Download 26 Generators and Relations

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Generators and Relations
Unless otherwise stated all groups will be nonabelian. When we describe a
group by “generators and relations” we are writing it as the quotient of the
free group F on a set of generators X by the normal subgroup R generated
by a set of relations ∆. In combinatorial group theory we keep track of this
particular choice of generators and relations. We begin with the definition
of a free group.
Definition 26.1. If X is a subset of a group F we say that F is a free group
with basis X if any set mapping φ : X → G from X to any group G extends
uniquely to a homomorphism φ : F → G (i.e., so that φ|X = φ).
This is a universal property so F is determined uniquely up to isomorphism by the set X provided that F exists. This definition has the advantage
that it is easy to state. Its disadvantage is that it does not show that free
groups exist. This definition also says that the free group functor is left
adjoint to the forgetful functor Ab → Sets. This matches our previous
characterization of free abelian groups so it easily leads us to the following
theorem.
Theorem 26.2. Let F be a free group with basis X and consider the quotient
map F → F/F 0 where F 0 E F is the commutator subgroup of F .
1. The elements of X map to distinct nontrivial elements of F/F 0 .
2. F/F 0 is the free abelian group generated by the image X of X in F/F 0 .
Proof. We will show that F/F 0 satisfies the universal property of free abelian
groups. Suppose that A is an additive group and φ : X → A is any set
function. Then φ extends uniquely to a homomorphism φ : F → A. Since A
is abelian, F 0 ≤ ker φ so we get an induced homomorphism ψ : F/F 0 → A.
This shows that F/F 0 is isomorphic to the free abelian group Z[X]. If we
let A = Z[X] then ψ : F/F 0 → Z[X] will be an isomorphism which sends
the image x ∈ F/F 0 of each x ∈ X to x ∈ Z[X]. This proves both (1) and
(2).
Construction of free groups
In order to show that every set X generates a free group we take the usual
description of the free group as the set of “reduced words” in the symbols xi
−1
and x−1
for x ∈ X (which we could formally
i where xi ∈ X. The symbols x
define by x−1 := (x, X)) form a set X −1 disjoint from X.
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Definition 26.3. A word on X is a sequence
` −1 ` w = (a1 , a2 , a3 , · · · ) where
each letter ai is an element of the set X X
{1} which has the property
`
`
Each word is a set functions w : N → X X −1 {1} where N is the set of positive
integers. (But not all such functions are words.)
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that
` for some (unique) integer n ≥ 0, the first n letters a1 , a2 , · · · , an lie in
X X −1 and all other letters are equal to 1. The integer n is called the
length of the word w.
The infinite string of 1’s at the end of a word are never shown. Thus an
arbitrary word of length n is written:
w = x²11 x²22 · · · x²nn
where xi is a sequence of elements of X, ²i = ±1 and x1 = x. There is always
a unique word of length 0 called the empty word. Another example is
w7 = abcc−1 b−1 a.
This is a words of length 7 in X = {a, b, c}. It is not a “reduced” word since
it contains cc−1 as a “subword.”
A subword of a word w is a sequence of consecutive letters in w. The
empty word is a subword of every word. Subwords can also be described in
terms of products. If w, u are words of length n, m then their product w · u is
the word of length n + m given by taking the n letters of w followed by the
m letters of u. The word u is a subword of w iff there are two other words
a, b so that w = a · u · b.
Definition 26.4. A word is called reduced if it does not contain two consecutive inverse letters, i.e., if it does not contain the subwords xx−1 or x−1 x
for any x ∈ X.
We want to show that the set of reduced words on X forms a group.
Multiplication on this set should be given by multiplying the words and
cancelling adjacent inverse letters. For example the reduced words u = abc
and v = c−1 b−1 a give u · v = w7 as above which can be reduced to the
two letter word a2 = aa. However, cancellation of adjacent inverse letters
is not a unique process. For example, the letters in w4 = aa−1 aa−1 can be
cancelled in two different ways. To make sure the result is unique we make
the following definition.
Definition 26.5. The reduced product of two reduced words w, u is defined
by wu = w0 · v 0 where w0 , v 0 are subwords of w, v so that
w = w0 · u,
v = u−1 · v 0
where u is a word of maximal length which occurs at the end of w so that its
inverse u−1 (given by taking the letters of u in reverse order with opposite
signs) occurs at the beginning of v.
Theorem 26.6. The set of reduced words in X forms a group under the
reduced product and this group is a free group with basis X.
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Proof. (van der Waerden) Instead of proving that reduced words form a
group, we first construct a group F (X) and then show that the elements of
F (X) are in 1-1 correspondence with the set W0 (X) of reduced words on X.
The group F (X) will be`a subgroup of the group of permutations of the
set W0 (X). For each x ∈ X X −1 let |x| : W0 (X) → W0 (X) be the mapping
defined as follows.
(
(a2 , · · · , an )
if a1 = x−1 ,
|x|(a1 , a2 , · · · , an ) =
(x, a1 , a2 , · · · , an ) otherwise.
Then we can easily show that |x| ◦ |x−1 | is the identity (since a1 , a2 are not
allowed to be inverse`
to each other). Consequently, |x| is a permutation of
W0 (X) for all x ∈ X X −1 . Let F (X) be the subgroup of the permutation
group of W0 (X) generated by all such permutations |x|.
A bijection ψ : F (X) → W0 (X) is given by ψ(f ) = f (∅). The inverse
φ : W0 (X) → F (X) is given by
φ(a1 a2 · · · an ) = |a1 | ◦ |a2 | ◦ · · · ◦ |an |.
(Clearly, ψ ◦ φ is the identity on W0 (X). To show that φ ◦ ψ is the identity
on F (X) note that:
1. The set of all f ∈ F (X) so that φψ(f ) = f forms a subgroup of F (X).
2. φψ(|x|) = |x|. Therefore this subgroup contains each generator |x| of
F (X).
Now we want to show (rigorously) that composition in F (X) corresponds
to reduced product in W0 (X), in other words that φ(wu) = φ(w) ◦ φ(v). This
is true from the definition of φ in the case when w · v is already reduced (so
that wu = w · v):
φ(a1 a2 · · · an b1 b2 · · · bm ) = |a1 |◦· · ·◦|an |◦|b1 |◦· · ·◦|bm | = φ(a1 · · · an )◦φ(b1 · · · bm ).
But this special case implies the general case using the fact that φ(u−1 ) =
φ(u)−1 :
φ(wu) = φ(w0 v 0 ) = φ(w0 )φ(v 0 ) = φ(w0 )(φ(u−1 )φ(u))φ(v 0 )
= (φ(w0 )φ(u−1 ))(φ(u)φ(v 0 )) = φ(w0 u−1 )φ(uv) = φ(w)φ(u)
Finally, we prove that F (X) ∼
= W0 (X) is the free group with basis X.
Let G be any group and let f : X → G be any set mapping. Then we define
f : W0 (X) → G by
f (a1 a2 · · · an ) = f (a1 ) ◦ · · · ◦ f (an ).
Since any homomorphism W0 (X) → G extending f must satisfy this formula,
the uniqueness is automatic. It suffices to show that f is a homomorphism,
i.e., that f (wv) = f (w) ◦ f (v). This is evidently true in the case when w · v is
reduced. Imitating the above argument we see that this special case implies
the general case.
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Relations
When we say that the group G has a presentation
G = hX|∆i
we mean that ∆ is a set of reduced words in X and G = F/R where R is
the normal subgroup of F generated by ∆. In other words, R is the set of
all products of conjugates of elements of ∆ ∪ ∆−1 .
We have the obvious fact:
Proposition 26.7. If H is another group then a homomorphism φ : G =
hX|∆i → H is given unique by any choice of φ(x) ∈ H for each x ∈ X with
the property that φ(r) = 1 for all r ∈ ∆.
There are some difficulties with presentations of groups given by the fact
that the word problem is unsolvable in general, i.e., given a word w in X there
is no algorithm to determine whether or not w can be written as a product
of conjugates of elements of ∆ ∪ ∆−1 (i.e., that it represents a nontrivial
element of G). A fortiori, this implies that the conjugacy problem is also
unsolvable, i.e., there is no algorithm to determine whether two given words
in X represent conjugate elements of G. However, every word in X does
represent a well-defined element of G and any algorithm which uses a solution
of the word problem as a subroutine will be a procedure with a well-defined
output.
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