easily

^{that the} group determined

by

^one

generator

a, and the relation

{a

²

},

^{has order 2.}

Thecanonical

homomorphism

cp:

F(S) F(S) /

^Nsatisfies the uni versal map-

ping property

^for

homomorphisms t/J

^of

F(S)

into groups G such that

t/J(x)

^{= e}

for allx E R. In view of

this,

^{one sometimes} calls the group

F(S)/N

the group determined

by

^the generators

S,

^and the relations x ⁼ e

(for

^{all x E}

R).

^For

instance,

the group in the

preceding example

^{would be}called the group determined

by

^the generatora, and the relation a2 =

e.

Let G be a group

generated by

^a finite number of

elements,

^and

satisfying

the relationx2 =

efor allx EG. What does G look like? It is easy^to show that G is commutative. Thenone can view G as a vectorspace ^over

Z/2Z,

^{so G is}

determined

by

^its

cardinality,

up ^to

isomorphism.

InExercises 34 and

35,

you willprove that there exist certain groups

satisfying

certain relations and with a

given ^order,

^sothat the group

presented

^{with these} generators and relations canbe

completely

determined. A

priori,

^{it is noteven}

clear if a group

given by generators

^{and relations is} finite. Even if it is

finite,

one does not know its order a

priori.

To show that a group of certain order

exists,

^{one has to use various} means, a common means

being

^to represent ^the group^{as a}group of

automorphisms

^ofsome

object,

for instance the

symmetries

ofa

geometric object.

This will be the method

suggested

for the groups in Exercises 34 and

35,

mentioned above.

Example.

^{Let G be a} group. For^x, y ^E G define

[x, y]

⁼

xyx-1y-1 (the commutator)

^and

Xy

⁼

xyx-

^l

(the conjugate).

^{Thenone has the}

cocycle

^relation

[x, yz]

⁼

[x, y]Y[x, z].

Furthermore,

suppose^x, y, ^{Z E} G and

[x, y]

⁼ y,

[y, z]

^{= Z,}

[z, x]

^{= x.}

Thenx ⁼ y ^{= z = e.} It is anexercise to prove these

assertions,

but one sees

that certain relations

imply

^{that a} group

generated by

^x, y, ^z

subject

^{to those}

relations is

necessarily

^trivial.

Next we

give

^{a somewhatmore}

sophisticated example.

^{We assumethatthe}

reader knows the basic

terminology

of fields and matrices as in

Chapter XIII,

but

applied only

^{to 2 x 2} matrices. Thus SL₂

(F)

^denotes the group of 2 ^x 2 matrices with

components

^{in a} field F and determinant

equal

^{to 1.}

Example.

^SL2

(F).

^{Let F be a} field. For b E Fand ^a E F, ^{a =t=}

0,

^we let

u(b)

⁼

( ), ^s(a)

⁼

( _}

^{and w=}

(_ ).

Then it is

immediately

^{verified that:}

SL O.

s(a)

⁼

wu(a-l)wu(a)wu(a-

^l

).

SL 1. u is an additive

homomorphism.