UNIVERZA NA PRIMORSKEMFakulteta za matematiko, naravoslovje in informacijske tehnologije

Algebraic Graph Theory

doc. dr. Ademir Hujdurovic

DRUGO UCNO GRADIVO

41 strani

Matematika, dodiplomski studijski program

PRVA IZDAJA

Koper, 2018

Contents

1 Preliminaries 41.1 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Cyclic and dihedral groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Direct and semidirect product of groups . . . . . . . . . . . . . . . . . . . . . . 61.4 Symmetric group Sn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Alternating group An . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.6 Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.7 Orbits and stabilizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.8 Transitive, semiregular and regular group actions . . . . . . . . . . . . . . . . . 151.9 Counting orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.10 Cycle index of permutation group . . . . . . . . . . . . . . . . . . . . . . . . . . 171.11 Polya enumeration theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2 Graphs 192.1 Number of non-isomorphic graphs . . . . . . . . . . . . . . . . . . . . . . . . . 202.2 Number of non-equivalent (proper) colourings . . . . . . . . . . . . . . . . . . . 21

3 Vertex-transitive graphs 233.1 Cayley graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 Normal Cayley graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3 Graph products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.4 Hamiltonicity of vertex-transitive graphs . . . . . . . . . . . . . . . . . . . . . . 30

4 Edge-transitive and arc-transitive graphs 33

5 s-arc-transitive graphs 355.1 Cubic-arc-transitive graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

6 Distance-transitive graphs 376.1 Eigenvalues of strongly regular graphs . . . . . . . . . . . . . . . . . . . . . . . 38

2

Introduction

This is set of lecture notes on undergraduate course ”Algebraic Graph Theory” at Facultyof Mathematics, Natural Sciences and Information Technologies of University of Primorska,Slovenia. As this is an undergraduate subject, the material gives only introduction to Alge-braic Graph Theory. The material presented in these lecture notes is collected from differentsources, books, lecture notes etc.

3

1 Preliminaries

1.1 Groups

In this section we review the background on group theory.

Definition 1.1. A group is a set G together with a binary operation ∗ such that:

(i) for every a, b ∈ G, a ∗ b ∈ G;

(ii) (∀a, b, c ∈ G) a ∗ (b ∗ c) = (a ∗ b) ∗ c, (∗ is associative,);

(iii) there exists an identity element e ∈ G, such that x ∗ e = e ∗ x = x, ∀x ∈ G;

(iv) given identity element e ∈ G, for every element x ∈ G, there exists its inverse x−1 ∈ G,such that x ∗ x−1 = x−1 ∗ x = e, for all x ∈ G.

If a subset H ⊆ G is also a group with the same binary operation ∗ then we say that H is asubgroup of G, and we write H ≤ G.

Identity element in G will sometimes be denoted with 1G or simply by 1. To simplify thenotation, we usually write gh instead of g ∗ h. We refer to the number |G| of elements in Gas the order of the group G.

If x∗y = y∗x then we say that x and y commute. If every pair of elements in G commutes,then G is said to be abelian.

Exercise 1.2. Let G be a group and H ⊆ G. Prove that H ≤ G if and only if the followingthree conditions hold:

(i) 1G ∈ H;

(ii) ∀h1, h2 ∈ H, h1h2 ∈ H;

(iii) ∀h ∈ H, h−1 ∈ H.

Exercise 1.3. If H and K are subgroups of G, then also H ∩K is a subgroup of G.

Definition 1.4. If H ≤ G, then the right coset of H in G is the set of the form Hg = hg |h ∈ H. Similarly, left coset of H in G is the set of the form gH = gh | h ∈ H.

Exercise 1.5. If Hg1 and Hg2 are cosets, then either Hg1 = Hg2 or Hg1 ∩Hg2 = ∅.

Exercise 1.5 implies that G is a disjoint union of right cosets of H.

Definition 1.6. If H ≤ G, then the index [G : H] of H in G is the number of cosets of H inG.

It is now easy to see that for a finite group G, |G| = [G : H]|H|, and in particular theorder of H divides the order of G. This is known as Lagrange’s theorem.

Note that the possible converse of the Lagrange’s whould be the following: If n is a divisorof |G|, then there exists H ≤ G such that |H| = n. We will see latter that this does not hold.

If g is an element of a group G, the order of g is the least integer n ≥ 1 such that gn = e,if such n exists, otherwise we say that g is of infinite order. In finite group all elements havefinite orders.

4

If G is any group and S is a subset of G, then with 〈S〉 we denote the smallest subgroupof G that contains S, and call it the subgroup generated by S. If S = g1, g2, . . . , gn theninstead of 〈g1, g2, . . . , gn 〉 we simply write 〈g1, g2, . . . , gn〉.

If a group G contains an element g such that G = 〈g〉, then we say that G is a cyclicgroup.

Exercise 1.7. Let p be a prime, and let G be a group of order p. Prove that G is cyclicgroup.

Definition 1.8. Let G be a group and let N ≤ G. If g−1Ng = N for every g ∈ G, then N issaid to be a normal subgroup of G, and we denote this with N G.

Exercise 1.9. Prove that the following claims are equivalent:

(i) N is a normal subgroup of G;

(ii) gN = Ng for every g ∈ G;

(iii) the set of all left cosets of N in G coincides with the set of all right cosets of N in G.

Exercise 1.10. Prove that in an abelian group every subgroup is normal.

Exercise 1.11. Let G be a group and H an index 2 subgroup of G. Prove that H is a normalsubgroup of G.

Exercise 1.12. Let G be a group and N1 and N2 be two normal subgroups of G. Prove thatN1 ∩N2 is a normal subgroup of G.

Let G be a group and H,K ≤ G. Let HK = hk | h ∈ H, k ∈ K. Set HK is not ingeneral a subgroup of G.

Exercise 1.13. If H and K are subgroups of G then we have that

|HK| = |H||K||H ∩K|

.

Exercise 1.14. Let G be a group and H,K ≤ G. If one of H or K is normal in G, then HKis a subgroup of G.

Definition 1.15. If G is a group and N G, then the factor group G/N is the group withelements gN : g ∈ G and the group operation (g1N) ∗ (g2N) = (g1g2)N .

If G is a finite group, then |G/N | = |G|/|N |.

Exercise 1.16. Prove that the factor group defined in Definition 1.15 is indeed a group.

Definition 1.17. If G and H are groups, a homomorphism ϕ : G → H is a map ϕ from Gto H such that ∀g1, g2 ∈ G:

ϕ(g1 ∗G g2) = ϕ(g1) ∗H ϕ(g2), (1)

where ∗G is the group operation in G and ∗H is the group operation in H.

5

Exercise 1.18. Let ϕ : G→ H be a homomorphism and let K ≤ G. Prove that ϕ(K) ≤ H.

Definition 1.19. Let ϕ : G→ H be a homomorphism and let eH denote the identity in H.Kernel Ker(ϕ) of ϕ is defined with Ker(ϕ) = g | g ∈ G,ϕ(g) = eH.

Exercise 1.20. Let ϕ : G→ H be a homomorphism. Prove that Ker(ϕ) G.

Definition 1.21. A bijective homomorphism ϕ : G → H is called isomorphism. If thereexists an isomorphism between G and H we say that G and H are isomorphic and writeG ∼= H.

1.2 Cyclic and dihedral groups

Let n be a positive integer. With Cn we will denote a cyclic group of order n, that isCn = 〈a〉, and an = 1. It can be shown that any two cyclic groups of order n are isomorphic.The following construction of cyclic group of order n will be useful later on.

Let Zn = 0, 1, . . . , n− 1 be the set of residues modulo n. Then it is not difficult to seethat Zn is a group with operation +n, that is addition modulo n. Identity element in thisgroup is 0, and x−1 = −x (mod n). This group is cyclic and of order n.

For a positive integer n, dihedral group D2n is generated by two elements ρ and τ , suchthat ρn = 1 and τ2 = 1. The remaining elements in D2n are given with

D2n = 1, ρ, . . . , ρn−1, τ, ρτ, . . . , ρn−1τ.

The operation in D2n is given with

ρi ∗ ρj = ρi+j ;

ρi ∗ ρjτ = ρi+jτ ;

ρiτ ∗ ρj = ρi−jτ ;

ρiτ ∗ ρjτ = ρi−j .

Dihedral group D2n is isomorphic to the group of all symmetries of a regular n-gone. Onestep rotation of the n-gon corresponds to the element ρ, and τ corresponds to a reflection ofthe n-gon.

1.3 Direct and semidirect product of groups

In this section we briefly present methods of constructing new groups from old ones. The firstand the simplest method is the direct product of groups.

Definition 1.22. Given two group G1 and G2, the direct product G1 × G2 of G1 and G2 isthe group with elements (g1, g2) : g1 ∈ G1, g2 ∈ G2 and the group operation is given with(x1, x2) ∗ (y1, y2) = (x1y1, x2y2).

Exercise 1.23. Prove that G1 ×G2 as defined above is indeed a group.

Theorem 1.24. Let G be a group and let H and K be a subgroups of G. Then G isisomorphic to the direct product of H and K if and only if the following three conditionshold:

6

(i) H and K are normal in G;

(ii) G = HK;

(iii) H ∩K = 1G.

Exercise 1.25. Determine which of the following groups are isomorphic: Z2×Z3, Z6, Z4×Z2,Z8.

A generalization of the notion of direct product is the notion of semidirect product.

Definition 1.26. Let G be a group and let N and K be a subgroups of G. Then G issemidirect product of N and K if and only if the following three conditions hold:

(i) N is a normal subgroup of G;

(ii) G = NK;

(iii) N ∩K = 1G.

We let N oK denote the semidirect product of N by K.

It is clear that every direct product is also a semidirect product, but semidirect productdoes not need to be a direct product. To get a better understanding of the structure of asemidirect product we first need to introduce some more terminology.

Definition 1.27. If G is a group then an automorphism of G is an isomorphism from G toG. The set of all automorphisms of a group G is denoted with Aut(G).

Exercise 1.28. The set Aut(G) together with the operation of composition of functions is agroup.

Definition 1.29. Two elements x and y in a group G are said to be conjugate if there existsan element g ∈ G such that gxg−1 = y. We often write gxg−1 = xg.

We can extend the notion of conjugate elements to the notion of conjugate subgroups. Wesay that H and K are conjugate subgroups of G if there exists g ∈ G such that gHg−1 = K.Again we will usually write Hg = gHg−1. It is not difficult to see that the relation ”isconjugate” is an equivalence relation in G. The equivalence classes of this relation are calledconjugacy classes.

Exercise 1.30. If G is a group and g ∈ G, then the mapping ϕ : G → G defined withϕ(x) = gxg−1 is an automorphism of G.

The automorphisms given in previous exercise are called inner automorphisms.

Exercise 1.31. The set Inn(G) of all inner automorphisms of G is a normal subgroup ofAut(G).

Exercise 1.32. Prove that N is normal subgroup of G if and only if ϕ(N) = N for everyϕ ∈ Inn(G).

Definition 1.33. A subgroup N of a group G is called characteristic if ϕ(N) = N , for everyϕ ∈ Aut(G). If N is characteristic subgroup of G we write N char G.

7

Exercise 1.34. If N char G then N G.

An important property of characteristic subgroup is the following:

Exercise 1.35. If G is a group and C and N are subgroups of G with C ≤ N ≤ G such thatC char N and N G then C G.

We now go back to the semidirect product of two groups.

Theorem 1.36. Let N and H be a groups and ϕ : H → Aut(N) be a homomorphism. LetG be the set N ×H with operation

(n1, h1) ∗ (n2, h2) = (n1 · ϕ(h−11 )(n2), h1h2).

Then G is isomorphic to a semidirect product of N by H.

Proof. We identify N with the set (n, 1H) | n ∈ N and H with the set (1N , h) | h ∈ H.It is now clear that N ∩ H = 1. We leave as an exercise to the reader to verify thatthe operation ∗ in G is associative, that (1N , 1H) is the identity element in G, and that(n, h)−1 = (ϕ(h)(n−1), h−1).

To see that N is normal in G we observe that

(n, h)−1(n′, 1H)(n, h) = (ϕ(h)(n−1), h−1)(n′, 1H)(n, h)

= (ϕ(h)(n−1)ϕ(h)(n′)), h−1)(n, h)

= (ϕ(h)(n−1)ϕ(h)(n′)ϕ(h)(n), 1)

= (ϕ(h)(n−1n′n), 1) ∈ N

It remains to prove that G = NH. Let g ∈ G be arbitrary. Then by the definition of G,it follows that g = (n, h) for some n ∈ N and h ∈ H. It is now easy to see that g = nh, wheren = (n, 1) and h = (1, h).

We are now going to show how the dihedral group D2n can be constructed as the semidirectproduct of groups Cn and C2. Let N = 〈ρ | ρn = 1〉 and H = 〈τ | τ2 = 1〉. In order toconstruct a semidirect product of groups N and H, we need a homomorphism ϕ : H →Aut(N). Let ϕ be defined with ϕ(1H) = idN and ϕ(τ) = ιN , where idN stands for identitymap in N , that is idN (x) = x (∀x ∈ N) and ιN stands for the inversion map in N , that isιN (x) = x−1 (∀x ∈ N).

Exercise 1.37. Prove that ϕ defined as above is indeed a homomorphism that maps H toAut(N).

Now using Theorem 1.36 we construct the group N o H ∼= Cn o C2. We have that theelements in the group N oH are given with N oH = (ρi, τk) : i ∈ 0, . . . , n− 1, k ∈ 0, 1.The group operation is given with

(ρi, 1H) ∗ (ρj , 1H) = (ρi · ϕ(1H)(ρj), 1H) = (ρi+j , 1H)

(ρi, 1H) ∗ (ρj , τ) = (ρi · ϕ(1H)(ρj), τ) = (ρi+j , τ)

(ρi, τ) ∗ (ρj , 1H) = (ρi · ϕ(τ)(ρj), 1H) = (ρi−j , τ)

(ρi, τ) ∗ (ρj , τ) = (ρi · ϕ(τ)(ρj), 1H) = (ρi−j , 1H)

8

Exercise 1.38. Prove that the mapping φ : N oH → D2n defined with

φ((ρi, 1H)) = ρi;

φ((ρi, τ)) = ρiτ ;

is isomorphism.

1.4 Symmetric group Sn

Let X be a nonempty set. A bijective mapping g from X to X is also called a permutationof X. Let Sym(X) denote the set of all permutations of the set X. If X has size n, then|Sym(X)| = n!.

Exercise 1.39. Let denote the symbol for composition of functions, that is

∀g, h ∈ Sym(X) (g h)(x) = g(h(x)),

Prove that Sym(X) together with operation forms a group.

The definition of the operation in Sym(X) means that for g1, g2 ∈ Sym(X) the productg1g2 is carried out by first applying g2 to x ∈ X , and then g1 to the image g2(x), that is weare multiplying permutations from right to left.

We denote the identity element of Sym(X) with idX . Let x ∈ X and g ∈ Sym(X). Wesay that x is a fix element of g, or x is fixed by g if g(x) = x. Otherwise, we say x is movedby g.

Definition 1.40. A permutation group is any subgroup G of Sym(X) for some nonemptyset X.

We say G is a finite permutation group when X is a finite set. In this note every set ismeant to be finite, unless it is said otherwise.

Notation. We denote by Sn the permutation group Sym(1, . . . , n). We write idn (or idwhen n is understood from the context) for id1,...,n.

There are two common ways in which permutation groups can be written. First is thetwo line notation for permutations, that was introduced by Cauchy. A permutation g of a setX = x1, x2, . . . , xn is written in the following way:

g =

(x1 x2 x3 · · · xn−1 xng(x1) g(x2) g(x3) · · · g(xn−1) g(xn)

)The other notation is to write g ∈ Sym(X) as a product of disjoint cycles.

Definition 1.41. A permutation g ∈ Sym(X) is called a k-cycle if there exists distinctelements x1, . . . , xk in X such that g(xk) = x1, and g(xi) = xi+1 for all i ∈ 1, . . . , k − 1,and g(x) = x for all x ∈ X \ x1, . . . , xk.

Notation. We denote by (x1 x2 . . . xk) the k-cycle g in Definition 1.41.

A 2-cycle is also called a transposition. Two cycles g1 = (x1 . . . xk) and g2 = (x′1 . . . x′l)

are disjoint if x1, . . . , xk ∩ x′1, . . . , x′l = ∅. Note that, this implies that g1g2 = g2g1.

9

Example 1.42. Let X = 0, 1, 2, 3, 4, 5, 6 and let g ∈ Sym(X) be given with g(x) = 4x+ 1(mod 7). Then the permutation g can be written as

g =

(0 1 2 3 4 5 61 5 2 6 3 0 4

)or as a product of disjoint cycles g = (0 1 5)(2)(6 4 3) = (6 4 3)(0 1 5) = . . . = (0 1 5)(3 6 4).

Observe that the order in which the disjoint cycles are written is not important.

Theorem 1.43. (Cyclic Decomposition) Every permutation g ∈ Sym(X) can be writtenas the product of disjoint cycles. Factorization is unique up to the order of factors.

Corollary 1.44. Every permutation can be written as the product of transpositions.

Proof. It is clear that

(a1 a2 . . . ar) = (a1 ar)(a1 ar−1) . . . (a1 a2).

By Theorem 1.43 it follows that every permutation can be written as a product of disjointcycles. On the other hand, every cycle can be written as the product of transposition, hencethe claim holds.

Exercise 1.45. Let g, h ∈ Sym(X), such that g has a cyclic decomposition

g = (a1 . . . ar) . . . (as . . . an).

Then hgh−1 has a cylic decomposition

hgh−1 = (h(a1) . . . h(ar)) · · · (h(as) . . . h(an)).

We call a cyclic decomposition of g to be maximal if it includes all possible 1-cycles. Apermutation g ∈ Sym(X) has cyclic structure [c1, . . . , cn] if a maximal cyclic decompositionof g consists of ck number of k-cycles, where k ∈ 1, . . . , n. Note that, the equality c1 +2c2 +· · ·+ ncn = n always holds. The following corollary of Theorem 1.45 is useful when studyingthe conjugacy of two permutations.

Corollary 1.46. The permutations g1 and g2 are conjugate in Sym(X) if and only if g1 andg2 have the same cyclic structure.

Exercise 1.47. Let g1, g2 ∈ S12 be given with their cyclic decompositions:

g1 = (1 3 5)(2 4 6)(7)(8 12)(9 10 11)

g2 = (1)(3 2 6)(7 4 5)(9 8 11)(10 12)

Find permutation h ∈ S12 such that hg1h−1 = g2.

10

1.5 Alternating group An

In this part we construct a subgroup of Sn of index 2. We saw that any permutation canbe written as the product of transpositions. However, the number of transpositions is notuniquely determine. On the other hand, we are now going to prove that the parity of thenumber of transpositions for a given permutation is fixed.

Theorem 1.48. If a permutation g ∈ Sn can be written as a product of r and s transpositions,then r ≡ s (mod 2).

Definition 1.49. A permutation g ∈ Sn is called even (odd) if g is the product of even(odd) number of transpositions.

Exercise 1.50. Prove that a cycle of length k is even permutation if and only if k is odd.

Notation. We denote by Alt(X) the subgroup of even permuations in Sym(X). In the casewhen |X| = n, then we denote by An the set all even permuations in Sn.

We show below that An, n ≥ 2 is in fact a subgroup of Sn of index 2 . The group An iscalled the alternating group of degree n.

Theorem 1.51. The set An, n ≥ 2 of all even permutations is a subgroup of Sn of index 2.

Proof. It is clear that id ∈ An.Let g, g′ ∈ An. The group Sn is generated by the set of all transpositions. Thus g = t1 · · · tr

and g′ = t′1 · · · t′s, where r and s are even. Then gg′ = t1 · · · trt′1 · · · t′s, r + s is even, and sogg′ ∈ An. Also, g−1 = tr · · · t1, and so g−1 ∈ An. We obtain that An is indeed a subgroup ofSn. Then (1, 2) /∈ An. It is not difficult to see that the mapping g 7→ (1, 2)g, where g ∈ An,is a bijective mapping from An to Sn \An. Thus |An| = n!/2.

Note that, since An has index 2 in Sn, An / Sn.

Exercise 1.52. Let n ≥ 2, and let H = id, (1 2) be a cyclic subgroup of Sn of order 2.Prove that Sn ∼= An oH.

Exercise 1.53. Let g ∈ Sn be a permutation and let [c1, c2, . . . , cn] be the cyclic structure ofg. Prove that g is odd permutation if and only if

n/2∑i=1

c2i ≡ 1 (mod 2).

Exercise 1.54. Prove that the group An (n ≥ 3) is generated by the set of all cycles of length3.

Exercise 1.55. Prove that A4 has no subgroup of order 6. (Hint: Use Exercises 1.11 and1.13).

11

Proof. Suppose that A4 has a subgroup N of order 6. Then N is an index 2-subgroup of A4

and by Exercise 1.11 N is normal in A4. Let V = id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3). ThenV is a subgroup of A4 of order 4. Since 4 does not divide 6 and V is not subgroup of N , andso N is a proper subgroup of V N . This implies that NV = A4. Now, using Exercise 1.13 weobtain

|N ∩ V | = |N ||V ||NV |

=6 · 412

= 2.

Since both N and V are normal in A4, it follows that N ∩V is also a normal subgroup of A4.However, the only subgroups of A4 of order 2 are of the form id, (a b)(c d). Let g = (1 2 3).Then (1 2)(3 4)g = (2 3)(1 4), (1 3)(2 4)g = (2 1)(3 4) and (1 4)(2 3)g = (2 4)(3 1), hence nosubgroup of order 2 in A4 is normal.

1.6 Group actions

In this part we introduce the concept of an action of a group. The motivation behind thisconcept is to study groups by representing them as permutation groups.

Definition 1.56. An action of a group G on a set X is a mapping f : G × X → X suchthat, writing the image f((x, g)) as g · x, the following properties hold.

A1. 1G · x = x for every x ∈ X.

A2. (g1g2) · x = g1 · (g2 · x) for every x ∈ X and for every g1, g2 ∈ G.

Given an action of a group G on a set X, we will also say that G acts on X. The cardinality|X| is called the degree of the action. Instead of g · x we will sometimes write gx or g(x).

Example 1.57. Let G = R \ 0 be the group of nonzero real numbers under the multipli-cation, and let X be the set of all vectors in R3. Then G acts on X via scalar multiplication,that is g · (x1, x2, x3) = (gx1, gx2, gx3), for (x1, x2, x3) ∈ R3 and g ∈ R \ 0.

Example 1.58. Let G ≤ Sym(X). Then G acts naturally on X with g · x = g(x), for x ∈ Xand g ∈ G.

Example 1.59. Let G be the group acting on a set X. Then G acts on the set X ×X withg · (x1, x2) = (g · x1, g · x2).

Theorem 1.60. Let G act on X and fix g ∈ G.

(1) The mapping πg : X → X, πg : x 7→ g(x) is in Sym(X).

(2) The mapping ρ : G→ Sym(X), ρ : g 7→ πg is a homomorphism.

(3) Let Φ : G→ Sym(X) be a homomorphism. Then this homomorphism defines action of agroup G on the set X with g · x := Φ(g)(x).

Definition 1.61. For an action of a group G on a set X we call the homomorphism ρ inTheorem 1.60 the corresponding permutation representation of G on X. The degree ofan action (or a premutation representation) is the size of X.

12

Definition 1.62. By the kernel and image of an action of G on a set X we mean the kerneland image, respectively, of the corresponding permutation representation ρ. We say that theaction is faithful if its kernel is trivial.

We illustrate the above definitions through the following example.

Example 1.63. Let G = D12 be the group of symmetries of a regular hexagon acting on itspoints 1, 2, 3, 4, 5, 6. Let d1 = 14, d2 = 25 and d3 = 36 be the three main diagonals of thehexagon (order in which the points are written is not important, for example d1 = 14 = 41).Let X = d1, d2, d3. Prove that G acts on the set X via g · x1x2 = g(x1)g(x2). Determinethe image and the kernel of this action. Is this action faithful?

Now we jump to a more theoretical example.

Theorem 1.64. Let G be a group, and let us take X to be G. Then G acts on X with

g · x = gx, x, g ∈ G.

This action is faithful.

Proof. We show first that we indeed have an action, i.e., the mapping defined above satisfiesAxioms A1 and A2.

A1: For every x ∈ X, 1G · x = 1Gx = x.

A2: For every x ∈ X and for every g1, g2 ∈ G, (g1g2) ·x = g1g2x = (g1)(g2x) = g1 · (g2 ·x).

Let g ∈ G be in the kernel of the action. This means g · x = x for all x ∈ X. But this justmeans gx = x for all x ∈ X, hence g = 1G, and the kernel is the trivial group. By definition,the action is faithful.

The permutation representation corresponding to the action of G in the previous theoremis called the left regular representation of G. We illustrate the usage of the left regularrepresentation in proving that every group is isomorphic to a permutation group.

Theorem 1.65 (Cayley). Every group is isomorphic to a permutation group.

Proof. Let G be a group. Let G act on X = G via left multiplication. By Theorem 1.60,it follows that ρ : G → Sym(X) is a group homomorphism (recall that ρ(g) = πg, andπg(x) = gx for x ∈ G). Then the image of G under this homomorphism ρ(G) is a subgroupof Sym(X). By the first isomorphism theorem, it follows that G/Ker(ρ) ∼= ρ(G). Since byTheorem 1.65 it follows that Ker(ρ) = 1G it follows that G ∼= ρ(G).

1.7 Orbits and stabilizers

In this part we introduce the two most basic concepts in connection with an action – theorbits and the stabilizers.

Definition 1.66. Let G act on X, and let x ∈ X. The orbit of G induced by x is the setg · x | g ∈ G.

13

Notation. We denote by OrbG(x) the orbit of G induced by x.

Definition 1.67. Let G act on X, and let x ∈ X. The stabilizer of x in G is the setg ∈ G | g · x = x.

Notation. We denote by Gx the stabilizer of x in G.

Example 1.68. Let X = Z10 = 0, 1, . . . , 9, and let G = Aut(Z10) = Z∗10 acting naturallyon X. Calculate the stabilizers G0, G1, and orbits 0G, 1G, 2G.

We have the following properties of orbits and stabilizers.

Theorem 1.69. Let G act on X and let x ∈ X.

(1) The set of all orbits OrbG(x) form a partition of X.

(2) The stabilizer Gx is a subgroup of G.

Proposition 1.70. Let G act on a set X, let g ∈ G, x ∈ X and let y = g · x. Then

Gy = gGxg−1.

Proof.

Gy = h ∈ G | h · y = y= h ∈ G | h · (g · x) = g · x= h ∈ G | (g−1hg) · x = x= h ∈ G | g−1hg ∈ Gx= h ∈ G | h ∈ gGxg−1= gGxg

−1.

Theorem 1.71 (Orbit-stabilizer). Let G be a group acting on a set X. Then for every x ∈ Xit holds |G| = |OrbG(x)| · |Gx|.

Proof. Let x ∈ X be fixed. Define the set Ω = (g, y) ∈ G×OrbG(x) | gx = y. We calculatethe cardinality of Ω with “double counting”. First, we obtain that

|Ω| =∑g∈G

∣∣ (g, y) ∈ G×OrbG(x) | y = gx∣∣ =

∑g∈G|y ∈ OrbG(x) | y = gx| =

∑g∈G

1 = |G|.

Second, we obtain that

|Ω| =∑

y∈OrbG(x)

∣∣ (g, y) ∈ G×OrbG(x) | y = gx∣∣ =

∑y∈OrbG(x)

|g ∈ G | y = gx|.

Let y ∈ OrbG(x) be fixed. Then there exists h ∈ G such that y = hx. Now we have

g ∈ G | y = gx = g ∈ G | hx = gx = g ∈ G | h−1gx = x = g ∈ G | h−1g ∈ Gx = h(Gx).

14

Since |hGx| = |Gx|, it follows that

|Ω| =∑

y∈OrbG(x)

|Gx| = |OrbG(x)| · |Gx|.

We conclude that |Ω| equals |G| = |OrbG(x)| · |Gx|, as required.

We are now going to present two more important types of group actions. First is theso-called conjugation action of G on G defined with

xg = gxg−1, x, g ∈ G.

Proposition 1.72. The conjugation action of a group G on G is well-defined group action.For g ∈ G, the orbit ORbG(g) is the conjugacy class of g in G. The stabilizer Gg is thecentralizer CG(g) of g in G. The kernel of this action is the center Z(G) of G.

Exercise 1.73. Does group G act on G with

xg = g−1xg, x, g ∈ G?

Next important example of group action is the so called coset action. Let G be a groupand H ≤ G. The coset action is the action of G on the set of left H-cosets in G definedwith

g · (xH) = gxH, x, g ∈ G.

Proposition 1.74. The coset action is indeed an action. The stabilizer of xH is GxH =xHx−1. The kernel is ∩x∈GxHx−1.

The kernel of the coset action is also called the core of H in G (notation: coreG(H)).

Exercise 1.75. Prove that coreG(H) is the largest normal subgroup of G which is containedin H.

Exercise 1.76. Let G be a p-group, that is a group of order pk, where p is a prime. Usingthe conjugation action, prove that Z(G) the center of G, is non-trivial.

1.8 Transitive, semiregular and regular group actions

Definition 1.77. Let G act on X. The action is transitive if it has only one orbits, and itis semiregular if |Gx| = 1 for all x ∈ X. An action which is both transitive and semiregularis called regular.

Example 1.78. Left regular representation of G, that is the action of G on G given withg · x = gx is regular action.

Proposition 1.79. Let G act transitively on X. Then cardinality |X| is a divisor of |G|.

Proposition 1.80. Let G act transitively on X. Then G is regular if and only if |G| = |X|.

Proposition 1.81. Let G be an abelian group which is transitive and faithful on the set X.Then G is regular on X.

15

Proof. Since G is transitive, it remains to prove that G is semiregular. Suppose that G is notsemiregular, that is, there exists x ∈ X, such that Gx 6= 1. Since G acts transitively on X,then for any x ∈ X we have OrbG(x) = X. Using Proposition 1.70 it follows that any twostabilizers are conjugate. However, since G is abelian, it follows that all stabilizers are thesame. Then the kernel of this action is equal to Gx, which contradicts the assumption thatG is faithful.

Exercise 1.82. Let p be a prime, and let G ≤ Sp be transitive. Prove that there exists acyclic subgroup H of G, such that H is regular.

Exercise 1.83. Let G be a group. Prove that the direct product G × G acts on G with(g1, g2) · x = g1xg

−12 . Prove that this action is transitive, and calculate the vertex stabilizer

(G×G)1. When is this action faithful?

Exercise 1.84. Let G act transitively on X and let H ≤ G. Prove that G = HGx if andonly if H is transitive.

1.9 Counting orbits

Definition 1.85. Let G be a group acting on a set X, and let g ∈ G. Then with fixX(g) wedenote the set of elements of X fixed by g, that is fixX(g) = x ∈ X | g · x = x. When theset X is clear from the context we simply write fix(g).

There is a simple relationship between the number of orbits of a finite group acting ona finite set and the number of fixed points of its elements. A wide range of applicationsin counting problems and combinatorics is based on elaborations of this relationship. Thetheorem itself has a long history and is often referred to (inaccurately) as the ”BurnsideLemma”; the simplest version is the following result.

Theorem 1.86 (Cauchy-Frobenius or Burnside). Let G be a finite group acting on a finiteset X, and let m be the number of orbits in this action. Then

m =1

|G|∑g∈G|fix(g)|.

Proof. Define the set Ω = (x, g) ∈ X × G | gx = x. We are going to count the number ofelements of Ω in two different ways. First suppose that the orbits of G are X1, . . . , Xm. Thenwe have

|Ω| =m∑i=1

|(x, g) ∈ Xi ×G | gx = x|. (2)

Suppose that x ∈ Xi is arbitrary. Then (x, g) ∈ Xi ×G | gx = x = xi ×Gx. Recall thatby orbit-stabilizer property, we have |Gx||Xi| = |G|. Hence we have

|Ω| =m∑i=1

∑x∈Xi

|Gx| =m∑i=1

∑x∈Xi

|G||Xi|

=

m∑i=1

|G| = m|G|. (3)

On the other hand we have

|Ω| =∑g∈G|(x, g) ∈ X ×G | gx = x| =

∑g∈G|fix(g)|. (4)

Combining (3) and (4) the result follows.

16

Corollary 1.87. If G is a finite transitive permutation group of degree n > 1 then G containsan element with no fixed points.

Proof. Since G acts transitively, we have that the number of orbits is m = 1. Then byCauchy-Frobenius theorem, we have |G| =

∑g∈G |fix(g)|. Since for 1G we have fix(1G) = X

and |X| ≥ 2, then it follows that there exists g ∈ G, such that fix(g) = ∅.

1.10 Cycle index of permutation group

Definition 1.88. Let G ≤ Sym(X), where X is a finite set of size n, and let g ∈ G. Let[c1(g), . . . , cn(g)] be the cyclic structure of permutation g, that is, ck(g) denotes the numberof cycles of length k in the cyclic decomposition of g. The cycle index ZG of permutationgroup G is the polynomial

ZG =1

|G|∑g∈G

xc1(g)1 · xc2(g)

2 · . . . · xcn(g)n

Exercise 1.89. Calculate ZS3 .

Exercise 1.90. Calculate ZS4 .

Exercise 1.91. Prove that ZCn = 1n

∑d|n ϕ(d)x

n/dd .

Exercise 1.92. Prove that ZDn = 12ZCn +

12x1x

(n−1)/22 , n odd,

14

(x2

1x(n−2)/22 + x

n/22

), n even.

Exercise 1.93. Prove that

ZSn =∑(c)

1∏nk=1 k

ckck!

n∏k=1

xckk ,

where the summation is over all integer partitions (c) of n.

1.11 Polya enumeration theorem

The result we are going to present now, and is usually referred to as Polya enumerationstheorem, was first published by John Howard Redfield in 1927. In 1937 it was independentlyrediscovered by George Polya, who then greatly popularized the result by applying it to manycounting problems, in particular to the enumeration of chemical compounds.

Let X be a finite set and let G ≤ Sym(X) be a group of permutations of X. The setX may represent a finite set of beads, and G may be a chosen group of permutations of thebeads. For example, if X is a necklace of n beads in a circle, rotations and reflections arerelevant so G is the dihedral group D2n of order 2n.

Definition 1.94. Let C be a finite set of colors. Let CX be the set of functions X → C, thatis CX is the set of all colorings of the set X with colors from C.

17

For f ∈ CX , and for g ∈ G, we define an induced action of G on the set CX , withg · f = f g−1. Two functions f1, f2 ∈ CX are called G-equivalent, if they belong to thesame orbit under the induced action of G on CX , or more precisely, if there exists g ∈ G suchthat f1(g(x)) = f2(x), for every x ∈ X. Polya’s enumeration theorem given below counts thenumber of orbits in this action, or equivalently the number of non-equivalent colorings of theset X with colors from C.

Theorem 1.95 (Polya enumeration theorem). Let X and C be finite sets, and let G ≤Sym(X). The number of orbits in the induced action of G on the set CX is equal to ZG withx1 = . . . = xi = . . . x|X| = |C|.

Proof. By Cauchy-Frobenius theorem, it follows that the number of orbits is

1

|G|∑g∈G|fixCX (g)|.

It remains to determine the size of fixCX (g) = f : X → C | f g−1 = f. Let f ∈ fixCX (g).Suppose that (x1 x2 . . . xk) is one cycle in the cyclic decomposition of g. Since f g = f , itfollows that f(x2) = (f g−1)(x2) = f(g−1(x2)) = f(x1). Similarly, we conclude that f(x1) =f(x2) = . . . f(xk). Therefore, all vertices in the same orbit of g have the same image underf . Since the number of possible images under f is |C|, it follows that |fixCX (g)| = |C|c(g),where c(g) is the number of cycles in the cyclic decomposition of g.

18

2 Graphs

A graph Γ consist of vertex set V (Γ) and edge set E(Γ), where an edge is an unordered pairof distinct vertices u, v. We will sometimes write uv instead of u, v. If u, v is an edge,we say that u and v are adjacent and that v is a neighbour of u, and we will denote this byu ∼ v. A set Γ(v) denotes the set of all neighbours of v, that is Γ(v) = u ∈ V (Γ) : u ∼ v.Degree of a vertex v ∈ V (Γ) is defined as the number of its neighbours, and is denoted bydΓ(v), that is dΓ(v) = |Γ(v)|. If the graph Γ is clear from context, then we simply write d(v).Order of a graph is the cardinality of its vertex-set.

Two graphs X1 = (V1, E1) and X2 = (V2, E2) are equal if V1 = V2 and E1 = E2. Althoughthis is a reasonable definition, for most purposes, the model of relationship is not changed ifwe rename the vertices of a graph. This naturally leads to the definition of isomorphic graphs.

Definition 2.1. For graphs X and Y , a function ϕ : V (X) → V (Y ) is called isomorphismfrom X to Y , if

1. ϕ is bijection;

2. ∀u, v ∈ V (X) u, v ∈ E(X)⇔ ϕ(u), ϕ(v) ∈ E(Y ).

Graphs X and Y are called isomorphic if there exist an isomorphism from X to Y , and wedenote this by X ∼= Y .

For a function ϕ : A→ B and for a subset S ⊆ A we define ϕ(S) = ϕ(s) : s ∈ S.

Lemma 2.2. Let ϕ be an isomorphism from X to Y . Then for every v ∈ V (X), we haveϕ(X(v)) = Y (ϕ(v)).

Corollary 2.3. Isomorphisms preserve degrees of vertices, that is for an isomorphism ϕ fromX to Y we have dX(v) = dY (ϕ(v)), ∀v ∈ V .

One can also consider isomorphism from a graph Γ to itself. In this case, instead of ”iso-morphism” we use word ”automorphism”. We repeat the definition to stress its importance.

Definition 2.4. Let Γ be a graph. An automorphism of Γ is a bijective function ϕ : V (X)→V (X) such that ∀u, v ∈ V (X) we have u, v ∈ E(Γ) ⇔ ϕ(u), ϕ(v) ∈ E(Γ). The set of allautomorphisms of a graph Γ is denoted by Aut(Γ).

Exercise 2.5. Determine all the automorphisms of the following graph.

1 2

3

4

5 6

7 8 9

Theorem 2.6. For a graph Γ, the set Aut(Γ) is group under the operation of compositionof functions, that is for f, g ∈ Aut(Γ), fg = f g, or equivalently (fg)(v) = f(g(v)) for everyv ∈ V (Γ).

19

Definition 2.7. The complement Γ of a graph Γ is the graph with V (Γ) = V (Γ) and forevery two distinct vertices u, v ∈ V (Γ) we have u, v ∈ E(Γ)⇔ u, v 6∈ E(Γ).

Lemma 2.8. For every graph Γ we have Aut(Γ) = Aut(Γ).

A graph Γ is called asymmetric if Aut(Γ) = id.

Exercise 2.9. For every positive integer n ≥ 6 there exists an asymmetric graph with nvertices.

2.1 Number of non-isomorphic graphs

In this section we will show how Polya’s enumeration theorem can be used for determiningthe number of non-isomorphic graph of given order. We will first demonstrate this on graphsof order 4.

Example 2.10. Prove that there are 11 non-isomorphic graphs with 4 vertices.

Proof. Let G(4) be the set of all graphs with vertex set V = 1, 2, 3, 4. Let X = i, j |1 ≤ i < j ≤ 4. We can think of X as the edge set of complete graph K4. Let C = w, b.Consider the set CX , that is set of all functions f : X → C. For any such function f , one candefine a graph Γf , with E(Γf ) = x ∈ X : f(x) = b, or in other words f , is colors edges ofK4 with two colors (w for white, and b for black), the graph Γf is the subgraph with all blackedges. Conversely, for any graph Γ with vertex set V , one can define a function fΓ : X → C,where fΓ(u, v) = 1 if and only if u, v ∈ E(Γ).

For a permutation g ∈ S4, with g(2) we denote the permutation of X induced by g, and

with S(2)n the corresponding permutation group of X. For example, if g = (1 2)(3 4) then

g(2) = (1, 2)(3, 4)(1, 3 2, 4)(1, 4 2, 3).

It is now easy to see that two graphs with vertex set V are isomorphic if and only if the

corresponding collorings are equivalent under the action of S(2)4 .

By Polya’s theorem, the number of non-equivalent collorings is ZS(2)4

(x1 = x2 = x3 = x4 =

2). It is left as exercise to calculate the cyclic index of S(2)4 .

Exercise 2.11. Let g ∈ Sn be a permutation with ck cycles of length k. The number ofcycles in the cyclic decomposition of g(2) is

γ(g) =n∑k=1

kc2k+1 +∑k

kc2k +∑k

kck(ck − 1)

2+∑r<t

(r, t)crct.

Exercise 2.12. Let V = 1, . . . , n, let X = i, j | 1 ≤ i < j ≤ n. For g ∈ Sn letg(2) denote the induced permutation of X and let γ(g) denote the number of cycles in cyclicdecomposition of g(2). Prove that the number of non-isomorphic graphs of order n is

1

n!

∑g∈Sn

2γ(g).

20

Definition 2.13. Let V be a finite set, and let P(V ) denote the power set of V . HypergraphH = (V, E) is given with the vertex set V and the hyperedge set E ⊆ P(V ) \ ∅. Twohypergraphs H1 = (V1, E1) and H2 = (V2, E2) are said to be isomorphic if there exists bijectivefunction f : V1 → V2 such that f(E1) = E2.

Exercise 2.14. Determine the number of non-isomorphic hypergraphs with n vertices, forn ∈ 3, 4.

2.2 Number of non-equivalent (proper) colourings

Suppose that we are given a graph Γ and a set of colors C, and we want to color the verticesof Γ with colors from C. For two such colourings f1 and f2 we will say that are equivalent, ifthere exist an automorphism ϕ ∈ Aut(Γ) such that f2 = f1ϕ. In other words, two colouringsare equivalent, if one can be obtained from the other after applying an automorphism of thegraph Γ. Note that here we are not talking about proper colourings, that is adjacent verticescan be coloured in the same way.

Theorem 2.15. The number of non-equivalent colourings of a graph Γ of order n with mcolors is given by ZAut(G)(x1 = . . . = xn = m)

Example 2.16. Determine the number of non-equivalent colourings of C8 (cycle of length 8)with 3 colors.

Definition 2.17. A proper colouring of a graph Γ with a set of colors S is a function f :V (Γ) → C such that adjacent vertices have different colors, that is ∀u, v ∈ V (Γ), u ∼ v ⇒f(u) 6= f(v). A chromatic polynomial χ(Γ, k) of a graph Γ counts the number of its propervertex-colourings with k colors, that is χ(Γ, k) is the number of proper colourings of Γ with kcolors.

Example 2.18. Let Pn be a path on n vertices. Then χ(Pn, k) = k(k − 1)n−1.

Example 2.19. Let Cn be a cycle of length n (C1 is a loop with single vertex, and C2∼= K2.

Then χ(Cn, k) = (k − 1)n + (−1)n(k − 1).

Proof. By deletion-contraction formula, it follows that χ(Cn, k) = χ(Cn − e, k)− χ(Cn/e, k),where Cn/e is the graph obtained by contracting edge e of Cn. Observe that Cn − e ∼= Pnand Cn/e ∼= Cn−1. Now use induction, and previous exercise.

Definition 2.20. Let Γ be a graph, and let P be a partition of V (Γ). We defined a quotientgraph of Γ with respect to P to be the graph with vertex set P and with two elementsP1, P2 ∈ P being adjacent if and only if there exist u ∈ P1 and v ∈ P2 such that u, v ∈ E(Γ).In this definition we allow loops as well, that is P1 = P2 is allowed. The obtainedgraph is denoted by ΓP . For an automorphism g ∈ Aut(Γ), let Pg be the set of orbits of 〈g〉.The corresponding quotient graph is denoted by Γg.

Remark 2.21. If a graph Γ contains a loop, then there is no proper colouring of Γ.

Theorem 2.22. The number of non-equivalent proper colourings of a graph Γ with k colorsis equal to

1

|Aut(Γ)|∑

g∈Aut(Γ)

χ(Γg, k).

21

In the case when Γ ∼= Cn, then we have the following result about the number of non-equivalent proper colourings.

Theorem 2.23. Let Cn be the cycle of length n. The number of non-equivalent propercolourings of Cn with k colors is equal to

1

2n

∑d|n

ϕ(d)χ(Cnd, k) if n is odd ,

1

2n

∑d|n

ϕ(d)χ(Cnd, k) +

n

2k(k − 1)n/2

if n is even

Exercise 2.24. Let X be the graph shown below.

• Find all automorphisms of X.

• Determine the number of non-equivalent proper colouring of X with k colors.

1 2 3 4 5 6

7

Exercise 2.25. Let X be the graph shown below.

1 2 3 4 5 6

7 8 9

• Find all automorphisms of X.

• Determine the number of non-equivalent proper colouring of X with k colors.

22

3 Vertex-transitive graphs

Definition 3.1. A graph Γ is said to be vertex-transitive if Aut(Γ) acts transitively on V (Γ),that is ∀u, v ∈ V (Γ), there exists ϕ ∈ Aut(Γ) such that ϕ(u) = v.

Example 3.2. The complete graph Kn of order n, is vertex-transitive, for every n ≥ 1.

Example 3.3. The graph Cn, cycle of length n is vertex-transitive graph, for every n ≥ 3.

We will now present some elementary properties of vertex-transitive graphs. Recall that agraph Γ is said to be d-regular if all the vertices of Γ have degree equal to d. If Γ is d-regularfor some d, then we say that Γ is regular, and d is called the valency of Γ.

Proposition 3.4. A graph Γ is vertex-transitive if and only if its complement Γ is vertex-transitive.

Proof. The result follows from the fact that Aut(Γ) = Aut(Γ).

Proposition 3.5. Let Γ be a vertex-transitive graph. Then Γ is regular.

Proof. Let u, v ∈ V (Γ). Since Γ is vertex-transitive, it follows that there exists ϕ ∈ Aut(Γ)such that ϕ(u) = v. Then by Corollary 2.3 it follows that dΓ(u) = dΓ(v), hence the resultfollows.

We saw that every vertex-transitive graph is regular. The converse is not true, a graphwhich is regular does not need to be vertex-transitive.

Example 3.6. The graph shown on Figure 3.6 is not vertex-transitive. In fact it is thesmallest (with respect to the number of vertices) connected regular graph which is not vertex-transitive. To see that this graph is not vertex-transitive, observe that its complement isdisjoint union of C4 and C3.

Figure 1: Example of a regular graph which is not vertex-transitive

3.1 Cayley graphs

Definition 3.7. Let G be a finite group, S ⊂ G such that S does not contain 1G, andS−1 = S (here S−1 = x−1 : x ∈ S. The Cayley graph of G with respect to S denotedby Cay(G,S) is the graph with vertex-set G, and two vertices x, y ∈ G are adjacent if andonly if x−1y ∈ S. In other words E(Cay(G,S)) = x, xs : x ∈ G, s ∈ S. Set S is calledthe connections set of Cay(G,S).

23

Example 3.8. Let n ≥ 3, and let G = Zn be the cyclic group of order n. Let S = 1, n− 1.Then Cay(G,S) ∼= Cn.

Theorem 3.9. Every Cayley graph is vertex-transitive.

Proof. Let Γ = Cay(G,S). For g ∈ G, let gL : G→ G be the mapping defined by gL(x) = gx.It is not difficult to see that for every g ∈ G, the mapping gL is an automorphism of Cay(G,S)(verify this!). Now for any x, y ∈ G, take g = yx−1. Observe that gL(x) = y. Hence Cay(G,S)is vertex-transitive graph.

In fact we proved something stronger in previous theorem. For a group G, let GL =gL : g ∈ G. GL is called left regular representation of G. Observe that GL is a group,GL ≤ Sym(G), and GL ∼= G. In previous theorem we proved that GL ≤ Aut(Cay(G,S)).

Exercise 3.10. Prove that for every Cayley graph Cay(G,S), the group GL is a regularsubgroup of Aut(Cay(G,S)).

Exercise 3.11. Prove that valency of Cay(G,S) is equal to |S|.

Exercise 3.12. Let X = Cay(Z10, ±2, 5) and Y = Cay(D10, ρ, ρ4, τ), where D10 isdihedral group of order 10, that is, D10 = 〈ρ, τ | ρ5 = τ2 = 1, ρτ = τρ−1〉. Representgraphically graphs X and Y and check if X and Y are isomorphic.

Lemma 3.13. Cayley graph Cay(G,S) is connected if and only if 〈S〉 = G (recall that 〈S〉denotes the group generated by S).

Proof. Suppose that 〈S〉 = G. Let x, y ∈ G, and g = x−1y. Then there exists s1, . . . , sk ∈ Ssuch that s1 · s2 . . . · sk = g. For 1 ≤ i ≤ k, let vi = x · s1 · . . . si. Observe that vk =x · s1 . . . sk = xg = x(x−1y) = y. This constructs a path from x to y, (x, v1, . . . , vk−1, vk = y)hence Cay(G,S) is connected.

Suppose now that Cay(G,S) is connected. Then for any g ∈ G, there exists a path1, v1, . . . , vk = g. Since vi ∼ vi+1 , it follows that vi+1 = visi for some si ∈ S. Therefore,g = vk = vk−1sk−1 = . . . = s1 . . . sk−1, hence 〈S〉 = G.

For a graph Γ with minimum degree at least 2, girth of Γ is defined as the minimumnumber of vertices of a cycle in Γ.

Example 3.14. Let Γ = Cay(G,S), where G is abelian group and |S| ≥ 3. Then girth of Γis at most 4.

Proof. Let a, b, c be three distinct elements from S. If a + b 6= 0, then we obtain a 4-cycle,0, a, a+ b, b. If a+ b = 0, then a+ c 6= 0, and hence we again obtain 4-cycle 0, a, a+ c, c.

Recall that an automorphism of a group G is a bijective function α : G → G such thatα(g1 · g2) = α(g1) · α(g2), for every g1, g2 ∈ G.

Lemma 3.15. Let G be a finite group, S an inverse closed subset of G, and α an automor-phism of the group G. Then Cay(G,S) ∼= Cay(G,α(S)).

Proof. Since α is automorphism of the group G, it is bijective mapping of the vertex set ofCay(G,S) into the vertex set of Cay(G,α(S)). We leave the details for the reader to verifythat α is also an isomorphism between these two graphs.

24

Figure 2: Petersen graph

Corollary 3.16. Let G be a group, S an inverse closed subset of G and α ∈ Aut(G). Thenα ∈ Aut(Cay(G,S)) if and only if α(S) = S.

We denote with Aut(G,S) = α ∈ Aut(G) | α(S) = S. The previous corollary showsthat Aut(G,S) is a subgroup of Aut(Cay(G,S). Hence we have the following.

Proposition 3.17. Let Γ = Cay(G,S). Then 〈GL,Aut(G,S)〉 ∼= GL o Aut(G,S) ≤ Aut(Γ).

We have seen that every Cayley graph is vertex-transitive. But it is not true that everyvertex-transitive graph is Cayley graph.

Example 3.18. Petersen graph shown on Figure 2 is vertex-transitive, but it is not a Cayleygraph.

Proof. Suppose that there exists a group of G order 10 such that Cay(G,S) is isomorphicto the Petersen graph. Since Petersen graph is 3-valent, then |S| = 3. There are two non-isomorphic groups of order 10, Z10 and D10. Observe that girth of the Petersen graph is 5.This shows that G cannot be abelian, hence it remains to consider the case when G = D10.

Recall that D10 = 1, ρ, ρ2, ρ3, ρ4, τ, τρ, τρ2, τρ3, τρ4. Since S is inverse closed, it followsthat it contains one or three involutions (elements of order 2). Suppose S contains exactlyone involution. Then without loss of generality, we may assume that S = ρ, ρ−1, τ (verifythis). Observe that the obtained graph is not isomorphic to the Petersen graph. The casewhen S contains three involutions is left for exercise.

Exercise 3.19. Prove that the automorphism group of the Petersen graph is isomorphic toS5.

Proof. We will represent Petersen graph in the following way. Let I = 1, 2, 3, 4, 5 and letV = A : A ⊂ I and |A| = 2. For A,B ∈ V we define that A ∼ B ⇔ A∩B = ∅. It is left forreader to verify that the graph obtained this way is isomorphic to the Petersen graph.

25

Let A be the automorphism group of the Petersen graph. For g ∈ S5, let g(2) denote thecorresponding permutation of V . It is easy to see that g(2) is automorphism of the Petersen

graph for every g ∈ S5. Hence S(2)5 ≤ A. We claim that we have equality. We will use orbit

stabilizer lemma several times to prove that |A| = 120, which will establish A = S(2)5∼= S5.

By the orbit stabilizer lemma, it follows that

|A| = |A1,2||OrbA(1, 2)| = 10 · |A1,2|, (5)

since Petersen graph is vertex-transitive. Let G = A1,2. We now use orbit-stabilizer lemmafor G to obtain:

|G| = |G3,4||OrbG(3, 4)| = 3 · |G3,4|, (6)

since |OrbG(3, 4)| = 3 (To see this we can use elements of S(2)5 ). Let H = G3,4. Use

orbit-stabilizer lemma for H to obtain:

|H| = |H4,5||OrbH(4, 5)| = 2 · |H4,5|, (7)

since OrbH(4, 5) = 4, 5, 3, 5. To see this one can use automorphism g(2) whereg = (3 4) ∈ S5. Let K = H4,5. Use orbit-stabilizer lemma for K to obtain

|K| = |K1,3||OrbK(1, 3)| = 2 · |K1,3|, (8)

since OrbK(1, 3) = 1, 3, 2, 3. To see this one can use automorphism g(2) whereg = (1 2) ∈ S5. Finally, we claim that K1,3 = 1A. Let α ∈ K1,3. Then α fixesvertices 1, 2, 3, 4, 4, 5 and 1, 3. Observe that 2, 4 is the unique common neighbourof 3, 4 and 1, 3. Hence it must be fixed by α. Now it easy to see that α fixes every vertex.Combining the above equalities implies that |A| = 120.

In the following theorem, we give necessary and sufficient conditions for a given graph tobe a Cayley graph. Recall that a permutation group G ≤ Sym(V ) is called regular if G istransitive, and Gv = 1G for every v ∈ V .

Theorem 3.20 (Sabidussi). Let Γ be a graph. Then Γ is isomorphic to Cayley graph ongroup G if and only if there exists a regular subgroup H ≤ Aut(Γ) such that H ∼= G.

Proof. We already proved that for a Cayley graph Cay(G,S), the groupGL ≤ Aut(Cay(G,S))is regular.

Suppose now that Γ is a graph, and that there exists a regular subgroup H ≤ Aut(Γ),such that G ∼= H. Suppose that V (Γ) = v1, . . . , vn. Then for each i ∈ 1, . . . , n, thereexists a unique element hi ∈ H such that hi(v1) = vi (uniqueness follows from the factthat H is regular). Observe that h1 = 1H . So we have H = h1 = 1H , h2, . . . , hn. Letα : H → G be an isomorphism, and let gi := α(hi). So we have G = g1 = 1G, . . . , gn. LetD = hi ∈ H : v1 ∼Γ hi(v1), and let S = α(hi) : hi ∈ D. We claim that Γ is isomorphicto the graph Cay(G,S). Let ϕ : V → G be defined with ϕ(vi) = gi. It is clear that ϕ isbijection.

26

Let vi, vj ∈ V (Γ). Observe that

vi ∼Γ vj ⇔ h−1i (vi) ∼Γ h

−1i (vj)

⇔ v1 ∼Γ (h−1i hj)(v1)

⇔ h−1i hj ∈ D

⇔ α(h−1i hj) ∈ S

⇔ α(h−1i )α(hj) ∈ S

⇔ g−1i gj ∈ S

⇔ gi ∼Cay(G,S) gj

⇔ α(vi) ∼Cay(G,S) α(vj)

This shows that α is indeed an isomorphism between Γ and Cay(G,S).

We have seen that Petersen graph is not Cayley graph. The followin result shows that allvertex-transitive graphs of prime order are Cayley graphs.

Lemma 3.21. Let Γ be a vertex-transitive graph of order p, where p is a prime. Then Γ isa Cayley graph on cyclic group Zp.

Proof. Let Γ be a vertex-transitive graph of order p, and let A = Aut(Γ). By orbit-stabilizertheorem, we have |A| = p|Av| for some vertex v. Hence, the order of group A is divisibleby p. By Cayley theorem, there exists an element g of order p in A. Let G = 〈g〉. ThenG is cyclic group of order p. Then using the orbit-stabilizer theorem for G, it follows thatp = |G| = |Gv||OrbG(v)|. Since p is prime, it follows that |Gv| = p, or |Gv| = 1. It is now easyto see that G is a regular subgroup of A, hence by Sabidussi theorem Γ is a Cayley graph onG.

Can it happen that Cay(G,S1) ∼= Cay(H,S2), where G and H are non-isomorphic groups?

Example 3.22. Let G = Z32, H = Z4 × Z2, S1 = (1, 0, 0), (0, 1, 0), (0, 0, 1), and S2 =

(1, 0), (3, 0), (0, 1). Observe that both Cay(G,S1) and Cay(H,S2) are isomorphic to the3-dimensional cube Q3.

Example 3.23. Prove that 3-dimensional cube has 48 automorphisms.

Proof. We will represent cube as Cay(Z32, (1, 0, 0), (0, 1, 0), (0, 0, 1)). Let A be the automor-

phism group of the cube. Then by orbit-stabilizer lemma, it follows that

|A| = 8|A(0,0,0)|. (9)

Now apply orbit-stabilizer lemma for A(0,0,0). We have

|A(0,0,0)| = |A(0,0,0),(1,0,0)||Orb|A(0,0,0)|(0, 1, 0)|. (10)

For a permutation g ∈ S3 define function g : Z32 → Z3

2 with g(x1, x2, x3) = (xg(1),xg(2),xg(3) .It is left for reader to prove that g is automorphism of the cube. Using these permutations,it can be easily seen that |Orb|A(0,0,0)|(0, 1, 0)| = 3. Now again use orbit-stabilizer lemma toobtain:

|A(0,0,0),(1,0,0)| = |A(0,0,0),(1,0,0),(0,1,0)||OrbA(0,0,0),(1,0,0)(0, 1, 0)|. (11)

27

Again using permutations g it is easy to see that |OrbA(0,0,0),(1,0,0)(0, 1, 0)| = 2. Furthermore,

it is not difficult to verify that |A(0,0,0),(1,0,0),(0,1,0)| = 1. Hence

|A| = 8|A(0,0,0)| = 8·3·|A(0,0,0),(1,0,0)| = 24·2·|A(0,0,0),(1,0,0),(0,1,0)||OrbA(0,0,0),(1,0,0)(0, 1, 0)| = 48.

(12)

Example 3.24. Let Cn be the cycle of length n. Find all groups G, such that Cn is cayleygraph on G.

Proof. Let Γ = Cn. We know that Aut(Γ) ∼= D2n = 〈ρ, τ | ρn = τ2 = 1, ρτ = τρ−1〉. If Γis Cayley graph on G, then by Sabidussi theorem, G is regular subgroup of D2n. Hence G isof order n. It is easy to see that the subgroup generated by ρ (which is cyclic of order n) isregular. Denote H = 〈ρ〉. Let G 6= H be another regular subgroup of D2n. Then GH = D2n.We have

|G ∩H| = |G||H||GH|

=n2

2n= n/2.

This means that n must be even, and G∩H is an index 2 subgroup of H, hence G∩H = 〈ρ2〉.So H contains n/2 rotations of form ρ2i, and n/2 reflections without fixed point. It can beseen that G is isomorphic to the dihedral group of order n.

3.2 Normal Cayley graphs

Let Γ = Cay(G,S). Recall that we proved that GL is regular subgroup of Aut(Γ), andalso that Aut(G,S) ≤ Aut(Γ). For such automorphisms we might say that they are obviousautomorphisms of Cayley graphs. In some cases, it turns out that these generate the fullautomorphism group of a given Cayley graph.

Definition 3.25. Let Γ = Cay(G,S). If GL is normal subgroup of Aut(G) then Cay(G,S)is said to be normal cayley graph on G with respect to S.

Theorem 3.26. Let Γ = Cay(G,S) and A = Aut(X). The following claims are equaivalent:

1. Cay(G,S) is normal cayley graph.

2. A1G = Aut(G,S);

3. A = 〈GL, Aut(G,S)〉 ∼= GL oAut(G,S);

4. |A| = |G| · |Aut(G,S)|.

Remark: If a graph Γ can be represented as Cay(G,S1) and Cay(H,S2) then it is possiblethat with respect to one representation it is normal Cayley graph, and with respect to theother one it is not normal Cayley graph.

Example 3.27. Let Γ be the cube. Then it can be represented as Cayley graph on Z32 and

Z4 × Z2, as we saw in Example 3.22. It can be seen that the first representation is normal,while the second is not.

28

Theorem 3.28. Let G be finite abelian group, and let S be an inverse closed generating set,such that 0 6∈ S. If S satisfies

∀s1, s2, s3, s4 ∈ S s1 + s2 = s3 + s4 6= 0⇒ s1, s2 = s3, s4,

then Cay(G,S) is normal Cayley graph.

Example 3.29. Let Γ = Cay(Z12, 1, 6, 11). Calculate Aut(Γ).

Proof. By Theorem 3.28 it follows that Γ is normal Cayley graph. Let us determine Aut(Z12, 1, 6, 11).We know that Aut(Z12) = ϕ1, ϕ5, ϕ7, ϕ11, where ϕk(x) = k · x. It is now easy to see thatAut(Z12, 1, 6, 11) = ϕ1, ϕ11. Observe that ϕ1 = id. Therefore, |Aut(Γ)| = 12 · 2 = 24and Aut(Γ) = 〈(Z12)L, ϕ11, ϕ11〉. Observe that elements of (Z12)L represent rotations, andϕ11 : x 7→ −x is reflection. Hence Aut(Γ) ∼= D24.

Example 3.30. Let Γ = Cay(Z29, 1, 13, 16, 28). Determine all automorphisms of Γ.

Example 3.31. LetG = Zn2 and let S = (1, 0 . . . , 0), . . . , (0, 0, . . . , 1). Prove that Cay(G,S)is normal, and determine Aut(Cay(G,S)). (Observe that Cay(G,S) is n-dimensional cube.)

Exercise 3.32. Let X = Cay(Z25, ±1,±7.

• Prove that X is normal Cayley graph.

• Calculate Aut(Z25, ±1,±7).

• Calculate |Aut(X)|.

3.3 Graph products

In this section we introduce notions of Cartesian, direct, lexicographic and strong product ofgraphs.

Definition 3.33. Let X and Y be graphs. Cartesian product XY of graphs X and Y is thegraph with vertex set V (X)× V (Y ) and (x1, y1) is adjacent to (x2, y2) if and only if x1 = x2

and y1, y2 ∈ E(Y ) or y1 = y2 and x1, x2 ∈ E(X).

Definition 3.34. Let X and Y be graphs. Direct product X × Y of graphs X and Y isthe graph with vertex set V (X) × V (Y ) and (x1, y1) is adjacent to (x2, y2) if and only ifx1, x2 ∈ E(X) and y1, y2 ∈ E(Y ).

Definition 3.35. Let X and Y be graphs. Strong product X Y of graphs X and Y is thegraph with vertex set V (X) × V (Y ) and (x1, y1) is adjacent to (x2, y2) if and only if one ofthe following holds

1. x1 = x2 and y1, y2 ∈ E(Y );

2. y1 = y2 and x1, x2 ∈ E(X);

3. x1, x2 ∈ E(X) and y1, y2 ∈ E(Y ).

Observe that the edge set of X Y is disjoint union of edge sets of XY and X × Y .

29

Definition 3.36. Let X and Y be graphs. Lexicographic product X[Y ] of X by Y is thegraph with vertex set V (X)× V (Y ) and (x1, y1) is adjacent to (x2, y2) if and only if x1 = x2

and y1, y2 ∈ E(Y ) or x1, x2 ∈ E(X).

The Cartesian, direct, strong and lexicographic products are called standard graph prod-ucts. If G ≤ Sym(A) and H ≤ Sym(B), then G × H acts naturally on A × B with(g, h) : (x, y) 7→ (g(x), h(y), for all g ∈ G and h ∈ H.

Proposition 3.37. Let G ≤ Aut(X) and H ≤ Aut(Y ). Then:

1. G×H ≤ Aut(XY );

2. G×H ≤ Aut(X × Y );

3. G×H ≤ Aut(X Y );

4. G×H ≤ Aut(X[Y ]).

Proof. It is clear that (g, h) is permutation of vertex set of any of the four standard graphproducts. Case by case analysis leads to the proof that (g, h) is automorphism of graphobtained as one of the four products. Details are left to the reader.

Proposition 3.38. Let X,Y be two graphs and Z is obtained as one of the four standardproducts of X and Y . If X and Y are vertex-transitive then Z is vertex-transitive. If X andY are Cayley graphs then Z is Cayley graph.

Proof. Follows by Proposition 3.37 and Theorem 3.20.

3.4 Hamiltonicity of vertex-transitive graphs

Definition 3.39. Hamilton path in a graph Γ is a path that visits each vertex exactly once.A Hamilton cycle (or Hamilton circuit) is a Hamilton path that is a cycle. A graph thatadmits a Hamilton cycle is called Hamiltonian graph.

The problem of determining whether or not a given connected vertex-transitive graphhas a Hamilton cycle or Hamilton path is one of the oldest and most well-known problemconcerning vertex-transitive graphs, and as we shall see many additional problems have arisenvia attempts to solve these problems. The problem began with what is usually referred to asa conjecture, posed by Lovasz.

Example 3.40. Let Γ = Cay(Z10, 2, 5, 8). Prove that Γ has a Hamiltonian cycle.

Example 3.41. Find a Hamilton cycle in graph Cay(Z12, ±3,±4).

Example 3.42. Petersen graph has a Hamiltonian path, but it has no Hamiltonian cycle.

Proof. It is easy to construct a Hamiltonian path in Petersen graph. To see that Petersengraph does not have a Hamilton cycle we will use the fact that the girth of Petersen graphis 5. Now suppose contrary, that there is a cycle v0 . . . v9 of length 10 in the Petersen graph.There are 5 more edges that are not lying on this cycle, and each vertex vi is incident with1 such edge. Since girth of the Petersen graph is 5, it follows that vi can be adjacent only

30

with vi+4, vi+5 or vi+6. If each vertex vi is adjacent with vi+5 then we obtain cycle of length4, for example v0, v1, v6, v5. So without loss of generality, we may assume that v0 is adjacentwith one of v4 or v6, and furthermore, without loss of generality we may assume that v0 isadjacent with v6. By the girth condition, we have that v1 can only be adjacent with v5, v6 orv7. Moreover, since v0 ∼ v6, it follows that v1 is not adjacent with v6. If v1 ∼ v5, we obtain4-cycle v0, v1, v5, v6 and if v1 ∼ v7 we obtain 4 cycle v0, v1, v7, v6, a contradiction with the factthat girth of Petersen graph is 5. The obtained contradiction shows that Petersen graph doesnot admit a Hamilton cycle.

The only known vertex-transitive graphs that don’t admit a Hamilton cycle are K1, K2,Petersen graph, truncation of Petersen graph, Coxeter graph (cubic graph of order 28) andtruncation of Coxeter graph. Of course graphs K1 and K2 can be considered as trivial exam-ples, as these graphs have no cycles at all. It turns out that none of the 4 non-trivial examplesis a Cayley graph. This has led to the following ”Folklore” conjecture.

Conjecture 3.43. Every connected Cayley graph of order at least 3 has a Hamilton cycle.

This conjecture has drawn a lot of attention of researchers working in the algebraic graphtheory.

We will prove that this conjecture is true for Cayley graphs on Abelian groups. Beforepresenting the proof we need the following lemma.

Lemma 3.44. For every n ≥ 1 and m ≥ 3 graph CmPn admits a Hamilton cycle.

Proof. Proof is left for reader as exercise.

Theorem 3.45. A connected Cayley graph of an abelian group of order at least 3 is Hamil-tonian.

Proof. Let Γ = Cay(G,S) be connected, where G is abelian and S = S−1. Since Γ isconnected, it follows that 〈S〉 = G. We proceed by induction on |S|. If |S| = 1, then S = a,a ∈ G, a is self-inverse and G is generated by a. This implies that G = 1, a, which is oforder 2, a contradiction. If |S| = 2, then either S = a, a−1 with a a generator of G, orS = a, b, where both a and b are self-inverse. In the former case, we have the Hamiltoncycle 1, a, a2, . . . , an−1, where an = 1 and n = |G| = |〈a〉|. In the later case, as G is abelianand G = 〈a, b〉, we see that G = Z2 × Z2. The only connected Cayley graphs of Z2 × Z2 areC4 and K4, which are both Hamiltonian. Assume the result holds for every generating set ofsize at most m− 1 ≥ 2, and let S be a self-inverse generating set of G of size m.

If m = 3, then S contains a self-inverse element a, and we define T := S \ a. If m > 3,then we take a to be arbitrary element of S, and define T := S \ a, a−1. Observe that inboth cases we have that 2 ≤ |T | < m. Let H = 〈T 〉. Then H is abelian group of order atleast 3. Let X = Cay(H,T ). Then we can apply induction hypothesis on X, since it is aconnected Cayley graph on abelian group of order at least 3 and its generating set has sizeless than m. Therefore, there exists a Hamilton cycle in the graph X. Suppose that thisHamilton cycle is h1, h2 . . . hk, where H = h1, . . . , hk. Let n be the smallest positive integersuch that an ∈ H. Then G is disjoint union of H,aH,a2H,. . . , an−1H. Observe now thatah1, ah2 . . . ahk is also a cycle, and that each x is adjacent with ax. Hence we obtain thatΓ contains a spanning subgraph isomorphic to CkPn. By previous Lemma, it follows thatthere is a Hamilton cycle in CkPn, hence there is also a Hamilton cycle in Γ.

31

Example 3.46. Let G = Z6 × Z4 in S = (3, 1), (3, 3), (2, 1), (4, 3). Find a Hamilton cyclein graph Cay(G,S).

32

4 Edge-transitive and arc-transitive graphs

Automorphism of a graph is defined as permutation of vertex-set, that preserves edges. Everyautomorphism of a graph can be also considered as a permutation of edges, that is we canconsider natural action of Aut(Γ) on E(Γ). For an automorphism ϕ ∈ Aut(Γ) and an edgee = x, y we define ϕ(e) = ϕ(x), ϕ(y).

Definition 4.1. A graph Γ = (V,E) is said to be edge-transitive if Aut(Γ) acts transitivelyon the edge set, that is for any two edges u, v, x, y ∈ E, there exists ϕ ∈ Aut(Γ) suchthat ϕx, y = u, v.

Example 4.2. Complete bipartite graph Km,n is edge-transitive. Furthermore, it is vertex-transitive if and only if m = n. Hence there exist edge-transitive graphs which are notvertex-transitive.

We have seen that every Cayley graph is vertex-transitive. Cayley graphs don’t need tobe edge-transitive.

Example 4.3. Cay(Z6, ±2, 3) is not edge-transitive.

Example 4.4. For every odd positive integer m ≥ 3 graph Cay(Zm ×Z2, (±1, 0), (0, 1)) isnot edge-transitive.

Theorem 4.5. Let Γ be an edge-transitive graph without isolated vertices. If Γ is not vertex-transitive, then Γ is bipartite. Moreover, Aut(G) has two orbits on vertices and these twoorbits form a bipartition of Γ.

Proof. Let u, v be an edge of Γ. Let U = ϕ(u) : ϕ ∈ Aut(Γ) and V = ϕ(v) : ϕ ∈ Aut(Γ).Since Γ is not vertex-transitive, it follows that U ∩ V = ∅. Let x be an arbitrary vertex of Γ.Since Γ has no isolated vertices it follows that there exist y ∈ V (Γ) such that x ∼ y. Since Γis edge-transitive, there exists ϕ ∈ Aut(Γ) such that ϕu, v = x, y. This shows that x ∈ Uor x ∈ V . Hence V (Γ) is disjoint union of U and V . Suppose now that there is an edge x, ywhere both x and y belong to U . Then there exists ϕ ∈ Aut(Γ) such that ϕu, v = x, y.This would imply that ϕ(v) = x or ϕ(v) = y and therefore x ∈ V or y ∈ V . However, this isimpossible, since U ∩ V = ∅. This shows that sets U and V form a bipartition of Γ.

Definition 4.6. Let Γ be a graph. Arc in Γ is a pair (u, v), where u and v are adjacentvertices of Γ.

Observe that for any graph Γ, the number of arcs is twice the number of edges, that is forevery edge u, v there are two arcs (u, v) and (v, u).

Definition 4.7. A graph Γ is said to be arc-transitive if Aut(Γ) acts transitively on the arcset of Γ, that is for any two pairs of adjacent vertices u ∼ v and x ∼ y, there exists anautomorphism ϕ ∈ Aut(Γ) such that ϕ(u) = x and ϕ(v) = y.

Proposition 4.8. Let Γ be a graph. If Γ is arc-transitive then Γ is edge-transitive. Moreover,if Γ is without isolated vertices, then Γ is vertex-transitive.

Proof is straightforward and is omitted.

33

Remark 4.9. There exist graphs which are vertex-transitive and edge-transitive but notarc-transitive. Such graphs are called half-arc-transitive. The smallest example is a 4-valentgraph of order 27 and it so called Doyle-Holt graph.

Proposition 4.10. Let Γ be a vertex-transitive graph. Then Γ is arc-transitive if and only iffor an arbitrary vertex v, the stabilizer Aut(G)v acts transitively on Γ(v), the set of neighbourof v in Γ.

Proof. Suppose first that Γ is arc-transitive. Let v ∈ V (Γ) and let a, b ∈ Γ(v) be arbitrary.Since Γ is arc-transitive, we can map arc (v, a) to arc (v, b), hence there exists automorphismϕ such that ϕ(v) = v and ϕ(a) = b. Hence ϕ ∈ Aut(Γ)v and maps a to b. This shows thatAut(G)v acts transitively on Γ(v).

Suppose now that Aut(G)v acts transitively on Γ(v). Let (x, y) and (a, b) be two arbitraryarc in Γ. Then there exist automorphism ϕ1 and ϕ2 such that ϕ1(x) = v and ϕ2(y) = v. Letc = ϕ1(y and d = ϕ2(b). Observe that c, d ∈ Γ(v). Then there exists ϕ3 ∈ Aut(Γ)v such thatϕ3(c) = d. It is now easy to verify that ϕ−1

2 ϕ3ϕ1 maps arc (x, y) to arc (a, b). This provesthat Γ is arc-transitive.

Example 4.11. Let Γ = Cay(Z17, 1, 2, 4, 8, 16, 15, 13, 9. Then Γ is arc-transitive.

Proof. Let S = 1, 2, 4, 8, 16, 15, 13, 9. Observe that Aut(Z17, S) = ϕ1, ϕ2, ϕ4, ϕ8, ϕ9, ϕ13, ϕ15, ϕ16and that it acts transitively on S. Observe also that Γ(0) = S. Recall that Aut(Z17, S) ≤Aut(Γ)0. This shows that Aut(Γ)0 acts transitively on neighbours of 0, and hence Γ is arc-transitive.

Proposition 4.12. A Cayley graph on Abelian group is arc-transitive if and only if it isedge-transitive.

Proposition 4.13. Let Γ be a vertex-transitive and edge-transitive graph which is not arc-transitive. Then Γ has even valency.

A Cayley graph on cyclic group is called circulant

Exercise 4.14. Find up to isomorphism all cubic edge-transitive circulants.

34

5 s-arc-transitive graphs

Definition 5.1. An s-arc in a graph is a sequence of vertices (v0, . . . , vs) such that consecutivevertices are adjacent and vi−1 6= vi when 0 < i < s. Note that an s-arc is permitted to usethe same vertex more than once. A graph Γ is s-arc transitive if its automorphism groupis transitive on s-arcs, that is for any two s-arcs (v0, . . . , vs) and (u0, . . . , us) there existsα ∈ Aut(Γ) such that α(vi) = ui for i = 0, . . . , s

Observe that 0-arcs are just the vertices and 1-arcs are the usual arcs.

Example 5.2. Cn is s-arc-transitive for every s

Example 5.3. Kn is s-arc-transitive only for s = 0, 1, 2.

Example 5.4. Cube Q3 is 2-arc-transitive but not 3-arc-transitive.

Theorem 5.5. Let Γ be a graph with minimum degree at least 2. Then Γ is (s + 1)-arc-transitive if and only if Γ is s-arc-transitive and the stabilizer in Aut(Γ) of any s-arc(v0, . . . , vs−1, vs) acts transitively on the Γ(vs) \ vs−1.

Example 5.6. Petersen graph is 3-arc-transitive but not 4-arc-transitive.

Proposition 5.7. If Γ is an s-arc-transitive graph with valency at least three and girth g,then g ≥ 2s− 2.

Proof. We may assume that s ≥ 3, since the condition on the girth is otherwise meaningless.It is easy to see that Γ contains a cycle of length g and a path of length g whose end-verticesare not adjacent. Therefore Γ contains a g-arc with adjacent end-vertices and a g-arc withnonadjacent end-vertices; clearly, no automorphism can map one to the other, and so s < g.Since Γ contains cycles of length g, and since these contain s- arcs, it follows that any s-arcmust lie in a cycle of length g. Suppose that u0, . . . , us is an s-arc. Denote it by A. Since us−1

has valency at least three, it is adjacent to a vertex w other than us−2 and us, and since thegirth of Γ is at least s, this vertex cannot lie in A. Hence we may replace us by w, obtaininga second s-arc B that intersects A in an (s− 1)-arc. Since B must lie in a circuit of length g,we thus obtain a pair of circuits of length g that have at least s− 1 edges in common. If wedelete these s − 1 edges from the graph formed by the edges of the two circuits of length g,the resulting graph still contains a cycle of length at most 2g− 2s+ 2. Hence 2g− 2s+ 2 > g,and the result follows.

5.1 Cubic-arc-transitive graphs

A graph is said to be cubic if it is 3-regular, that is if each vertex has degree 3. In 1947 Tutteshowed that for any s-arc transitive cubic graph, s ≤ 5. This was, eventually, the stimulus fora lot of work. One outcome of this was a proof, by Richard Weiss, that for any s-arc transitivegraph, s ≤ 7. This is a very deep result, the proof of which depends on the classification ofthe finite simple groups.

Definition 5.8. A graph Γ is s-arc regular if it is s-arc-transitive and stabilizer in Aut(Γ) ofany s-arc is trivial, or equivalently, the automorphism group Aut(Γ) acts regularly on the setof all s-arcs of Γ.

35

Lemma 5.9. Let Γ be a connected cubic graph that is s-arc transitive, but not (s + 1)-arctransitive. Then Γ is s-arc regular.

Theorem 5.10 (Tutte). If Γ is an s-arc regular cubic graph, then s ≤ 5.

Corollary 5.11. If Γ is cubic arc-transitive graph then |Aut(Γ)| = |V (Γ)| · 3 · 2s−1 for somes ∈ 1, 2, 3, 4, 5.

Example 5.12. Petersen graph is 3-arc-regular.

36

6 Distance-transitive graphs

Definition 6.1. A connected graph Γ is distance-transitive if given any two ordered pairs ofvertices (u, u′) and (v, v′) such that d(u, u′) = d(v, v′), there is an auto- morphism α of Γ suchthat α(v, v′) = (u, u′).

Example 6.2. Complete graph Kn is distance-transitive.

For a connected graph Γ and a vertex v ∈ V (Γ) let Γi(v) be the set of vertices at distancei from v, that is Γi(v) = u ∈ V (Γ) | d(u, v) = i. Observe that for any connected graphΓ with diameter d we have that V (Γ) is partitioned into sets Γ0(v), Γ1(v), . . ., Γd(v). Thefollowing theorem gives a characterization of distance-transitive graphs, based on the actionof automorphism group on sets Γi(v). This partition is called distance-partition of Γ.

Theorem 6.3. Let Γ be a connected graph. Then Γ is distance transitive if and only if thefollowing conditions hold:

1. Γ is vertex-transitive;

2. Aut(Γ)v acts transitively on each of sets Γi(v) (i = 1, . . . , diam(Γ)), for any vertexv ∈ V (Γ).

Corollary 6.4. Let Γ be a connected distance-transitive graph. Then Γ is arc-transitive.

Example 6.5. Petersen graph is distance-transitive.

We would like to note that distance-transitive graphs are not necessarily s-arc-transitivefor higher values of s.

Example 6.6. A cube graph Q3 is distance-transitive, it has diameter 3, but it is not 3-arc-transitive.

Example 6.7. Let G = Z17 and S = 1, 2, 4, 8, 9, 13, 15, 16 and let Γ = Cay(G,S). Then Γis distance-transitive with diameter 2.

Example 6.8. Let p ≡ 1 (mod 4) be a prime, and let G = Zp. Let S = x2 : x ∈ 1, . . . , p− 1.Observe that S = −S and that |S| = (p − 1)/2. Let Γ = Cay(G,S). Then Γ is distance-transitive with diameter 2.

Example 6.9. Let p be a prime, and q = pn such that q ≡ 1 (mod 4). Let Fq be the field oforder q. Let S be the set of all non-zero squares in Fq, that is S = x2 | x ∈ Fq \ 0. Let Gbe the additive group of the field Fq, that is G = (Fq,+). The graph Γ = Cay(G,S) is calledPaley graph. Prove that Paley graphs are distance-transitive.

Suppose that Γ is a connected distance-transitive graph and u ∈ V (Γ). Since the cells ofthe distance partition Γi(u) are orbits of Aut(Γ)u, every vertex in Γi(u) is adjacent to thesame number of other vertices, say ai, in Γi(u). Similarly, every vertex in Γi(u) is adjacentto the same number, say bi, of vertices in Γi+1(u) and the same number, say ci, of vertices inΓi−1(u). The graph X is regular, and its valency is given by b0, so if the diameter of Γ is d,we have

ci + ai + bi = b0, i = 0, 1, ..., d.

37

These numbers are called the parameters of the distance-transitive graph, and determinemany of its properties.

Distance transitivity is a symmetry property in that it is defined in terms of the existenceof certain automorphisms of a graph. These automorphisms impose regularity properties onthe graph, namely that the numbers ai, bi, and ci are well-defined. There is an importantcombinatorial analogue to distance transitivity, which simply asks that the numerical regu-larity properties hold, whether or not the automorphisms exist. Given any graph Γ we cancompute the distance partition from any vertex u, and it may occur ”by accident” that everyvertex in Γi(u) is adjacent to a constant number of vertices in Γi−1(u),Γi(u) and Γi+1(u),regardless of whether there are any automorphisms that force this to occur. Such graphs arecalled distance-regular graphs.

Proposition 6.10. Every distance-transitive graph is distance-regular.

A special case are distance-regular graphs of diameter 2. Such graphs are called stronglyregular graphs.

Exercise 6.11. Let Γ be a distance-regular graph with diameter 2. Prove that there existparameters (n, k, λ, µ) such that Γ has order n, it is k-regular, every edge lies on λ triangles(in other words every pair of adjacent vertices have λ common neighbours), and every pair ofdistinct non-adjacent vertices have µ common neighbours. A graph Γ is said to be stronglyregular with parameters (n, k, λ, µ).

Example 6.12. Prove that the complement of a strongly regular graph is strongly regular.

Example 6.13. Let Γ be a connected vertex-transitive graph, and suppose that Aut(G)vhave exactly three orbits in its action on V (Γ). Prove that Γ is distance-transitive.

Example 6.14. The graph L(Kn) (that is the line graph of Kn) is strongly regular withparameters (n(n− 1)/2, 2n− 4, n− 2, 4).

Example 6.15. The graph L(Kn,n) is strongly regular with parameters (n2, 2n− 2, n− 2, 2).

Exercise 6.16. Let G = Z4 × Z4 and let S = ±(1, 0),±(0, 1),±(1, 1). The graph Γ =Cay(G,S) is called Shrikhande graph. Prove that it is strongly regular, but not distance-transitive.

6.1 Eigenvalues of strongly regular graphs

Let Γ be a graph of order n with vertex set v1, . . . , vn. Adjacency matrix A = A(Γ) is n-by-n matrix with value Ai,j equal to 1 if and only if vi, vj ∈ E(Γ). Observe that adjacencymatrix is a symmetric matrix, that is AT = A. Such matrices have several nice properties.

Lemma 6.17. All eigenvalues of a real symmetric matrix are real.

Proof. Suppose that α ∈ C is an eigenvalue of a symmetric matrix A. Let x 6= 0 be acorresponding eigenvector. Then Ax = αx. Then x∗Ax = x∗αx = αx∗x. On the other handαx∗x = x∗αx = x∗Ax = (x∗Ax)∗ = αx∗x. Therefore, α = α, hence α is real.

Lemma 6.18. Let A be a symmetric real matrix. Then A is diagonalizable.

38

Exercise 6.19. Let Γ be a connected k-regular graph. Then k is an eigenvalue of A(Γ) withmultiplicity one.

Proof. Let Γ be a connected k-regular graph of order n, and let v1, . . . , vn be the vertex setof Γ. Let A be its adjacency matrix. Then it is easy to see that A · 1 = k1, hence k is aneigenvalue. Suppose that x is an eigenvector of A corresponding to k. Let x = (x1, . . . , xn)T .Since x is eigenvector corresponding to eigenvalue k, it follows that Ax = kx. Let xj be themaximum of x1, . . . , xn, that is xi ≤ xj for every i ∈ 1, . . . , n. Then

kxj = (kx)j = (Ax)j =∑

Aj,ixi =∑

vi∈Γi(vj)

xi ≤ kxj .

Therefore, we conclude that xi = kj for every i such that vi ∈ Γ(vj). The connectedness of Γnow implies that all xi must be equal to xj , hence x = xj1. This shows that the multiplicityof k as an eigenvalue if 1.

Suppose A is the adjacency matrix of the (n, k, λ, µ) strongly regular graph Γ. We candetermine the eigenvalues of the matrix A from the parameters of Γ and thereby obtain somestrong feasibility conditions. The uv-entry of the matrix A2 is the number of walks of lengthtwo from the vertex u to the vertex v. In a strongly regular graph this number is determinedonly by whether u and v are equal, adjacent, or distinct and nonadjacent. Therefore, we getthe equation

A2 = kI + λA+ µ(J − I −A)

Where J denotes all 1 matrix. The above equation can be rewritten as

A2 − (λ− µ)A− (k − µ)I = µJ.

We can use this equation to determine the eigenvalues of A. Since Γ is regular with valencyk, it follows that k is an eigenvalue of A with eigenvector (1, . . . , 1). Any other eigenvector ofA is orthogonal to (1, . . . , 1) (this follows since A is symmetric matrix, that is AT = A). Letz be an eigenvector for A with eigenvalue θ 6= k. Then

A2z − (λ− µ)Az − (k − µ)Iz = µJz = 0,

soθ2 − (λ− µ)θ − (k − µ) = 0.

Therefore, the eigenvalues of A different from k must be zeros of the quadratic

x2 − (λ− µ)x− (k − µ).

If we set ∆ = (λ−µ)2 +4(k−µ) (the discriminant of the quadratic) and denote the two zerosof this polynomial by τ and θ, we get

τ =λ− µ−

√∆

2

and

θ =λ− µ+

√∆

2

39

Now, τθ = (µ−k), and so, provided that µ < k, we get that τ and θ are nonzero with oppositesigns. We see that the eigenvalues of a strongly regular graph are determined by its parameters(although strongly regular graphs with the same parameters need not be isomorphic). Themultiplicities of the eigenvalues are also determined by the parameters.

Exercise 6.20. Determine the multiplicities of eigenvalues of an (n, k, λ, µ) strongly regulargraph.

Theorem 6.21. A connected regular graph with exactly three distinct eigenvalues is stronglyregular.

Proof. Suppose that Γ is connected and regular with eigenvalues k, θ, and τ , where k is thevalency of Γ, and let n be the order of Γ. Let A be the adjacency matrix of G. SinceA is symmetric, the sum of multiplicities of its eigenvalues equals n. Moreover, since Γ isconnected, eigenvalue k has multiplicity 1. Define matrix M with

M :=1

(k − θ)(k − τ)(A− θI)(A− τI)

Observe that kernel of M consists precisely of eigenvectors of A corresponding to θ or τ , hencekernel of M has dimension n− 1. Moreover, we have

M · [1, . . . , 1]T =1

(k − θ)(k − τ)(A− θI)(A− τI) · [1, . . . , 1]T

=1

(k − θ)(k − τ)(A− θI)((A · [1, . . . , 1]T − τ · [1, . . . , 1]T ))

=1

(k − θ)(k − τ)(A− θI)((k − τ) · [1, . . . , 1]T ))

=1

(k − θ)(A− θI) · [1, . . . , 1]T = [1, . . . , 1]T .

This implies that M = 1nJ (explain why).

We have shown that J is a quadratic polynomial in A, and thus A2 is a linear combinationof I, J , and A. Accordingly, Γ is strongly regular.

Exercise 6.22. Let Γ be a Moore graph with valency k and diameter 2. Prove that k ∈2, 3, 7, 57.

Proof. Observe that a Moore graph with valency k and diameter 2 is (k2 + 1, k, 0, 1) stronglyregular. Calculate the eigenvalues and the corresponding multiplicities. Use the fact thatmultiplicities of eigenvalues are integers.

Exercise 6.23. Let Γ be a connected graph of order 25 and let A be its adjacency matrix. IfA2 +A− 6I = 6J , prove that Γ is strongly regular, and determine its parameters (n, k, λ, µ).

Proof. Observe that we can rewrite the given equality as:

A2 = 12I + 5A+ 6(J − I −A).

This means that for vertex v ∈ V (Γ), the number of walks of length 2 between v and v is 12.Hence Γ is regular with valency 12. Similarly we see that for two adjacent vertices u and vthe number of common neighbours is 5 and for two non-adjacent vertices, the number of theircommon neighbours is 6. Therefore, Γ is (25, 12, 5, 6)-strongly regular graph.

40

Exercise 6.24. Let Γ be a connected graph of order 21 and let A be its adjacency matrixsuch that A2 −A− 6I = 4J .

(a) Prove that Γ is strongly regular graph;

(b) Determine parameters (n, k, λ, µ) of Γ;

(c) Determine eigenvalues of A and their multiplicities.

41