1.8 Free Groups; Generators and Relations

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Let S be a nonempty set and let F be a group. Say that F is free on S if

there exists a function _ : S ! F such that if G is any group and _ : S ! G

is any function then there is a unique homomorphism f : F ! G such that

the diagram

S

_ - F

@

@

@

@

@

_

R         􀀀

􀀀

􀀀

􀀀

􀀀

f

G

commutes.

The following is routine, but important (see Exercises 1, 2 and 3):

Proposition 1.8.1 Let F be free on the set S, with mapping _ : S ! F.

(i) _ : S ! F is injective.

(ii) F = h_(S)i.

(iii) Via the map _ : {s} ! Z, _(s) = 1, the additive group Z is free on

one generator.

(iv) If F is free on a set with more than one generator, then F is nonabelian.

The following is absolutely fundamental.

Proposition 1.8.2 If S is nonempty, then a free group exists on S, and is

unique up to isomorphism.

Now let G be an arbitrary group. It is clear that G is the homomorphic

image of some free group F. Indeed, let F be the free group on the set G;

the map F ! G is then that induced by 1G : G ! G. More generally (and

economically), if G = hSi, and if F is free on S, then the homomorphism

F ! G induced by the inclusion S ! G is surjective.

1.8. FREE GROUPS; GENERATORS AND RELATIONS 37

The following notation, though not standard, will prove useful. Let H be

a group, and let R be a subset of H. Denote by hhRii the smallest normal

subgroup of H which contains R. This normal subgroup hhRii is called

the normal closure of R. The reader should note carefully the difference

between hRi and hhRii. Now assume that G = hSi is a group, and that

F is free on S, with the obviously induced homomorphism F ! G. Let

K = ker(F ! G), and assume that K = hhRii. Then it is customary to say

that G has generators S and relations R, or that G has presentation

G = hS| r = e, r 2 Ri.

A simple example is in order here. Let D be a dihedral group of order

2k; thus D is generated by elements n, k 2 D such that nk = h2 = e, hnh =

n−1. Let F be the free group on the set S = {x, y}. The kernel of the

homomorphism F ! D determined by x 7! n, y 7! h can be shown to be

hhxk, y2, (yx)2ii (we’ll prove this below). Thus D has presentation

D = hx, y| xk = y2 = (xy)2 = ei.

One need not always write each “relation” in the form r = e. Indeed, the

above presentation might just as well have been written as

D = hx, y| xk = y2 = e, yxy = x−1i.

The concept of generators and relations is meaningful in isolation, i.e.,

without reference to a given group G. Thus, if one were to write “Consider

the group

G = hx, y| x4 = y2 = (xy)2 = ei, ”

then one means the following. Let F be the free group on the set S = {X, Y }

and let K = hhX4, Y 2, (XY )2ii. Then G is the quotient group F/K, and

the elements x, y are simply the cosets XK, Y K 2 G/K.

Presented groups, i.e., groups of the form hS|Ri are nice in the sense

that if H is any group and if _ : S ! H is any function, then _ determines

a uniquely defined homomorphism hS|Ri ! H precisely when _ “kills all

elements of R.” This fact is worth displaying conspicuously.

Theorem 1.8.3 Let hS| Ri be a presented group, and let _ : S ! H be a

function, where H is a group. Then _ extends uniquely to a homomorphism

hS| Ri ! H if and only if

_(s2)_(s2) · · · _(sr) = e

whenever s1s2 · · · sr 2 R.

38 CHAPTER 1. GROUP THEORY

To see this in practice, consider the presented group G = hx, y|x2 = y3 =

(xy)3 = ei, and let H = A4, the alternating group on 4 letters. Let _ be

the assignment

_ : x 7! (1 2)(3 4); y 7! (1 2 3);

since ((1 2)(3 4))2 = (1 2 3)3 = ((1 2)(3 4)(1 2 3))3 = e, the above theorem

guarantees that _ extends to a homorphism _ : hx, y|x2 = y3 = (xy)3 =

ei ! A4. Furthermore, since A4 = h(1 2)(3 4), (1 2 3)i, we conclude that

_ is onto. That _ is actually an isomorphism is a little more difficult (see

Exercise 11); we turn now to issues of this type.

Consider again the dihedral group D = hn, hi of order 2k, where nk =

h2 = e, hnh = n−1. Set G = hx, y| xk = y2 = (xy)2 = ei. We have

immediately that the map x 7! n, y 7! h determines a surjective homomorphism

G ! D. Since |D| = 2k, we will get an isomorphism as soon

as we learn that |G| _ 2k. This isn’t too hard to show. Indeed, note that

the relation (xy)2 = e implies that yx = x−1y. From this it follows easily

that any element of G can be written in the form xayb. Furthermore, as

xk = e = y2, we see also that every element of G can be written as xayb,

where 0 _ a _ k −1, 0 _ b _ 1. Thus it follows immediately that |G| _ 2k,

and we are done.

The general question of calculating the order of a group given by generators

and relations is not only difficult, but, in certain instances, can be

shown to be impossible. (This is a consequence of the unsolvability of the

so-called word problem in group theory.) Consider the following fairly simple

example: G = hx, y| xy = y2x, yx = x2yi. We get

y−1xy = y−1y2x = yx = x2y = xxy,

so that y−1 = x. But then

e = xy = y2x = y(yx) = y,

so y = e, implying that x = e. In other words, the relations imposed on

the generating elements of G are so destructive that the group defined is

actually the trivial group.

Exercises 1.8

1. Show that if F is free on the set S via the map _ : S ! F, then

1.8. FREE GROUPS; GENERATORS AND RELATIONS 39

(a) _ is injective.

(b) F = h_(S)i.

2. Let |S| = 1, and let F be free on S. Prove that F _= (Z, +).

3. Let |S| _ 2, and let F be free on S. Prove that F is not abelian.

4. Let F be free on the set S, and let F0 be the subgroup of F generated

by S0 _ S. Prove that F0 is free on S0.

5. Prove that hx, y, z| yxy2z4 = ei is a free group. (Hint: it is free on

{y, z}. )

6. Prove that hx, y| yx = x2y, xy3 = y2xi = {e}.

7. Prove that hx, y| xy2 = y3x, x2y = yx3i = {e}.

8. Let G be a free group on a set of more than one element. Prove that

G/G0 is infinite.

9. Compute the structure of G/G0 for each finitely presented group below.

(i) hx, y| x6 = y4 = e, x3 = y2i,

(ii) hx, y| x3 = y2 = ei,

(iii) hx, y| x2 = y3 = (xy)3 = ei,

(iv) hx, y| x2 = y3 = (xy)4 = ei,

(v) hx, y| x2 = y3 = (xy)5 = ei.

10. Prove that

hx, y| x4 = e, y2 = x2, yxy−1 = x−1i _= hr, s, t|r2 = s2 = t2 = rsti.

11. Show that |hx, y| x2 = y3 = (xy)3 = ei| _ 12. Conclude that A4

_=

hx, y| x2 = y3 = (xy)3 = ei.

12. (a) Show that |hx, y| x2 = y3 = (xy)4 = ei| = 24.

(b) Show that |hx, y| x2 = y3 = (xy)5 = ei| = 60.

13. Let D = D8, the dihedral group of order 8. Prove that Aut(D) _= D8.

40 CHAPTER 1. GROUP THEORY

14. Let k, l,m be positive integers and set

D = D(k, l,m) = h_, _| _k = _l = (__)m = ei,

_ = _(k, l,m) = ha, b, c| a2 = b2 = c2 = (ab)l = (bc)l = (ac)m = ei.

Prove that D is isomorphic with a subgroup of index 2 in _.

15. Let Q2n+1 be the generalized quaternion group of order 2n+1. Show

that Q2n+1 has presentation

Q2n+1 = hx, y| x2a = e, y4 = x2a−1

, yxy−1 = x−1i.

(Hint: see Exercise 11.)

16. The Free Product of Groups. Let Ai, i 2 I be a family of groups. A

free product of the groups Ai, i 2 I is a group P, together with a

family of homomorphisms μi : Ai ! P, such that if _i : Ai ! G is any

family of homomorphisms of the groups Ai into a group G, then there

exists a unique homomorphism f : P ! G making the diagram below

commute for each i 2 I:

Ai

μi - P

@

@

@

@

@

i

R         􀀀

􀀀

􀀀

􀀀

􀀀

f

G

Prove that the free product of the groups Ai, i 2 I exists and is unique

up to isomorphism. (Hint: the uniqueness is just the usual categorical

nonsense. For the existence, consider this: let Xi, i 2 I be a family of

pairwise disjoint sets, with Xi in bijective correspondence with Ai, i 2

I, say with bijection _i : Xi ! Ai. Now form the free group F(X) on

the set X = [i2IXi. Similarly, for each i 2 I, we let F(Xi) be the

free group on the set Xi, and set Ki = ker F(Xi) ! Ai, i 2 I. Set

K = hhKi| i 2 Iii, set P = F(X)/K and define μi : Ai ! P via the

composition

Ai

_!

F(Xi)/Ki ! P,

1.8. FREE GROUPS; GENERATORS AND RELATIONS 41

where F(Xi)/Ki ! P is induced by Xi ,! X. Now prove that P

satisfies the necessary universal mapping property. The group P, so

constructed, is generally denoted _i2IAi. The free product of two

groups A and B is denoted A _ B.)

17. Let A _= Z3 and let B _= Z2. Prove that

A _ B _= hx, y| x3 = y2 = ei.

(In fact, it turns out that the above group is isomorphic with PSL2(Z).)

18. Let S be a set, and define groups indexed by S by setting As = Z

(the additive group of the integers) for each s 2 S. Prove that the free

group on S is isomorphic with _s2SAi.

Chapter 2 Field and Galois Theory

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