Combinatorial Group Theory [Lecture notes]

54 Pages·2011·0.443 MB·English

by  Martin Edjvet, with A Surdina

Combinatorial Group Theory
Dr Martin Edjvet, The University of Nottingham, Spring semester 2011
LATEX by Alexandra Surdina, last changes: April 18, 2011.
These are the lecture notes for the module Combinatorial Group Theory at
the University of Nottingham by Dr Martin Edjvet, based on handwritten
notes. I’m happy to correct any mistakes you find, just send me an email:
[e-mail]
1 Free groups
By way of introduction consider the dihedral group D of order 10, the group of
10
isometries of the regular 5-gon. It consists of five rotations and five reflections.
1
5 2
4 3
Let x = (12345),y = (25)(34). Then |x| = 5,|y| = 2, and
y−1xy = (25)(34)(12345)(25)(34) = (15432) = x−1
Therefore,
yx = y−1x = x−1y−1 = x−1y
and y−1x−1 = yx−1 = xy
(rewrite rules). Thus, all elements of D are of the form xiyj.
10
For example,
w = xyx2y3x3y−1x−2y11 = x(yx)xyx3y−1x−2y
= xx−1yxyx3y−1x−2y = (yx)yx3y−1x−2y
= (x−1y)yx3y−1x−2y = x2y−1x−2y
= x2(y−1x−1)x−1y = x2xyx−1y
= x3(yx−1)y = x4y2
= x4
⇒ D = {e,x,x2,x3,x4,y,xy,x2y,x3y,x4y}
10
(Does D really contain ten elements? In other words: Could some of the ele-
10
ments above still be equal? No! Suppose for example x3 = x4y ⇒ x−1 = y .)
(cid:32)
2
To introduce some language:
x and y are generators for D .
10
{x,y} is a generating set.
x5 = y2 = e,y−1xy = x−1 are relations.
w is a word in {x,x−1,y,y−1}.
This module is essentially about studying groups in terms of generators and re-
lations. Now suppose we are given a group G and we know that G is generated by
a non-empty subset S, we write G = (cid:104)S(cid:105).
This means that every g ∈ G can be expressed as a finite product g = sε1 ·...·sεk
1 k
where ε = ±1,s ∈ S,1 ≤ i ≤ k. (That is, each g ∈ G is a word in S ∪S−1.)
i i
This product is said to be reduced (with respect to S) if the following condition is
satisfied:
s = s ⇒ ε +ε (cid:54)= 0
i i+1 i i+1
This simply says that s−1s and ss−1(s ∈ S) are forbidden subwords.
Example. S = {x,y}
not reduced reduced
(stillnotreduced!)
xyy−1x2y → x3y
x−1y4xx−1y−2x−1x2 →x−1y4xx−1y−2x
→ x−1y4y−2x
→ x−1y3y−1x → x−1y2x
The fundamental idea is that of a free group. Roughly speaking, a free group F is
one for which any two reduced words that look different are not equal in F.
Example. In D we have y−1xy = x−1 so D is not a free group (with respect to
10 10
{x,y}).
Again: In a free group, the only relations amongst the generating set are the obvious
ones: xx−1 = x−1x = e.
3
Definition. A group F is said to be free on X ⊆ F if given any group G and any
mapping θ : X → G there exists a unique homomorphism θ(cid:48) : F → G extending θ,
that is, having the property that xθ(cid:48) = xθ (∀x ∈ X). That is, the diagram
ι (cid:47)(cid:47)
X F
θ
(cid:15)(cid:15) (cid:126)(cid:126) θ(cid:48)
G
commutes. Here ι : X → F is the inclusion determined by xι = x (∀x ∈ X).
Remarks. (1) ι and θ are mappings; θ(cid:48) is a homomorphism.
(2) This definition allows us to distinguish between words in F.
x x ,x x ∈ F (x ∈ X)
1 2 1 3 i
x(cid:95)1 x(cid:95)2 x(cid:95)3 X (cid:47)(cid:47)F
(cid:127)
(cid:127)
(cid:127)
(cid:127)
(cid:127)
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)θ(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) θ(cid:48)
(12) (123) (124) S
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(x x )θ(cid:48) = (x )θ(cid:48)(x )θ(cid:48) = (12)(123)
1 2 1 2
(x x )θ(cid:48) = (x )θ(cid:48)(x )θ(cid:48) = (12)(124)
1 3 1 3
(3) If we replace “group” by “abelian group” in the above definition, we get a free
abelian group on X.
Exercise. Show that every free abelian group A is a direct sum of copies of Z:
A ∼= ⊕Z
Lemma 1.1. If F is free on X then X generates F.
Proof. Let H = (cid:104)X(cid:105) ⊆ F be the subgroup generated by X, and let θ : X → H be
the mapping xθ = x (∀x ∈ X). Let θ(cid:48) : F → H be the corresponding extension. Let
ι : H → F be defined by hι = h (∀h ∈ H). Then θ(cid:48)ι extends θι . But so does
2 2 2 2
id , and so θ(cid:48)ι = id by uniqueness. Therefore F = Im(id ) = Im(θ(cid:48)ι ) ⊆ H
F 2 F F 2
⇒ F = H = (cid:104)X(cid:105).
If F is free on X then X is a (free) basis for F and |X| = r(F) is the rank of F.
4
∼
Theorem 1.2. If F is free on X (i = 1,2) and |X | = |X | then F = F . (Free
i i 1 2 1 2
groups of the same rank are isomorphic.)
Proof. Assume |X | = |X |. Let φ : X → X be a bijection. Let α,β be the
1 2 1 2
following extensions (in which ι and ι are inclusions):
1 2
X ι1 (cid:47)(cid:47)F X ι2 (cid:47)(cid:47)F
1 1 2 2
(cid:14) (cid:14)
(cid:14) (cid:14)
(cid:14) (cid:14)
φ (cid:14) φ−1 (cid:14)
(cid:15)(cid:15) (cid:14)(cid:14) (cid:15)(cid:15) (cid:14)(cid:14)
(cid:14) (cid:14)
X2 (cid:14)(cid:14)(cid:14)α X1 (cid:14)(cid:14)(cid:14)β
(cid:14) (cid:14)
(cid:14) (cid:14)
ι2(cid:14)(cid:14) ι1(cid:14)(cid:14)
(cid:15)(cid:15) (cid:6)(cid:6)(cid:14) (cid:15)(cid:15) (cid:6)(cid:6)(cid:14)
F F
2 1
Now ι αβ = φι β = φφ−1ι = ι , so αβ : F → F extends ι .
1 2 1 1 1 1 1
X ι1 (cid:47)(cid:47)F
(cid:15)(cid:15)1ι1(cid:126)(cid:126)(cid:124)(cid:124)(cid:124)(cid:124)(cid:124)α(cid:124)(cid:124)β(cid:124) 1
F
1
By uniqueness, αβ = id , since id also extends ι . Similarly βα = id . This says
F1 F1 1 F2
that α : F → F and β : F → F are bijective homomorphisms and so are inverse
1 2 2 1
isomorphisms.
Let F be a group and let X ⊆ F. For the group G let Hom(F,G) denote the set of
homomorphisms from F → G, and let Map(X,G) denote the set of mappings from
X (cid:55)→ G.
Define ρ : Hom(F,G) → Map(X,G) by φρ = ιφ where ι : X → F is inclusion.
Exercise. (1) ρ is surjective iff for all maps X → G there exists θ(cid:48) as in the defini-
tion of F being free on X.
(2) ρ is injective iff θ(cid:48), if it exists, is unique.
(3) F is free on X iff ρ is bijective ∀ groups G.
∼
Theorem 1.3. If F is free on X (i = 1,2) and F = F then |X | = |X |. (Free
i i 1 2 1 2
groups of different rank are not isomorphic.)
5
∼
Proof. Since F = F we have |Hom(F ,G)| = |Hom(F ,G)| for any group G.
1 2 1 2
Therefore |Map(X ,G)| = |Map(X ,G)| for any group G. Let G = C be the cyclic
1 2 2
group of order two. Then |Map(X ,G)| = 2|X1| = |Map(X ,G)| = 2|X2|
1 2
⇒ |X | = |X |.
1 2
Free groups are (isomorphic to): Z,F ,F ,...,F ,... and F where F is the free
2 3 n ∞ n
group of rank n and F is the free group of X where X is infinite. Let X be an
∞
arbitrary non-empty set. We construct a free group F(X) with X as a free basis.
Step 1. First form another copy of X,
ˆ
X = {xˆ : x ∈ X}
ˆ
where the elements of X will later become the inverses of the elements of X. Let
X±1 := X ∪Xˆ. Now form the sets of words W of length n ≥ 0 in X±1 which are
n
n-tuples of elements of X±1. Thus
(cid:136)
W consists of ( ), the empty word (sometimes denoted by e).
0
(cid:136)
W consists of (x),(xˆ),x ∈ X
1
(cid:136) W consists of (x,y),x,y ∈ X±1
2
and so on. Now discard all words containing an adjacent pair: (...,x,xˆ,...) or
(...,xˆ,x,...) where x ∈ X. The remaining words are called reduced words. Let W(cid:102)
n
(cid:83)
denote the set of reduced words of length n. Finally let F(X) = W(cid:102).
n≥0 n
Notation: We are writing:
(cid:136) x−1 for xˆ
(cid:136)
x x ...x for (x ,x ,...,x )
1 2 k 1 2 k
(cid:136) xn for (x,...,x) ∈ W(cid:102)
n
(cid:136) x−n for (xˆ,...,xˆ) ∈ W(cid:102)
n
(for x,x ∈ X).
i
6
Example. IfX = {x,y},W(cid:102) = {x2,y2,x−2,y−2,xy,xy−1,x−1y,x−1y−1,yx,y−1x,y−1x−1,yx−1}.
2
Step 2. “Juxtaposition plus cancellation” is the binary operation.
Given a = (x ,...,x ) ∈ W(cid:102),b = (y ,...,y ) ∈ W(cid:103),
1 l l 1 m m
ab := (x ,...,x ,y ,...,y )
1 l−r r+1 m
where r is the largest integer k such that none of (x ,y ),(x ,y ),...,(x ,y )
l 1 l−1 2 l−k+1 k
is reduced. Thus ab ∈ W(cid:102) . Now we need to check the group axioms.
l+m−2r
Closure: Immediate.
Identity: The empty word ( ).
Inverses: (x ,...,x )−1 = (xˆ ,...,xˆ ) with the understanding that xˆˆ := x
1 l l 1
for x ∈ X.
Associativity: Let c = (z ,...,z ) ∈ W(cid:102) and let bc = (y ,...,y ,z ,...,z ) ∈
1 n 1 1 m−s s+1 n
W(cid:102) . We want to show: (ab)c = a(bc).
m+n−2s
“We always prove (ab)c = a(bc). Why is this enough to prove that it doesn’t matter
how to bracket an arbitrary expression? We always take it for granted, don’t we?
This is a hidden horrible exercise in group theory.”
If a,b or c is the empty word the result is obvious. So, assume that l,m,n ≥ 1.
Then there are three cases to consider:
Case 1.
r+s < m (The cancellations in ab and bc are disjunct).
Then both sides of (ab)c = a(bc) are equal to
(x ,...,x ,y ,...,y ,z ,...,z ) ∈ W(cid:102)
1 l−r r+1 m−s s+1 n l+m+n−2(r+s)
Example. r = 2,s = 3,m = 6
a = xyxy−1x−1 (which is equal to (x,y,x,yˆ,xˆ) but we won’t use this notation.)
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b = xyxy−1xy−1 and c = yx−1yx
(ab)c = (xyxy−1x−1xyxy−1xy−1)yx−1yx
= (xyxxy−1xy−1)yx−1yx
= xyxxx
a(bc) = xyxy−1x−1(xyxy−1xy−1yx−1yx)
= xyxy−1x−1(xyxx)
= xyxxx
Case 2. r+s = m (The cancellations in ab and bc meet in exactly one place so that
b is completely cancelled out.)
Here both sides equal (x ,...,x ,z ,...,z ) ∈ W .
1 l−r s+1 n l+n−m
Example. r = 2,s = 2,m = 4
a = xyx−1y−1 (l = 4)
b = yxyx (m = 4)
c = x−1y−1y−1x (n = 4)
(ab)c = (xyx−1y−1yxyx)x−1y−1y−1x
= (xyyx)x−1y−1y−1x
= xyy−1x ∈ W
4
a(bc) = xyx−1y−1(yxyxx−1y−1y−1x)
= xyx−1y−1(yxy−1x)
= xyy−1x ∈ W
4
(Of course this word can be further reduced to xx ∈ W(cid:102) but we stop the reduction
2
as soon as we arrive in W(cid:102) . It’s sufficient to know that we end up with the
l+m−n
same irreduced word at some point since it would lead us to the same reduced word
as well.)
Case 3. r+s > m (The cancellations overlap).
In this case put β = (y ,...,y ),γ = (y ,...,y ),δ = (y ,...,y ), where
1 m−s m−s+1 r r+1 m
8
γ describes the overlapping part.
By hypothesis, γ has length r−m+s−1+1 = r−m+s > 0 and b = βγδ,
a = αγ−1β−1 with α = (x ,...,x ), c = δ−1γ−1ε with ε = (z ,...,z )
1 l−r s+1 n
Then,
(ab)c = (αγ−1β−1βγδ)(δ−1γ−1ε)
= (αδ)(δ−1γ−1ε)
= α(γ−1ε)
a(bc) = (αγ−1β−1)(βγδδ−1γ−1ε)
= (αγ−1β−1)(βε)
= (αγ−1)ε
Since α and γ−1 are adjacent in the reduced word a there is no cancellation in
forming their product. Similarly this is also the case for γ−1 and ε since they are
adjacent in the reduced word c. Therefore:
a(bc) = aγ−1ε = (ab)c
Example. Take r = 4,s = 4,m = 5.
a = xyxy−1x2y = (xyx)(y−1x)(xy) = αγ−1β−1
b = y−1x−2yx = (y−1x−1)(x−1y)(x) = βγδ
c = x−1y−1xyx3 = (x−1)(y−1x)(yx3) = δ−1γ−1ε
Then (ab)c = αγ−1ε = (xyx)(y−1x)yx3 = xyxy−1xyx3 = a(bc).
Step 3. Get rid of the brackets and commas.
x → x
(x ,...,x ) = x ...x
1 l 1 l
and identify xˆ with x−1. Note that (cid:104)X(cid:105) = F(X).
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Step 4. Finally we show that F(X) is free on X.
For a given group G and a mapping θ : X → G define θ(cid:48) : F(X) → G as follows:
e θ(cid:48) = e
F(X) G
xθ(cid:48) = xθ ∀x ∈ X
x−1θ(cid:48) = (xθ)−1 ∀x ∈ X
and (x ,...,x )θ(cid:48) = (x θ(cid:48))(x θ(cid:48))...(x θ(cid:48)) for x ,...,x ∈ W(cid:102).
1 l 1 2 l 1 l l
It is clear that θ(cid:48) extends θ and if φ : F(X) → G is another extension of θ then θ(cid:48)
and φ (given that θ(cid:48) is a homomorphism) agree on the generating set X so θ(cid:48) = φ
(uniqueness). (See Exercise Sheet 1, Question 1.)
It remains to show that θ(cid:48) is homomorphism: Let a = x ...x ∈ W(cid:102) and b =
1 l l
y ...y ∈ W(cid:103) and ab = x ...x y ...y ∈ W(cid:102) . By definition of ab we
1 m m 1 l−r r+1 m l+m−2r
see that y = x−1 for 1 ≤ i ≤ r (to get cancellations) and so by definition of θ(cid:48),
i l−i+1
y θ(cid:48) = (x θ)−1 = (x θ(cid:48))−1.
i l−i+1 l−i+1
(To see this if y ∈ X,x−1 ∈ X−1 then y θ(cid:48) = y θ = (y−1θ(cid:48))−1 = (x θ(cid:48))−1; and
i l−i+1 i i i l−i+1
if y ∈ X−1,x−1 ∈ X then y θ(cid:48) = x−1 θ(cid:48) = (x θ)−1 = (x θ(cid:48))−1 .)
i l−i+1 i l−i+1 l−i+1 l−i+1
Therefore,
(aθ(cid:48))−1((ab)θ(cid:48))(bθ(cid:48))−1 = [x θ(cid:48)...x θ(cid:48)][x θ(cid:48)...x θ(cid:48)y θ(cid:48)...y θ(cid:48)][y θ(cid:48)...y θ(cid:48)]−1
1 l 1 l−r r+1 m 1 m
= (x θ(cid:48))−1...(x θ(cid:48))−1(y θ(cid:48))−1...(y θ(cid:48))−1
l l−r+1 r 1
= (y θ(cid:48))...(y θ(cid:48))(y θ(cid:48))−1...(y θ(cid:48))−1 = e
1 r r 1
So (aθ(cid:48))(bθ(cid:48)) = ((ab)θ(cid:48)) ⇒ θ(cid:48) ∈ Hom(F(X),G).
We have proved:
Theorem 1.4. The group F(X) of reduced words in X±1 is free on X.
∼
Lemma 1.5. If F is free of rank r and H = F then H is free of rank r.
Proof. Let F be free on X with |X| = r. Let φ : F → H be an isomorphism and
let Y = {xφ : x ∈ X} ⊆ H.
10