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Combinatorial Group TheoryDr 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:
1 Free groups
By way of introduction consider the dihedral group D10 of order 10, the group of
isometries of the regular 5-gon. It consists of five rotations and five reflections.
1
2
34
5
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 D10 are of the form xiyj.
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
⇒ D10 = {e, x, x2, x3, x4, y, xy, x2y, x3y, x4y}
(Does D10 really contain ten elements? In other words: Could some of the ele-
ments above still be equal? No! Suppose for example x3 = x4y ⇒ x−1 = y .)
2
To introduce some language:
x and y are generators for D10.
{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 = 〈S〉.
This means that every g ∈ G can be expressed as a finite product g = sε11 · . . . · sεkk
where εi = ±1, si ∈ S, 1 ≤ i ≤ k. (That is, each g ∈ G is a word in S ∪ S−1.)
This product is said to be reduced (with respect to S) if the following condition is
satisfied:
si = si+1 ⇒ εi + εi+1 6= 0
This simply says that s−1s and ss−1(s ∈ S) are forbidden subwords.
Example. S = {x, y}
not reduced (still not reduced!) reduced
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 D10 we have y−1xy = x−1 so D10 is not a free group (with respect to
{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′ : F → G extending θ,
that is, having the property that xθ′ = xθ (∀x ∈ X). That is, the diagram
X
�
ι // F
θ′~~
G
commutes. Here ι : X → F is the inclusion determined by xι = x (∀x ∈ X).
Remarks. (1) ι and θ are mappings; θ′ is a homomorphism.
(2) This definition allows us to distinguish between words in F .
x1x2, x1x3 ∈ F (xi ∈ X)
x1_
��
x2_
��
x3_
��
X
�
// F
θ′����������
(12) (123) (124) S4
(x1x2)θ′ = (x1)θ
′(x2)θ′ = (12)(123)
(x1x3)θ′ = (x1)θ
′(x3)θ′ = (12)(124)
(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 = 〈X〉 ⊆ F be the subgroup generated by X, and let θ : X → H be
the mapping xθ = x (∀x ∈ X). Let θ′ : F → H be the corresponding extension. Let
ι2 : H → F be defined by hι2 = h (∀h ∈ H). Then θ′ι2 extends θι2. But so does
idF , and so θ′ι2 = idF by uniqueness. Therefore F = Im(idF ) = Im(θ′ι2) ⊆ H
⇒ F = H = 〈X〉.
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 Fi is free on Xi (i = 1, 2) and |X1| = |X2| then F1∼= F2. (Free
groups of the same rank are isomorphic.)
Proof. Assume |X1| = |X2|. Let φ : X1 → X2 be a bijection. Let α, β be the
following extensions (in which ι1 and ι2 are inclusions):
X1
φ
��
ι1 // F1
α
�����������������
X2
ι2��
F2
X2
φ−1
��
ι2 // F2
β
�����������������
X1
ι1��
F1
Now ι1αβ = φι2β = φφ−1ι1 = ι1, so αβ : F1 → F1 extends ι1.
X1
ι1��
ι1 // F1
αβ~~||||||||
F1
By uniqueness, αβ = idF1 , since idF1 also extends ι1. Similarly βα = idF2 . This says
that α : F1 → F2 and β : F2 → F1 are bijective homomorphisms and so are inverse
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 7→ 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′ as in the defini-
tion of F being free on X.
(2) ρ is injective iff θ′, if it exists, is unique.
(3) F is free on X iff ρ is bijective ∀ groups G.
Theorem 1.3. If Fi is free on Xi (i = 1, 2) and F1∼= F2 then |X1| = |X2|. (Free
groups of different rank are not isomorphic.)
5
Proof. Since F1∼= F2 we have |Hom(F1, G)| = |Hom(F2, G)| for any group G.
Therefore |Map(X1, G)| = |Map(X2, G)| for any group G. Let G = C2 be the cyclic
group of order two. Then |Map(X1, G)| = 2|X1| = |Map(X2, G)| = 2|X2|
⇒ |X1| = |X2|.
Free groups are (isomorphic to): Z, F2, F3, . . . , Fn, . . . and F∞ where Fn is the free
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,
X = {x : x ∈ X}
where the elements of X will later become the inverses of the elements of X. Let
X±1 := X ∪ X. Now form the sets of words Wn of length n ≥ 0 in X±1 which are
n-tuples of elements of X±1. Thus
� W0 consists of ( ), the empty word (sometimes denoted by e).
� W1 consists of (x), (x), x ∈ X
� W2 consists of (x, y), x, y ∈ X±1
and so on. Now discard all words containing an adjacent pair: (. . . , x, x, . . .) or
(. . . , x, x, . . .) where x ∈ X. The remaining words are called reduced words. Let Wn
denote the set of reduced words of length n. Finally let F (X) =⋃n≥0 Wn.
Notation: We are writing:
� x−1 for x
� x1x2 . . . xk for (x1, x2, . . . , xk)
� xn for (x, . . . , x) ∈ Wn
� x−n for (x, . . . , x) ∈ Wn
(for x, xi ∈ X).
6
Example. IfX = {x, y}, W2 = {x2, y2, x−2, y−2, xy, xy−1, x−1y, x−1y−1, yx, y−1x, y−1x−1, yx−1}.
Step 2. “Juxtaposition plus cancellation” is the binary operation.
Given a = (x1, . . . , xl) ∈ Wl, b = (y1, . . . , ym) ∈ Wm,
ab := (x1, . . . , xl−r, yr+1, . . . , ym)
where r is the largest integer k such that none of (xl, y1), (xl−1, y2), . . . , (xl−k+1, yk)
is reduced. Thus ab ∈ Wl+m−2r. Now we need to check the group axioms.
Closure: Immediate.
Identity: The empty word ( ).
Inverses: (x1, . . . , xl)−1 = (xl, . . . , x1) with the understanding that ˆx := x
for x ∈ X.
Associativity: Let c = (z1, . . . , zn) ∈ W1 and let bc = (y1, . . . , ym−s, zs+1, . . . , zn) ∈
Wm+n−2s. We want to show: (ab)c = a(bc).
“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
(x1, . . . , xl−r, yr+1, . . . , ym−s, zs+1, . . . , zn) ∈ Wl+m+n−2(r+s)
Example. r = 2, s = 3,m = 6
a = xyxy−1x−1 (which is equal to (x, y, x, y, x) but we won’t use this notation.)
7
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 (x1, . . . , xl−r, zs+1, . . . , zn) ∈ Wl+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 ∈ W4
a(bc) = xyx−1y−1(yxyxx−1y−1y−1x)
= xyx−1y−1(yxy−1x)
= xyy−1x ∈ W4
(Of course this word can be further reduced to xx ∈ W2 but we stop the reduction
as soon as we arrive in Wl+m−n. It’s sufficient to know that we end up with the
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 = (y1, . . . , ym−s), γ = (ym−s+1, . . . , yr), δ = (yr+1, . . . , ym), where
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 = (x1, . . . , xl−r), c = δ−1γ−1ε with ε = (zs+1, . . . , zn)
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
(x1, . . . , xl) = x1 . . . xl
and identify x with x−1. Note that 〈X〉 = F (X).
9
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′ : F (X)→ G as follows:
eF (X)θ′ = eG
xθ′ = xθ ∀x ∈ X
x−1θ′ = (xθ)−1 ∀x ∈ X
and (x1, . . . , xl)θ′ = (x1θ
′)(x2θ′) . . . (xlθ
′) for x1, . . . , xl ∈ Wl.
It is clear that θ′ extends θ and if φ : F (X) → G is another extension of θ then θ′
and φ (given that θ′ is a homomorphism) agree on the generating set X so θ′ = φ
(uniqueness). (See Exercise Sheet 1, Question 1.)
It remains to show that θ′ is homomorphism: Let a = x1 . . . xl ∈ Wl and b =
y1 . . . ym ∈ Wm and ab = x1 . . . xl−ryr+1 . . . ym ∈ Wl+m−2r. By definition of ab we
see that yi = x−1l−i+1 for 1 ≤ i ≤ r (to get cancellations) and so by definition of θ′,
yiθ′ = (xl−i+1θ)
−1 = (xl−i+1θ′)−1.
(To see this if yi ∈ X, x−1l−i+1 ∈ X−1 then yiθ′ = yiθ = (y−1i θ′)−1 = (xl−i+1θ
′)−1; and
if yi ∈ X−1, x−1l−i+1 ∈ X then yiθ′ = x−1l−i+1θ
′ = (xl−i+1θ)−1 = (xl−i+1θ
′)−1 .)
Therefore,
(aθ′)−1((ab)θ′)(bθ′)−1 = [x1θ′ . . . xlθ
′][x1θ′ . . . xl−rθ
′yr+1θ′ . . . ymθ
′][y1θ′ . . . ymθ
′]−1
= (xlθ′)−1 . . . (xl−r+1θ
′)−1(yrθ′)−1 . . . (y1θ
′)−1
= (y1θ′) . . . (yrθ
′)(yrθ′)−1 . . . (y1θ
′)−1 = e
So (aθ′)(bθ′) = ((ab)θ′)⇒ θ′ ∈ 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
X� � ι1 //
ρ~~~~~~~~~~
F
φ~~~~~~~~~
θ′
||
��
��
��
z
Y� � ι2 //
θ
''PPPPPPPPPPPPPPP H
φ′
�����
G
We claim that H is free on Y . Since |Y | = r the result follows.
Let ρ : X → Y be the bijection xρ = xφ (∀x ∈ X). Let G be any group and θ any
mapping from Y to G. Since F is free on X there exists a unique homomorphism
θ′ : F → G extending ρθ. Let φ′ : H → G be the homomorphism defined by
φ′ = φ−1θ′. Clearly ρι2 = ι1φ. Therefore ρθ = ι1θ′ = ι1φφ
′ = ρι2φ′. Now ρ is
a bijection, so θ = ι2φ′, that is, φ′ extends θ. Suppose now that θ = ι2φ. Then
ρθ = ρι2φ.
⇒ ρθ = ι1φφ ⇒ φφ extends θ. Since θ′ also extends θ and is unique, φ = φ−1θ′ =
φ′.
(What we have proved here is that “being free” is an isomorphism invariant.)
Theorem 1.6. A group F is free on a subset X ⊆ F iff
(i) X generates F , and
(ii) no reduced word in X±1 of positive length is equal to the identity in F .
Proof. Let θ′ be the unique extension of the inclusion ι2 : X → F .
X
ι2
��
ι1 // F (X)
θ′||y
yy
yy
F
(⇐): If (i) holds then θ′ is surjective. If (ii) holds then θ′ is injective. Therefore θ′
is an isomorphism and so F is free on X by 1.5.
(⇒): If F is free on X and |X| = r, then both F and F (X) are free groups of
rank r. Therefore by theorem 1.2, F ∼= F (X). Since (i) and (ii) hold in F (X) it is
clear that they also hold in F .
Theorem 1.7. Every group is isomorphic to a quotient group of a free group.
11
Proof. Let G be a given group and let X be any generating set for G. Such an X
always exists (take X = G if necessary). Let θ′ be the unique extension given by:
X
ι2
��
ι1 // F (X)
θ′||y
yy
yy
G
Then (first isomorphism theorem) F (X)/kerθ′ ∼= Imθ′.
But X ⊆ Imθ′ and 〈X〉 = G. Thus, Imθ′ = G.
We investigate torsion in free groups.
A reduced word
a = x1 . . . xl (xi ∈ X±1, a ∈ F (X)
is called cyclically reduced if x1 6= x−1l .
Example. The word x2y−1xyx−1 is reduced but not cyclically reduced.
The word yx2y3x−1y is cyclically reduced.
Suppose a = x1 . . . xl is reduced and a2 = x1 . . . xl−rxr+1 . . . xl ∈ W2l−2r. So l(a) =
2l − 2r. Question: How big can r be?
Example. Let’s assume l = 9 and r = 5.
a = x1x2x3x4 x5 x6 x7 x8 x9x1 x2 x3 x4 x5
?
x6x7x8x9
But x5x5 is a reduced word of length 2, so r = 5 is impossible!
Example. What about l = 8 and r = 5?
a = x1x2x3x4 x5 x6 x7 x8x1 x2 x3 x4
?
x5x6x7x8
If x5x4 = e then x4 = x−15 ⇒ x4x5 = e which contradicts the fact that a is reduced.
12
More generally, if l = 2k + 1 and r > k then we get x2k+1 = e (contradiction), and if
l = 2k and r ≥ k then we get “a not reduced” (contradiction). Therefore r < l(a)/2.
Now let a = u−1au where a is cyclically reduced.
a = (x1 . . . xr)−1xr+1 . . . xl−r(x1 . . . xr) l(a) > 0 for a 6= e
Example.
a = x−1y−1x2y4xyx−2yx
a = y4xy
Note that l(a2) = 2l(a)− 2r > 2l(a)− l(a) = l(a).
More generally,
an = (u−1au)n = u−1anu
⇒ l(an) = l(u−1anu) = nl(a) + 2r > (n− 1)l(a) + 2r = l(u−1an−1u) = l(an−1).
We have proved:
Theorem 1.8. F (X) is torsion free (the only element of finite order is e).
Remark. 〈Q\{0}, x〉 has elements +1 and −1 of finite order so it is not torsion free
– therefore it’s not free.
If g ∈ G then the centralizer of g in G is the subgroup CG(g) = {h ∈ G : hg =
gh} ≤ G.
Theorem 1.9. For any w ∈ F (X)\{e}, Cw = CF (X)(w) ∼= Z (infinite cyclic group).
We will need some Lemmas:
Lemma 1.10. Let a, b ∈ F (X). If ab = ba then ∃c ∈ F (X) such that a = ck and
b = ch where k, h ∈ Z.
Proof. The proof is by induction on l(a) + l(b). The result is clear if a = e or b = e,
in particular, the result holds for l(a) + l(b) = 1.
13
Assume a 6= e and b 6= e. Let a = x1 . . . xl ∈ Wl, b = y1 . . . ym ∈ Wm and assume
without loss of generality that l ≤ m. Suppose in reduced form we have:
ab = x1 . . . xl−ryr+1 . . . ym
ba = y1 . . . ym−rxr+1 . . . xl
where 0 ≤ r ≤ l.
(If we have r cancellations in ab we also have r cancellations in ba since ab = ba so
ab and ba have the same length.)
� If r = 0 then xi = yi for 1 ≤ i ≤ l and so b = au where l(u) = m − l < l(b).
Now au = b ⇒ au = a−1ab = a−1ba = a−1aua = ua. So by induction we see
that a and u are powers of a common element. But then so is au = b.
� If r = l then b = a−1v with l(v) < l(b). Now a−1v = b ⇒ a−1v = aa−1b =
aba−1 = aa−1va−1 = va−1 (since ab = ba ⇔ a−1b = ba−1). ⇒ a−1 and v are
powers of a common element. But then, so is a−1v = b.
� 0 < r < l: See Exercise Sheet, question 5.
Lemma 1.11. (i) If a, b ∈ F (X) and an = bn then a = b.
(ii) If w ∈ F (X) then |{c ∈ F (X) : cn = w for some n ∈ N}| <∞.
Proof. (i) Write:
a = u−1au ⇒ l(a) = l(a) + 2l(u)
b = v−1bv ⇒ l(b) = l(b) + 2l(v)
14
an = bn ⇒ u−1anu = v−1bnv
⇒ nl(a) + 2l(u) = nl(b) + 2l(v)
an = bn ⇒ a2n = b2n
⇒ u−1a2nu = v−1b2nv
⇒ 2nl(a) + 2l(u) = 2nl(b) + 2l(v)
⇒ l(a) = l(b)⇒ l(u) = l(v)
Therefore u−1anu = v−1bnv (reduced). ⇒ u = v and a = b⇒ a = b.
(ii) Let cn = w. If w = e then c = e by Theorem 1.8. Write c = u−1cu. If w 6= e
then c 6= e and so c 6= e. Now l(w) = l(cn) = l(u−1cu) = nl(c) + 2l(u) ≥ n.
But l(w) ≥ n for only finitely many n. Therefore, since w is an nth power of
at most one element for each fixed n by (i), the result follows.
Lemma 1.12. If akbh = bhak for a, b ∈ F (X), h, k ∈ Z\{0} then ∃c ∈ F (X) such
that a, b ∈ 〈c〉 (⇒ ab = ba since they’re both powers of a common element).
Proof. Exercise. [Hint: Apply Lemma 1.11 (i) twice.]
Lemma 1.13. If a ∈ F (X)\{e} then C = CF (X)(a) is Abelian.
Proof. Let u, v ∈ C. Then ua = au and va = av. Assume without any loss that
u 6= e and v 6= e. By Lemma 1.10 there exist b, d ∈ F (X) and p, q, r, s ∈ Z\{0} such
that:
u = bp, a = bq
v = dr, a = ds
⇒ bqds = dsbq ⇒ ∃c ∈ F (X) such that b = ch, d = ck (by 1.12).
Therefore u = chp and v = ckr commute.
15
Proof of Theorem 1.9. Observe that wn ∈ Cw for all n ∈ Z. By Theorem 1.7 it
follows that Cw is infinite. Pick d ∈ Cw\{e} such that d has minimal length. Let
v ∈ Cw. We claim that v is a power of d. Since d and v commute (by 1.13) we have
d = uh, v = uk for some u ∈ F (X). Now uk ∈ Cw implies: uk commutes with w and
so u commutes with w by 1.12, that is, u ∈ Cw. Write u = a−1ua. Then:
l(d) = l(uh) = |h|l(u) + 2l(a)
= (|h| − 1)l(u) + l(u) + 2l(a)
= (|h| − 1)l(u) + l(u)
By minimality we have |h| = 1 therefore d = u±1 ⇒ v = d±h and we have shown
that Cw = 〈d〉.
Remark. 〈w〉 ≤ CF (X)(w), but equality does not always hold!
Take for example X = {x, y}, w = x4. Here Cw 6= 〈x4〉.
2 Schreier’s Method
Dedekind’s Theorem tells us that if A is free Abelian and B ≤ A then B is free
Abelian and rankB ≤ rankA. In this section, we will prove that a subgroup of a
free group is free. However there is no bound on the rank of a free subgroup.
Let F = F (X) be the free group on X. Let H ≤ F . We show that H is a free group.
1. The well-ordering of F
A partial ordering on a set S is a binary relation < on S such that < is
(01) irreflexive: ¬(s < s) (∀s ∈ S)
(02) transitive: a < b ∧ b < c⇒ a < c
and < is called a total ordering if also
(03) (∀s, t ∈ S) s < t or s = t or t < s
and < is called a well-ordering if also
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(04) any nonempty set T in S has a least element. That is, ∀T ⊆ S : ∃t ∈ T such
that t < t ∀t ∈ T\{t}.
Notes. 1. (04) ⇒ (03)
2. Theorem. Any set can be well-ordered.
Let < be a well-ordering of X±1. Let a = x1 . . . xl, b = y1 . . . ym ∈ F (X) where
xi, xj ∈ X±1 and so that l(a) = l and l(b) = m. Then write a < b if either l < m or
l = m and xr < yr and xi = yi for 1 ≤ i ≤ r − 1.
Exercise. This yields a well-ordering on F (X).
Example. Let X = {x, y}. Well-ordering of X±1: x < y < x−1 < y−1 (Question:
What is the 10th word in F (X)? — Answer: e < x < y < x−1 < y−1 < xx < xy <
xy−1 < yx < yy < . . .)
Lemma 2.1. Let w = x1 . . . xn be a reduced word in X±1 with n > 1 and let v be
any element of F = F (X). Then v < x1 . . . xn−1 ⇒ vxn < w.
Proof. If l(v) < n−1 then l(vxn) < n = l(w) (Caution: this is the usual “less than”,
not our ordering!) and vxn < w (this on the other hand is our well-ordering “less
than”). Now suppose that v = x1 . . . xr−1yr . . . yn−1 where yr < xr.
If xn = y−1n−1 then l(vxn) = n− 2 < l(w) and the result follows.
If xn 6= y−1n−1 then vxn = x1 . . . xr−1yr . . . yn−1xn < x1 . . . xn = w.
2. Schreier Transversal (1927)
We have H, our fixed subgroup of F . Let U be a right transversal for H in F .
Then for any w ∈ F , Hw ∩ U consists of a single element that we denote w.
Example (What is a transversal?). C20 = 〈x〉, H = 〈x5〉 = {1, x5, x10, x15}. Then
the other partitions we get are Hx,Hx2, Hx3 and Hx4. Either two cosets are the
same or their intersection is empty (since the partitions can be considered as equiv-
alence classes with a ∼ b⇔ a = b mod H).
A transversal is a set of coset representatives. There are 45 transversals, e.g.
{x5, x11, x12, x3, x4}. Here, x16 = x11.
17
A subset S of F has the Schreier property (SP) if it contains all initial segments of
all its elements, that is, w = x1 . . . xn ∈ S ⇒ x1 . . . xn−1 ∈ S, where l(w) ≥ 1.
A Schreier transversal for H in F is a transversal U with (SP). Note that this implies
e ∈ U .
Lemma 2.2. Every subgroup H of F has a Schreier transversal.
Proof. Choose the least element of each right coset of H in F in the above well-
ordering of F . Let U denote the resulting transversal. Suppose that x1 . . . xn−1 /∈ U
but w = x1 . . . xn ∈ U . Let v be the least element of Hx1 . . . xn−1.
Then v < x1 . . . xn−1 and this implies vxn < x1 . . . xn−1xn = w by Lemma 2.1. But
now Hv = Hx1 . . . Hxn−1 ⇒ Hvxn = Hw ⇒ vxn ∈ Hw and this contradicts the
minimality of w in Hw.
Example. X = {x, y}, F = F (X) ordered as above, x < y < x−1 < y−1. Let
H be the normal closure of S = {x3, y2, x−1y−1xy} in F . Then H is the inter-
section of all the normal subgroups of F containing S. For example, H E F and
S ≤ N E F ⇒ H E N .
We show that |F : H| = G.
Let C6 = {e, a, a2, a3, a4, a5} = 〈a〉 and define θ : X → C6 by xθ = a2, yθ = a3. Let
θ′ : F → C6 be the unique extension of θ.
Then (yx−1)θ′ = (yθ′)(xθ′)−1 = a ⇒ θ′ is onto. x3θ′ = y2θ′ = x−1y−1xyθ′ = e ⇒
S ≤ ker θ′ ≤ F ⇒ H ≤ ker θ′ ⇒ |F : H| = |F : ker θ′| · | ker θ′ : H| ≥ |F : ker θ′| =
|Im θ′| = 6 by the first isomorphism theorem.
On the other hand the quotient group F/H is generated by Hx and Hy and there
(Hx)3 = (Hy)2 = H, Hx−1y−1xy = H ⇒ HxHy = HyHx.
(If N is a normal subgroup of G and a ∈ G, then Na = aN,Nk = N, (Na)k = Nak.
Normal subgroups make life much easier.)
⇒ F/H is Abelian of order ≤ 6 ⇒ |F : H| ≤ 6 ⇒ |F : H| = 6 since we al-
ready know that |F : H| ≥ 6.
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(The rearranging works like this:
Hx−1y−1xy = H
⇒ x−1y−1Hxy = H
⇒ Hxy = yxH = Hyx
⇒ H2xy = H2yx
⇒ HxHy = HyHx
General observation: N E G normal, [a, b] ∈ N ⇔ NaNb = NbNa.)
The transversal {e, x, x2, y, xy, x2y} has SP.
In general, take the least-element approach:
Take a look at the ordered elements.
e x y x−1 y−1 xx xy xy−1 yx yy yx−1 . . .
Now for each class we search for the element representing it. Clearly, e ∈ H, x ∈
Hx, y ∈ Hy. The next element is x−1 and it’s in Hx2 because x3 ∈ H iff Hx2 = H,
y−1 ∈ Hy with a similar arguument involving y2 ∈ H. Of course the next ele-
ments xx ∈ Hx2 and xy ∈ Hxy. Now, Hxy−1 = HxHy−1 = HxHy = Hxy and
Hyx = HyHx = HxHy = Hxy, and Hyy = Hy2 = H. But Hyx−1 = HyHx−1 =
HyHx2 = Hx2Hy = Hx2y and so it completes our transversal.
Transversal: {e, x, y, x−1, xy, yx−1}.
3. The Schreier generators
Recall that Hw∩U = w where U is a right Schreier transversal for H in F = F (X).
Properties.
1. w = w
2. Hw = Hw
19
3. w = w iff w ∈ U
4. (∀u ∈ U, x ∈ X±1) Hux = Hux⇒ Hu = Huxx−1 ⇒ uxx−1 = u.
Lemma 2.3. H is generated by A = {ux(ux)−1|u ∈ U, x ∈ X±1}
Proof. Since Hux = Hux we have Huxux−1 = H, that is, ux(ux)−1 ∈ H, so A ⊆ H.
Now let h ∈ H and write h as a reduced word h = x1 . . . xn (xi ∈ X±1). Define the
following sequence of elements of U :
u1 = e
ui+1 = uixi (1 ≤ i ≤ n)
and now put ai = uixiu−1i+1 = uixi(uixi)
−1 ∈ A (1 ≤ i ≤ n). Then
a1a2 . . . an = (u1x1u−12 )(u2x2u
−13 ) . . . (un−1xn−1u
−1n )(unxnu
−1n+1)
= u1x1x2 . . . xnu−1n+1 = u1hu
−1n+1
= hu−1n+1
Observe that u−1n+1 = h−1(a1 . . . an) ∈ H ⇒ u−1n+1 ∈ H ∩ U = {e} ⇒ h = a1 . . . an.
Returning to the previous example we obtain:
HHHHHH
HHHU
X±1x y x−1 y−1
e e e e y−2
x x3 e e xy−2x−1
x−1 e x−1yxy−1 x−3 x−1y−1xy−1
y yxy−1x−1 y2 e e
xy xyx2y−1 xy2x−1 xyx−1y−1 e
yx−1 e yx−1yx yx−2y−1x−1 yx−1y−1x
The entries are: uxux−1, e.g. exex−1 = e,Hx3 = Hy2 = Hx−1y−1xy = H,Hxy =
Hyx. (In example, for x ∈ U, y−1 ∈ X±1 we get xy−1(xy−1)−1 = xy−1(xy)−1 =
xy−1y−1x−1.)
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4. Decomposition of the set A.
Notice that in the previous table there are redundancies. In fact, {column 1} ∪
{column 2} = ({column 3} ∪ {column 4})−1. Also, e appears at entry (u, x) iff
ux ∈ U . It turns out that these are the only redundancies.
LetB = {b1, b2, b3, b4, b5, b6, b7} := {x3, yxy−1x−1, xyx2y−1, x−1yxy−1, y2, xy2x−1, yx−1yx}.
More generally put B = {uxux−1 : u ∈ U, x ∈ X, ux /∈ U}.
Lemma 2.4. The sets B = {uxux−1 : u ∈ U, x ∈ X−1, ux /∈ U} and B−1 = {b−1 :
b ∈ B} coincide. Moreover A = B ∪ B ∪ {e}. (In other words, B generates H.)
Proof.
A\{e} = {uxux−1 : u ∈ U, x ∈ X±1, uxux−1 6= e}
= {uxux−1 : u ∈ U, x ∈ X±1, ux 6= ux}
= {uxux−1 : u ∈ U, x ∈ X±1, ux /∈ U}
= {uxux−1 : u ∈ U, x ∈ X, ux /∈ U} ∪ {uxux−1 : u ∈ U, x ∈ X−1, ux /∈ U}
= B ∪ B
Now for u ∈ U, x ∈ X±1,
(uxux−1)−1 = uxx−1u−1
= uxx−1uxx−1−1
= u′x−1u′x−1−1
where u′ = ux ∈ U . (Here we used the identity u = uxx−1 which we noted earlier.)
Also:
ux /∈ U iff ux 6= ux
iff u 6= uxx−1
iff uxx−1 6= uxx−1
iff uxx−1 /∈ U
iff u′x−1 /∈ U
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In conclusion 〈B〉 = H. Let x ∈ X to get B−1 ⊆ B and let x ∈ X−1 to get B−1 ⊆ B
and result.
5. Freeness of the generators B.
Let b = uxux−1 and b′ = vyvy−1 be members of B ∪ B−1 = A\{e}. So bb′ =
uxux−1vyvy−1.
Lemma 2.5. (i) ux when reduced retains the final x.
vy when reduced retains the final y.
(ii) xux−1 when reduced retains the inital x.
yvy−1 when reduced retains the inital y.
(iii) xux−1v when reduced retains the inital x.
ux−1vy when reduced retains the final y.
(iv) bb′ = e precisely when v = ux, x = y−1 and u = vy, that is, b′ = uxx−1u−1
(i.e. b′ is the free inverse of b).
Proof. (i) Let v = y1 . . . ym be reduced where yi ∈ X±1. If there is cancellation
in forming vy then ym = y−1 and so vy = y1 . . . ym−1. If m = 1 then vy = e
and if m > 1 then vy ∈ U (by (SP)) and in both cases b′ = vyvy−1 = e, a
contradiction. Similarly ux retains the final x.
(ii) Let ux = x1 . . . xl (xi ∈ X±1) be reduced. If x is cancelled in xux−1 =
xx−1l . . . x−11 then x = xl. But then
ux = uxx−1x = x1 . . . xl−1x
= x1 . . . xl−1x by (SP)
= x1 . . . xl−1xl = ux
⇒ b = e. a contradiction. Similarly yvy−1 retains the initial y.
(iii) If ux−1vy does not retain the final y then since vy retains y it follows that vy
is an initial segment of ux. But by (SP) vy ∈ U and so vy = vy ⇒ b′ = e, a
22
contradiction.
If xux−1v does not retain the initial x then (xux−1)−1 = uxx−1 must be an
initial segment of v. So it is in U by (SP). But then ux = uxx−1x = uxx−1x =
ux⇒ b = e, a contradiction.
(iv) If bb′ = e then x must cancel with y. For this to happen we must have ux = v,
then x = y−1 and u = vy. So b′ = vyvy−1 = uxx−1u−1 = b−1 is freely the
inverse of b.
Example.
z :=
b1︷ ︸︸ ︷(u1x1u1x1
−1)
b2︷ ︸︸ ︷(u2x2u2x2
−1)
b3︷ ︸︸ ︷(u3x3u3x3
−1)
b4︷ ︸︸ ︷(u4x4u4x4
−1)
= w1 x1u1x1−1u2x2 u2x2
−1b3b4
= w1x1w2 x2u2x2−1u3x3 u3x3
−1u4x4u4x4−1
= w1x1w2x2w3 x3u3x3−1u4x4 u4x4
−1
= w1x1w2x2w3x3w4 x4u4x4−1
= w1x1w2x2w3x3w4x4w5
where l(wi) ≥ 0 with relation to X±1.
Note that lenghts are relative: l(z) = 4 relative to B but l(z) ≥ 4 relative to X±1.
Important: When we talk of a reduced word we always mean reduced relative to
some given set of generators.
Example. X = {x, y}, B ≤ F (X), B = 〈bi : 1 ≤ i ≤ 5〉 where
b1 = x2yx−2
b2 = x−1yx3y
b3 = y4
b4 = y−1x−1y−1x−2
b5 = y−1x
23
Here b1b2b−12 b3 is not reduced relative to B or to X.
The word b1b22b3 = x2yx−2x−1yx3yx−1yx3yy4 is reduced relative to B and to X.
However, the word
b2b3b4 = x−1yx3y y4y−1 x−1y−1x2
is reduced relative to B but not reduced relative to X.
Also:
b2b4b1b5 = x−1 y x3yy−1 x−1 y−1 x−2x2 y x−2 y−1 x
⇒ {b1, . . . , b5} is not a free basis for B.
(Question: Is {b1, . . . , b4} a free basis for B?)
6. Proof of Theorem
Theorem 2.6 (Nielsen-Schreier Theorem). Let F be a free group and H ≤ F . Then
H is free. Moreover if |F : H| = g and rank F = r are both finite then
rank H = (rank F − 1)|F : H|+ 1 = (r − 1)g + 1
Proof. Let X be a set of free generators for F . Let U be a Schreier transversal
for H and let B be the set of generators for H as constructed above. It follows
from Lemma 2.5 that if w = b1 . . . bn is a word reduced relative to B then after
reducing we get a reduced word in X of length at least n and so w 6= e. Therefore
B is a free basis for H. For the last statement, observe that B is indexed by the
pairs (u, x) ∈ U ×X with ux /∈ U and so rank H = |U | × |X| − b = gr − b where
b = |{(u, x, v) ∈ U×X×U : ux = v}| It remains to prove b = g−1. Let T denote the
graph with g vertices labelled by the elements of U and having a directed edge from
u to v labelled x iff ux = v where x ∈ X. By (SP) every vertex of T is connected by
a path to e. In particular, T is connected. Also T has no circuits since F is free on
X. Therefore T is a tree with g − 1 edges. Since there is a bijection between edges
of T , E(T ), and {(u, x, v) ∈ U ×X × U : ux = v} it follows that b = g − 1.
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Example. Let F = F (X) be the free group on X = {x, y}, N the normal closure
of {x5, y2, (yx)2} in F . Then |F/N | = 10, |Nx| = 5 and |Ny| = 2 (see later).
Now let H = 〈N, y〉 ≤ F (X). Then N E H and H = N ∪ Ny ⇒ |H : N | = 2 so
|F : H| · |H : N | = |F : N | ⇒ |F : H| = 5. Thus rank H = (2− 1)5 + 1 = 6.
However H is not a normal subgroup of F .
We obtain Schreier generators for H. As before x < y < x−1 < y−1. Observe that a
Schreier transversal for H in F is U = {e, x, x−1, x2, x−2}. Since |Nx| = 5 in F/N
it follows that H,Hx,Hx−1, Hx2, Hx−2 are distinct.
If Hxi = Hxj for i 6= j it follows that xj−i ∈ H = N ∪Ny, a contradiction.
Certainly xj−i /∈ N and xj−i ∈ Ny ⇒ Nxj−i︸ ︷︷ ︸order5
= Ny︸︷︷︸order2
, a contradiction.
u x ux ux uxux−1
e x x x e
e y y e ?
x x x2 x2 e
x y xy x−1 xyx
x−1 x e e e
x−1 y x−1y ? ?
x2 x x3 x−2 x5
x2 y x2y ? ?
x−2 x x−1 x−1 e
x−2 y x−2y ? ?
yxyx ∈ N ⇒ yxyx ∈ H ⇒ Hyxyx = H
⇒ Hxyx = H ⇒ Hxy = Hx−1
In general, if N is a normal subgroup then every cyclic permutation of elements
and inverses belongs to N . (If x1x2 . . . xk is in a normal subgroup then so is
(xi . . . xkx1 . . . xi−1)±1.) For example, yxyx ∈ N
⇒ Nyxy = Nx−1
⇒ (Nyxy)2 = (Nx−1)2
⇒ Nyxy2xy = Nx−2
25
Nyx2y = Nx−2 since N is normal. Therefore yx2yx2 ∈ N ≤ H ⇒ Hyx2yx2 = H
⇒ Hx2yx2 = H ⇒ Hx2y = Hx−2 ⇒ x2y = x−2.
3 Presentations
Let X be a set, F = F (X) the free group on X, R ⊆ F , N the normal closure of R
in F (denoted by R or RF
). Put G ∼= F/N .
When we have this situation we write
G = 〈x|R〉
and call this a presentation of the group G. The elements of X are called generators,
and the elements of R are called defining relators. A group G is said to be finitely
presented if G = 〈X|R〉 where |X| <∞ and |R| <∞.
Important: G = 〈X|R〉. The elements of G are cosets Nw of N in F where
w ∈ F (X). However we often write w for Nw. Thus “w = 1 in G” iff w ∈ N iff
N = Nw. G = 〈Nx : x ∈ X〉 but we usually use G = 〈X〉 to mean this.
The identity in G is denoted by: 1, e, 1G, eg.
Exercise. Let g ∈ F (X). Show that g ∈ N = R iff g =∏k
i=1 h−1i rεii hi where
hi ∈ F (X), ri ∈ R, εi = ±1.
Notation: 〈x, y|x2y−2︸ ︷︷ ︸relator
〉 = 〈x, y|x2y−2 = 1︸ ︷︷ ︸relation
〉 = 〈x, y|x2 = y2︸ ︷︷ ︸relation
〉
Example. F (X) = 〈X| 〉 is the free group on X.
G = 〈X|X〉 is the trivial group.
If G = 〈x, y|x3, y2, x−1y−1xy〉 = 〈x, y|x3 = y2 = 1, xy = yx〉 then (we get an Abelian
group since the two generators commute and) G ∼= C3 × C2 = C6, the cyclic group
of order 6.
Example. Let G = 〈x|xn〉 where n ≥ 1. Then g = 1 in G iff g ∈ {xn} iff
g =∏k
i=1wix±nwi where wi ∈ F ({x}).
26
⇒ g = xqn for some q ∈ Z.
It follows that the elements of G are 1, x, x2, . . . , xn−1.
⇒ G ∼= Cn.
Moreover every finite cyclic group is a homomorphic image of the infinite cyclic
group Z = 〈x| 〉. The kernel is again cyclic and is the normal closure of xn for some
n ≥ 1.
Cyclic groups: Z = 〈x| 〉, Cn = 〈x|xn〉, n ≥ 1.
Theorem 3.1. Every group has a presentation and every finite group can be finitely
presented.
Proof. Let G be any group and X ⊆ G be a set of generators for G. Then G =
〈x| ker θ′〉 where θ′ : F (X)→ G is the unique homomorphism extending ι : X → G.
If G is finite, |G| = l <∞, say, then so is X, |X| = r, say. Then ker θ′ is generated
by a set B of cardinality (r − 1)l + 1 by Nielsen-Schreier.
Since 〈B〉 = ker θ′ E F (X it follows that:
B = 〈B〉 (since B ⊆ 〈B〉 by definition of normal closure but 〈B〉 ⊆ B because it’s
the smallest subgroup containing B).
⇒ G = 〈X|B〉, a finite presentation.
Remarks. 1. Every group has infinitely many presentations.
(For example, 〈x|B〉 = 〈x, y|B, y = 1〉 and so on.)
2. Some infinite groups can be finitely presented. (Uncountable groups can’t.)
3. If G = 〈X|K〉 where K E F (X) and 〈S〉 = K then G = 〈X|S〉.
Lemma 3.2. If F,G,H are groups and ν : F → G,α : F → H are homomorphisms
such that
(i) Im ν = G
(ii) ker ν ⊆ kerα
27
then there exists a homomorphism α′ : G→ H such that να′ = α.
Fν
��@@@@@@@α // H
G
α′>>~
~~
~
“α factors through G”
Proof. Given g ∈ G, (i) allows us to pick f ∈ F such that fν = g. Define gα′ := fα.
well-defined: If fν = f ′ν = g then f−1f ′ ∈ ker ν ⊆ kerα.
⇒ fα = f ′α.
commuting: f(να′) = (fν)α′ = gα′ = fα
homomorphism: Let fν = g and f ∗ν = g∗ so that g∗α′ = f ∗α.
Then (gg∗)α′ = (fνf ∗ν)α′ν hom.
= (ff ∗)να′ = (ff ∗)αα hom.
= fαf ∗α = gα′g∗α′.
Theorem 3.3 (von Dyck). If G = 〈X|R〉 and H = 〈X|S〉 where R ⊆ S ⊆ G(X)
then ∃ epimorphism φ : G → H fixing X elementwise (∀x ∈ X) and such that
kerφ = S\R. Conversely every quotient group of G = 〈X|R〉 has a presentation
〈X|S〉 where R ⊆ S.
Proof. Let ν : F (X) → G and α : F (X) → H denote the natural homomorphisms
(wν7→ Rw,w
α7→ Sw). Since ν is onto and ker ν = R ⊆ S = kerα we get α′ : G→ H
such that να′ = α.
Fα //
ν
��@@@@@@@ H
Gα′=φ
>>~~
~~
Since ν and α fix x ∈ X, so does α′. Moreover α′ is onto since να′ = α which is
onto; and kerα′ = (kerα)ν. To see this let w ∈ kerα′. Then w ∈ G and so there
exists f ∈ F (X) such that fν = w ⇒ fνα′ = wα′ = eH ⇒ fα = eH ⇒ f ∈ kerα⇒
w ∈ (kerα)ν.
Conversely if u ∈ (kerα)ν then u = fν where fα = eH . But u = fν ⇒ uα′ =
fνα′ ⇒ uα′ = fα = eH ⇒ u ∈ kerα′.
But (kerα)ν = Sν = S\R as required.
28
For the converse: If H is a quotient group of G let θ be the composite of the
natural maps F (X)→ G→ H so that ker θ ⊇ R and H = 〈X| ker θ〉.
Remark. This theorem says that “adding relators to a presentation for G” is the
same as “taking quotients of G”.
Example. C6 = 〈x|x6〉, H = 〈x|x6, x4〉.
(x6 = 1, x4 = 1⇒ x2 = 1. Or: C6 = 〈x〉, C3 = 〈x4〉, C6/C3).
Then: H ∼= C2.
Theorem 3.4 (Substitution Test). Suppose we are given a presentation G = 〈X|R〉,
a group H and a mapping θ : X → H. Then θ extends to a homomorphism
θ′′ : G → H iff (∀x ∈ X, ∀r ∈ R) the result of substituting xθ for x in r yields the
identity in H.
Proof. Consider the commutative diagram where ι1, ι2 are inclusions and ν : F (X)→
G is the natural map (wν→ Rw). Let θ′ : F (X)→ H extend θ.
R
ι2��
Xι1 //
θ
""EEEEEEEEE F (X)
θ′
�����
ν // Gθ′′
||yy
yy
yx � // x � // xR
H
(⇐): The substitution condition can be rephrased as R ⊆ ker θ′ E F (X). Now
ker ν = R ⊆ ker θ′ = ker θ′. Now apply (3.2) to get θ′′ : G → H. Observe that
θ = ι1θ′ = ι1νθ
′′ ⇒ θ′′ extends θ.
(⇒): For the converse the existence of such a θ′′ entails that R ⊆ R = ker ν ⊆
ker(νθ′′) = ker θ′.
Note: θ′′ is onto iff 〈Xθ〉 = H.
Example. G = 〈a, b, c|a2b2a−2b−1c−1b−1c〉,
H = 〈x, y, z|[x, y], [y, z], [z, x]〉
θ : {a, b, c} → H, aθ = x, bθ = y, cθ = z
29
Check: x2y2x−2y−1z−1y−1zgroup Abelian
= x2−2y2−1−1z−1+1 = 1.
⇒ θ extends to a homomorphism θ′′ : G→ H, aθ′′ = x, bθ′′ = y, cθ′′ = z.
Theorem 3.5. If G = 〈X|R〉 and H = 〈Y |S〉 then G × H = 〈X, Y |R, S, [X, Y ]〉
where [X, Y ] = {[x, y] : x ∈ X, y ∈ Y }
Proof. Let D = 〈X, Y |R, S, [X, Y ]〉. We must show that D ∼= G × H. The in-
clusions X ↪→ D, Y ↪→ D induce homomorphisms θ : G → D,φ : H → D
by substitution test. Define α : G × H → D by (g, h)α = gθhφ. Notice that
(x, 1)α = xθ1φ = x and (1.y)α = y. Moreover (g1, h1)(g2, h2)α = (g1g2h1h2)α =
(g1g2)θ(h1h2)φ = g2θg2θh1φh2φ = g1θh1φg2θh2φ (since [X, Y ] are relators in D)
= (g1, h1)α(g2, h2)α⇒ α ∈ Hom(G×H,D).
On the other hand the mapping of X ∪ Y into G×H sending x to (x, 1) and y to
(1, y) extends by the substitution test to a homomorphism β : D → G ×H. Since
[x, y] in D rewrites to [(x, 1), (1, y)] in G×H and (x, 1)(1, y) = (x, y) = (1, y)(x, 1).
Finally αβ : G × H → G × H sends (x, 1) to (x, 1)αβ = xβ = (x, 1) and sends
(1, y) to (1, y) and xβα = x, yβα = y. It follows that α and β are mutually inverse
isomorphisms.
Example. C3 = 〈x〉, C4 = 〈y〉.
C3 × C4 = 〈x, y|x3, y4, [x, y]〉
C2 × C5 × C10 = 〈x, y, z|x2, y5, z10, [x, y], [y, z], [z, x]〉
C2 × C2 × C2 = 〈a, b, c|a2, b2, c2, [a, b], [b, c], [c, a]〉
= 〈a, b, c|a2, b2, c2, (ab)2, (bc)2, (ca)2〉
Recall that the commutator subgroup or derived subgroup G′ or [G,G] of a given
group G is the subgroup of G generated by {[g, h] : g, h ∈ G}.
Remarks. (1) G′ E G
(2) Gab := G/G′ is Abelian.
(3) G/N is Abelian iff G′ ⊆ N , N normal subgroup.
(4) In General, G′ 6= {[g, h] : g, h ∈ G}. (But the equality often holds.)
30
Theorem 3.6. If G = 〈X|R〉 then Gab = 〈X|R, [X,X]〉 where [X,X] = {[x, y] :
x, y ∈ X}.
Proof. Let D = 〈X|R, [X,X]〉. We must prove that D ∼= Gab. By von Dyck there
exists an epimorphism θ : G→ D with ker θ = [X,X]. It remains to show that G′ =
[X,X]. Since the generators of D ∼= G/ker θ is Abelian it follows that G′ ⊆ ker θ.
On the other hand [X,X] ⊆ G⇒ [X,X] ⊆ G′ = G′ and so G′ = [X,X].
Example. 1. G = 〈x, y, z|x2yz−1yz−1〉
Gab = 〈x, y, z|[x, y], [y, z], [z, x], x2y2z−2〉.
2. G = 〈x, y|xyx−2yxy−2〉
Gab = 〈x, y|[x, y]〉 infinite
3. G = 〈x, y|x4y−2x−2y, y3x−1y−2x〉
Gab = 〈x, y|[x, y], x, y〉 trivial
4. G = 〈x, y, z, t|x2y2zy−2, x3t4x−2t−1, txyt−1x−5, yt2y−1t−1〉
Gab = 〈x, y, z, t|[X,X], x2z, xt3, x−4y, t〉
[= 〈x, y, z|[X,X], x2z, x, yx−4〉 = 〈y, z|[y, z], z, y〉 = {1}].
Since G � Gab it follows that if Gab is infinite G is also infinite. We describe the
so-called Tietze transformations which allow us to pass from one presentation of
a group G to another.
Lemma 3.7. Let F = 〈X| 〉, G =〉X|R〉 and suppose that w ∈ F and r ∈ R\R ⊆ F .
Let y be a symbol not in X. Then both of the inclusions
α1 : X → 〈X|R, r〉
β1 : X → 〈X, y|R, y−1w〉
extend to isomorphisms with domain G.
Proof. The fact that these mappings extend to homomorphisms with domain G is
immediate from the substitution test. Now observe that the maps α2 : X → 〈X|R〉
and β2 : X → 〈X|R〉 which fix elementwise and where yβ2 = w extend, again by
the substitution test, to homomorphisms.
31
Now α1α2 and β1β2 (where ˆ denotes extension) fix the generating set X, and so
both equal idG. Therefore αi, βi are isomorphisms.
Note 1.
We say that r ∈ R\R is a consequence of the relators R.
Note 2.
The four isomorphisms of 3.7 are the so-called Tietze transformations :
R+: adjoining a relator: 〈X|R〉 → 〈X|R, r〉 where r ∈ R\R
R−: removing a relator: 〈X|R〉 → 〈X|R− {r′}〉 where r′ ∈ R ∩R\{r′}
X+: adjoining a generator: 〈X|R〉 → 〈X, y|R, y−1w〉 where y /∈ X,w ∈ F .
X−: removing a generator: 〈X|R〉 → 〈X − {y}|R − {y−1w}〉 where y ∈ X,w ∈
F (X\{y}) and y−1w is the only relator of R that involves y.
Example.
R+ : 〈x, y, z|z−1xyz = y, y3 = 1〉
= 〈x, y, z|z−1xyz = y, y3 = 1, (xy)3 = 1〉
R− : 〈a, b, c|a4, a8, abc〉
= 〈a, b, c|a4, abc〉
X+ : 〈x, y, z|x2 = y2 = z4 = 1, (xyz)4 = 1〉
= 〈x, y, z, t|y2 = y2 = z4 = 1, (xyz)4 = 1, t = xyz2xyz〉
(or choose any other word in x, y, z)
X− : 〈x, y, z, t|t = z3x, y4 = 1〉
= 〈x, y, z|y4 = 1〉
32
In practice we do many transformations at the same time.
〈x, y|xl, ym, (xy)n〉
= 〈x, y, a|xl, ym, (xy)n, a−1xy〉
= 〈x, y, a|xl, ym, an, a−1xy〉
= 〈y, a|(ay−1)l, ym, an〉
= 〈y, a, b|(ay−1)l, ym, an, b−1y−1〉
= 〈a, b|(ab)l, bm, an〉
Example.
〈x, y, z|x = yzy−1, y = zxz−1, z = xyx−1〉
= 〈x, z|x = yxyx−1y−1, y = xyx−1xxy−1x−1〉
= 〈x, y|xyx = yxy〉
= 〈x, y, a|ax = ya, a = xy〉
= 〈x, a|ax = x−1a2〉
= 〈x, a, b|ax = x−1a2, b = ax〉
= 〈a, b|b2 = a3〉
Remark. Let G = 〈Y |S〉 and suppose that s ∈ S is of the form s1y±1s2 where y ∈ Y
and y is not included in either s1 or s2. Then y = (s−11 s−12 )±1, a word not involving
y. So G =⟨Y − {y}|S
⟩where S consists of all relators in S − {s} in which y is
replaced by (s−11 s−12 )±1 wherever it appears.
Example. 〈x, y, z, t|t3ytx−1z, z2yty, x3ytz2, x5〉
t3ytx−1z = 1⇒ y = t−3z−1xt−1
G = 〈x, z, t|z2(t−3z−1xt−1)t(t−3z−1xt−1), x3(t−3z−1xt−1), x5〉
Theorem 3.8. Given two presentations of the same group, one can be obtained
from the other by a finite sequence of Tietze transformations.
Proof. Suppose G = 〈x|R(x) = 1〉 = 〈Y |S(y) = 1〉. Let X = X(Y ) and Y = Y (X)
be two systems of equations expressing the X in terms of Y and the Y in terms of
33
X. We can do this since G = 〈X〉 = 〈Y 〉.
Now
〈X|R(X) = 1〉
X+ : = 〈X, Y |R(X) = 1, Y = Y (X)〉
R+ : = 〈X, Y |R(X) = 1, Y = Y (X), X = X(Y )〉
R+ : = 〈X, Y |R(X) = 1, Y = Y (X), X = X(Y ), R(X(Y )) = 1〉
R− : = 〈X, Y |Y = Y (X), X = X(Y ), R(X(Y )) = 1〉
R+ : = 〈X, Y |Y = Y (X), X = X(Y ), R(X(Y )) = 1, Y = Y (X(Y ))〉
R− : = 〈X, Y |X = X(Y ), R(X(Y )) = 1, Y = Y (X(Y ))〉
X− : = 〈Y |R(X(Y )) = 1, Y = Y (X(Y ))〉
R+ : = 〈Y |R(X(Y )) = 1, Y = Y (X(Y )), S(Y ) = 1〉
= 〈Y |S(Y ) = 1〉
Example. If A = 〈x|x12, x30〉 then |A| = 6 = gcd(12, 30) (If x12 = 1 then x24 = 1.
If x24 = x30 = 1 it follows that x6 = 1. ⇒ A = 〈x|x12, x30, x6〉 = 〈x|x6〉.
Example. Cm × Cn ∼= Cmn iff gcd(m,n) = 1.
(⇒): Let p be prime, p|m and p|n. Then Cm×Cn = 〈a〉×〈b〉 contains the subgroup⟨am/p
⟩×⟨bn/p
⟩ ∼= Cp × Cp a non-cyclic group (since every element has order p).
This contradicts our assumption that Cm × Cn is cyclic (all subgroups of a cyclic
group are cyclic).
(Cp = 〈x〉 , Cp = 〈y〉 ⇒ (x, y)m = (xm, ym) ⇒ |(x, y)| ≤ p < p2. If |G| = n and
g ∈ G then gn = 1 so here either ap or bp equals 1.)
34
(⇐): (m,n) = 1 ⇐⇒ ∃u, v ∈ Z s.t. um+ vn = 1.
Cmn = 〈a|amn = 1〉
= 〈a, x, y|amn = 1, x = am, y = an〉
=⟨a, x, y|amn = 1, x = am, y = an, [x, y] = 1, xn = 1, ym = 1, xuyv = aum+vn = a
⟩= 〈x, y|(xuyv)mn = 1, x = (xuyv)m, y = (xuyv)n, [x, y] = xn = ym = 1〉
= 〈x, y|(xn)um(ym)vn = 1, x = xumyvm, y = xunyvn, [x, y] = xn = ym = 1〉
= 〈x, y|x = xum, y = yvn, [x, y] = xn = ym = 1〉
But xn = 1⇒ xvn = 1⇒ x1−um = 1⇒ x = xum and ym = 1⇒ y = yvn.
So we get Cmn = 〈x, y|xn = ym = [x, y] = 1〉 = Cm × Cn.
Strategy:
abstract presentations oo subst. test // concrete groups
Example (The dihedral group). Let D2n denote the dihedral group of order 2n,
the 2n symmetries of the regular n-gon. This n rotations and n reflections. Let
H = 〈x, y|xn = y2 = 1, y−1xy = x−1〉. Define a mapping θ : {x, y} → D2n by xθ =
rotation of order n and yθ = any reflection. Then H = 〈xθ, yθ〉 and (xθ)n =
1, (yθ)2 = 1, and (yθ)−1(xθ)(yθ) = (xθ)−1. Therefore θ extends to a homomorphism
θ : H → D2n by the substitution test. Now θ is onto so it remains to show that θ is
injective.
Example (The quaternions). Let H denote the subgroup of GL(2,C) generated by
A =
ω 0
0 ω
, B =
0 1
−1 0
where ω = eiπ/n
Claim: H ∼= Q2n = 〈x, y|xn = y2, x2n = 1, y−1xy = x−1〉
Note that B /∈ 〈A〉 ⇒ |H| ≥ 4n
H = {I, A . . . A2n−1, B,BA . . . BA2n−1}.
Define θ : {x, y} → 〈A,B〉 by xθ = A, yθ = B.
Since An = B2 =
−1 0
0 −1
, A2n =
1 0
0 1
, A−1BA = B−1, we get θ : Q2n
onto� H
extending θ by the substitution test. It remains to show that |Q2n| ≤ 4n.
Now y−1xy = x−1 ∈ 〈x〉, yxy = x−1 ∈ 〈x〉 =⇒ 〈x〉 E Q2n.
35
But Q2n/〈x〉 = 〈y|y2 = 1〉.
=⇒ |Q2n| = |〈y|y2 = 1〉||〈x〉| ≤ 2 · 2n = 4n.
Note that Q2n = 〈x, y|yn = y2, y−1xy = x−1〉 since y−1xy = x−1 ⇒ (y−1xy)n =
x−n ⇒ y−1xny = x−n ⇒ xn = x−n.
Example (Groups of order 8). Let G be a group of order 8 and let x ∈ G have
maximal order so that |x| ∈ {2, 4, 8}.
If |x| = 8 then G ∼= C8.
If |x| = 2 then G ∼= C2 × C2 × C2.
If |x| = 4 let y ∈ G\〈x〉. Then y−1xy ∈ 〈x〉 E G so y−1xy ∈ {e, x, x2, x3}. Clearly
y−1xy /∈ {e, x2}. If y−1xy = x then G ∼= C2 × C4 so let y−1xy = x3. If y2 = 1 then
we get D8. If y2 6= 1 then y2 ∈ 〈x〉 (otherwise we could write down more than 8
distinct elements). This forces y2 = x2 and G ∼= Q4.
Thus the groups of order 8 are:
〈x|x8〉, 〈x, y|x4, y2, [x, y]〉, 〈x, y, z|x2, y2, z2, [x, y], [y, z], [z, x]〉,
D8 = 〈a, b|a4 = b2 = (ab)2 = 1〉, Q4 = 〈a, b|a2 = b2, b−1ab = a−1〉
Example (The Heisenberg group).
H is the group of matrices of the form
1 r s
0 1 t
0 0 1
, r, s, t ∈ Z.
Put A =
1 1 0
0 1 0
0 0 1
, B =
1 0 0
0 1 1
0 0 1
, C =
1 0 1
0 1 0
0 0 1
Then Ak =
1 k 0
0 1 0
0 0 1
, Bl =
1 0 0
0 1 l
0 0 1
, Cm =
1 0 m
0 1 0
0 0 1
and Ak ·Bl · Cm =
1 k m+ kl
0 1 l
0 0 1
.
=⇒ H = 〈A,B,C〉.
Let G = 〈a, b, c | [a, b] = c, [c, a] = 1, [c, b] = 1〉 and define θ : {a, b, c} → H by
36
aθ = A, bθ = B, cθ = C. Now [A,B] = C, [C,A] = [C,B] = 1. So θ extends to
θ ∈ Hom(G,H) by the substitution test. Clearly θ is surjective.
Now let L = {n = akblcm : k, l,m ∈ Z} ⊆ G. Observe that eG ∈ L (k = l = m = 0).
We claim that Lg ⊆ L (∀g ∈ G). It would then follow that L = G.
To prove the claim it is enough to consider a±1, b±1 and c±1 and show that each of
ua, ua−1, ub, ub−1, uc, uc−1
is in L.
Now
uc±1 = akblcm±1 ∈ L
ub±1 = akblcmb±1 = akbl±1cm ∈ L ([c, b] = 1)
Observe that
[a, b] = c⇔ a−1b−1ab = c
⇔ a−1b−1a = cb−1
⇔ a−1ba = bc−1
and so
a−1bla = (bc−1)l = blc−l
Similarly abla−1 = blcl.
Therefore
ua±1 = akblcma±1
= akbla±1cm ([c, a] = 1)
= aka±1blc±1cm ∈ L
But now distinct members of L are sent by θ to distinct matrices in H which means
θ is injective, so G ∼= H.
Note. akblcm is a normal form for G.
Symmetric groups.
37
Let Sn denote the symmetric group of degree n so that |Sn| = n! and Sn = 〈(i, i+1) :
1 ≤ i ≤ n〉.
Put
Gn = 〈x1, . . . , xn|R, S, T 〉
R = {x2i = 1 : 1 ≤ i ≤ n− 1}
S = {(xi xi+1)3 = 1 : 1 ≤ i ≤ n− 2}
T = {[xi, xj] = 1 : 1 ≤ i < j − 1 < n− 1}
Claim: Sn ∼= Gn.
Define θ : {x1, . . . , xn−1} → Sn by xiθ = (i i + 1). Then θ extends to a homomor-
phism θ : Gn → Sn by the substitution test and θ is injective so |Gn| ≥ |Sn| = n!.
We show by induction on n that |Gn| ≤ n! and so θ is an isomorphism.
Proceed by induction on n.
If n = 1 then Gn = {e} and |Gn−1| ≤ (n − 1)!. Let H be the subgroup of Gn
generated by x1, . . . , xn−2 and define
y0 = 1
yi = yn−1 . . . yn−i
Consider the subset A = {hyi : h ∈ H, 0 ≤ i ≤ n− 1} of Gn. Observe that H ≤ A
and eG ∈ A.
We show that hyixj ∈ A.
There are six possibilities:
(i) i = 0, j < n− 1 : hyixj = hxj ∈ H ⊆ A
(ii) i = 0, j = n− 1 : hyixj = hxn−1 = hyi ∈ A
(iii) i > 0, j > n− i : exercise
(iv) i > 0, j = n− i : hyixj = hxn−1 . . . xn−ixn−i(R)= hxn−1 . . . xn−i+1 = hyi−1 ∈ A
(v) i > 0, j = n− i− 1 : hyixj = hyi+1 ∈ A
(vi) i > 0, j < n− i− 1 : hyixj(T )= (hxj)yi = h′yi ∈ A.
38
It follows that A = Gn and |Gn| = |A| ≤ n × |H|. Now since the relations in
Gn that involve x1, . . . , xn−2 are precisely those in Gn−1 we get the homomorphism
φ : Gn−1 → Gn where xiφ = xi (1 ≤ i ≤ n− 2).
Now Im φ = H ⇒ (n− 1)! ≥ |Gn−1| (induction) ≥ |Im φ| = |H|.
⇒ n! = n(n− 1)! ≥ n · |H| ≥ |Gn|.
Example (The rationals 〈Q,+〉). If ab∈ Q then a
b= a(b−1)!
b=
1
b!+
1
b!+ . . .+
1
b!︸ ︷︷ ︸a(b−1)! times
Hence 〈Q,+〉 = 〈 1n!
: n ≥ 1〉
Put G = 〈tn(n ≥ 1)|tnn = tn−1 (n ≥ 2)〉.
Claim: 〈Q,+〉 ∼= G.
For convenience we write additively so that
G = 〈xn (n ≥ 1)|nxn = xn−1 (n ≥ 2)〉
Define θ : {xn : n ≥ 1} → Q by xnθ = 1n!
. By the substitution test θ extends to a
surjective homomorphism θ : G→ Q. We must check that θ is injective.
Let w ∈ G. Then w = a1x1 + a2x2 + . . . + aNxN where ai ∈ Z for some N ≥ 1. So
θ(w) =∑N
n=1ann!
.
We say that w has normal form if:
1) aN 6= 0
2) 0 ≤ an ≤ n− 1 for 2 ≤ n ≤ N
3) a1 is arbitrary.
Any word can be put into normal form. To do this we work from aN . Suppose that
we have got as far as k where N ≥ k ≥ 2.
Then ak = qk + a′k where 0 ≤ a′k < k. Then ak · xk = (qk + a′k)xk = qkxk + a′kxk =
qxk−1 + a′kxk and w = a′NxN + . . .+ a′kxk + (ak−1 + q)xk−1 + a1x1.
Suppose now that w,w′ are in normal form and θ(w) =∑N
n=1ann!
and θ(w′) =∑Mn=1
bnn!
. Suppose that aj = bj for all 1 ≤ j ≤ k− 1 but ak > bk. (If k = 1 then this
is strictly speaking the empty set.)
Thenakk!
+N∑
n=k+1
ann!
=bkk!
+M∑
n=k+1
bnn!
39
1
k!≤ ak − bk
k!≤ ak − bk
k!+
N∑n=k+1
ann!
=M∑
n=k+1
bnn!≤ 1
k!− 1
M !
(for the last step see below) which gives a contradiction. It follows that θ is injective.
Note. To prove∑M
n=k+1bnn!≤ 1
k!− 1
M !we fix k and proceed by induction on
M ≥ k + 1.
If M = k + 1 we get bk+1
(k+1)!on the left hand side. But bk+1 < k + 1 and this implies
bk+1
k+1≤ 1
k!− 1
(k+1)!.
Assume that is true for M − 1. Then
M∑n=k+1
bnn!
=M−1∑n=k+1
bnn!
+bMM !≤(
1
k!− 1
(M − 1)!
)+M − 1
M !=
1
k!− 1
M !
4 Finitely generated Abelian groups
Recall that if G = 〈X|R〉 and H = 〈Y |S〉 then
G×H = 〈X, Y |R, S, [X, Y ]〉
Gab = G/G′ = 〈X|R, [X,X]〉
Proposition 4.1. Let F = F (X) be free of rank r.
(i) Fab = 〈X|[X,X]〉
(ii) Fab ∼= Zr
(iii) Fab ∼= free Abelian group of rank r.
Proof. (i) is known and (ii) ⇒ (iii).
We prove (ii) by induction on r. If r = 1 then Fab = 〈x| 〉 = Z. Assume the claim
is true for 1 ≤ k < r. Then
Fab = 〈x1, . . . , xr|[xi, xj] (i 6= j)〉
= 〈x1, . . . , xr−1|[xi, xj] (i 6= j)〉 × 〈xr| 〉
= Zr−1 × Z = Zr
40
Notation. A = A(X) is free Abelian group on X. We will write additively.
Theorem 4.2. If X generates an Abelian group G then there exists an epimorphism
(that is: a surjective homomorphism) A(X) → G fixing X elementwise: Every
Abelian group is the homomorphis image of some free Abelian group.
Proof. If G = 〈X|R〉 is Abelian then G = Gab = 〈X|R, [X,X]〉. By von Dyck G is
a factor of A(X) = 〈X|[X,X]〉 by the normal closure of R.
Theorem 4.3 (Dedekind). If A = A(X) is free Abelian of rank r and B ≤ A then
B is free Abelian of rank ≤ r. (Think subspaces of vector spaces.)
Proof. If r = 1 then A ∼= Z and the result is known. Assume r > 1 and result true
for 1 ≤ k < r. Let X = {x1, . . . , xr} and define the subgroups H = 〈x1, . . . , xr−1〉
and C = 〈xr〉 of A. Then H is free Abelian of rank r − 1 and A ∼= H ⊕ C. By
induction B ∩H ≤ H is free Abelian on y1, . . . , ys, say, where s ≤ r − 1. Also
B/(B ∩H) ∼= (B +H)/H ≤ A/H ∼= C
(the product of Abelian groups is the same as the sum). So B/(B ∩ H) is either
trivial or infinite cyclic.
If trivial then B = B ∩H and result follows.
So assume B/(B ∩H) = 〈b+B ∩H〉 where b ∈ B\H.
Now b = h+ l · xr where h ∈ H, l ∈ Z\{0}.
We claim that B is free Abelian on {y1, . . . , ys, b} and so has rank s + 1 ≤ r, as
required. Clearly B = 〈Y 〉. Now suppose that∑s
i=1 kiyi +kb = 0 (ki, k ∈ Z). Then
k · l · xr = k(b− h) = −∑s
i=1 kiyi − kh ∈ H ⇒ k · l · xr ∈ H ∩ C = {0}.
Since l 6= 0 this forces k = 0 ⇒∑s
i=1 kiyi = 0 ⇒ ki = 0 (1 ≤ i ≤ s) since
{y1, . . . , ys} is a basis.
Thus every element of B is uniquely a Z-linear combination of the elements of Y ,
so Y is a free basis for B.
(This is equivalent to saying B ∼= Zs+1.)
Change of generators.
41
We know that a finitely generated Abelian group is of the form A(X)/B where A =
A(X) is free Abelian on X = {x1, . . . , xr} and B is free Abelian on Y = {y1, . . . , ys}
where s ≤ r. Let Y = Y (X) be the elements of Y written as words in X.
Then A(X)/B = 〈X|[X,X], Y (X)〉.
Suppose that {u1, . . . , un} is another set of generators for A. Then we get X = X(U)
and U = U(X), in which
(1) xi =∑n
j=1 pijuj for 1 ≤ i ≤ r
(2) uj =∑r
k=1 qjkxk for 1 ≤ j ≤ n
(pij, qjk ∈ Z). Substituting (2) into (1) yields
r∑j=1
pijqjk = δij =
1 i = j
0 i 6= j
by uniqueness.
So if P = [pij]r×n and Q = [qjk]n×r then PQ = Ir.
Substituting (1) into (2) yields QP = In.
If n ≥ r then n = rank (In) = rank (QP ) ≤ rank Q ≤ r. So n = r and Q = P−1.
Conversely any transformation of type (2) with [qjk] invertible over Z will yield a
new set of generators for A.
Now let B ≤ A. So B is free on Y = {y1, . . . , ys} where s ≤ r and we get
(3) yk =∑4
i=1mkixi for 1 ≤ k ≤ r
Thus B is determined by the matrix M = [mki]s×r relative to Y and X.
If we change to generators U of A instead of X then we must substitute (1) into (3).
In matrix terms we change from M to MP = MQ−1. If Y is changed to another
set V of free generators for B using an s × s-matrix T invertible over Z then B is
now determined by
TM relative to V,X
TMQ−1 relative to V, U
42
Theorem 4.4. The subgroup B = 〈Y 〉s of A = 〈X〉r is determined by the s × r
coefficient matrix M = [mki]s×r. Changing the free generators X and Y corresponds
to post- and pre-multiplication of M by matrices over Z. Conversely if T and Q are
invertible over Z, the coefficient matrix TMQ−1 determines the same subgroup of A
as does M .
Example.
G = 〈x, y, z, t|(xyz)6 = 1, t2 = (xz)2, (xy3zt2)2 = 1, (yt2)2 = x2z3, (xyz)4(yt)2 = 1〉
Gab = 〈x, y, z, t|[x, y], [x, z], [x, t], [y, z], [y, t], [z, t], x6y6z6 = 1, x2z2t−2 = 1,
x2y6z2t4 = 1, x−2z−3y2t4 = 1, x4y6z4t2 = 1〉
Gab∼= A(X)/B(Y )
where X = {x, y, z, t}, Y = {6x + 6y + 6z, 2x + 2z − 2t, 2x + 6y + 2z + 4t,−2x −
3z + 2y + 4t, 4x+ 6y + 4z + 2t}
M =
6 6 6 0
2 0 2 −2
2 6 2 3
−2 2 −3 4
4 6 4 2
d1 0 0 0
0 d2 0 0
0 0 d3 0
0 0 0 d4
0 0 0 0
Relation matrix: By performing row and column operations (which correspond to
pre- and post-multiplication by invertible matrices) we can reduce M to a canonical
form (Smith normal form) from which
(i) free generators for the subgroup B can be read off
(ii) A(X)/B(Y ) can be identified as a product of cyclic groups.
Row operations.
P: permuting rows
M: multiplying a row by ±1
A: adding an integer multiple of one row to another
(Column operations are similarly defined.)
43
We now describe an algorithm (see handout) for reducing any s × r matrix M
over Z to the canonical form D = diag(d1, . . . , dk) where k = min(r, s) and di is a
non-negative integer (1 ≤ i ≤ k) such that di|di+1 for 1 ≤ i ≤ k − 1.
Remarks. 1. The divisibility condition implies that any 1’s that occur amongst the
di occur at the beginning and any 0’s occur at the end.
2. d1 = hcf(entries of M)
3. Those di’s not equal to 0 or 1 are called the invariant factors or torsion coeffi-
cients.
4. The number of 0’s is the rank.
5. The uniqueness of the invariant factors and rank follows from linear algebra.
Returning to Example (*): See Handout (d1 = 1, d2 = 2, d3 = 6, d4 = 0).
Exercise. Reduce this matrix to Smith normal form:
132 68 68
78 76 40
78 112 40
Translating all this into group theory we obtain:
A(X)/B =A({x1, . . . , xr})〈d1x1, . . . , dkxk〉
=A({x1, . . . , xr})〈d1x1, . . . , dlxl〉
where l ≤ k, di > 0
=〈x1〉 × . . .× 〈xr〉〈d1x1〉 × . . .× 〈dlxl〉
= 〈x1|d1x1〉 × 〈x2|d2x2〉 × . . .× 〈xl|dlxl〉 × Zr−l
Returning to Example (*)
1 0 0 0
0 2 0 0
0 0 6 0
0 0 0 0
0 0 0 0
44
we get:
Gab = 〈x|x〉 × 〈y|2y〉 × 〈z|6z〉 × Z ∼= C2 × C6 × Z
rank = 1, invariant factors = {2, 6}.
Examples. (i)
1 0 0 0 0 0
0 2 0 0 0 0
0 0 16 0 0 0
0 0 0 16 0 0
C2 × C16 × C16 × Z2
〈x1, . . . , x6|x1, x22, x163 , x164 , [xi, xj]〉
(ii)
1 0 0 0 0
0 2 0 0 0
0 0 4 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
C2 × C4 × Z2
(iii) 3 0 0
0 3 0
0 0 9
C3 × C3 × C9
Theorem 4.5 (Basis theorem for finitely generated Abelian groups). Given a finitely
generated Abelian group G there are integers k,m ≥ 0 and integers di ≥ 2 (1 ≤ i ≤
k) such that di|di+1 for 1 ≤ i ≤ k − 1, and
G ∼= Cd1 × . . .× Cdk × Zm
45
Moreover each (d1, . . . , dk,m) uniquely determines G.
Example. C6 × C10∼= (C2 × C3)× C10
∼= C2 × (C3 × C10) ∼= C2 × C306 0
0 10
→6 6
0 10
→ 6 6
−12 −2
→−30 0
−12 −2
→2 0
0 30
Remark. Let P (n) denote the number of partitions of n ≥ 1. For example P (5) = 7:
5, 1 + 4, 2 + 3, 1 + 1 + 3, 1 + 2 + 2, 1 + 1 + 1 + 2, 1 + 1 + 1 + 1 + 1.
It follows from the theorem that the number of Abelian groups of order n (= pk11 ·
. . . · pkrk , pi distinct primes, ki ≥ 1) equals∏r
i=1 P (ki).
Example. How many Abelian groups are there of order n = 71906968240509600?
Answer: n = 253352112317, the number of groups is P (5) ·P (3) ·P (2) ·P (2) ·P (7) =
7 · 3 · 2 · 2 · 15 = 1260.
Example. List the Abelian groups of order 144.
Suppose that the relation matrix M is a square matrix. Then the row and column
operations only alter detM up to a factor of ±1.
Corollary. If G = 〈X|R〉 is a finite presentation and |X| = |R| then
|Gab| = ± detM
Corollary. If G = 〈X|R〉 is a finite presentation and |X| > |R| then G is an infinite
group.
Proof. Since |X| > |R| the relation matrix has more columns than rows. So the
normal form of M must contain at least |X| − |R| columns of zeroes. It follows that
Gab is infinite. But Gab is a quotient of G so G is infinite.
Example.
Cn = 〈x|xn = 1〉
Q2n = 〈x, y|xn = y2, y−1xy = x−1〉
G = 〈x, y, z|y−1xy = xa, z−1yz = ya, x−1zx = za〉
then |G| <∞ for a ≥ 3. Also: G cannot be 2-generated.
46
Example. List the Abelian groups of order 144.
Open problem: Is there a group G having a presentation G = 〈X|R〉 where
(1) |G| <∞,
(2) |X| = |R| = 4, and
(3) G can not be generated by fewer than 4 generators?
5 Group Extensions
The group G is said to act on the group A if for each g ∈ G, a ∈ A there exists a
unique element ag ∈ A such that
aeA = a (∀a ∈ A)
(ag1)g2 = ag1g2 (∀a ∈ A, g1, g2 ∈ G)
(a1a2)g = ag1a
g2 (∀a1, a2 ∈ A, g ∈ G)
Example. If A ⊆ G and we know g−1ag ∈ A (∀a ∈ A, g ∈ G) then
ag := g−1ag
is an action of G on A (conjugation).
aeG = e−1G aeG = a
(ag1)g2 = (g−11 ag1)g2 = g−12 (g−11 ag1)g2 = ag1g2
(a1a2)g = g−1a1a2g = g−1a1gg
−1a2g = ag1ag2
Definition. An automorphism of A is an isomorphism A→ A. Aut(A) is the group
of all automorphisms of A with composition of maps.
Theorem 5.1. Let G act on A (as groups). Then for each g ∈ G there corresponds
a mapping φg : A→ A defined by φg(a) = ag (∀a ∈ A), and moreover φg ∈ Aut(A).
The mapping φ : G → Aut(A) defined by gφ = φg (∀g ∈ G) is a homomorphism
and we call φ the action.
Conversely let φ ∈ Hom(G,Aut(A)). Then G acts on A (with action φ) if we define
ag := a(gφ) ∈ A.
47
Proof. We prove only the last statement.
aeG = a(eGφ) = a(eAut(A)) = a(iddA) = a
(a1a2)g = (a1a2)(gφ)
= a1(gφ)a2(gφ) (since gφ ∈ Hom(A,A))
= ag1 ag2
ag1g2 = a((g1g2)φ)
= a((g1φ)(g2φ))
= (a(g1φ))(g2φ)
= (ag1)g2
Suppose G acts on A (with action φ). Consider K = G × A (Cartesian product)
with the binary operation
(x, a)(y, b) = (xy, ayb) (∀x, y ∈ G)(∀a, b ∈ A).
(Note: ay = a(yφ).)
Then we get a group K called the semi-direct product of G on A; and we write
K = G×φ A or A ��φG
For example
(x, a)(x−1, (a−1)x−1
) = (xx−1, ax−1
(a−1)x−1
) = (eG, (aa−1)x
−1
)
= (eG, ex−1
a ) = (eG, ea)
So (x, a)−1 = (x−1, (a−1)x−1
).
Note. eyA = eA(yφ) = eA.
Note. Direct product of G,A is obtained when ay = a (∀a ∈ A,∀y ∈ G), that is,
a(yφ) = a (∀a ∈ A, ∀y ∈ G), that is, yφ = idA (∀y ∈ G), that is, φ : G→ Aut(A)
is defined by yφ = idA (∀y ∈ G).
48
The mappings
A→ K a 7→ (eG, a)
G→ K g 7→ (g, eA)
give injective homomorphisms. (Check!)
It is customary to identify A,G with their isomorphic images in K, that is, re-
gard them as subgroups of K.
Observe that
(x, b)−1(eG, a)(x, b) = (x−1, (b−1)x−1
)(x, axb)
= (eG, ((b−1)x
−1
)xaxb) = (eG, b−1axb) ∈ A
Thus: A E K,G ≤ K,A ∩G = {(eG, eA)} and K/A ∼= G (∗)
Pictorially:
1 // A // K // G // 1
Definition. An extension of a group G by a group A is a group G having a normal
subgroup N such that A ∼= N and G/N ∼= G.
Example. If θ : G→ G is an epimorphism then G is an extension of G by ker θ.
Example. K = A ��φG is an extension of G by A.
Pictorially:
1 // Aβ1
// Gβ2
// G // 1
where ker β1 = {eA}, Im β1 = ker β2, Im β2 = G.
Definition. A sequence of groups Ai and homomorphisms αj
A0α0 // A1
α1 // A2α2 // . . . αn−1
// An
49
is called exact if Im (αi−1) = ker(αi) for each i.
A short exact sequence is an exact sequence where n = 4 and A0 = A4 = 1.
Therefore group extensions ≡ short exact sequences.
1 // A1 inj.// A2 surj.
// A3// 1
A diagram is a directed graph whose edges are homomorphisms between the end
points and whose vertices are groups.
A diagram is called commutative if given any two vertices and any two paths between
them the corresponding composite homomorphisms coincide.
Lemma 5.2 (The Five Lemma).
Let
A0
φ0��
α0 // A1
φ1��
α1 // A2
φ2��
α2 // A3
φ3��
α3 // A4
φ4��
B0 β0// B1 β1
// B2 β2// B3 β3
// B4
be a commutative diagram with exact rows. If φ0, φ1, φ3 and φ4 are isomorphisms
then so is φ2.
Proof.
(φ2 injective)
Let a ∈ kerφ2. We show that a = 0.
Then aα2φ3 = aφ2β2 = eβ2 = e
⇒ aα2 ∈ kerφ3 = e (φ3 injective)
⇒ a ∈ kerα2 = Im α1 (exact)
⇒ a = a1α1 for some a1 ∈ A1
⇒ e = aφ2 = a1α1φ2 = a1φ1β1
⇒ a1φ1 ∈ ker β1 = Im β0(exact)
⇒ a1φ1 = b0β0 for some b0 ∈ B0
Since φ0 is surjective ∃a0 ∈ A0 s.t. a0φ0 = b0
⇒ a1φ1 = a0φ0β0 = a0α0b1
⇒ a1 = a0α0 (φ1 injective)
50
⇒ a1 ∈ Im α0 = kerα1 (exact)
⇒ a = a1α1 = e
⇒ kerφ2 = {e}.
(φ2 surjective)
Exercise.
Suppose now that we are given an extension
1 // Aι // G
ν // G // 1 exact
and presentations G = 〈X|R〉 and A = 〈Y |S〉. Our aim is to obtain a presentation
for G.
Let Y = {yι = y : y ∈ Y } ⊆ G and let S = {s : s ∈ S} ⊆ G where s is obtained from
s ∈ S by replacing each occurence of y in s by y. Now let X = {x : x ∈ X±1} ⊆ G
be members of a transversal for the image of ι in G such that xν = x.
Note.
G ∼= G/ ker ν = G/Im ι
and
x1ν = x2ν iff x−11 x2 ∈ ker ν
iff x−11 x2 ∈ Im ι
iff x1Im ι = x2Im ι.
For each r ∈ R let r be the word in X obtained from r by replacing each x with x.
Now r ∈ ker ν = Im ι so each r can be written as a word vr in the y.
Put R = {rv−1r : r ∈ R} ⊆ G. Finally Im ι E G and so each conjugate x−1yx, xyx−1 ∈
Im ι and so is a word wx,y, wx−1,y in Y . Put T = {x−1yxw−1x,y : x ∈ X±1, y ∈ Y } ⊆ G.
Theorem 5.3.
G = 〈X, Y |R, S, T 〉
Proof. Let D = 〈X, Y |R, S, T 〉. We must show that G ∼= D.
Define θ : X ∪ Y → G by
xθ = x yθ = y
51
Then θ extends to a homomorphism θ : D → G. The restriction of θ to 〈Y 〉 ≤ D
gives rise to the composite homomorphism θ1 : 〈Y 〉 → Im ι ∼= A where yθ1 = y.
Since all the defining relators S of A with each y replaced by y yield the identity
in 〈Y 〉 then using the substitution test the mapping Y → 〈Y 〉, y 7→ y extends to a
homomorphism A→ 〈Y 〉 inverse to θ1. So θ1 is an isomorphism.
The presence of T in the presentation for D implies that 〈Y 〉 E D. So we have the
commutative diagram
Dθ //
nu1""EEEEEEEEE G
ν // G
D/〈Y 〉
θ2
<<zz
zz
z
where Im ν1 = D/〈Y 〉, ker ν1 = 〈Y 〉 and since 〈Y 〉θν ⊆ (Im ι)ν = (ker ν)ν = eG
⇒ ker ν1 = 〈Y 〉 ⊆ ker(θν) and so we can apply Lemma 3.2 to obtain
θ2 : D/〈Y 〉 → G
x〈Y 〉θ2 = xν1θ2 = xθν = xν = x
But the relators R of G with each x replaced by x〈Y 〉 all held in D/〈Y 〉 so applying
the substitution test we obtain a homomorphism G → D/〈Y 〉, x 7→ x〈Y 〉, inverse
to θ2. So θ2 is an isomorphism.
We now have:
1 // A // G // G // 1
1
θ0
OO
// 〈Y 〉
θ1
OO
// D
θ
OO
// D/〈Y 〉
θ2
OO
// 1
θ3
OO
where the rows are exact and each θi is an isomorphism.
Corollary. Let G = 〈X|R〉 and A = 〈Y |S〉. Let α : G → Aut(A) be a homomor-
phism such that
y(xα) = wx,y ∈ 〈Y 〉 = A (∀x ∈ X, ∀y ∈ Y )
Then the semi-direct product A ��αG has a presentation
A ��αG = 〈X, Y |R, S, x−1yxw−1x,y (x ∈ X, y ∈ Y )
52
Proof. Apply 5.3 to the extension 1 // A // A ��αG // G // 1 . Since G ≤
A ��αG it follows that all the vr are trivial. Remove the tildas to get result.
Also note that we use X instead of X±1 for x−1yxw−1x,y. Since xalpha ∈ Aut(A) it
follows that {wx,y : y ∈ Y } generates A.
Suppose we know that x−1yx = wx,y then xwx,yx−1 = y ∈ A. So y = wn1
x,y1. . . wnk
x,yk
and xyx−1 = (xwx,y1x−1)n1(xwx,y2x
−1)n2 . . . (xwx,yk)nk = y1 . . . yk ∈ A = 〈Y 〉
Example.
A = Cn = 〈y|yn = 1〉, G = Cm = 〈x|xm = 1〉
When does G act on A?
We want G to act with action α, say, α : G→ Aut(A), where
(∗) y(xα) = yl
for some l ∈ {1, . . . , n− 1}.
Since xα ∈ Aut(A) we must have 〈yl〉 = A⇔ hcf(l, n) = 1.
Also since (xα)m = xmα = eGα = eAut(A) = idA we have y(xα)m = y ⇔ ylm
= y
⇔ lm ≡ 1 mod n (∗∗) (⇒ hcf(l, n) = 1).
Subject to (∗∗), (∗) completely determines α since
yr(xsα) = (y(xα)s)r = yr·ls
By corollary 5.4
A ��αG = 〈x, y|xm, yn, x−1yx = yl〉
Example. A = C5 = 〈y|y5 = 1〉, G = 〈x|x4 = 1〉, l ∈ {1, 2, 3, 4},
m = 4, n = 5, l4 ≡ 1 mod 5.
In fact this is true for all values of l.
〈x, y|x4, y5, x−1yx = y〉 = C4 × C5∼= C20
〈x, y|x4, y5, x−1yx = y2〉
〈x, y|x4, y5, x−1y = y3〉
〈x, y|x4, y5, x−1y = y4〉.
53
Example. A = 〈y|y6 = 1〉, G = 〈x|x2 = 1〉, l ∈ {1, 2, 3, 4, 5},
m = 2, n = 6, l2 = 1 mod 6
12 ≡ 1, 22 ≡ 4, 32 ≡ 3, 42 ≡ 4, 52 ≡ 1
〈x, y|x2, y6, x−1yx = y〉 = C2 × C6
〈x, y|x2, y6, x−1yx = y−1〉 = D12
54
