10. EL OFICIO DE MOLINERO.
10.2. CONSIDERACION SOCIAL DEL MOLINERO.
The knots GKa,b and SKa,b are analogues of square and granny knots built from (a, b)-torus knots, whereaand bare co-prime. We defineGKa,b to be the connect sum of two right-handed (a, b)-torus knots or two left-handed (a, b)-torus knots, and SKa,b to be the connect sum of a right-handed (a, b)-torus knot and a left-handed (a, b)-torus knot. Recall from Lemma 3.1 that the meridian of Ga,b is equal to xcy−d, where c and d ∈ N are a solution to bc−ad = 1. We can write the presentations for π1(GKa,b) andπ1(SKa,b) as follows:
π1(GKa,b)∼=hx, y, w, z|xa=yb, wa=zb, xcy−d =wcz−di, π1(SKa,b)∼=hx, y, w, z|xa=yb, wa=zb, xcy−d =wcz−di, where both have the same meridian xcy−d = wcz−d, but different longitudes λ
GKa,b = x
awa and λSKa,b =x
aw−a. As we can see both of GK
a,b and SKa,b have the same knot group. The presentation for Gn(GKa,b) and Gn(SKa,b) can be obtained in exactly the same way as those for Gn(GK) and Gn(SK):
Gn(GKa,b)∼=hx, y, w, z, ν|xa=yb, wa=zb, νn=xcy−d =wcz−d, xawaν =νxawai, Gn(SKa,b)∼=hx, y, w, z, ν|xa=yb, wa=zb, νn=xcy−d =wcz−d, xaw−aν =νxaw−ai. By Nelson and Neumann [21], we know that Gn(GKa,b) is not isomorphic to Gn(SKa,b). The goal of this study is to show that the difference between these groups can be detected by counting homomorphisms into suitably chosen finite groups. We will give the proof in Chapter 6.
ski [4], Chirikjian [6], Humphreys [10], Ledermann [14], Meldrum [19], Rose [24], Rotman [25], Scott [27] and Tuffley [29].
5.1
Dihedral groups
In this section, we will outline the dihedral groups with some examples.
Definition 5.1. For n ≥ 3 the dihedral group Dn is the group of symmetries of a regular polygon with n sides. For n = 1 or n = 2, Dn is the group of symmetries of an interval or rectangle respectively. The order of the dihedral group Dn is 2n, for n ≥1.
There are two types of symmetries of the n-gon; exactly half of them are rotations and the other half are reflections as follows:
Rotations ρ0, ρ2π n, ρ 4π n, . . . , ρ 2(n−1)π n
, where ρθ is an anti-clockwise rotation of angleθ. Reflections µ0, µπn, µ2π
n, . . . , µ
(n−1)π n
, where µθ is reflection about the line through the origin meeting the vertical axis with an angle θ.
The dihedral group is generated by a rotation through 2π/n and a reflection. The dihedral group Dn has the presentation ha, b|bn =a2 = e, aba−1 =b−1i, where a represents a reflection
and b a rotation.
We will clarify the above discussion in the following example.
Example 5.1. Let ρ0, ρ2π
3 and ρ
4π
3 be the rotations of D3 and let µ0, µ
π
3 and µ23π be the
reflections of the group. Then the group of symmetries of the triangle can be written as D3 ={ρ0, ρ2π 3 , ρ 4π 3 , µ0, µ π 3, µ23π}.
As we can see in Figure 5.1 rotating the triangle gives three symmetries: ρ0 is the identity
rotation, ρ2π
3 is the 120
◦ anti-clockwise rotation and ρ
4π
3 is the 240
◦ anti-clockwise rotation.
Figure 5.1: The elements ofD3, the symmetry group of an equilateral triangle. (i) is a diagram
of the three rotations. (ii) is a diagram of the three reflections of D3.
The reflections across the diagonal and vertical lines of symmetry give three more symmetries which are µ0, the reflection through the vertical line; µπ3, the reflection through the diagonal
running northwest to southeast; andµ2π
3 , the reflection through the diagonal running northeast
to southwest.
We will conclude this section with the following remark which showsDnmay be realised as a subgroup of Sn.
Remark 5.1. By labeling the vertices of the regular polygon with 0,1, . . . , n−1, Dn can be realised as a subgroup of Sn. It is generated by the rotation ρ(i) = i+ 1 and the reflection σ(0) = 0, σ(i) = n−i, for 1 6 i6 n−1. Also, both ρ and σ can be written in cycle form as follows:
ρ= (0 1 . . . n−1)
σ= (0)(1n−1)· · ·(bn/2c dn/2e).
Note that (bn/2c dn/2e) represents the two cycle (k k + 1) when n = 2k+ 1 is odd, and the one cycle (k) when n= 2k is even.
Let us illustrate the above remark by an example.
Example 5.2. The group S5 has a subgroup that is isomorphic to D5 (Figure 5.2), which is
generated by ρ and σ, where
ρ= (0 1 2 3 4), σ= (0)(1 4)(2 3).
Figure 5.2: A pentagon with labelled vertices which realises D5 as a subgroup of S5.
5.2
Semidirect products
In this section, we will describe the semidirect product and some of its properties. To be able to define the semidirect product, the direct product needs to be defined first.
Definition 5.2. Let Gbe a group. Then G is to said to be a direct product of two groups H and K if and only if they are normal subgroups of Gsuch thatH∩K ={e} and G=HK.
Definition 5.3. Gis said to be the semidirect product ofH andK if and only if H and K are subgroups of G, which satisfy the following conditions:
H CG, H∩K ={e}, HK =G.
LetG=HK, whereHandK are subgroups ofGand satisfy the conditions in Definition 5.3. Each element ofGcan be uniquely expressed in the form hk, sincehk =h0k0 implies (h0)−1h= k0k−1 ∈ H∩K = {e}. Since H C G, for each k ∈ K there is an automorphism of H which
is the inner automorphism of k restricted to H given by h 7→ khk−1. Moreover, the map
K →Aut(H) given byk 7→(h7→khk−1) is a group homomorphism, as we now show. Suppose ϕk is the inner automorphism of k restricted toH. Then, for t, k ∈K and h∈H,
ϕtk(h) = tkh(tk)−1 =tkhk−1t−1 =t(khk−1)t−1 =ϕt(khk−1) =ϕt(ϕk(h)) = (ϕt◦ϕk)(h).
For h ∈ H and k ∈ K we associate the ordered pair (h, k) ∈ H × K with the element hk ∈ G = HK. For hk, h0k0 ∈ G and the homomorphism ϕ, the multiplication on G can be
This is what called a semidirect product, and it is denoted by HoϕK, or it can be written as HoK, when ϕis understood.
We now give an example of the semidirect product.
Example 5.3. Let H = hx|xn = 1i ∼=
Zn, and K = hy|y2 = 1i ∼= Z2. Let ϕ : K → Aut(H)
be the homomorphism such that ϕy(x) = x−1. Then the semidirect product of the normal subgroup H and the subgroup K is
G=HoϕK.
We now show thatG∼=Dn, the dihedral group of order 2n. The element (x,1) has order n and (1, y) has order 2. So,
(1, y)(x,1)(1, y) = (ϕy(x), y)(1, y) = (ϕy(x), y2) = (ϕy(x),1) = (x−1,1) = (x,1)−1.
Hence if we setv = (x,1) and u= (1, y) then v and u satisfy vn =u2 = 1 and uvu =v−1. We can see that v and u generate G and |G| =|K||H| = 2n. So the relations of the group G are exactly the same as those of the dihedral group Dn and that impliesGis isomorphic toDn.
Remark 5.2. The example above shows that the dihedral group Dn is isomorphic to the semidirect product G=Zno Z2.
Next, we will use the semidirect product to construct the groupDp,q;θ, which will form the building blocks for our target groups in Chapter 6.
5.3
The construction of
D
p,q;θWe will begin with the definition of Tuffley’s Dp,q [29]. For p and q distinct primes the group Dp,q is a semidirect product
Dp,q =Zqp−1o Zq.
To define multiplication in Dp,q we regard Zqp−1 as the additive group of the finite field Fpq−1.
The multiplicative groupF×pq−1 is cyclic of orderpq
−1−1, and so contains an elementζ of order
q, asq dividespq−1−1 by Fermat’s Theorem. The multiplication inD
p,q can be defined by the action of i ∈Zq onZpq−1 given by multiplication by ζi. We remark that Dp,q may be regarded as a generalised dihedral group, in the sense that Dp,2 ∼=Dp.
IfF(x) = 1+x+x2+· · ·+xq−1factors over
Zp, then the isomorphism type ofDp,qdepends on the choice of rootζ, and therefore to avoid this ambiguity we need to defineDp,q;θ. To construct
Thus Dp,q;θ is a subgroup of Tuffley’s Dp,q [29] for a suitably chosen element ζ of order q, and Dp,q;θ∼=Dp,q whenθ is irreducible over Zp.
Remark 5.3. In the special case where q = 2, we have F(x) =x+ 1 which is irreducible over
Zp. Then ζ =−1 and we see that Dp,2;x+1 ∼=Dp, the dihedral group.
SinceVp;θ is normal, there is a map fromDp,q;θ intoZq, which we will use in Section 5.5. If f = (v, i)∈Dp,q;θ, then we will write [f] =i for its image inZq.
The next section proves some properties of the group Dp,q;θ.
5.3.1
Properties of
D
p,q;θIn this section, we will introduce some properties of Dp,q;θ and show that it is generated by any element of order p together with any element of order q. Also, we will define cyclic and noncyclic solutions to xa=yb, in a group G, and show when D
p,q;θ has a noncyclic solution to xa =yb.
The following lemma will present some facts about the elements of Dp,q;θ. For a proof see Tuffley [29].
Lemma 5.1 (Tuffley, [29, Lemma 3.2]). Let Dp,q;θ =Vp;θo Zq, such that Vp;θ = (Zp)degθ. Then 1. Elements in Dp,q;θ are of order 1, p, or q.
2. If two elements f and g ∈Dp,q;θ commute with each other, then they belong to Vp;θ or the same cyclic subgroup of order q.
3. If an element f = (v,0)∈Vp;θ has order p, then the conjugacy class off is(ζiv,0), where 06i6q−1.
4. Iff = (v, i)has orderq, then the conjugacy class of f is{g : [g] =i}={(ω, i) :ω ∈Vp;θ}.
Remark 5.4. Since p and q are prime, as a consequence of Lemma 5.1 we note that if f is a nontrivial element of Dp,q;θ, then f has an nth root in Dp,q;θ if and only if ord(f) is co-prime to n. Moreover, any such nth root belongs to hfi.
(v,1) |j = 0, . . . , q−1}. Then letting hβi act on α= (u,0) by conjugation we see that βjαβ−j = (vj, j)(u,0)(vj, j)−1 = (vj, j)(u,0)(−ζ−jvj,−j) (as (vj, j)−1 = (−ζ−jvj,−j)) = (vj, j)(u−ζ−jvj,−j) = (vj+ζj(u−ζ−jvj), j−j) = (vj−vj+ζju,0) = (ζju,0),
so (ζju,0) belongs to hα, βi for j = 0,1, . . . , q−1.
Let d= degθ. We claim that (u,0),(ζu,0), . . . ,(ζd−1u,0) are linearly independent, and there-
fore generate Vp;θ. We regard Vp;θ = (Zp)d as a vector space over Zp. Suppose that
c0u+c1ζu+· · ·+cd−1ζd−1u= 0, (5.1)
where ci ∈Zp fori= 0, . . . , d−1. Factoring (5.1) we get "d−1 X i=0 ciζi # u= 0,
so Pdi=0−1ciζi = 0 ∈ Fp;θ. But then ζ is a root of Pd
−1
i=0 cix
i, a polynomial of degree less than the degree of θ, so we must have ci = 0 for all i since θ is irreducible over Zp. Then (ζiu,0), i= 0, . . . , d−1 are linearly independent and therefore generate
Vp;θ as claimed. Since we are interested in homomorphisms from Ga,b, we are interested in solutions to xa =yb in a groupG. The following definition gives more explanation.
Definition 5.4. IfG has a solution to xa =yb, then the solution is cyclic if hx, yi is a cyclic subgroup of Dp,q;θ. Otherwise, the solution is noncyclic.
Similarly, we will say that ρ : Ga,b → G is cyclic if hρ(x), ρ(y)i is cyclic. Otherwise, ρ:Ga,b→G is noncyclic.
We now characterise whenDp,q;θ has a noncyclic solution to xa=yb.
Lemma 5.3. Let a and b be co-prime. Any noncyclic solution to xa = yb in D
p,q;θ satisfies xa =yb = 1. Consequently, such a solution exists if and only if p|a and q|b, or q|a and p|b. Proof. Suppose xand y are a solution to xa=yb such thatxa=yb =g 6= 1. Theng has order either porq. The element g belongs tohxi andhyi, wherehxi andhyiare groups of orderp or q which are primes. Therefore,g generates hxiand hyi, andhxiand hyiare cyclic subgroups of hgi. This implies thathx, yi=hgi is cyclic, so any noncyclic solution must satisfyxa =yb = 1. We show that a noncyclic solution exists only if p|a and q|b, or q|a and p|b. Let xa =yb be a noncyclic solution satisfying xa =yb = 1. Since a and b are co-prime and p and q are distinct
It is important in the classification of permutation groups. For our purposes we may define the wreath product as follows.
Definition 5.5. LetGbe a group, and let X be a set, then GX =Q
i∈XGi, where Gi =Gfor i ∈X. Suppose H is a group that acts on X on the right. Then there is a natural left action of H onGX defined by
(h(g))i =gi·h,
whereg = (gi)i∈X. Then thewreath productGoH is the semidirect productGXoϕH, where ϕis the homomorphism ϕ:H →Aut(GX) defined as follows: for (g
i)i∈X ∈GX and h∈H, we have (ϕh(g))i =gi·h.
We will illustrate the multiplication in the wreath product with a simple remark followed by an example.
Remark 5.6. Suppose (g1, g2, . . . , gn),(k1, k2, . . . , kn) ∈ GX, X = {1, . . . , n} and h1, h2 ∈ H.
Then the multiplication rule in GoH =GX oϕH can be defined by (g1, g2, . . . , gn), h1 (k1, k2, . . . , kn), h2 = (g1k1·h1, g2k2·h1, . . . , gnkn·h1, h1h2). If g = (g1, . . . , gn), h
, then we define ˆg = h. In general, let g, k ∈ GoH and ˆg ∈ H. Then (gk)i = giki·gˆ, and the multiplication in H is given by gkc = ˆgk. The multiplication inˆ permutations is composed from left to right, as we will see in the following example.
Example 5.4. Let (g1, g2, g3),(k1, k2, k3)∈G3 and let (1 2),(1 3)∈S3. Then, the product of
g = (g1, g2, g3),(1 2) and k = (k1, k2, k3),(1 3) in G3oS3 is gk = (g1, g2, g3),(1 2) · (k1, k2, k3),(1 3) , = (g1k2, g2k1, g3k3),(1 2 3) .
In the following section we will construct the group Wh,ks,t;;%τ by modifying the construction of Tuffley’s Hq,r
Figure 5.3: When ˆω has ordert = 5, its action onzhas a single 1-cycle (p) and all other cycles are of length 5.
5.5
The construction of
W
h,k;τs,t;%To construct our target group Wh,ks,t;%;τ we will modify the construction of Tuffley’s Hq,r
p , which is a wreath productDq,roP SL(2, p). We will do this by replacingP SL(2, p) withDs,t;%as follows. Given distinct primes h, k, s and t and suitable polynomials τ ∈ Zk[x] and % ∈ Zt[x], the group Wh,ks,t;;%τ is a wreath product of Dh,k;τ over Ds,t;%,
Wh,ks,t;%;τ =Dh,k;τ oDs,t;%= (Dh,k;τ)zoDs,t;%,
where z = {g ∈ Ds,t;%|[g] = 1} is a conjugacy class of Ds,t;% and we let Ds,t;% act on z by conjugation.
Elements of Wh,ks,t;;%τ have the form
ω= ((ωi)i∈z,ω),ˆ
where ωi ∈Dh,k;τ for each i∈z, so there will be |Vs;%|=sdeg% entries. The element ˆω ∈Ds,t;% has order 1, s or t.
Remark 5.7. We use Lemma 5.1 to describe the cycle structure of an element ˆω∈Ds,t;%acting onz by conjugation. Note that all elements ofz have ordert. If ˆω has orders, then ω ∈Vs;%. Elements of order s and t do not commute, and therefore ˆω has no fixed point and all cycles are of length s, because s is prime. If ˆω has order t, then it commutes only with hωi. There isˆ a unique element ˆω ∈ hωiˆ such that ˆωr ∈
z, 06 r 6t−1. Then, ˆω has a unique fixed point
in z and therefore all other cycles have length t, because t is prime. Figure 5.3 pictures the above discussion, with t= 5.
The map Dh,k;τ →Zk leads to a map W h,k;τ
s,t;% →ZkoDs,t;% which is given by [ω] = (([ωi])i∈z,ω).ˆ
It is convenient to factor the map Wh,ks,t;%;τ →ZkoDs,t;%into two maps to distinguish a subgroup of Wh,ks,t;;%τ isomorphic to ZkoDs,t;%. Choose ξ inDh,k;τ of order k such that [ξ] = 1, sohξi ∼=Zk. We define Ak
To be able to use Ws,t;%, we need to restrict h, k, s and t to be distinct primes which will allow some flexibility in dealing with the processes in the following chapter in order to prove the main theorem.
We conclude this section with an example which illustrates the group operations in the wreath product GoD5.
Example 5.5. Consider the wreath product GoD5,2;x+1. Since D5,2;x+1 is isomorphic to D5,
we therefore have GoD5. The groupD5 is generated by ρ= (0 1 2 3 4) and σ = (0)(1 4)(2 3)