finite simple groups in geometry and topology · finite simple groups in geometry and topology...

FINITE SIMPLE GROUPS IN GEOMETRY AND TOPOLOGY

LINDSAY WHITE

Contents

Introduction 11. Classification of Finite Simple Groups of Rank 2 12. Knot Theory 103. Graph Theory 164. Fundamental Group 295. Algebraic Geometry 306. Actions on Spheres 32Conclusions 33References 33

Introduction

The classification of all finite simple groups was an enormous accomplishment (the proofwas done by many different people, consisting of tens of thousands of pages). However,these groups frequently show up in other areas of mathematics such as geometry andtopology. There are numerous finite simple groups, but we will focus our attention onfinite simple groups of rank 2. This list is much shorter, so our examples will be moreconsistent because there are only a few families of groups to look at, as opposed to lookingat all finite simple groups. We will see the use of these rank 2 finite simple groups in theareas of knot theory, graph theory and algebraic geometry. In knot theory, their use inknot colouring polynomials allows us to distinguish the Kinoshita-Terasaka and Conwayknots. One of the families of rank 2 finite simple groups is the set of PSL2(Fp) groups (forp ≥ 5), which show up as quotient groups of the fundamental groups of some Brieskornspheres, as well as certain knot groups. It also turns out that a lot of the PSL2(Fp)groups admit a certain presentation which guarantees that the Cayley graph of the grouphas a Hamiltonian cycle. We also look at which of these groups can act on spheres.

1. Classification of Finite Simple Groups of Rank 2

First, let’s recall some group theory definitions:

Definition 1.1. H is a normal subgroup of G if

∀h ∈ H,∀g ∈ G, ghg−1 ∈ H1

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or equivalently,gH = Hg ∀g ∈ G

This means that a normal subgroup is invariant under conjugation by elements of thelarger group G.

Example 1.2. Consider G = S3, the symmetric group on 3 elements, consisting of thepermutations {1, (12), (13), (23), (123), (132)}. Let H be the subgroup

H = {1, (123), (132)}Then, it can easily be shown (by direct computation) that ghg−1 ∈ H ∀g ∈ G, h ∈ H.

Definition 1.3. A simple group G has no proper (nontrivial) normal subgroup.

Definition 1.4. The Rank of a Group G, denoted

rank(G) = max{rankpG | p | |G|}where p is a prime and

rankpG ≤ k

ifZp × Zp × ...× Zp 6⊂ G

where there are k + 1 copies of Zp. From this definition, if there are k copies of Zpcontained in G, this in turn means that rankpG ≥ k. So we can alternatively say thatrankpG = k if k is the largest number such that Zp × Zp × ...× Zp ⊂ G, where there arek copies of Zp.

We are looking specifically at groups with rank 2, which means we are looking at groupsthat do not contain Zp×Zp×Zp for any prime p, but do contain Zp×Zp for some primep. The classification of finite simple groups of rank 2 is a rather complicated problem -we will not get into the proof of the classification here, but use the result to see wherethese groups show up in other areas such as Knot Theory, Graph Theory and AlgebraicGeometry.

Theorem 1.5. Classification of Finite Simple Groups of Rank 2[Adem-Smith [1]]A finite simple group of rank 2 must be one of the following:

(1) PSL2(Fp) p odd prime, p ≥ 5(2) PSL2(Fp2) p odd prime(3) PSL3(Fp) p odd prime(4) PSU3(Fp) p odd prime(5) PSU3(F4)(6) A7

(7) M11

Here, Fp refers to the finite field consisting of p elements, where p is a prime. Let’s recallwhat these groups are. We will need to define some of these in terms of other groups,so we will start there. All of the fields we will be dealing with (these show up in theprojective special linear and projective special unitary groups) are of the form F = Fq,for q a power of some prime p (specifically, we had either q = p, q = p2 or q = 4). We will

FINITE SIMPLE GROUPS IN GEOMETRY AND TOPOLOGY 3

define these groups using Fq, but note that there are definitions for arbitrary fields F aswell.

Definition 1.6. [Dummit-Foote [6, p35]] The General Linear Group of degree n(over a finite field Fq) is the set of n × n invertible matrices, with group operation asmatrix multiplication. We denote this as:

GLn(Fq) = {M |M is an n× n matrix with entries from Fq, and det(M) 6= 0}

It has order:

|GLn(Fq)| = (qn − 1)(qn − q)(qn − q2)...(qn − qn−1)[Coxeter-Moser [5, p92]]

Definition 1.7. [Dummit-Foote [6, p48]] The Special Linear Group of degree n(over a finite field Fq) is the set of n × n invertible matrices with determinant 1 (withgroup operation as matrix multiplication). We denote this as:

SLn(Fq) = {M ∈ GLn(Fq) | det(M) = 1}

where the matrices have entries in Fq. It has order:

|SLn(Fq)| =(qn − 1)(qn − q)(qn − q2)...(qn − qn−1)

q − 1

[Coxeter-Moser [5, p92]]

Note that SLn(Fq) is a subgroup of GLn(Fq) - specifically, the subgroup of matriceswith determinant 1.

The groups that we had in our classification theorem are the projective special lineargroups. Recall that points in projective space can be thought of as lines through theorigin in R2. In this way, if two points in standard Euclidean space lie on the same line(through the origin), then they are considered to be the same point in projective space.Lines through the origin have the form y = mx (where m is the slope of the line), so inorder for two points to be on the same line of this form (for a fixed m), they must bescalar multiples of each other. Keeping this in mind, the definitions for the projectivegeneral linear and projective special linear groups are as can be expected.

Definition 1.8. [Coxeter- Moser [5, p92]] The Projective General Linear Group(over a finite field Fq) is the quotient of the general linear group by all nonzero scalartransformation matrices.

PGLn(Fq) ∼= GLn(Fq)/{λI}where λ is a scalar in Fq. It has order:

|PGLn(Fq)| =(qn − 1)(qn − q)(qn − q2)...(qn − qn−1)

q − 1

This is |GLn(Fq)|q−1 . We divide by q − 1 because there are q − 1 non-zero scalars in Fq, so

|{λI}| = q − 1.

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Definition 1.9. [Coxeter-Moser [5, p93]] The Projective Special Linear Group (overa finite field Fq) is the quotient of the special linear group by all nonzero scalar transfor-mation matrices λI.

PSLn(Fq) = SLn(Fq)/{λI}where λ is a scalar in Fq. The order of this group is:

|PSLn(Fq)| =|GLn(Fq)|d(q − 1)

where d = gcd(n, q − 1).

Going back to the classification of finite simple groups of rank 2, we had the followingprojective special linear groups show up: PSL2(Fp) for p ≥ 5 and p odd, PSL2(Fp2) forp odd, and PSL3(Fp) for p odd. We then have:

(1) PSL2(Fp): 2 × 2 matrices (with entries in the finite field of p elements, Fp), upto scalar multiplication (that is, up to multiplication by λI, where I is the 2 × 2identity matrix, λ a scalar), for p ≥ 5 and p odd. PSL2(Fp) has a presentation as:

PSL2(Fp) = 〈a, b|ap = b2 = (ab)3 = (a2ba(p+1)

2 b)3 = 1〉

[Coxeter-Moser [5, p94]](2) PSL2(Fp2): 2× 2 matrices (with entries in the finite field of p2 elements, Fp2), up

to scalar multiplication (that is, up to multiplication by λI , where I is the 2× 2identity matrix, λ a scalar), for p odd.

(3) PSL3(Fp): 3 × 3 matrices (with entries in the finite field of p elements, Fp), upto scalar multiplication (that is, up to multiplication by λI, where I is the 3 × 3identity matrix, λ a scalar), for p ≥ 5 and p odd.

We continue with the definitions of the groups in the list. The next group we needto define is the projective special unitary group over a finite field. First, we recall theanalogous definitions over the complex numbers:

Definition 1.10. [Nomizu, [14, p237]] A unitary matrix is a matrix M such that

MM∗ = M∗M = I

where M∗ is the conjugate transpose. The unitary group, denoted U(n), is the set ofall n×n unitary matrices. The special unitary group, denoted SU(n) is the subgroupof unitary matrices with determinant 1. In both cases, the group operation is matrixmultiplication, and the matrix entries are in C.

Before definining the unitary group for finite fields, we note that in this definition forC, we have that C is a quadratic extension over R, and that the conjugate transposeoperation is an involution. Recall that an involution is an automorphism of order 2. Ifwe apply ∗ to M twice, we get M , so that the conjugate transpose is an involution. Thiswill show us the similarity between the usual unitary definition over C, and the one overfinite fields, defined below.


Definition 1.11. The Special Unitary Group (of degree n) over a finite field Fq,denoted SUn(Fq), is the set of n × n unitary matrices with determinant 1 (with groupoperation as matrix multiplication), with entries in Fq2 . Note that some authors will referto this as SUn(Fq2) instead of SUn(Fq). Here a unitary matrix is a matrix M such that

MM∗ = M∗M = I

where M∗ is the tranpose of M , with α applied to each entry, where α is the automorphismwhich sends x 7→ xq (in Fq2). By[ Prop 5.1, http://www.maths.qmul.ac.uk pjc/class gps/ch5.pdf],we have that:

|SUn(Fq)| =|Un(Fq)|q + 1

where

|Un(Fq)| = qn(n−1)

2

n∏i=1

(qi − (−1)i)

Here, we have the quadratic extension Fq2 over Fq (in the definition above we wereconsidering entries in C, which is a quadratic extension over R). We can also see that αis an involution (that is, it has order 2). This is because the quadratic extension we have

here (Fq2 over Fq) is the splitting field of xq2 − x, so that α2(x) = α(xq) = xq

2= x. In C

we used conjugate transpose as our involution, and here we use α.

Definition 1.12. The Projective Special Unitary Group (of degree n) over a finitefield Fq, denoted PSUn(Fq), is the quotient of the special unitary group by scalar unitarymatrices, for scalars whose order divides n. By [Prop 5.1, http://www.maths.qmul.ac.uk/pjc/class gps/ch5.pdf],we have:

|PSUn(Fq)| =|SUn(Fq)|

d

where d = gcd(n, q + 1).

We need the scalars to have order dividing n in order to have determinant 1 (a require-ment for being a special unitary matrix). We will illustrate this by looking at the 3 × 3case. Consider the determinant of a scalar matrix M:

detM = det

λ 0 00 λ 00 0 λ

= λ3

In order for us to have λ3 = 1, the order of λ must divide 3. It is clear then that ingeneral, for an n × n scalar matrix, in order for it to have determinant 1, we must haveλn = 1, so that the order of λ must divide n.

Now we will look at the groups we had in our list. We had PSU3(Fp) for p odd, as wellas PSU3(F4). In both cases, we are dealing with 3× 3 unitary matrices with determinant1, with entries in Fp2 and F16, respectively. We are quotienting out by scalar unitarymatrices (for scalars whose order divides 3). The requirement that a scalar matrix M is

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unitary (that is, MM∗ = M∗M = I) means we have:

M∗M =

λq 0 00 λq 00 0 λq

λ 0 00 λ 00 0 λ

=

λq+1 0 00 λq+1 00 0 λq+1

=

1 0 00 1 00 0 1

with λ ∈ Fq2 . (Recall that M∗ was the transpose of M , with α applied to each entry, whereα(x) = xq). In order for this matrix condition to be satisfied, we need to have λq+1 = 1.So a scalar unitary matrix must satisfy λq+1 = 1. We are trying to determine whichmatrices we are quotienting out by. We have just stated the condition necessary for ascalar matrix to be unitary. We also need our matrices to have determinant 1. Therefore,as explained earlier, we must have λ3 = 1, which means the order of λ must divide 3.Clearly, if λ = 1, then the order is 1, which divides 3, and we will have λ3 = λq+1 = 1. Ifλ 6= 1, then λ cannot have order one and therefore must have order exactly 3, since thereare no other divisors of 3. Since we also have the requirement that λq+1 = 1 as well, wemust have that the order of λ divides q + 1 as well. If λ 6= 1, in order to satisfy both ofthese, we must have that 3|(q + 1) (since 3 ≤ q + 1 for q ≥ 2). This makes sense becauseif λ 6= 1 then λ ∈ Fq2∗, so the order of λ (which we stated must be 3) must divide |Fq2∗|.|F∗q2| = q2 − 1 = (q + 1)(q − 1). This means the order of λ (for any λ ∈ Fq2) must divideeither q + 1 or q − 1. So our condition that the λ divides q + 1 makes sense. Therefore,the matrices we are quotienting out by are of the form:

{λI | λ3 = λq+1 = 1}

In order for this to be nontrivial, we must have that 3|(q + 1).

Example 1.13. Consider the group PSU3(F5). Here, we have matrices with entries inF25. We want to determine which scalar matrices we quotient out by in SU3(F5) to getPSU3(F5). The conditions above tell us that we are looking for scalars λ ∈ F25 such thatλ3 = λ6 = 1. (Here, q = 5, so q + 1 = 6). We will consider F25 as F25

∼= F5(α), where αis a root of x2 + x+ 1. (It can easily be checked that x2 + x+ 1 is irreducible over F5, byverifying it has no roots in F5).Then, our elements in F25 are of the form:

{a+ αb | a, b ∈ F5}

We now need to find out which ones satisfy λ3 = λ6 = 1.

We see that the elements satisfying λ3 = λ6 = 1 are 1, λ and 4λ + 4. Therefore,to get PSU3(F5) from SU3(F5), we quotient out by the scalar matrices {I, λI, (4λ+ 4)I}.

We also compute the orders for these two groups. From the formulas above, we have:

|PSUn(Fq)| =|SUn(Fq)|

d


λ λ2 λ3 λ4 λ5 λ6

0 0 0 0 0 01 1 1 1 1 12 4 3 1 2 43 4 2 1 3 44 1 4 1 4 1α 4α + 4 1 α 4α + 4 1

α + 1 α 4 4α + 4 4α 1α + 2 3α + 3 α + 3 4α 4α + 1 3α + 3 3 3α + 4 4 4α + 2 2α + 3α + 4 2α α + 3 2 2α + 3 4α + 1

2α α + 1 3 α 3α + 3 42α + 1 2 4α + 2 4 3α + 4 32α + 2 4α 2 4α + 4 3α 42α + 3 3α 3α + 4 α + 1 3α + 1 22α + 4 2α + 2 3α + 4 4α 3α + 2 2

3α α + 1 2 α 2α + 2 43α + 1 2α + 2 2α + 1 4α 2α + 3 23α + 2 3α 4α + 1 4α α + 3 3α + 33α + 3 4α 3 4α + 4 2α 43α + 4 2 α + 3 4 2α + 1 3

4α 4α + 4 4 α α + 1 14α + 1 2α 4α + 2 α + 1 α + 2 34α + 2 3 2α + 1 4 α + 3 24α + 3 3α + 3 4α + 2 4α α + 4 34α + 4 α 1 4α + 4 α 1

where d = gcd(n, q + 1) and where

|SUn(Fq)| =|Un(Fq)|q + 1

where

|Un(Fq)| = qn(n−1)

2

n∏i=1

(qi − (−1)i)

In our list, we have PSU3(Fp) (for p an odd prime), and PSU3(F4). For PSU3(Fp), wehave n = 3, and q = p (an odd prime), so we have:

|U3(Fp)| = p3(3−1)

2

3∏i=1

(pi − (−1)i)

|SU3(Fp)| =|Un(Fp)|p+ 1

=p3(p+ 1)(p2 − 1)(p3 + 1)

p+ 1= p3(p2 − 1)(p3 + 1)

|PSU3(Fp)| =|SU3(Fp)|

d=p3(p2 − 1)(p3 + 1)

d

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where d = gcd(3, q + 1). There are two possibilities - either d = 1 or d = 3.For PSU3(F4), we have n = 3 and q = 4, so we have:

|U3(F4)| = 43(3−1)

2

3∏i=1

(4i − (−1)i) = 43(4 + 1)(42 − 1)(43 + 1)

.

|SU3(F4)| =|Un(F4)|

4 + 1=

43(4 + 1)(42 − 1)(43 + 1)

4 + 1= 43(42 − 1)(43 + 1)

|PSU3(F4)| =|SU3(F4)|

dwhere d = gcd(3, 4 + 1) = 1. Then we have:

|PSU3(F4)| = 43(42 − 1)(43 + 1) = 43 · 15 · 65 = 43 · 3 · 52 · 13

The next group we had in the classification theorem was the alternating group A7.

Definition 1.14. The Alternating Group A7, is a subgroup of the symmetric groupS7 (the group of all permutations on 7 objects), consisting of all even permutations on 7objects (that is, permutations which can be decomposed into an even number of transpo-sitions). The order of the symmetric group is 7! and the order of the alternating group is7!2

. The alternating group A7 has a presentation as:

A7 = 〈a, b|a5 = b3 = (ab)7 = (ba−1ba)2 = (ba−2ba2)2 = 1〉where a = (34567) and b = (123). [Coxeter-Moser [5, p67]]

Definition 1.15. [http://en.wikipedia.org/wiki/Mathieu group M11] The Mathieu groupM11 is a group (of order 7920) of permutations on 11 objects, and is sharply 4-transitive.This means that if M11 acts on a set X (containing at least 4 elements), then for any twodisjoint subsets of pairwise distinct elements of X, say {x1, x2, x3, x4} and {y1, y2, y3, y4}(where no element appears more than once in all 8 elements), there is an element g ∈ Gsuch that

g · xi = yifor i = 1, 2, 3, 4. M11 has a presentation as:

M11 = 〈a, b, c|a11 = b5 = c4 = (ac)3 = 1, b−1ab = a4, c−1bc = b2〉where a = (012345678910), b = (14593)(281076) and c = (1543)(26107) [Coxeter-Moser[5, p99]]. M11 has elements of order 1, 2, 3, 4, 5, 6, 8 and 11. There is one element oforder 1, 5 of order 11 and 11 of each of the remaining possible orders.

Proposition 1.16. All of the groups in Theorem 1.5 have 2-rank 2.

By [Gorenstein, [8, Thm 1.86]], all of the groups from Theorem 1.4 have 2-rank ≤ 2.For a group to have 2-rank 1, it cannot be simple [Gorenstein, [8, p13]]. Since all of thegroups in Theorem 1.4 are simple, they therefore all must have 2-rank exactly equal to 2.

For the groups to have rank 2 (above we only stated they have 2-rank 2), this meanswe must have that

Zp × Zp × Zp 6⊂ G

for any prime p. We will prove this for a few of the groups listed in Theorem 1.5.


Proposition 1.17. All of the groups in Theorem 1.5 have 2-rank 2. That is,

Zp × Zp × Zp 6⊂ G

for any prime p.

Proof. The order of a subgroup of G must divide the order of G, and we know that|Zp × Zp × Zp| = p3, so let’s first look at the orders of the groups:

|A7| = 2520 = 23 · 32 · 5 · 7Here, the possible primes p dividing the order of A7 are 2, 5, 7 and 9. We already knowthat the 2-rank is 2. Now, the order of a subgroup has to divide the order of a group, sowe could only have 5-rank ≤ 1, 7-rank ≤ 1 and 3-rank ≤ 2. This is because we only have1 power of both 5 and 7 showing up in the prime decomposition of the order of A7, andfor 3 we have a 2-power. Since the rank of the group is the maximum of all p-ranks, andwe have that all p-ranks are ≤ 2, with the 2-rank equal to 2, we have that A7 has rank 2.

|M11| = 7920 = 24 · 32 · 5 · 11

Similarly, we have that the 5-rank and 11-rank have to be ≤ 1, and the 3-rank has to be≤ 2. We know that the 2-rank is exactly 2, so M11 has rank 2.

For the projective special linear groups, we cannot determine that they have rank 2from the orders alone, as we will see below. From above, we have the formula:

|PSLn(Fq)| =(qn − 1)(qn − q)(qn − q2)...(qn − qn−1)

d(q − 1)

The first case we consider is PSL2(Fp) (for p odd, ≥ 5). Here, d = gcd(2, p−1) = 2 (sincep− 1 is even). After expanding and simplifying, we have:

|PSL2(Fp)| =p(p2 − 1)

2=p(p− 1)(p+ 1)

2

Since p is odd, we have that both p− 1 and p+ 1 are even, and also that p cannot divideeither p + 1 or p − 1. Additionally, since the difference between p + 1 and p − 1 is 2,p + 1 and p − 1 have no common prime factors aside from 2. However, we do have thepossibility that there are powers higher than 2 of a prime showing up in either p + 1 orp− 1 (or possibly both, but with different primes). For example, if we take p = 109, then

p− 1 = 108 = 22 · 33

So in this case, we cannot see that the |PSL2(Fp)| groups have rank 2 just from lookingat the orders, as it is possible that the 3-rank could be 3.

Similarly, for PSL2(Fp2), we get:

|PSL2(Fp2)| =p2(p2 + 1)(p2 − 1)

2

In our case, p is odd, which means both p2 − 1 and p2 + 1 are even. We also have thatp cannot divide either p2 + 1 nor p2 − 1, and p2 + 1 and p2 − 1 have no common prime

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factors (aside from 2). Again, we run into the problem that p2 + 1 or p2− 1 could containpowers greater than 2 of a prime. For example, if we take p=53, then

p2 − 1 = 2808 = 23 · 33 · 13

So we cannot see that the |PSL2(Fp)| groups have rank 2 just from looking at the orderseither. We take another approach for the projective special linear groups: The groups withp-rank 1 (for all primes p) are precisely those groups with periodic cohomology. [Carlson-Chebolu-Minac, [4]]. We have that SL2(Fq) has periodic cohomology [Wolf, [19]], so thatit has p-rank 1 for all primes p. We know that PSL2(Fq) has 2-rank 2 (for the ones thatshowed up on our list), and need to show that for all other primes pi, the pi-rank is 1. Wehave that PSL2(Fq) is a quotient of SL2(Fq). From above, we know that it is:

PSL2(Fq) ∼= SL2(Fq)/{λI}where the order of λ must divide 2, which means that λ = ±1. {λI|λ = ±1} ∼= Z/2, sowe have:

PSL2(Fq) ∼= SL2(Fq)/Z/2So we have a surjection, say f , from SL2(Fq) onto PSL2(Fq) with kernel Z/2. Supposethat PSL2(Fq) had Z/pi×Z/pi contained in it for some prime pi. Then when we lift thisgroup back to SL2(Fq), it would have to have come from Z/pi×Z/pi(×ker(f)) ∈ SL2(Fq).Since the kernel does not contain Z/pi (unless pi = 2), we need both copies of Z/pi to bein SL2(Fq). But this contradicts SL2(Fq) having pi-rank 1 for all primes pi. Therefore,PSL2(Fq) must have pi-rank 1 for all odd primes pi. On the other hand, since thekernel of this map is Z/2, we only need one copy of Z/2 in SL2(Fq) in order to haveZ2 × Z2 ∈ PSL2(Fq), so we can (and in fact do, as stated above) have 2-rank 2 inPSL2(Fq).

Lastly, we have the groups PSU3(Fp), and PSU3(F4). From above, we have the orders:

|PSU3(Fp)| =p3(p2 − 1)(p3 + 1)

d

where d = gcd(3, q+1). There are two possibilities - either d = 1, or d = 3. For PSU3(Fp),we have:

|PSU3(F4)| = 43(42 − 1)(43 + 1) = 43 · 15 · 65 = 43 · 3 · 52 · 13

In PSU3(F4), the only prime with a power greater than 2 is p = 2, but we already knowthe 2-rank is exactly 2. Since we cannot have p-rank greater than 2 for any other prime,this group must have rank 2. However, for PSU3(Fp), we have the possibility of the p-rankbeing 3, so we cannot say that this group has rank 2 just from looking at the order. Werefer the reader to [Gorenstein, [8]] for the proof that the PSU3(Fp) groups have rank 2.

�

2. Knot Theory

The finite simple groups of rank 2 provide some interesting applications in knot theory.One of the more significant applications is their use in being able to distinguish betweenthe Kinoshita-Terasaka and Conway knots, as will be seen below. First, we start with amore basic application.


Theorem 2.1. [Johnson [9]] Every finite simple group is a quotient of a knot group.

Looking at finite simple groups of rank 2 is just a special case of this theorem, and wetherefore know that every group listed in Theorem 1.5 is the quotient of a knot group forsome knot K.

Example 2.2. The (3, 5)-torus knot has knot group:

π1(K) = 〈a, b|a3 = b5〉[Rolfsen, [15, p53]] On the other hand, it was stated above that A7 (one of our finitesimple groups of rank 2) has a presentation as:

A7 = 〈a, b|a5 = b3 = (ab)7 = (ba−1ba)2 = (ba−2ba2)2 = 1〉So, we can see that A7 is a quotient of π1(K) since quotient groups can be formed by

adding extra relations to the starting group (By [Dummit-Foote [6, Exercise 16, p86]], anyquotient group H of G is of the form H ∼= G/N , where N is the smallest normal subgroupgenerated by a set of elements {a1, ..., an}. From this, we can see that the presentation ofH is the same as the presentation of G, with the extra relations a1 = ... = an = 1.) Moreexplicitly,

A7∼= π1(K)/N

where N is the smallest normal subgroup containing the group elements (ab)7, (ba−1ba)2,(ba−2ba2)2, 1. More generally, we have that a (p, q)-torus knot has knot group:

π1(K) = 〈a, b|ap = bq〉[Rolfsen, [15, p53]] From above, we have that PSL2(Fp) has a presentation as:

PSL2(Fp) = 〈a, b|ap = b2 = (ab)3 = (a2ba(p+1)

2 b)3 = 1〉We can see that PSL2(Fp) is a quotient of the knot group of a (p, 2)-torus knot. A(p, 2)-torus knot has knot group:

π1(K) = 〈a, b|ap = b2〉PSL2(Fp) is clearly a quotient of this, since PSL2(Fp) just has the extra relations (ab)3 =

(a2ba(p+1)

2 b)3 = 1, ap = 1.

Another use of the groups in Theorem 1.5 is in Knot Colouring Polynomials, wheresome of the finite simple groups of rank 2 provide some interesting results.

Definition 2.3. [Eisermann, [7]] Let G be a group and α ∈ G. Then (G,α) is a colouringgroup if G is generated by the conjugacy class of α.

From the steps taken to find a Wirtinger presentation of a knot group, it is easily seenthat all elements of a knot group are in the same conjugacy class. Recall that in theWirtinger presentation of a knot group, each crossing gives rise to a relation of eitherc = aba−1 or c = a−1ba, depending on the orientations at the crossing. There are 3 arcsat each crossing - an overcrossing, and two under crossings (hence the 3 letters a, b andc which correspond to the 3 arcs). We start with 2 labelled arcs, and then look at onecrossing at a time, using these relations to label the new arcs we come across. When we

12 LINDSAY WHITE

come across arcs that have already been labelled, we need to make sure the labelled arcagrees with what our current crossing’s relation is giving us. Setting the two equal, weget a relation. These relations give us our knot group and hence, every element is in thesame conjugacy class (because of our labeling relations). We also have that every simplegroup is a colouring group. By definition of a simple group, there is no normal subgroup,meaning that the group is generated by a single conjugacy class. So we have that everyfinite simple group is a colouring group, and in particular, every finite simple group ofrank 2 is a colouring group.

To define a knot colouring polynomial, we consider the set of representations (homo-morphisms):

ρ : π1(K)→ G

with ρ(mK) = α, where mK is a (fixed) meridian of K, and the colouring group G (inour case, a finite simple group of rank 2) and element α ∈ G are fixed. From the theoremabove, we know that for all finite simple groups G, such a homomorphism will exist.

Definition 2.4. [Eisermann, [7]] The colouring polynomial is defined as:

PGα(K) =

∑ρ

ρ(lK)

where lK is a (fixed) longitude of K, and G is a finite colouring group.

Since each element in the summation is an element of G, we get a linear combinationof elements in G with integer coefficients (more particularly natural numbers), so in factwe have an element of the group ring NG ⊂ ZG (by definition of group ring). As well,we note that because G is finite, there can only be a finite number of homomorphisms ρfrom π1(K) onto G, so the sum is finite.

Recall now some notation/terminology regarding knots:

Definition 2.5. [Eisermann, [7]] The obverse of an oriented knot K, denoted Kx, is themirror image of the knot K.

Definition 2.6. [Eisermann, [7]] The reverse of an oriented knot K, denoted K!, is theknot K with orientation reversed.

Definition 2.7. [Eisermann, [7]] The inverse of an oriented knot K, denoted K∗, is themirror image of the knot K with orientation reversed.


[http://charles.hamel.free.fr/knots-and-cordages/Geometric-transformations.html]We now generalize these ideas to automorphisms:

Definition 2.8. [Eisermann, [7]] An obversion of the pair (G,α) is an automorphismx : G→ G with αx = α−1. (α ∈ G is fixed a fixed element).

Definition 2.9. [Eisermann, [7]] A reversion of the pair (G,α) is ananti-automorphism ! : G→ G with α! = α. (α ∈ G is fixed a fixed element). Note: inorder for a function f : G→ G to be an anti-automorphism, we must have thatf(x)f(y) = f(yx) (in contrast to an automorphism, which would havef(x)f(y) = f(xy)).

Definition 2.10. An inversion of the pair (G,α) is morphism ∗ = x·! : G→ G, where !

is a reversion and x is an obversion. (α ∈ G is fixed a fixed element).

Example 2.11. Consider the group PSL2(F5) and let

α =

[1 10 1

]Then, for an obversion (which is not necessarily unique), we need xx = x−1, so in thiscase we need: [

1 10 1

]x=

[1 −10 1

]We could then have: [

a bc d

]x=

[a −b−c d

]An obversion also needs to be an automorphism, so we have to have that(g1g2)

x = g1xg2

x.We verify that it is a homomorphism:

14 LINDSAY WHITE

(

[a bc d

] [e fg h

])x

=

[ae+ bg af + bhce+ dg cf + dh

]x=

[ae+ bg −af − bh−ce− dg cf + dh

]On the other hand, we have:[

a bc d

]x[e fg h

]x=

[a −b−c d

] [e −f−g h

]=

[ae+ bg −af − bh−ce− dg cf + dh

]These are equal, so we have that x is a homomorphism. To check that it is anautomorphism, we need to verify that the resulting matrix still has determinant 1. Sincewe only changed the sign of both entries on the same diagonal, the determinant will notchange, and hence x is an automorphism.For a reversion, we need: [

1 10 1

]!=

[1 10 1

]We could then have: [

a bc d

]!=

[d bc a

]To show this is an anti-homomorphism, we need to check (g2g1)

! = g1!g2

!:

(

[e fg h

] [a bc d

])!

=

[ea+ fc eb+ fdga+ hc gb+ hd

]!=

[gb+ hd eb+ fdga+ hc ea+ fc

]On the other hand, we have:[

a bc d

]![e fg h

]!=

[d bc a

] [h fg e

]=

[dh+ bg df + bech+ ag cf + ae

]=

[gb+ hd eb+ fdga+ hc ea+ fc

]Therefore, we have that ! is an anti-homomorphism. To check that it is ananti-automorphism, we need to verify that the resulting matrix still has determinant 1.Since we only interchanged entries on the same diagonal, the determinant will notchange, and hence ! is an anti-automorphism.For an inversion, we have: [

a bc d

]∗=

[d −b−c a

]2.1. The Kinoshita-Terasaka and Conway Knots. Where the Alexander, Jones,Homfly and Kaufmann polynomials fail to distinguish the Kinoshita-Terasaka andConway knots, knot colouring polynomials are able to distinguish these. We will showtwo examples of this, both using finite simple groups of rank 2.First, let K denote the Kinoshita-Terasaka knot, and C, the Conway knot.


[http://en.wikipedia.org/wiki/Mutation %28knot theorys%29]Here, the Kinoshita-Terasaka knot is pictured on the left, and the Conway knot on the

right. As we can see in the picture, these two knots are almost identical except for onesmall section - this section in C is just a rotation of the corresponding section in K.Despite this difference, the polynomials mentioned above were still unable to detect adifference between the two knots.

Example 2.12. Let G = PSL2(F7). Then, choosing any element α of order 3, by[Eisermann [7]], we have:

PGα(K) = 1 + 6α

PGα(K∗) = 1 + 6α2

PGα(C) = 1 + 12α

PGα(C∗) = 1 + 12α2

This shows that K 6= C, and also that both K and C do not equal their inverse.

Example 2.13. Let G = M11. Then, by [Eisermann [7]], we have:

PGα(K) = 1 + 11α3 + 11α7

PGα(K∗) = 1 + 11α4 + 11α8

PGα(Kx) = 1 + 11α4 + 22α8

PGα(K !) = 1 + 22α3 + 11α7

PGα(C) = 1 + 11α3 + 11α7

PGα(C∗) = 1 + 11α4 + 11α8

PGα(Cx) = 1 + 11α4 + 11α6 + 11α8

PGα(C !) = 1 + 11α3 + 11α5 + 11α7

Here, just looking at PGα(K), PG

α(C), PGα(K∗) and PG

α(C∗) is not enough to showthat K and C are distinct, as it was in Example 2.12. But, when we look at PG

α(Kx),PG

α(Cx), PGα(K !) and PG

α(C !) as well, we can conclude not only that K and C aredistinct (since PG

α(Kx) 6= PGα(Cx)), but also that both K and C are not equal to their

own inverse, reverse or obverse.

16 LINDSAY WHITE

3. Graph Theory

Every group can be represented as a graph, so this leads to some applications of thefinite simple groups of rank 2 in graph theory. Specifically, we will look at Cayleygraphs, prime graphs and Hamiltonian cycles.

Definition 3.1. [Sjerve, Cherkassoff, [17, p2]] Given a group G = 〈g1, ...., gn〉, a Cayleygraph is a graph with the vertex set consisting of elements of G, and edges definedbetween two distinct vertices s and t if and only if there exists a generator of G, gi, suchthat s = tgi.

Example 3.2. To get the idea of a Cayley Graph, we will look at an easy example withsmaller size before looking at examples of finite simple groups of rank 2. Consider forexample the additive group Z/6Z. From group theory, we know that there are only 2possible generators for this group, 1 or 5. (For any group Z/nZ, the generators are thenumbers relatively prime to n). Let’s say we’re given the presentation of this group with5 as our generator. Our vertices will be the 6 elements of the group {0, 1, 2, 3, 4, 5}. Todetermine the edges, we need to find out which pairs of elements satisfy s = tgi forgi = 5. We can write out the list of multiples of 5 (our only generator):

(1) 0 · 5 = 0(2) 1 · 5 = 5(3) 2 · 5 = 10 = 4 mod 6(4) 3 · 5 = 15 = 3 mod 6(5) 4 · 5 = 20 = 2 mod 6(6) 5 · 5 = 25 = 1 mod 6

The only pairs of distinct elements s and t satisfying s = tgi, for gi = 5, are {1, 5} and{2, 4}, so we have an edge between 1 and 5 and between 2 and 4.

•2 •4

•3

•1 •5We now recall some terminology from graph theory:

Definition 3.3. [Sjerve, Cherkassoff, [17, p2]] First, recall that a path is a sequence ofvertices alternating with edges, {v1, e1, v2, e2, ..., vn}, where each edge ei (in thesequence) is an edge between the vertices vi−1 and vi. A cycle is a path which ends onthe same vertex it started at (that is, v1 = vn). A Hamiltonian path is a path whichreaches each vertex exactly once. Similarly, a Hamiltonian cycle is a cycle whichreaches each vertex exactly once.

Theorem 3.4. [Sjerve, Cherkassoff, [17]] Let G be a group which has a presentation as:

G ∼= 〈g1, g2, g3|g12 = g22 = g3

2 = 1, g1g2 = g2g1, ...〉where the ”...” can be any other relations, aside from g1g3 = g3g1 and g2g3 = g3g2.Then the Cayley graph of this presentation has a Hamiltonian cycle.


It turns out that the rank 2 PSL2(Fp) groups (that is, those with p ≥ 5) all have apresentation of the form above, except when p = 7. [Sjerve, Cherkassoff, [17, Prop 4.4]]

Example 3.5. Consider PSL2(F5) ∼= A5.

18 LINDSAY WHITE

As we will see below, it turns out that this group satisfies the conditions of the theorem,and hence has a Hamiltonian cycle. We can see the Hamiltonian cycle along the outercircle in the graph above.(The Hamiltonian cycle was found by a program created by Cameron White, ComputerScience Student at University of Waterloo).Now we will show that the group satisfies the conditions of the theorem. Let:

g1 = (12)(34)

g2 = (13)(24)

g3 = (12)(45)

Then, we need to show that g12 = g2

2 = g32 = 1 and that g1g2 = g2g1, as well as

g1g3 6= g3g1 and g2g3 6= g3g2.Any disjoint pair of 2-cycles will have order 2, since:

[(ab)(cd)]2 = (ab)(cd)(ab)(cd) = id

Since g1, g2 and g3 are all disjoint pairs of 2-cycles, we then have:

g12 = g2

2 = g32 = 1

Next, we need to show that g1g2 = g2g1, g1g3 6= g3g1 and g2g3 6= g3g2.:

g1g2 = (12)(34)(13)(24) = (14)(23)

g2g1 = (13)(24)(12)(34) = (14)(23)

=⇒ g1g2 = g2g1

g1g3 = (12)(34)(12)(45) = (345)

g3g1 = (12)(45)(12)(34) = (354)

=⇒ g1g3 6= g3g1

g2g3 = (13)(24)(12)(45) = (14523)

g3g2 = (12)(45)(13)(24) = (13254)

=⇒ g2g3 6= g3g2

It is easy to see that these 3 elements will generate the entire group A5. So, we see thatPSL2(Fp) ∼= A5 has a presentation as in the theorem. Then, our theorem tells us thatthe Cayley graph of A5 must have a Hamiltonian cycle.In order to draw the Cayley graph, we need to determine which pairs of elements satisfythe relation s = tgi for the 3 generators above (there is an edge between s and t ifs = tgi for some i, i ∈ {0, 1, 2}). Then we can draw the graph. Each ”s” will be joinedby an edge to the 3 elements in it’s row in the table below.


a a · g1 a · g2 a · g3 a a · g1 a · g2 a · g3(id) (12)(34) (13)(24) (12)(45) (243) (142) (123) (14532)

(12)(34) (id) (14)(32) (345) (245) (14352) (13)(25) (142)(12)(35) (345) (15324) (354) (253) (15342) (12453) (15432)(12)(45) (354) (13254) (id) (254) (15432) (13)(45) (152)(13)(24) (14)(23) (id) (14523) (345) (12)(35) (14253) (12)(34)(13)(25) (15234) (245) (15423) (354) (12)(45) (15423) (12)(35)(13)(45) (12354) (254) (123) (12345) (135) (14325) (134)(14)(23) (13)(24) (12)(34) (13245) (12354) (13)(45) (15432) (135)(14)(25) (15243) (13452) (15)(24) (12435) (145) (15)(23) (14)(35)(14)(35) (12453) (15342) (12435) (12453) (14)(35) (253) (143)(15)(23) (13425) (12435) (13254) (12534) (153) (14532) (15)(34)(15)(24) (14325) (135) (14)(25) (12543) (154) (23)(45) (153)(15)(34) (125) (14235) (12534) (13245) (14235) (125) (14)(23)(23)(45) (13542) (12543) (132) (13254) (15423) (12)(45) (15)(23)(24)(35) (14532) (153) (14352) (13425) (15)(23) (145) (15234)(25)(34) (152) (14523) (15342) (13452) (235) (14)(25) (234)

(123) (134) (243) (13)(45) (13524) (14523) (152) (14235)(124) (143) (132) (145) (13542) (23)(45) (154) (235)(125) (15)(34) (13245) (154) (14235) (13245) (15)(34) (13524)(132) (234) (124) (23)(45) (14253) (15324) (345) (15243)(134) (123) (142) (12345) (14325) (15)(24) (12345) (15324)(135) (12345) (15)(24) (12354) (14352) (245) (15234) (22)(35)(142) (243) (134) (245) (14523) (13524) (25)(34) (13)(24)(143) (124) (234) (12453) (14532) (24)(35) (12534) (243)(145) (12435) (13425) (124) (15234) (13)(25) (14352) (13425)(152) (25)(34) (13524) (254) (15243) (14)(25) (235) (14253)(153) (12534) (24)(35) (12543) (15324) (14253) (12)(35) (14325)(154) (12543) (13542) (125) (15342) (253) (14)(35) (25)(34)(234) (132) (143) (13452) (15423) (13254) (354) (13)(25)(235) (13452) (15243) (13542) (15432) (254) (12354) (253)

Example 3.6. We now consider an example using matrices instead of permutations, asnot many PSL2(Fp) are isomorphic to alternating or symmetric groups. Let’s considerG = PSL2(F13). We will show that this also satisfies the conditions of the theorem, sothat we can conclude its Cayley graph has a Hamiltonian cycle.Let:

g1 =

[0 1−1 0

]g2 =

[5 00 −5

]g3 =

[5 90 −5

]First, we need to show that g1

2 = g22 = g3

2 = 1 (first condition of the theorem). Thenwe have:

g12 =

[0 1−1 0

]2=

[−1 00 −1

]= [

[1 00 1

]]

20 LINDSAY WHITE

g22 =

[5 00 −5

]2=

[−1 00 −1

]mod 13 = [

[1 00 1

]] mod 13

g32 =

[5 90 −5

]2=

[−1 00 −1

]mod 13 = [

[1 00 1

]] mod 13

(since we are in PSL2(F13), a scalar multiple of the identity is equivalent to theidentity).Now, we need to verify that g1g2 = g2g1, as well as g1g3 6= g3g1 and g2g3 6= g3g2.

g1g2 =

[0 1−1 0

] [5 00 −5

]=

[0 −5−5 0

]

g2g1 =

[5 00 −5

] [0 1−1 0

]=

[0 55 0

]Since we are in PSL2(F13), scalar multiples of matrices are equivalent, meaning thesetwo matrices are the same, so g1g2 = g2g1.

g1g3 =

[0 1−1 0

] [5 90 −5

]=

[0 −5−5 −9

]

g3g1 =

[5 90 −5

] [0 1−1 0

]=

[−9 55 0

]=⇒ g1g3 6= g3g1

g2g3 =

[5 00 −5

] [5 90 −5

]=

[−1 60 −1

]

g3g2 =

[5 90 −5

] [5 00 −5

]=

[−1 −60 −1

]=⇒ g3g2 6= g2g3

To see that these 3 elements generate all of PSL2(F13) (which is not obvious in thiscase), we look at the possible subgroups of PSL2(F13) and eliminate all of the propersubgroups as possibilities. For this, we need the following theorem:

Theorem 3.7. [Sjerve, Cherkassoff, [17, p3, Thm 2.7]] If H is a subgroup of PSL2(Fp),it must be one of the following:

(1) Projective: PSL2(Fp) (the entire group)(2) Affine: Consider the following subgroups of PSL2(Fp2). Any subgroup of

PSL2(Fp) that is isomorphic to one of the following is called affine:


a. {[a b0 a−1

]: a, b ∈ Fp, a 6= 0}

b. {[a 00 a

]: a ∈ Fp2 , ap+1 = 1}

where a 7→ a = ap is the involution in Fp2.

Note that this is an involution since Fp2 is the splitting field of xp2 − x so that

a 7→ a = ap has order 2, as ap2

= a.c. Exceptional Subgroups: D2t, t ≥ 2, A4, A5, S4

Continuing with our example, we need to eliminate all of the possible subgroups asidefrom the entire group itself. First, we consider the possibility that

H = 〈g1, g2, g3|g12 = g22 = g3

2 = 1, g1g2 = g2g1, ...〉with g3g2 6= g2g3, g3g1 6= g1g3, is dihedral.Recall that if H is dihedral (of order 2n), it can be presented as:

H = 〈r, s|sn = 1, r2 = 1, sr = rs−1〉[Coxeter-Moser, [5, p6]].The relation sr = rs−1 implies that skr = rs−k (by a simple induction). In order for D2n,for n odd, to have a presentation as in H, we need to have 3 involutions, two of whichcommute, and the third of which does not commute with either of the other two. If wehave that n is odd, then our elements of order 2 (that is, our involutions) consist of:

{r, rs, ...., rsn−1}Note that none of the powers of s will have order 2 since sn = 1 (and n is odd), and theorder of sk for k ≤ n has to divide n. All of these elements are of the form rsk (fork = 0, ..., n− 1). We show that for n odd, none of these will commute. Consider two ofthese elements, say rsj and rsk. Then we have:

(rsj)(rsk) = r(sjr)sk = r(rs−j)sk = r2s−j+k = s−j+k

(rsk)(rsj) = r(skr)sj = r(rs−k)sj = r2s−k+j = s−k+j

In order for these two to be equal, we must have that −k+ j ≡ −j + k mod n. But thisis equivalent to 2j ≡ 2k mod n, which is only possible if n is even (this further reducesthe problem to j ≡ k mod n

2. Therefore, if H is dihedral, it is of even order.

Going back to the generators of H, we have that g3 does not commute with g1 or g2. Wealso have that g3 does not commute with g1g2:

g1g2g3 =

[0 −5−5 0

] [5 90 5

]=

[0 11 1

]mod 13

On the other hand, we have:

g3g1g2 =

[5 90 5

] [0 −5−5 0

]=

[−6 11 0

]mod 13

Therefore, g3 does not commute with g1, g2 or g1g2. The following proposition allows usto conclude that H cannot be dihedral:

22 LINDSAY WHITE

Proposition 3.8. Given two involutions g1, g2 in D2n (the dihedral group of order 2n,where n is even), such that g1g2 = g2g1, a third involution g3 must commute with at leastone of g1, g2 or g1g2.

Proof. Consider the dihedral group D2n, where n is even. Then our involutions consistof {r, rs, ..., rsn−1, sn

2 }. We need to show that for any pair of commuting involutions g1and g2, there does not exist a third which commutes with either g1, g2 or g1g2. We beginby determining which(distinct) involutions commute, and which ones do not. The pairsof commuting (distinct) involutions are of the form:

(1) {r, rsn2 }

(2) {sn2 , rsk} for k = 0, 1, ...n− 1

(3) {rsk, rsj} for k ≡ j mod n2}

The pairs of non-commuting (distinct) involutions are of the form:

(1) {r, rsk} for k 6= n2

(2) {rsk, rsj} for k 6≡ j mod n2}

We need to show that for any pair of commuting involutions g1 and g2, a third willcommute with either g1, g2 or g1g2. In the first set of commuting involutions, we haveg1 = r and g2 = rs

n2 . Then we have:

g1g2 = r(rsn2 ) = r2s

n2 = s

n2

From our list of commuting involutions, we see that sn2 commutes with every involution,

so a third involution will commute with g1g2. In the second set of commutinginvolutions, we have g1 = s

n2 and g2 = rsk for some k = 0, ..., n− 1. Any other

involution will also commute with g1, so we are done. In the third set of commutinginvolutions, we have g1 = rsk and g2 = rsj, for k ≡ j mod n

2. Then we have:

g1g2 = (rsk)(rsj) = r(skr)sj = r(rs−k)sj = r2s−k+j = s−k+j

Here, we have that k ≡ j mod n2, which means that −k + j ≡ 0 mod n

2. This means

−k + j is either n or n2. If −k + j = n, then we have sn = 1, and all elements commute

with 1. If we have −k + j = n2, then we have s

n2 , and all involutions commute with s

n2 .

Therefore, given two involutions g1, g2 in D2n (the dihedral group of order 2n, where nis even), such that g1g2 = g2g1, a third involution g3 must commute with at least one ofg1, g2 or g1g2. �

Next we need to show that H cannot be affine. In H, we have 3 elements of order 2. Weclaim that in the affine subgroups, there are no elements of order 2. Recall, that thesewere subgroups of PSL2(Fp2), and our subgroups of Fp are isomorphic to these. In thefirst case for affine subgroups, we had:

{[a b0 a−1

]: a, b ∈ Fp, a 6= 0}

We look for an element of order 2.[a b0 a−1

]2=

[a2 ab+ ba−1

0 a−2

]


In order for this to be the identity, we would need a2 = a−2 = 1 mod p2, for a ∈ Fp (forp an odd prime). But we cannot have a2 = 1 for a ∈ Fp unless a = 1, which would forceb = 0, and then our original matrix would be the identity, which has order 1. Therefore,there is no element of order 2 in this subgroup. In the second case for affine subgroups,we had:

{[a 00 a

]: a ∈ Fp2 , ap+1 = 1}

where a 7→ a = ap is the involution in Fp2 . Again, we look for the possibility of anelement of order 2: [

a 00 a

]2=

[a2 00 a2

]=

[a2 0

0 ap2

]=

[a2 00 a

]This can only be the identity if a = a2 = 1 for a ∈ Fp2 , where ap+1 = 1. The requirementthat a = 1 means our original matrix must be the identity (since a = ap = 1). Therefore,we have no elements of order 2 in this subgroup either. Therefore, H cannot be affine.Now we eliminate the possibility of G being either A4, A5, or S4 by noting that anelement in any of these groups cannot have order greater than 5.Consider the element g1g3, which we show has order 6.

g1g3 =

[0 −5−5 −9

](g1g3)

2 =

[0 −5−5 −9

] [0 −5−5 −9

]=

[−1 66 2

](g1g3)

3 =

[−1 66 2

] [0 −5−5 −9

]=

[9 33 4

](g1g3)

4 =

[9 33 4

] [0 −5−5 −9

]=

[−2 66 1

](g1g3)

5 =

[−2 66 1

] [0 −5−5 −9

]=

[9 −5−5 0

](g1g3)

6 =

[9 −5−5 0

] [0 −5−5 −9

]=

[−1 00 −1

]which is equivalent to the identity matrix in PSL2(F13), so g1g3 has order 6.=⇒ G cannot be isomorphic to A4, A5 or S4

We now only have the possibility that G is PSL2(13), as desired.

Definition 3.9. [Khosravi-Fakhraei [10]] A prime graph is a graph, denoted Γ(G),with vertex set

π(G) = {p | p | |G|}where p is a prime. Two distinct elements p and q are joined by an edge if and only ifthe group G contains an element of order pq.Denote by {πi(G)} the connected components of Γ(G). By convention, the firstconnected component, π1(G) is the component containing the vertex 2.

24 LINDSAY WHITE

Definition 3.10. Recall that (similarly) for a natural number n, we have π(n) = {pprime | p | n}.

Example 3.11. A7:

|A7| =7!

2= 7 · 6 · 5 · 4 · 3 = 23 · 32 · 5 · 7

π(A7) = {2, 3, 5, 7}

•2 •7

•3

•5

��

Example 3.12. M11:|M11| = 7920 = 2·32 · 5 · 11

π(M11) = {2, 3, 5, 11}M11 has elements of order 1, 2, 3, 4, 5, 6, 8 and 11 (as mentioned earlier), so the onlyedge is 2 · 3.

•2 •5

•3

•11

��

Definition 3.13. Recall that a graph is 2-regular if every vertex in the graph is anendpoint of exactly two edges.

Theorem 3.14. [Khosravi-Fakhraei [10]] Let G be a finite simple group of rank 2. Thenthe first connected component, π1(G), of Γ(G), is 2-regular if and only if G is one of thefollowing groups:

(1) PSL(2, p), 4|p− 1, |π(p− 1)| = 3(2) PSL(2, p2), 4|p2 − 1, |π(p2 − 1)| = 3(3) PSL(2, p), 4|p+ 1, |π(p+ 1)| = 3

This theorem was adapted to finite simple groups of rank 2. In the general case, therewas a fourth option, and we will show that it can be eliminated for our groups. Thefourth possibility was:

PSL(2, p2) , 4|p2 + 1, |π(p2 + 1)| = 3

To see why this is not an option in our case, let’s determine which of our groups satisfythe 4 possible conditions (see table below).Our smallest example satisfying the first option is 61, 121 for the second, 59 for the 3rd,and no example for the fourth. In our examples, we always had 4|(p+ 1) or 4|(p− 1),which means we always have 4|(p2 − 1). This in turn means 4 cannot divide p2 + 1, asthe difference between the 2 numbers is only 2. We show that in general, we can neverhave 4|(p2 − 1) by showing that 4 always has to divide one of either p+ 1 or p− 1 for p


p 4|(p+ 1) 4|(p− 1) π(p− 1) π(p+ 1)5 X X 1 27 X X 2 111 X X 2 213 X X 2 217 X X 1 219 X X 2 223 X X 2 229 X X 2 331 X X 3 137 X X 2 241 X X 2 343 X X 3 247 X X 2 253 X X 2 259 X X 2 361 X X 3 267 X X 3 271 X X 3 273 X X 2 2

p2 4|(p2 − 1) 4|(p2 + 1) π(p2 − 1) π(p2 + 1)25 X X 2 249 X X 2 2121 X X 3 2169 X X 3 3289 X X 2 3361 X X 3 2841 X X 4 2961 X X 3 31369 X X 2 21681 X X 4 21849 X X 4 32209 X X 3 72809 X X 3 33481 X X 4 23721 X X 4 24489 X X 4 35041 X X 4 25329 X X 3 4

26 LINDSAY WHITE

odd: Let p = 2k + 1 for some k. Then p− 1 = 2k and p+ 1 = 2k + 2 = 2(k + 1). Ifk = 2m, then p− 1 = 2(2m) = 4m and hence 4|(p− 1) and we are done. If k is insteadodd, then k + 1 = 2l for some l and we have p+ 1 = 2(k + 1) = 2(2l) = 4l, and hence4|(p+ 1). Thus, 4 must always divide either p+ 1 or p− 1, which means 4 alwaysdivides p2 − 1. As explained above, this implies 4 can never divide p2 + 1.Before we draw some graphs as examples, we need the following theorem:

Theorem 3.15. [McCullough,Wanderley [12]] For PSL2(Fq), where q = pk for some k,the orders of group elements are: p and the divisors of q+1

d, q−1

d, where d = gcd(2, q − 1).

Going back to Theorem 3.12, we had that the first prime p satisfying the first option is61, so let’s consider the prime graph for PSL2(F61). To determine the vertices for theprime graph of PSL2(F61), we need to determine the prime divisors of the order of thegroup.

|PSL2(F61)| =61(62)(60)

2= 113460

= 22 · 3 · 5 · 31 · 61

So our vertex set is π(G) = {2, 3, 5, 31, 61}.To determine the edges, we need to know what orders of elements show up in this group.Using the theorem above, we have:

p = 61(= q)

q + 1

2=

62

2= 31

q − 1

2=

60

2= 30 = 2 · 3 · 5

Therefore, our orders are 2, 3, 5, 2 · 3, 2 · 5, 3 · 5, and 2 · 3 · 5. This means our edges are2 · 3, 2 · 5 and 3 · 5.

•61 •3

•2

•31 •5

OOOOOOOOO

��//////////

As we can see in the graph, the first connected component (that is, the component with2 in it), is 2-regular.The first prime p satisfying the second set is 11 (p2 = 121), so let’s consider the primegraph for PSL2(F112). To determine the vertices for the prime graph of PSL2(F112), weagain need to determine the prime divisors of the order of the group.

|PSL2(F112)| =(112)(112 + 1)(112 − 1)

2= 885720

= 23 · 3 · 5 · 112 · 61



p = 11

q + 1

2=

122

2= 61

q − 1

2=

120

2= 60 = 22 · 3 · 5

Therefore, our orders are 2, 3, 5, 22, 2 · 3, 2 · 5, 3 · 5, 2 · 3 · 5, 22 · 3, 22 · 5, and 22 · 3 · 5.This means our edges are 2 · 3, 2 · 5 and 3 · 5.

•61 •3

•2

•11 •5

OOOOOOOOO

��//////////

Again we see that the first connected component is 2-regular.Since our first two examples are almost identical, let’s try another example satisfyingthe second set to try to get something different for our graph. We have p = 37 satisfyingthe second set (p2 = 1369), so let’s consider the prime graph for PSL2(F372). Todetermine the vertices for the prime graph of PSL2(F372), we need to determine theprime divisors of the order of the group.

|PSL2(F372)| =(372)(372 + 1)(372 − 1)

2

= 1282862520

= 23 · 32 · 5 · 19 · 372 · 137

So our vertex set is π(G) = {2, 3, 5, 19, 37, 137}.To determine the edges, we need to know what orders of elements show up in this group.Using the theorem above, we have:

p = 37

q + 1

2=

1370

2= 685 = 5 · 137

q − 1

2=

1368

2= 684 = 22 · 32 · 19

Therefore, our edges are 5 · 137, 2 · 3, 2 · 19 and 3 · 19.

28 LINDSAY WHITE

•19 •3

•2

•37 •5

•137

OOOOOOOOO

ooooooooo

oooooooo

As we can see in the graph, the first connected component is 2-regular, but in this graphwe now have a second component (which is not 2-regular).We have p = 59 satisfying the third set, so let’s consider the prime graph for PSL2(F59).To determine the vertices for the prime graph of PSL2(F59), we need to determine theprime divisors of the order of the group.

|PSL2(F59)| =59(59 + 1)(59− 1)

2= 102660

= 22 · 32 · 5 · 29 · 59


p = 59q + 1

2=

60

2= 30 = 2 · 3 · 5

q − 1

2=

58

2= 29

Therefore, our edges are 2 · 3, 2 · 5 and 3 · 5.

•59 •3

•2

•29 •5

OOOOOOOOO

��//////////

As we can see in the graph, the first connected component (ie the component with 2 init), is 2-regular.Lastly, we show an example of a group that did not satisfy the conditions of thetheorem. We have p2 = 412 = 1681 not satisfying any of the options, so let’s considerthe prime graph for PSL2(F412). To determine the vertices for the prime graph ofPSL2(F412), we need to determine the prime divisors of the order of the group.

|PSL2(F412)| =(412)(412 + 1)(412 − 1)

2= 2375051280


= 24 · 3 · 5 · 7 · 292 · 412

So our vertex set is π(G) = {2, 3, 5, 7, 29, 41}.To determine the edges, we need to know what orders of elements show up in this group.Using the theorem above, we have:

p = 41

q + 1

2=

1682

2= 841 = 292

q − 1

2=

1680

2= 840 = 23 · 3 · 5 · 7

Therefore, our edges are 2 · 3, 2 · 5, 2 · 7, 3 · 5, 3 · 7 and 5 · 7.

•41 •3

•2

•29 •5

•7

OOOOOOOOO

ooooooooo

444444444444444444

Here, we see that the first connected component is 3-regular, rather than 2-regular.

4. Fundamental Group

Recall that the fundamental group of a space is the set of homotopy classes of loops.As we are looking at finite simple groups of rank 2, it is natural to consider when thesegroups show up as fundamental groups of a space, or if there is any other way to relatethem to the fundamental group of a space.It turns out that the PSL2(Fp) groups can be related to the fundamental groups ofhyperbolic manifolds - we will not discuss hyperbolic geometry here, but use this resultto relate the knot group (the fundamental group of the knot exterior) of certain knots tothe PSL2(Fp) groups.

Theorem 4.1. [Long, Reid, [11]] Let M be a finite volume hyperbolic 3-manifold. ThenPSL2(Fp) is a quotient group of π1(M), for some p.

There is an interesting application of this theorem in knot theory - it turns out that allknot exteriors S3/K, for K a knot other than a torus or satellite knot, are (finitevolume) hyperbolic 3-manifolds [Menasco-Thistlethwaite, [13]]. A knot with hyperbolicexterior is called a hyperbolic knot. Some classic examples of hyperbolic knots are thefigure eight, Kinoshita-Terasaka and Conway knots. We recall the definitions of satelliteand torus knots:

Definition 4.2. A (p, q) - torus knot is a knot embedded in a torus that wrapsaround the torus p times meridianally, and q times longitudinally.

30 LINDSAY WHITE

Definition 4.3. [Marc Lackenby, ”The Crossing Number of Satellite Knots”] Let L be anontrivial knot in S3. Then a knot K is a satellite knot (with companion knot L) ifK lies in a regular neighbourhood, N(L) ,of the knot L. K also cannot lie in a 3-ball inN(L), nor be the core curve of N(L).

From the theorem above, we immediately have that if K is a knot other than a satelliteor torus knot, then the knot group of K, π1(S

3/K), has PSL2(Fp) as a quotient groupfor some p. In fact, when K is a (p, 2)-torus knot, we also have that π1(S

3/K) hasPSL2(Fp) as a quotient group. We can see this as follows: A (p, q)-torus knot has knotgroup presentation:

π1(S3/K) = 〈x, y|xp = yq〉

[Rolfsen, [15, p53]]. On the other hand, PSL2(Fp) has presentation as:

PSL2(Fp) = 〈x, y|xp = y2 = (xy)3 = (x2yxp+12 y)2 = 1〉

[Coxeter-Moser [5, p94]]. So when q = 2, clearly the knot group of the (p, 2) torus knothas the presentation:

π1(S3/K) = 〈x, y|xp = y2〉

As mentioned earlier, we can get a quotient of this by adding extra relations. Therefore,PSL2(Fp) is a quotient group of the knot group of a (p, 2)-torus knot. Note that wehave the same result for a (2, q)-torus knot by interchanging the roles of x and y in theknot group presentation.We then have the following theorem:

Theorem 4.4. Let K be a knot other than a satellite knot or (p, q) torus knot, withneither p = 2 or q = 2 (In other words, K can be a (2, q)- or (p, 2)-torus knot). Thenthe knot group of K, π1(S

3/K), has PSL2(Fp) as a quotient group for some p.

5. Algebraic Geometry

The main theme of this paper has been relating the finite simple groups of rank 2, anarea of algebra, to areas of geometry and topology. Algebraic geometry is a natural areato look at to illustrate this connection further. The Brieskorn Spheres are a constructionfrom algebraic geometry, and as we will see shortly, their fundamental groups can berelated to the PSL2(Fp) groups as well. This ties together the previous section onfundamental groups with both algebra and geometry. First we recall the defintion of ahomology 3-sphere, before defining the Brieskorn spheres.

Definition 5.1. [Rolfsen, [15, p244]] An integral homology 3-sphere is a3-dimensional manifold, X, having the same homology groups as the 3-sphere. That is,

H0(X,Z) = H3(X,Z) = Z

and

H1(X,Z) = H2(X,Z) = 0

Now we are ready to define Breiskorn spheres:


Definition 5.2 (Saveliev, [16, p4]). A Brieskorn 3-sphere, denoted Σ(p, q, r), wherep, q, r are pairwise relatively prime positive integers, is a homology 3-sphere, defined by:∑

(p, q, r) = V ∩ Sε

where V = {(x, y, z)|xp + yq + zr = 0} ⊂ C3 and Sε is a small 5-sphere.

We note that this construction does in fact produce a 3-dimensional manifold. Thevariety V induces one complex relation, which brings complex dimension down by 1,which is equivalent to reducing the real dimension by 2. Intersecting V with a 5-spherethen produces something with real dimension 3.

Example 5.3. Let X = Σ(2, 3, 7). It turns out that the fundamental group hasPSL2(F7) as a quotient group.We have:

π1(X) = 〈x, y, z, h|h central, x2h = y3h−1 = z7h−1 = xyz = 1〉

We have a presentation of PSL2(F7) as:

PSL2(F7) = 〈x, y, z|x2 = y3 = z7 = xyz = [y, x]4 = 1〉

Recall that the centre of a group G is the set of elements which commute with allelements of G. h central means that h is in the centre. We therefore always have h = 1in the centre, as 1 commutes with all elements. Now, if we choose h = 1 in ourpresentation of π1(X), we would get:

〈x, y, z|x2 = y3 = z7 = xyz = 1〉

which almost gives us PSL2(F7), except we need the additional relation [y, x]4 = 1.Therefore, we have PSL2(F7) as a quotient of π1(X):

PSL2(F7) = π1(X)/〈h, [y, x]4〉

[Boden, Herald, Kirk [3]]

It turns out that we always have PSL2(Fp) as a quotient of certain Brieskorn spheres:

Theorem 5.4. Any PSL2(Fp) group is a quotient of a Brieskorn (p, 2, 3) sphere.

Proof. The fundamental group of a Brieskorn (p, q, r) sphere has a presentation as:

π1(Σ(p, q, r)) = 〈x, y|xp = yq = (xy)r = 1〉

[Coxeter-Moser, [5, p67]] So a (p, 2, 3) sphere has presentation as:

π1(Σ(p, 2, 3)) = 〈x, y|xp = y2 = (xy)3 = 1〉

On the other hand, PSL2(Fp) has a presentation as:

PSL2(Fp) = 〈x, y|xp = y2 = (xy)3 = (x2yxp+12 y)2 = 1〉

[Coxeter-Moser, [5, p67]] So we can see that PSL2(Fp) is again a quotient ofπ1(Σ(p, 2, 3)). �

32 LINDSAY WHITE

6. Actions on Spheres

Theorem 6.1. [Main Result][Zimmermann, [20]] Of the finite simple groups of rank 2,only PSL2(F5) can act (orientation-preservingly and local linearly) on S2, none on S3

or S4. PSL2(Fp) admits a linear action on Sp, and A7, PSL2(F7) and PSU3(F3) canall act on a homology 5-sphere.

We will look at the S2 case. First, recall that by a group (G) action on Sn, we meanan injective homomorphism from G into the group of orientation preservinghomeomorphisms of Sn. We may also consider G to be a group of orientation-preservinghomeomorphisms of Sn. [Zimmermann [20]]There are a few different kinds of group actions, including:

Definition 6.2. [Zimmermann [20]] A topologogical action is a group action where Gacts by homeomorphism.

Definition 6.3. [Zimmermann [20]] A smooth action is a group action where G actsby diffeomorphism.

Definition 6.4. [Zimmermann [20]] A linear (orthogonal) action is a group actionwhere G acts on Sn by orthogonal maps; that is, G is a subgroup of SO(n+ 1).

Now, let’s consider finite groups acting on S2. We will consider G to be a finite group oforientation-preserving homeomorphisms of S2.

Theorem 6.5. [Zimmermann, [20]] Let G be a finite group of orientation-preservinghomeomorphism of S2. Then the action of G on a 2-manifold is locally linear.

Definition 6.6. A group action is locally linear if for each fixed point of a nontrivialelement of G (that is, for any x such that g(x) = x for some g ∈ G, g 6= id), there is aregular neighbourhood (since we’re looking at 2-manifolds, the neighbourhood will behomeomorphic to a 2-disk) around x on which 〈g〉 acts as a standard orthogonalrotation of the 2-disk. Note that 〈g〉 fixes x as g2(x) = g(g(x)) = g(x) = x, and so on.

Since we know (by the theorem above) that a group action on S2 is locally linear, let’sfirst look at the possible groups that could act on S2 that are linear. By definition of alinear action, these are subgroups of SO(3).The classification of all finite subgroups of SO(3) is as follows:

(1) Zn for any n (cyclic group of order n)(2) Dn for any n (dihedral group of order 2n)(3) A4, alternating group of order 12(4) S4, symmetric group of order 24(5) A5, alternating group of order 60

[Armstrong, [2, p105]]Of this list, we see immediately one finite simple group of rank 2 - A5

∼= PSL2(F7). Wenote that due to the classification of simple groups (see [Solomon, [18]]), the only othersimple groups in this list are Zp for p a prime. But these groups are not isomorphic toany of our finite simple groups of rank 2, so we have that the only finite simple group ofrank 2 that can act on S2 is PSL2(F7).


Conclusions

We have seen that the finite simple groups of rank 2 are useful in various areas withingeometry and topology. In knot theory, they help to distinguish the Kinoshita-Terasakaand Conway knots. They also are quotients of the fundamental groups of knot groupsand certain Brieskorn spheres. Some of these groups’ Caley graphs admit a Hamiltoniancycle, and some of these groups can also act on spheres.

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