Patrones de las respuestas fenotípicas

Let X = {x1, . . . , xr} be a set of elements, called the alphabets, where

the cardinality r may not be finite. Let X−1 _{= {x}−1

1 , . . . , x−r1} be

another set with X ∩X−1 _{= ∅. A sequence w = a}

1a2· · ·an, ai ∈

X ∪X−1_{, is called a word in X of length n. The empty set is also} viewed as a word, called the empty word. The set of all words in X of finite length is denoted byW. Letw1 =a1a2· · ·anandw2 =b1b2· · ·bm

be in W. We define their product w1w2 = a1a2· · ·anb1b2· · ·bm ∈ W.

Thus W with this multiplication is a monoid, the empty word is the identity element of W.

Free groups, generators and relations 45

Two wordsw1 andw2(alsow2andw1) are calledadjacentifw1 =uv andw2 =uxix−i 1v oruxi−1xiv, whereu, v ∈W andxi ∈X. Two words

w1 and w2 are called equivalent, denoted by w1 ∼ w2, if there is a sequence of words w1 = u1, u2, . . . , un−1, un = w2, such that ui and

ui+1 are adjacent words for i = 1,2, . . . , n−1. It is easy to see that “∼” is an equivalence relation, and if w1 ∼ w10 and w2 ∼ w20, then

w1w2 ∼ w10w02. We use [w] to denote the equivalence class containing

w. Let F be the set of all equivalence classes. Define a multiplication of the equivalence classes by

[w1][w2] = [w1w2],

then the multiplication is well defined, and F is a group with respect to this multiplication, called the free group generated by X.

A free group F is uniquely determined by the free generator set X, and it is uniquely determined by the cardinality r of X up to isomor- phism. We callr the rankof the free group F. From now on we use Fr

to denote the free group of rank r.

Theorem 1.7.1 Every group withr generators is isomorphic to a fac- tor group of Fr.

Proof: Let G = ha1, . . . , ari. Consider the free group Fr of rank

r with free generator set {x1, . . . , xr}. Define a map η : [xi] 7→ ai,

i = 1,2, . . . , r, and extend it to Fr. It is easy to see that η is an

epimorphism and hence

G∼=Fr/Ker η.

¤

Let again G = ha1, . . . , ari and let Fr = hx1, . . . , xri be the free

group of rank r. By Theorem 1.7.1, G ∼=Fr/K, where K is a normal

subgroup of Fr. Consider a word

f(x1, . . . , xr) =xni11· · ·x

ns

is ∈K, i1, . . . , is∈ {1, . . . , r}

inK. Then we have a word

f(a1, . . . , ar) =ani11· · ·a

ns

in G. We call the equation f(a1, . . . , ar) = 1 or the word f(x1, . . . , xr)

inFr which belongs toK a relation inG.

A set R = {fi(a1, . . . , ar) = 1 | i ∈ I} of relations in G is called

a set of defining relations of G if the normal closure of a set M =

{fi(x1, . . . , xr) ∈ Fr | i ∈ I} in Fr is the group K, i.e., MFr = K. In

this case, we write

G=ha1, . . . , ar |fi(a1, . . . , ar) = 1, i∈Ii,

or

G=ha1, . . . , ar |fi(a1, . . . , ar), i∈Ii,

and call it a presentation of a group G. If r is finite G is said to be finitely generated, and bothr and the index setI are finiteG is said to be finitely presented.

Now we give some examples to show how to determine a finite group by generators and defining relations.

Example 1.7.1 Let G = ha, bi with defining relations a2 _{= b}2 ₌ (ab)n _{= 1. Then G}∼_=D

2n.

Solution: Since G=ha, bi=hab, bi and

b−1(ab)b =b−1a−1 = (ab)−1 ∈ habi,

we have habiEG. HenceG=hab, bi=habi · hbi, and every element in

Ghas the form (ab)i_bj _{wherei= 0,1, . . . , n−1;j = 0,1. It follows that}

|G| ≤ 2n. On the other hand, the dihedral group D2n = hx, y | xn =

y2 _{= 1, y}−1_xy=x−1_{i. As another generating set of D}

2n,{xy−1, y} has

relations (xy−1₎2 _=y2 _{= (xy}−1 _·y)n _{= 1; henceD}

2n is a homomorphic

image of G. However, |D2n| ≤ |G|, implying that |G| = 2n and G ∼=

D2n. (See Example 1.3.2.) ¤

Example 1.7.2 Let G = ha, bi with defining relations a3 _{= b}2 ₌ (ab)3 _{= 1. Then G}∼_=A

Free groups, generators and relations 47

Solution: By the relationsa3 _{= 1 andb}2 _{= 1, every element of Gcan} be expressed as either ai_{, or a}i_ba±1_{b· · ·a}±1_baj_{, where i, j = 0,±1. By}

(ab)3 _{= 1 again, ababab = 1, implying bab = (aba)}−1 _{= a}−1_ba−1_{, and}

aba= (bab)−1 _=ba−1_{b. It follows that ba}±1_{b =a}∓1_ba∓1_{. By using this} relation, we may transform the elementai_ba±1_{b· · ·a}±1_baj _{into the form}

ai_baj _{with only one b in the product. Therefore}

G={ai, aibaj |i, j = 0,±1}.

It follows that |G| ≤ 3 + 3 × 3 = 12. On the other hand, in the alternating group A4, letting x = (123) and y = (12)(34), we have

xy = (243), hence we have x3 _{= y}2 _{= (xy)}3 _{= 1. Obviously, A} 4 =

hx, yi, implying thatA4 is a homomorphic image of G. Since |A4|= 12 and |G| ≤12, it forces that |G|= 12 and G∼=A4. ¤

As shown in Example 1.7.1, a group can have more than two dif- ferent presentations. Hence, one can ask naturally whether any two groups defined by the presentations are isomorphic or not. To answer it, we first introduce the Tietze transformations.

Definition 1.7.1 For a givenG=ha1, . . . , am |f1(ai), . . . , fn(ai)i, the

Tietze transformations R and G are defined as follows:

R: If f1(ai) = · · ·=fn(ai) = 1 impliesf(ai) = 1, then

ha1, . . . , am |f1(ai), . . . , fn(ai)i

R ↓ ↑ R−1

ha1, . . . , am |f1(ai), . . . , fn(ai), f(ai)i

G: Ifais a word on a1, . . . , am, saya=w(ai) = w(a1,. . . , am), then

ha1, . . . , am |f1(ai), . . . , fn(ai)i

G ↓ ↑ G−1

ha1, . . . , am, a |f1(ai), . . . , fn(ai), a=w(ai)i

Here, R−1 _{and G}−1 _{are the inverse Tietze transformations of Rand G,} respectively.

Theorem 1.7.2 (Tietze) Any two finite presentations

hai |fj(ai)i=ha1, . . . , am |f1(ai), . . . , fn(ai)i,

ha0

i |fk0(a0i)i=ha01, . . . , a0p |f10(a0i), . . . , fq0(a0i)i

of a group G can be convertible into each other by a finite sequence of Tietze transformations. Proof: hai |fj(ai)i−→ haR i |fj(ai), fk0(wi0(ai))i, wherea0 i =w0i(a1, . . . , am) is a word on ai G −→ hai, a0i |fj(ai), fk0(wi0(ai)), a0i =w0i(ai)i R −→ hai, a0i |fj(ai), fk0(a0i), fk0(w0i(ai)), ai0 =w0i(ai)i R−1 −→ hai, a0i |fj(ai), fk0(a0i), a0i =wi0(ai)i R −→ hai, a0i |fj(ai), fk0(a0i), a0i =wi0(ai), ai =wi(a0i)i, where ai =wi(a01, . . . , a0p) is a word on a0i

−→ (by the reverse order) ... −→ ha0 i |fk0(a0i)i. _¤ Example 1.7.3 Show Zpq ∼=Zp×Zq if (p, q) = 1. Solution: Clearly, Zpq =ha| apq = 1i ; Zp×Zq =hx, y |xp, yq, xy = yxi. ha|apq _{= 1i by setting x=a}q_{, y =a}p −→ ha, x, y |apq_{, x=a}q_{, y =a}p_i

(p, q) = 1⇒αp+βq= 1 for some α, β,and thena1 _=aαp_aβq _=yα_xβ

−→ ha, x, y |(yα_xβ₎pq_{, x= (y}α_xβ₎q_{, y = (y}α_xβ₎p_{, a=y}α_xβ_{, x}p_{, y}q_{, xy =yxi}

−→ hx, y |(yα_xβ₎pq_{, x= (y}α_xβ₎q_{, y = (y}α_xβ₎p _xp_{, y}q_{, xy =yxi}

because the first three relations come from the last three relations

−→ hx, y |xp_{, y}q_{, xy =yxi.}

Graphs of groups and Groups of graphs 49

Example 1.7.4 Show ha, b|aabb = 1i ∼=hc, d|cdcd−1 _{= 1i.}

Solution: ha, b|aabb= 1i

−→ ha, b, x |x=ab, axb= 1i −→ ha, b, x |x=ab, a=b−1_x−1_i

−→ hb, x |x=b−1_x−1_bi

−→ hc, d|cdcd−1 _{= 1i, by setting c=x}−1_{, d=b}−1_{. ¤}

Exercises

1.7.1. Let G = ha, bi with defining relations a3 _{= b}3 _{= (ab)}2 _{= 1. Then} G∼=A4.

1.7.2. Let G = ha, bi with defining relations a4 _{= b}2 _{= (ab)}3 _{= 1. Then} G∼=S4.

1.7.3. Let G = ha, bi with defining relations a5 _{= b}2 _{= (ab)}3 _{= 1. Then} G∼=A5.

1.7.4. LetG=ha1, a2, . . . , an−1i with defining relationsa2i = 1, (aiai+1)3= 1,(a_ia_j)2 _{= 1, wherei, j= 1,2, . . . , n−1 andj−i >1. ThenG}∼_=Sn. 1.7.5. Let G=ha3, a4, . . . , ani with defining relationsa3i = (aiaj)2 = 1, for

3≤i, j≤n andi6=j. ThenG∼=An.

1.7.6. Show D_2n ∼= ha, b | a2 _{= b}n _{= 1, aba = b}−1_{i ∼= ha, b | a}2 _{= b}n ₌

(ab)2 _{= 1i.}

1.7.7. Showha, b|an_=bm _{= (ab)}` <sub>= 1i ∼=hc, d|c</sub>` _=dm_{= (cd)}n_{= 1i.}