THE STRUCTURE OF UNITARY CAYLEY GRAPHS

MEGAN BOGGESSCOLUMBIA UNION COLLEGE

TIFFANY JACKSON-HENDERSON

ST. AUGUSTINE’S COLLEGEISIDORA JIMENEZ

MILLS COLLEGE

RACHEL KARPMANSCRIPPS COLLEGE

Abstract. In this paper we explore structural properties of unitary Cayley graphs, including clique and

chromatic number, vertex and edge connectivity, planarity, and crossing number.

1. Introduction

A graph is a pair G = (V,E). V (G) is the set of vertices of G, and E(G) is the set of edges. An edge {u, v}is an unordered pair of distinct vertices; the vertices u and v are the endpoints of the edge. The vertices ofa graph can be pictured as dots, and the edges as line segments whose endpoints are vertices. An edge e issaid to be incident at a vertex v if v is an endpoint of e. Two vertices are said to be adjacent if there is anedge between them. Adjacent vertices are sometimes referred to as neighbors

A graph is simple if there is at most one edge between any two vertices, and the endpoints of each edge aredistinct. All graphs discussed in this paper are simple graphs. A graph H is a subgraph of a graph G,denoted H ⊆ G, if V (H) ⊆ V (G) and E(H) ⊆ E(G). A graph H is an induced subgraph of G ifV (H) ⊆ V (G), and if E(H) contains every edge of G whose endpoints are in H.

In this paper, we investigate the properties of unitary Cayley graphs. A unitary Cayley graph, GR is thegraph whose vertex set is a ring R, where {i, j} is an edge of GR if and only if (i− j) is a unit in R. If R isthe cyclic ring of order n, we denote its unitary Cayley graph Gn.

Figures 1 and 2 show some examples of unitary Cayley graphs.

Date: 17 July 2008.

Figure 1. The GR graph for R = Z2 × Z4.

1

Figure 2. The Gn graphs for Z6, Z7, and Z8

The motivation for this project is derived from the problem of finding representation numbers of generalgraphs. A representative labeling modulo n is a labeling of each vertex of the graph with an element of thecyclic ring of order n, such that two vertices are adjacent if and only if their difference is a unit in the ring.The representation number for an arbitrary graph is the smallest integer n for which we can create arepresentative labeling of the graph modulo n. Finding the representation number of an arbitrary graphremains an open problem [2].

If a graph is representable modulo n, then it must be an induced subgraph of the unitary Cayley Graph ofof the cyclic ring Zn [2].

Unitary Cayley graphs of rings which are the direct product of cyclic rings relate to the generalizedrepresentation number of a graph. The definition of a generalized representation number of a graph issimilar to that of the (ordinary) representation number. Let R be a ring that is a direct product of cyclicrings. A representative labeling of a graph in R is a labeling which associates each vertex with an elementin R, such that two vertices are adjacent if and only if the difference of their labels is a unit in R. Thegeneralized representation number of an arbitrary graph is the order of the smallest ring in which thegraph is representable. If a graph is representable in R, then it must be an induced subgraph of the unitaryCayley graph of R [2].

In this paper, we explore the properties of unitary Cayley graphs. The properties we examine includechromatic number, clique number, connectivity, and planarity.

2. Definitions

Let G be a graph and v ∈ V (G). The degree of v, denoted deg(v), is number of edges incident at v. Themaximum degree of G, which we denote ∆(G), is degree of the vertex of largest degree. Similarly, theminimum degree δ(G) is the degree of the vertex of smallest degree. A graph is said to be k-regular, orsimply regular, if every vertex of the graph has degree k.

A walk is a sequence of pairwise adjacent vertices of a graph. A walk is closed if its initial and terminalpoints are equal. A path is a walk in which no vertex is repeated. A closed path is called a cycle A trail isa walk in which no edge is repeated. An Euler circuit is a closed trail containing all the edges of a graph.

A Hamiltonian cycle is a cycle which contains every vertex of a graph.

A graph is connected if there exists a path between any two vertices. Otherwise, the graph is disconnected2

The distance between two vertices of a graph is the number of edges of the shortest path between them.The diameter of a connected graph is the maximum distance between two vertices of the graph. Adisconnected graph is said to have infinite diameter.

A graph is complete if every vertex is adjacent to every other vertex. The complete graph with n vertices isdenoted Kn.

An independent set is a set of pairwise non-adjacent vertices.

A graph G is multipartite if its vertex set is a union of disjoint independent sets, which are known aspartite classes. A bipartite graph is a multipartite graph with 2 partite classes. A multipartite graph iscomplete multipartite if each vertex in each partite class is connected to every vertex in every other parititeclass. We denote the complete multipartite graph with m partite classes of size n1, n2 . . . , nm Kn1,n2,...,nm

.A bipartite graph is complete bipartite if every vertex is adjacent to every vertex outside its partite class.

An automorphism is a mapping from α : V (G)→ V (G), which preserves both adjacency and non-adjacency.

3. Vertex Degree

Proposition 1. Let n = pe11 . . . pes

s for distinct primes p1 . . . ps, where p1 < p2 . . . < ps. The degree of avertex v ∈ V (Gn) is given by deg(v) = φ(n), where φ is Euler’s totient function.

Proof. For any v ∈ V (Gn). Using the notation of Evans et. al., we assign each vertex v to a setX(v1, . . . vs), where 0 ≤ vi ≤ pi − 1 and vi ≡ v (mod pi) [2].

A vertex v in Gn is adjacent to a vertex w ∈ X(w1, . . . ws) if only if vi 6= wi for all i with 1 ≤ i ≤ s [2]. Forif vi = wi for some i, then v ≡ w (mod pi), which implies that v − w ≡ 0 (mod pi). Hence, pi dividesv − w, and v − w is not a unit of n. This implies v and w are non-adjacent.

Conversely, if vi 6= wi for all i, then v − w 6≡ 0 (mod pi) for all i. Hence, none of the prime divisors of ndivide v − w. This implies that v − w is a unit of Zn, and hence v and w are adjacent.

By construction, each set X(v1 . . . vs) has cardinalitys∏

i=1

pei−1i . Finding the degree of v thus reduces to

finding the number of sets whose vertices are adjacent to v, then multiplying by the number of elementsper set.

In the collection of sets X(v1 . . . vs), each vi ranges from 0 to pi, and thus takes on one of pi possiblevalues. Suppose the vertices of X(w1, . . . ws) are adjacent to those of X(v1, . . . vs), the set containing v.Then each wi must satisfy the conditions wi 6= vi, 0 ≤ wi ≤ pi − 1. Each wi thus takes on one of pi − 1possible values. The total number of sets X(w1, . . . ws) whose vertices are adjacent to those of any

X(v1, . . . vs) is given bys∏

i=1

(pi − 1). Hence, the degree of v is given by

deg(v) =

(s∏

i=1

(pi − 1)

)(s∏

i=1

pei−1i

)

=s∏

i=1

(pi − 1)(pei−1i )

=s∏

i=1

φ(peii )

= φ(n)

3

�

This proposition has a number of useful consequences.

Corollary 1. Gn is φ(n)-regular for all n.

Proof. This follows immediately from Proposition 1, which gives φ(n) as the degree of an arbitrary vertexof Gn.

�

Corollary 2. The total number of edges of Gn is |E(G)| = 12

s∏i=1

p2ei−1i (pi − 1) = n

(φ(n)

2

).

Proof. The total number of edges of Gn is half the total degree of Gn. Since all vertices of Gn have the

same degree, we simply multiply the degree of a vertex of Gn byn

2. Thus |E(G)| = n

2

s∏i=1

pei−1i (pi − 1).

Expanding n ass∏

i=1

peii gives |E(G)| = 1

2

s∏i=1

p2ei−1i (pi − 1) = n

(φ(n)

2

).

�

Corollary 3. Gn is Eulerian for all n ≥ 3.

Proof. Gn is Eulerian if and only if every vertex of Gn is of even degree [3]. We will show that if n ≥ 3, thedegree of each vertex of Gn is even.

Expand n as n = pe11 . . . pes

s , where the pi are distinct primes.If n is not a power of 2, then pi is odd for some i, so pi − 1 is even. For any

v ∈ V (G), deg(v) =s∏

i=1

pei−1i (pi − 1) is even. Therefore, Gn is Eulerian.

If n = 2e and n ≥ 3, then e > 1. For any v ∈ V (G), we have deg(v) = 2e−1, again by proposition 1. Sincek − 1 > 0, deg(v) is even. Hence Gn is Eulerian, and the proof is complete.

�

4. Vertex and Edge Coloring

Definition: The chromatic Nnumber, denoted χ(Gn), is the least number of colors that can be used tocolor vertices in such a way that adjacent vertices are not colored the same.

Definition: The Clique Number, denoted ω(Gn) is the size of largest complete subgraph that can be foundin a graph.

It is known in Graph Theory that ω(Gn) ≥ χ(Gn because if there exist a clique subgraph the smallest thechromatic number can be is the clique number.

Theorem 1. Gn is a Graph. Let n = pe11 p

e22 ...p

ess , where p1 < p2 < ... < ps and ei ≥ 1 for all i = 1, ..., s.

Then ω(Gn) = χ(Gn) = p1.

Proof. Let m represent the vertices and km is the coloring. For each m, 0 ≤ m ≤ n− 1, there is a uniquekm, such that m ≡ km(modp1). The vertex m is assigned the color km. If two vertices m and m′ receivethe same color then km ≡ km′(modp1), so m ≡ m′(modp1). This allows m to not be adjacent to m′. Thus,this coloring is proven and therefore, χ(Gn ≤ p1)

We observe that in Gn there exist a p1-clique. The clique in Gn is {0, 1, ..., p1 − 1} which impliesp1 ≤ ω(Gn).

4

Based upon the proof the following can be stated, p1 ≤ ω(Gn) ≤ χ(Gn) ≤ p1 and therefore we concludethat ω(Gn) = χ(Gn) = p1. �

Figure 3. G12 where ω(G12) = χ(G12) = p1 = 2.

Figure 3 shows that ω(G12) = χ(G12) = p1 = 2. For the chromatic number, the vertices that are coloredyellow are 0modulo2 and the vertices that are colored pink are 1modulo2. The clique number isrepresented by the vertices {0,1}.

Next, we consider the edge chromatic number of Gn.A proper edge coloring of a graph G assigns a color to each edge of G such that no edges of like color areincident at the same vertex.

The edge chromatic number of G, denoted Λ(G), is the smallest number of colors which can be used togenerate a proper coloring of G.

To find the edge chromatic number of Gn, we use the following lemma.

Lemma 1. Let n ≥ 3 be given. Then Gn can be decomposed intoφ(n)

2Hamiltonian cycles whose edge sets

are disjoint.5

Proof. Let u1,−u1, . . . ur,−ur be the units of Zn. Clearly, for any ui, the vertices labeled 0, ui, 2ui, . . . nui,where multiplication is modulo n, form a Hamiltonian cycle of Gn. We denote this cycle Cui

n . We will show

that Cu1n , Cu2

n , . . . Curn form a set of

φ(n)2

Hamiltonian cycles in Gn whose edge sets are disjoint.

Let {a, b} be an edge of Cuin . Let l(a) and l(b) be the labels assigned to a and b, respectively. Without loss

of generality, l(a) > l(b). By definition of Cuin , l(a)− l(b) = ui. But this implies that l(b)− l(a) = −ui, so

{a, b} is also an edge of C−uin . Thus, Cui

n and C−uin have the same edge set. By the definition of a

Hamiltonian cycle, they also have the same vertex set. So Cuin = C−ui

n . However, {a, b} cannot be an edgeof any cycle Cuj

n , where ui 6= ±uj . For if this were so, we would have l(a)− l(b) = ±uj , which cannotoccur. Therefore, the edge set of Cui

n is disjoint from that of every other cycle Cujn , except for C−ui

n .

Cu1n , Cu2

n , . . . Curn thus form a set of

φ(n)2

edge-disjoint Hamiltonian cycles in Gn.

Next, we will show that the union of these Hamiltonian cycles is the entire graph Gn. By definition of a

Hamiltonian cycle,ur⋃i=1

Cuin has the same vertex set as Gn. Clearly, the edge set of

ur⋃i=1

Cuin is a subset of

E(Gn). Each Cuin has exactly n edges, so the total number of edges in

ur⋃i=1

Cuin is n

(φ(n)

2

). But this is

precisely the number of edges in E(G). Hence, the edge set ofur⋃i=1

Cuin is equal to that of Gn, and

ur⋃i=1

Cuin = Gn.

�

Figure 4. G8 is a union of 2 disjoint Hamiltonian cycles.

Figure 4 shows G8 as a union of 2 disjoint Hamiltonian cycles. Note thatφ(8)

2= 2. The solid lines form a

cycle consisting of edges the difference of whose endpoints is 1; the dashed lines, a cycle consisting of edgesthe difference of whose endpoints is 3.

Our derivation of the edge chromatic number of Gn requires Vizing’s theorem.6

Vizing’s Theorem. Let G be a simple graph, and let ∆(G) be the maximum degree of G. Then eitherΛ(G) = ∆(G) or Λ(G) = ∆(G) + 1 [5].

Proposition 2. If n is even, Λ(Gn) = φ(n). If n is odd, Λ(Gn) = φ(n) + 1

Proof. Case 1: Suppose n is even. By lemma 1, Gn can be decomposed intoφ(n)

2edge-disjoint

Hamiltonian cycles. Since n is even, each of these cycles has even length. Each cycle can thus be coloredwith two colors, simply by assigning the two colors to alternate edges of the cycle. We color eachHamiltonian cycle of Gn in this manner, such that no two cycles have a color in common. The result is aproper coloring of Gn using φ(n) colors. This implies that Λ(Gn) ≤ φ(n) Since φ(n) is the maximumdegree of Gn, Vizing’s theorem shows that Λ(Gn) ≥ φ(n). Combining these results gives Λ(Gn) = φ(n),and the proof is complete in this case.

Case 2: Suppose n is odd. We will show that Gn cannot be colored by only φ(n) colors, and thus, byVizing’s theorem, a proper coloring of n requires φ(n) + 1 colors.

In a proper coloring, no two edges of like color can share an endpoint. Thus, for odd n, a proper coloring of

Gn contains no more thann− 1

2edges of a given color. Using φ(n) colors, we can color at most

φ(n)(n− 1

2

)edges of Gn. But Gn has a total of φ(n)

(n2

)edges. So it is impossible to color all the edges

of Gn using only φ(n) colors. Thus Λ(Gn) is greater than φ(n), the maximum degree of Gn. By Vizing’stheorem, Λ(G)n = φ(n) + 1.

�

5. Partite Classes of Gn

Proposition 3. : Gn is multipartite with p1 partite sets, where n = pe11 p

e22 ...p

ess and p1 < p2 < . . . < ps are

primes.

Proof. : For 0 ≤ i ≤ p1 we define each partite set as Ai = {x ∈ V (Gn) : n ≡ i(modp1)}. TheA1 ∪ ... ∪Ap1−1 = V (Gn) and Ai forms an independent set because they are equivalence classes modulop1.Therefore the number of partite sets is less than or equal to p1.

The set {0, 1, . . . , p1 − 1} shows that there exist a clique with p1 elements. The clique shows that thenumber of partite sets is greater than or equal to p1. This concludes that p1 is equal to the number ofpartite sets.

�

6. Connectivity

We now consider the edge connectivity of Gn.Let G be a connected graph. A disconnecting set of edges in G is a subset D ⊆ E(G), such that removingthe edges in D from G yields a disconnected graph.

The edge connectivity, denoted κ′, of a connected graph G, is the order of the smallest disconnecting set ofedges in G.

Let x and y be distinct vertices of a graph G. An x, y-disconnecting set of edges in G is a subset C of E(G)such that removing the edges of C from G yield a graph with no paths from x to y.

To find the edge connectivity of Gn, we make use of Menger’s theorem for edge connectivity.

Menger’s Theorem Suppose x and y are distinct vertices of a graph G. Then the minimim size of anx, y-disconnecting set of edges equals the maximum number of pairwise edge-disjoint x, y-paths in G.

7

Proposition 4. The edge connectivity κ′(Gn) of Gn is equal to φ(n).

Proof. For n < 3, the proposition is trivial. Let n ≥ 3 be given. First, we show κ′(Gn) ≥ φ(n).

Let v and w be vertices of Gn. By lemma 1, Gn can be decomposed intoφ(n)

2disjoint Hamiltonian cycles.

Hence, there are at least φ(n) paths from v to w whose edge sets are disjoint. By Menger’s theorem, todisconnect v and w, we must remove at least φ(n) edges. Since v and w are arbitrary vertices of Gn,κ′(Gn) ≥ φ(n).

To see that κ′(Gn) = φ(n), note that we can isolate a vertex v of Gn, simply by removing all edges incidentat v. Since the degree of any vertex of Gn is φ(n), we can isolate a vertex by removing φ(n) edges. Hence,κ′(Gn) ≤ φ(n). Since we have already shown the other inequality, κ′(Gn) = φ(n), and the proof is complete.

�

Now we consider the vertex connectivity of Gn. κ(Gn)=min{|S|: G− S is disconnected or only has onevertex}.

We use Whitney’s theorem as an upperbound for vertex connectivity, since we have proven thatκ′(G) = φ(n). We use Menger’s Theorem as the basis for the vertex connectivity proof.

Whitney’s Theorem[4]. κ(G) ≤ κ′(G) ≤ φ(n).

Menger’s Theorem[1]. If two vertices x, y ∈ V (G) are nonadjacent, the maximum number of internallydisjoint xy-paths in G is equal to the minimum number of vertices whose deletion destroys all xy-paths.

Proposition 5. κ(Gn) = κ′(Gn) = φ(Gn).

Proof. We know from Proposition 4 that κ′(Gn)=φ(n). In this proof, we will show that Gn has φ(n)internally disjoint paths from any two nonadjacent vertices. Thus the minimum number of vertices whosedeletion destroys all xy-paths will be φ(n), by Menger’s Theorem.

Consider Gn where n=pr11 p

r22 · · · prm

m and p1, . . . , pn are distinct primes. Let there be two nonadjacentvertices labeled (0, 0, . . . , 0) and x = (p1a1, p2a2, . . . , ptat, bt+1, . . . , bm), where pi - bi. Then there existcommon neighbors of the form (c1, . . . , ct, ct+1, . . . , cm), where pi - ci and pi - (bi − ci) for all i. Thisconnects (0, 0, . . . , 0) and x by internally disjoint paths of length two.

However, there may be vertices adjacent to (0, 0, . . . , 0) which are not adjacent to x. Choose a vertex thatis adjacent to (0, 0, . . . , 0) but not to x. Let this vertex be u, where u = (u1, . . . , um), pi - bi for all i andpi|(ui − bi) for some t+ 1 ≤ i ≤ m. Thus, u is adjacent to (0, 0, . . . , 0) and nonadjacent to x.

Case 1: All pi are odd or pi = 2 for some t+ 1 ≤ i ≤ m.Now we choose a vertex, v = (v1, . . . , vm), that is adjacent to both u and x. We define this vertex as

follows. If pi|(ui − bi), then vi = ui − bi. For 1 ≤ i ≤ t, let vi ={ui + 2 if ui ≡ (pi − 1) (mod pi)ui + 1 otherwise

If pi - (ui − bi) and t+ 1 ≤ i ≤ m, then

vi ={ui + 2 if (ui − bi) ≡ (pi − 1) (mod pi)ui + 1 otherwise

So v is adjacent to x. Thus (0, 0, . . . , 0) is connected to x by internally disjoint paths of length 3. We know(0, 0, . . . , 0) has φ(n) neighbors. We have now shown there are also φ(n) internally disjoint paths from(0, 0, . . . , 0) to x for the case where all primes are odd or pi = 2 for some t+ 1 ≤ i ≤ m.

Case 2: Let pi = 2 for some 1 ≤ i ≤ t. Without loss of generality, let p1 = 2. Let there be two nonadjacentvertices labeled (0, 0, . . . , 0) and x. Preserving the notation of the previous case, choose a vertex u which isadjacent to x.

8

Notice that p1a1 = 2a1 is even and u1 is odd, so there is no vertex v = (v1, . . . , vm) that is adjacent to bothx and u.

We create a path from u to x as follows:

Let w = (w1, . . . , wm) be defined by wi = ui + 1 for all 1 ≤ i ≤ t and when pi - (ui − bi) for t+ 1 ≤ i ≤ m.If pi|(ui − bi) for t+ 1 ≤ i ≤ m, then let

wi ={bi + 2 if bi ≡ (pi − 1) (mod pi)bi + 1 otherwise

Notice that for t+ 1 ≤ i ≤ m, pi 6= 2.

Let v = (v1, . . . , vm) by defined by vi = ui for all 1 ≤ i ≤ t and when pi - (ui − bi) for t+ 1 ≤ i ≤ m. Ifpi|(ui − bi) for t+ 1 ≤ i ≤ m then vi = ui − bi.

Then u,w, v, x is a path from u to x and all such paths are vertex disjoint.

Again, we have shown that there are still φ(n) internally disjoint paths from (0, . . . , 0) to(p1a1, p2a2, . . . , ptat, bt+1, . . . , bm), even if pi = 2 for some 1 ≤ i ≤ t. Thus, we have proven that there areφ(n) internally disjoint paths between any nonadjacent vertices of the graph Gn.

κ(Gn) = κ′(Gn) = φ(n). �

7. Connectedness and Diameter

Proposition 6. Let R = Zn1 × . . .× Znr . Then GR is connected if and only if ni is even for at most onevalue of i.

Proof. (⇒) Suppose there exists i 6= j such that ni and nj are both even. Let z be the vertex (0, . . . , 0) ofGR. We will show that GR has a vertex w such that there is no path in GR from z to w.

Let v = (v1, . . . , vr) be any vertex of GR not equal to z, and suppose there exists a path of length l betweenv and z. Then vi is the sum of l− 1 units in Zni

, and vj is the sum of l− 1 units in Znj. Since n1 and n2 are

even, each unit of Zn1 or Zn2 must be odd. Thus, vi and vj are each the sum of l − 1 odd numbers, whichneed not be distinct. This implies that either vi and vj are both even (if l − 1 is even), or they are bothodd (if l− 1 is odd). Consider a vertex w = (w1, w2, . . . , wr), where wi is even and wj is odd. By the aboveargument, there is no path from this vertex to (0, 0, . . . , 0). This implies that GR is not a connected graph.

(⇐) Next, suppose ni is even for at most 1 value of i. We will show that there exists a path between anytwo vertices of Gn. By translation, it suffices to show that there exists a path between the vertexz = (0, 0, . . . , 0) and each other vertex of Gn.

Assume there exists a non-empty subset D of V (GR) such for any d ∈ D, no path exists between z and d.

Consider the following partial ordering of R. For any a, b ∈ R, we say a < b if ai ≤ bi for all i. Choose avertex d = (d1 . . . , dr) ∈ D such that d is minimal with respect to the partial ordering described above.

Case 1: Suppose ni is odd for all i. By construction, dj 6= 0 for some 1 ≤ j ≤ s. If d = (d1, d2, . . . , dr),

where di ={

0 if i 6= j1 if i = j

. Then (0, 0, . . . , 1, . . . , 0), (1, 1, . . . , 2, . . . , 1), (0, 0, . . . , 0) is a path from z to d.

This path is valid because nj is odd, and hence 1 and 2 are units of Znj.

If d has some other value, we define f = d− (0, 0, . . . 1, 0). Because d is a minimal element of d, f /∈ D, andthere exists a path from z to f . Hence, it suffices to show there exists a path from f to d. By translation,this reduces to finding a path from z to (0, 0, . . . 1, 0). Such a path is given above. Hence, we can find apath between f and d

Concatenating this path with a path from z to f gives a path from z to d. This contradicts the claim thatno path exists between z and d.

9

Therefore, our initial assumption was false. For any vertex v of GR, there exists a path between z and v.Thus, if ni is odd for all i, GR is a connected graph.

Case 2: Suppose ni is even for exactly one value of i. Choose d a minimal element of D. If di is 0, thendj 6= 0 for some j 6= i. Then nj is odd, and we employ the argument used in the odd case.

If di 6= 0, we must modify our argument slightly. If d = (di, d2, . . . , dr), where dk ={

0 if k 6= i1 if k = i

.

Then (0, 0, . . . , 1, . . . , 0), (1, 1, . . . , 0, . . . , 1), (2, 2, . . . , 1, . . . 2), (0, 0, . . . , 0)is a path from z to d.

Otherwise, we define f = d− (0, 0, . . . , 1, . . . , 0). As in the odd case, there exists a path between z and f ,and it suffices to show that there exists a path between f and d. But this reduces to finding a pathbetween (0, 0, . . . , 0) and (0, 0, . . . , 1, . . . , 0), which we have already accomplished. Hence, there exists apath between f and d.

Concatenating this path with the path from z to f gives a path from f to d. This contradictions ourassumption that no path exists between d and z. Hence, our initial assumption is false, and GR is aconnected graph. �

Note that the above result implies that Gn is connected for all n. Expanding n as pe11 . . . pes

s , where the pi

are distinct primes, we see that Zn is isomorphic to R, where R = Zp1e1 × . . .×Zps

es . But peii is even for at

most one value of i, because 2 is the only even prime. Hence, the above result shows that GR is aconnected graph. Since Zn is isomorphic to R, Gn is isomorphic to GR. This implies that Gn is connected.

Using an argument similar to that used in in the proof of proposition 6, we now derive the diameter of Gn

for all n.

Proposition 7. The diameter of Gn is 1 if n is prime, 2 if n is an odd composite number, and 3 if n is aneven composite number.

Proof. Case 1: Suppose n is prime. Claim: Gn = Kn, and the diameter of Gn is 1. [2]

Let n be prime, and let v and w be vertices of Gn. Then v − w ∈ Zn, and v − w < n. Because n is prime,this implies that v − w is relatively prime to n. Hence v − w is a unit of the ring Zn, and v and w areadjacent. Since v and w are arbitrary, each vertex of Gn is adjacent to every other vertex, and Gn = Kn.Case 2: Next, suppose n is an odd composite number, where n = pe1

1 . . . pess . As in Proposition 1, we assign

each vertex v of Gn to a class X(h1 . . . hs), where each hi ranges from 0 to pi−1. Let v, w ∈ V (G) be given,where v ∈ X(v1, . . . , vs) and w ∈ X(w1, . . . , ws). If vi 6= wi for all 1 ≤ i ≤ s, then v and w are adjacent.

Suppose vi = wi for at least one value of i. Then v and w are non-adjacent, which implies that thediameter of Gn is at least 2. If n is odd, pi ≥ 3 for all i. Hence, each hi has at least three possible values.Thus, we can construct a class X(u1, . . . un) such that ui 6= vi and ui 6= wi for all 0 ≤ i ≤ n. Letu ∈ X(u1, . . . , un) be given. Then u is adjacent to both v and w, so v, u, w is a path of length two betweenv and w. Since v and w are arbitrary nonadjacent vertices of Gn, the diameter of Gn is 2.

Case 3: Suppose n is even, and n ≥ 2. We again let v, w ∈ V (G) be given, where v ∈ X(v1, . . . , vs) andw ∈ X(w1, . . . , ws). If vi 6= wi for all 1 ≤ i ≤ s, then v and w are adjacent. If vi = wi for some i, thesituation is more complicated. We have n = pe1

1 . . . pess , with p1 = 2. Thus, for each X(h1 . . . hs), h1 has

only two possible values. If v1 = w1 for v and w, we can again construct a class X(u1, . . . un) such thatui 6= vi and ui 6= wi for all 0 ≤ i ≤ n. Thus, any vertex in l is adjacent to both v and w, and the minimumlength for a path between v and w is 2.

If, v1 6= w1, but vi = wi for some i ≥ 2, the above argument fails. Clearly, v and w are non-adjacent.However, we cannot construct a class whose vertices are adjacent to both v and w. The first coordinate ofany class will be equal either to v1 or w1. Thus, there is no path of length two between v and w. Since

10

there exist pairs of vertices in Gn which are not connected by any path of length 1 or 2, the diameter of Gn

is at least 3.

We now show that the diameter of Gn is exactly 3. We construct a class of vertices X(u1, . . . , us), asfollows. If vi 6= wi, let ui = wi. If vi = wi, let ui be any integer such that ui 6= wi, 0 < ui < pi − 1. Letu ∈ X(u1, . . . us) be given. Then u is adjacent to v. Also, u and w agree in their first coordinate, so thereexists a vertex y adjacent to both u and w. Hence v, u, y, w is a path of length three between v and w. Sothe minimun length for such a path is three, so the diameter of Gn is at most 3. Combining this result withthe lower bound for diameter given above, this implies that the diameter of Gn is exactly 3.

�

8. Planarity

A graph is planar if it can be drawn in the plane with no edge crossings.

A planar graph is to be embedded in the plane if it is drawn in the plane with no edge crossings. Such adrawing is called a planar embedding of the graph or, alternatively, a plane graph.

We use several well-known theorems to find the values of n for which Gn is planar. The first is aconsequence of Euler’s formula.

A face of a plane graph is a region of the plane enclosed by edges of the graph. The space surrounding thegraph is considered a face.

The face length of a face of a plane graph is the number of edges that form the boundary of that face.

Euler’s formula. [5] Let G be a plane graph with v vertices, e edges, and f faces, then v − e+ f = 2 .

Corollary 4. [5] Let G be a plane graph with v vertices, e edges, and f faces, where v ≥ 3. Thene ≤ 3v − 6 .

Proof. It suffices to show that the result holds when G is connected. If G is disconnected, we simply addedges to G to form a connected graph, G′. Then G′ has e′ edges, where e′ > e. If we can prove thate′ ≤ 3v − 6, it follows that e ≤ 3v − 6.

In a simple graph, the length of each face is at least 3. When we sum the lengths of the faces of G, eachedge is counted twice. Let {li} be a list of lengths of the faces of G. Then 2e =

∑fi=1 li ≥ 3f . Substituting

into Euler’s formula, v − e+ f = 2, we obtain e ≤ 3v − 6. �

Corollary 5. Any planar graph has a vertex of degree less than 6 [3].

Proof. Let G be a plane graph with v vertices, e edges, and f faces. If v ≤ 6, the result follows. If v > 6,

we note that by Corollary 4, e ≤ 3v − 6, which implies2ev≤ 6− 12

v, and thus

2ev< 6. However, 2e is the

sum of the degrees of the vertices of G, because each edge has exactly two endpoints. So the inequality2ev< 6 is equivalent to the statement that the average degree of G is less than 6. This implies that G has

at least one vertex of degree less than 6, and the proof is complete.

�

The corollaries to Euler’s formula give a simple way to show that certain graphs are non-planar. However,a graph which meets the conditions for planarity given by Corollies 4 and 5 may still be non-planar.Kuratowski’s theorem, stated below, provides a criterion that can be used to determine the planarity ofany graph.

Kuratowski’s Theorem. A graph is planar if and only if it contains no subgraph which is a subdivisionof K5, the complete graph on 5 vertices, or K3,3, the complete bipartite graph with partite sets of 3 verticeseach [5].

11

Note that both K5 and K3,3 are themselves non-planar.

Figure 5. The non-planar graphs K5 (left) and K3,3 (right)

Given a graph G, we obtain a subdivision of G by inserting an arbitrary number of vertices whichsubdivide the edges in a subset of E(G)

Figure 6. An example of a subdivision.

In figure 6, the graph on the right is a subdivision of the graph on the left. The edge {u, v} is subdividedby the addition of the vertex d; the edge {u, v} is replaced by the path u, d, v.

Proposition 8. Gn is planar if and only if n ∈ {1, 2, 3, 4, 6}.

Proof. For any of the given values of n, we can easily produce a planar drawing of Gn, as shown in figure 7.12

Figure 7. The planar Gn graphs.

We will now show that Gn is not planar for any other value of n. By the corollary to Euler’s formula, aplanar graph must have a vertex with degree less than six. Earlier, we showed that Gn is a φ(n)-regular.Thus, if φ(n) ≥ 6, Gn cannot be planar. This implies that Gn is nonplanar for n = 7, n = 9, and n ≥ 13. Itremains to be shown that G5, G8, G10 and G12 are nonplanar.

G5 is equal to K5, the complete graph on five vertices, and is thus nonplanar. G8, G10, and G12 eachcontain subgraphs that are subdivisions of K3,3, as shown in figure 8.

Figure 8. Subgraphs of G8, G10 and G12 which are subdivisions of K3,3.

By Kuratowski’s theorem, G8, G10, and G12 are non-planar. Therefore, the only values of n for which Gn

is planar are 1, 2, 3, 4 and 6.�

We now seek to extend this result to GR graphs, where R is an arbitrary direct product of cyclic rings.13

Lemma 2. Let R = Zn1 × . . .Znr. Then GR is k-regular, and deg(v) =

r∏i=1

φ(ni) for all v ∈ V (G).

Proof. Let v be a vertex of GR, and suppose w is a vertex of GR adjacent to v. Let l(v) = (v1, v2, . . . vr) bethe label assigned to v, and let l(w) = (w1, w2, . . . wr) be the label assigned to w. Then wi must be thelabel of a vertex adjacent to the vertex labeled vi in the graph Gni

. Since the degree of any vertex of Gniis

φ(ni), wi has φ(ni) possible values. Since v is an arbitrary vertex of GR, this argument implies that GR is

k-regular, with degreer∏

i=1

φ(ni).

�

Proposition 9. Let R be a ring of the form Zn1 × . . .× Znr. Then GR is planar if and only if R has the

form Z2 × . . .Z2 × Zn, where n ∈ {1, 2, 3, 4, 6}.

Proof. (⇒) Let R = Z2 × . . .× Z2 × Zn, where n ∈ {1, 2, 3, 4, 6}. We will show that GR is a planar graph.When n = 1, R = Z2 × . . .Z2. We note that the degree of any vertex of GR is one. This follows from thelemma, and the fact that φ(2) = 1. GR is thus a disjoint union of K2 graphs, which can easily be drawn inthe plane without edge crossings.

The cases R = Z2 × . . .Z2 ×Z4 and R = Z2 × . . .Z2 ×Z6 require a little more work. We begin by observingthat both G4 and G6 are even cycles.

Choose n such that Gn is an even cycle, and suppose m ∈ N. Let R be the ring obtained from Zn bytaking the direct product with Z2 m− 1 times. We will show that GR is a disjoint union of even cycles,which implies that GR is a planar graph.

Choose a vertex v of Gn. We give v a new label l(v), an ordered m-tuple, as follows. Set the lastcoordinate of l(v) equal to v, and let the remaining coordinates be an arbitrary list L of ones and zeroes.We then proceed around the cycle, assigning an m-tuple l(vi) to every vertex vi, according to the followingrule: we set the last coordinate l(vi) equal to vi. If the distance from v to vi is even, we fill in theremaining coordinates with the ordered list L. If the distance from v to vi is odd, we fill in the first m− 1coordinates with a list of values obtained from L by interchanging the 1′s and 0′s

The resulting labeling shows that Gn is a subgraph of GR, where R is the ring obtained from Zn by takingthe direct product with Z2 m− 1 times. We repeat this process a total of 2m−1 times, constructing 2m−1

copies of Gn and labeling the vertices of each in a manner analogous to that described above. For eachcopy of Gn, we let the first m− 1 coordinates of the label assigned to v be a distinct list of ones and zeroes.We will show that the resulting union of disjoint cycles is equal to GR, with R as above.

The scheme described assigns each element of R to a unique vertex of the resulting graph. By construction,this graph is isomorphic to GR. Since this graph is a disjoint union of cycles, it is clearly planar. Hence, forR = Z2 × . . .Z2 × Z4 and R = Z2 × . . .Z2 × Z6, GR is a planar graph.

Figure 9 provides an example of how this process works.14

Figure 9. GR, with R = Z2 × Z4 is isomorphic to the disjoint union of two G4 graphs.

For the remaining case, n = 3, we simply note that Z6 is isomorphic to Z2 × Z3 by the Chinese RemainderTheorem. Hence, for R = Z2 × Z3, GR is isomorphic to G6. The result for G6 thus implies that GR,R = Z2 × . . .Z2 × Z3is a planar graph.

(⇐) Next, we show that no other direct product of cyclic rings yields a planar GR graph. Again, we caneliminate any ring whose graph has degree greater than 5. However, this is not sufficient to yield a finitelist. Taking the direct product of a ring R with Z2 gives a graph with different structure, but the samedegree. Thus, we may obtain rings whose graphs have sufficiently small degree by multiplying any of thefollowing rings by Z2 an arbitrary number of times:

Z5, Z7, Z8, Z10, Z12, Z3 × Z3, Z3 × Z6, Z4 × Z4, Z4 × Z6, and Z6 × Z6.

We have already shown that the cyclic rings listed above are non-planar. Consider the non-cyclic rings inthe above list. In Figure 11, we demonstrate that each of their respective GR graphs contains a subgraphwhich is a subdivision of K3,3, and is thus non-planar. More specifically, the GR graphs of all except forZ3 × Z3 contain a subgraph of the form shown in figure 10. This graph is a subdivision of K3,3, where anypath between vertices in different partite classes has odd length. In addition, G8, G10 and G12 havesubgraphs of this form, as shown in the planarity proof for Gn graphs. G7 is the complete graph on 7vertices, and therefore has K3,3 as a subgraph.

Figure 10. This graph is a subdivision of K3,3. Note that each edge in the underlying K3,3

has been replaced by a path of odd length.

15

Figure 11. subgraphs of non-cyclic rings which are subdivisions of K3,3.

Consider a GR graph which has a subgraph of the form shown in Figure 10. Suppose we take directproduct of R with Z2 an arbitrary number of times. We will prove that the resulting ring has a non-planarGR graph.

Let R be a ring, and let H be a subgraph of GR such that H is a subdivision of K3,3, and each pathcorresponding to an edge of the underlying K3,3 has odd length. We label the vertices of this subgraph, asfollows. Let H0 denote one of the partite sets in H, and let H1 denote the other. We append an additionalcoordinate, with a value of zero, to the label of each vertex in H0. We then proceed along each path whichcorresponds to an edge in K3,3, appending a coordinate to the label of each. If the distance between a givenvertex and the nearest vertex of H0 is odd, let this additional coordinate have a value of 1; if the distanceis even, let it have a value of 0. The resulting labeling shows that H is a subgraph of R×Z2. Thus, the GR

graph coresponding to R× Z2 is nonplanar. By repeating this process an arbitrary number of times, wecan show that the ring obtained by taking the direct product of R with Z2 an arbitrary number of timeshas a GR graph with H as a subgraph. Thus, the GR graph corresponding to such a ring is nonplanar.

16

Finally, we examine the case of Z3 × Z3, which has no subgraph of the form H. We note that Z2 × Z3 × Z3

is isomorphic to Z6 × Z3, which does have such a subgraph. The above argument thus implies that Z3 × Z3

with Z2 an arbitrary number of times will yield a ring whose graph is nonplanar.

We have shown that the only non-cyclic rings yielding planar GR graphs are those obtained by taking thedirect product of Z2 with a ring Zn such that Gn is planar.

�

9. Crossing Number

Proposition 10. : For any n ≥ 5, the complete graph Kn is non-planar.

Proof. Assume for contradiction that Kn for some n ≥ 5 is a planar graph. Since Kn is a complete graph,

then the number of vertices is n and e=kn

2, where k represents the degree of each vertex. In this case k =

n− 1. By replacing k by n− 1 we get: e=(n− 1)n

2. Now by using Euler’s Formula we get that

e ≤ 3(n)− 6 which implies that(n− 1)n

2≤ 3(n)− 6. By simple algebra this reduces to n2 − 7n+ 12 ≤ 0.

Thus n ≤ 4, which is a contradiction of our assumption that n ≥ 5.

�

The minimum number of edge crossings required to represent a graph G in the plane is said to be thecrossing number denoted as γ(G).

Theorem 2. γ(K6) = 3.

Proof. Figure 12 demonstrates that γ(K6) ≤ 3.

Figure 12.

We prove γ(K6) ≥ 3.

Case 1: There exists a vertex v such that adding v to K6 − v note K6 − v ∼= K5 yields no more edgecrossings.

This implies that if we start with a K5 drawing and add another vertex to get K6, the edges incident atthe vertex do not increase the number of edge crossings. This vertex must reside inside a face of K5 whereit can connect to all the other vertices without any more edge crossings. Without loss of generality, assumethis is the inside face, then the other vertices would have to use the outer face to connect. Start by usingthe vertex labeled 0 and connect it to the vertex labeled 2. Then start use the vertex labeled 1 and connectit to the vertex labeled 3. No matter what side one chooses, this will create one edge crossing. Repeat this

17

procedure to connect the vertex labeled 2 to the vertex labeled 4 and this will create at least one morecrossing. Then use the same method to connect the vertices 3 and 0.This yields at least 5 crossings.

Case 2: For every vertex v, adding v to K7 − v yields at least one more edge crossing.

Subcase 1:If there exists a v such that K6-v has at least two edge crossings, then we are done becauseadding v to the drawing has to add at least 1 more edge crossing (If not, we would be back in the first case).

Subcase 2: If not, then for every v K6-v has exactly 1 edge crossing. The resulting drawing is anisomorphism of K5. Also we know that this drawing has one form, which uses a subgraph of K4. Thus,there is only a unique way of drawing it. By doing an exhaustive check, we can verify that regardless wherewe put the vertex, it creates at least two more edge crossings. Therefore, γ(K6) ≥ 3. Combining the twoproofs we get that the γ(K6) = 3. �

Proposition 11. : 5 ≤ γ(G7) ≤ 9.

Proof. We prove that γ(G7) ≥ 5. Note: G7∼= K7.

Case 1: There exists a vertex v, such that adding v to K7 - v yields no more edge crossings.

This implies that if we start with a K6 drawing and add another vertex to get K7, this vertex does notincrease the number of edge crossings. Insert the vertex inside a face of K6 where the vertex can connect toall the other vertices without any edge crossing. Without loss of generality, assume this is the inside face,then the other vertices would have to use the outer face to connect. Start by using the vertex labeled 0 andconnect it to the vertex labeled 2. Then start with the vertex labeled 1 and connect it to the vertex labeled3. No matter what side you choose, this will create one edge crossing. Repeat this procedure to connectthe vertex labeled 2 to the vertex labeled 4 and this will create at least one other crossing. Then use thesame method to connect the vertices 3 and 5 and 4 and 6 and 5 and 0. This yields six edge crossings.

Case 2:For every vertex v, adding v to K7 − v yields at least one more edge crossing.

Subcase 1:If there exists a v such that K7-v has at least 4 edge crossings, then we are done because v hasto add at least one more edge crossing (If not, we would be in the first case).

Subcase 2:If not, then K7 − v ∼= K6 has exactly 3 edge crossings. By our result that K6 has a crossingnumber of 3 and doing an exhaustive check to verify that regardless of where we insert the vertex v, wehave two extra edge crossings. This yields five edge crossings. Therefore, γ(G7) ≥ 5.

We prove that γ(G7) ≤ 9 by looking at Figure 13.

Figure 13.

�18

Theorem 3. γ(G8) = 4.

Proof. We prove γ(G8) ≤ 4 by using Figure 14

Figure 14.

We prove γ(G8) ≥ 4 using the following Lemma.

Lemma 3. [5]. If G is a triangle-free planar graph with n vertices and e edges, then e ≤ 2n− 4.

Note: G8 = K4,4 is bipartite, and therefore any subgraph of it is triangle free. Now suppose that G8 can bedrawn in the plane with 3 or fewer crossings. This means that by removing 3 edges (the edges that createthe crossings) from G (but keeping all vertices), we may produce a planar subgraph G′. However, G′ isthen triangle-free with 8 vertices and 13 edges, so this contradicts Lemma 3 because we have 13 ≤ 12.Therefore, γ(G8) ≥ 4. Combining the two statements above we get that γ(G8) = 4. �

Proposition 12. γ(G9) ≤ 16.

Proof. Figure 15 yields this result.

Figure 15.

�

Theorem 4. γ(G10) = 4.19

Proof. First, we prove γ(G10) ≤ 4 by using Figure 16.

Figure 16.

Next, we prove γ(G10) ≥ 4.

Note: G10 is bipartite. Using the same technique, we used in Theorem 3 G′, is triangle-free with 10 verticesand 17 edges, which leads us to the contradiction of Lemma 3 because we have 17 ≤ 16. Therefore,γ(G10) ≥ 4. Combining the two statements above we get that γ(G10) = 4. �

Proposition 13. : 4 ≤ γ(G12) ≤ 6.

Proof. First, we prove γ(G12) ≤ 6 by observing Figure 17

Figure 17.

We prove γ(G12) ≥ 4.

Note: G12 is bipartite. Using the same technique we used in Theorem 3, G′ is triangle-free with 12 verticesand 21 edges, which leads us to the contradiction of Lemma 3 because we have 21 ≤ 20. Therefore,γ(G12) ≥ 4.

�20

10. Automorphisms

Finally, we look at the automorphisms of Gn. An automorphism of a graph is a bijective mapping from thevertex set of the graph to itself which preserves adjacency and non-adjacency. The automorphisms of agiven graph form a group under the operation composition of functions.

Theorem 5. Let n =s∏

i=1

peii , where the pi are distinct primes, and p1, p2, ...ps. Then a bijective mapping

α : V (Gn)→ V (Gn) induces an automorphism of Gn if and only if α preserves congruence modulo pi forall i.

Proof. (⇐) Suppose α preserves congruence modulo pi for all i. Let v and w be vertices of Gn. If v and ware non-adjacent, then we have v ≡ w (mod pi) for some i. Because α preserves congruence modulo pi,α(v) ≡ α(w) (mod pi). Hence, by the rules of modular representation, α(v) is adjacent to α(w). So α mapsnonadjacent vertices of Gn to nonadjacent vertices of Gn.

Next, suppose that v and w are adjacent. We will show that α(v) and α(w) are adjacent. Consider a primedivisor pi of n. Suppose α maps a vertex u into a residue class r mod pi. Since α preserves congruencemodulo pi, every vertex whose label is congruent to u modulo pi maps into r. Gn contains

n

pivertices

whose labels are congruent to u modulo pi, which is exactly the number of vertices with labels in r. By thepigeonhole principle, no two vertices which are not congruent modulo pi can map to vertices whose labelsare in the same residue class modulo pi. Because v and w are adjacent, we have v 6≡ w (mod pi) for all i.By the above argument, this implies that α(v) 6≡ α(w) (mod pi) for all i. Hence, α(v) is adjacent to α(w).So α maps adjacent vertices of Gn to adjacent vertices of Gn. Hence, α is an automorphism of Gn.

(⇐) Next, suppose that α is an automorphism of Gn. We will show that Gn preserves congruence modulopi for all i.

We first show that α preserves congruence modulo p1, the smallest prime divisor of n. Consider the set Aof vertices of Gn whose labels belong to a residue class a modulo p1. A is an independent subset of V (Gn).Since α is an automorphism of Gn, α maps A to an independent subset S of V (Gn). We will show that theelements of S are congruent modulo p1.Let t and u be vertices in S. Then clearly, |t− u| > p1. For otherwise, we would have t 6≡ u (mod pi) for alli, so t and u would be adjacent, contradicting the definition of S. Hence, S contains

n

p1elements between 0

and n, with the difference between any two elements greater than or equal to p1. We arrange the elementsof S in a list from smallest to largest, and consider consecutive elements of S. Therefore, the differencebetween consecutive elements of S must be exactly p1. This implies that the elements of S are congruentmodulo p1.

Next, suppose α preserves congruence modulo p1, p2, . . . pm. We will show that α preserves congruencemodulo pm+1.

Case 1: Suppose n is odd. Let v and w vertices of Gn such that v ≡ w (mod pm+1). Assume|α(v)− α(w)| < pm+1. Then α(v) and α(w) are in different residue classes modulo pi for all i ≥ m.

The number of common neighbors of v and w in Gn is given bys∏

i=1

pei−1i di where di = pi − 1 if v ≡ w

(mod pi) and di = pi − 2 if v 6≡ w (mod pi). To see this, we assign each vertex v to a set X(v1, . . . , vs),where 0 ≤ vi ≤ p− 1 and vi ≡ l(v) (mod pi).

Let u be a vertex adjacent to both v and w, and let X(u1, . . . , us) be the class containing u. Then each ui

must differ from both vi and wi. Hence, ui can take on either pi − 1 possible values, if ui = vi, or pi − 221

possible values, if ui 6= vi. Hence, u belongs to one ofs∏

i=1

di sets, each of which contains∏

i=1

pei−1i elements.

Hence, the number of common neighbors of v and w iss∏

i=1

pei−1i di. Because n is even, pi ≥ 3 for all i, and

hence di ≥ 1 for all i. Therefore, any two vertices have at least one common neighbor.

Since α is an automorphism of Gn, the number of common neighbors of α(v) and α(w) must be equal to

that of v and w. Since α preserves congruence modulo p1, p2, . . . , pm, we know thatm∏

i=1

pmi−1i di is identical

for the two pairs v, w and α(v), α(w). For α(v) and α(w), di = pi − 2 for all i ≥ m+ 1. Since v ≡ w

(mod q), we have dm + 1 = pm+1 − 1 for v and w. The value ofs∏

i=1

pki−1i di is thus greater for v and w than

for α(v) and α(w). This implies that v and w have more common neighbors than do α(v) and α(w), whichis a contradiction. Hence, |α(v)− α(w)| ≥ pm+1.

Consider the subset Q of V (Gn), consisting of all vertices of Gn whose labels are congruent to v modulopm+1. Q is an independent set. Since α is an automorphism, α maps Q to an independent set, which wedenote R. R contains exactly

n

pm+1elements between 0 and n− 1. By the above argument, any two

elements of R must differ by at least pm+1. Arrange the elements of R in a list from smallest to largest. Bythe pigeonhole principle, successive elements on this list must differ by exactly pm+1. But this implies thatthe elements of R are congruent modulo pm+1. Hence, α preserves congruence modulo pm+1.

Case 2: Suppose n is even. In this case, the proof is more complicated. Again, we consider two verticeslabeled v and w, where v ≡ w (mod pm+1). If, in addition, v ≡ w (mod 2), we can apply the commonneighbors argument used above, and show that |α(v)− α(w)| ≥ pm+1. If, v 6≡ w (mod 2), this argumentfails, because v and w have no common neighbors. We therefore use a variation on this argument. Insteadof considering common neighbors of v and w, we examine paths of length 3 between v and w.

— Claim: the number of paths of length 3 between two vertices of Gn which are not congruent modulo 2 is

given bys∏

i=1

p2(ei−1)i ci, where the value of ci is given by the following:

If v ≡ w (mod pi), then ci = p2i − 3pi + 2.

If v 6≡ w (mod pi), then ci = p2i − 3pi + 3.

To prove this, we again assign each vertex of Gn to a set X, as described above. Suppose v, t, u, w is a pathin Gn. Let X(v1, . . . , vs) be the set containing v, X(t1, . . . , ts) the set containing t, X(u1, . . . , us) the setcontaining u, and X(w1, . . . , ws) the set containing w.

Suppose vi = wi for a given value of i. Then ti can take on any value between 0 and pi − 1 not equal to vi,or one of pi − 1 different values. Similarly, ui can take on any value between 0 and pi − 1 not equal to wi orui, or one of pi − 2 values. Hence, the number of possible combinations of values of ti and ui is given byci = (pi − 1)(pi − 2) = p2

i − 3pi + 2.

On the other hand, suppose vi 6= wi. Then ti can again take on one of pi − 1 possible values. If ti = wi,then ui can take on any value not equal to ti, or one of pi − 1 possible values. If ti 6= wi, then we againhave pi − 2 possible values for ui. Hence, the number of possible combinations of values of ti and ui isgiven by ci = 1(pi − 1) + (pi − 2)(pi − 2) = p2

i − 3pi + 3.

Multiplying the values of ci for 1 ≤ i ≤ s gives the number of possible combination of coordinates for thesets X(t1, . . . , ts) and X(u1, . . . , us). To obtain the total number of paths of length 3 between v and w, we

22

multiply this expression by the square of the number of elements in each class, and obtain a total of∏si=1 p

2(ei−1)i ci such paths between any two vertices.

Since α is an automorphism of Gn, the number of paths of length 3 between α(v) and α(w) is equal to the

number of such paths between v and w. Since α preserves congruence modulo p1, . . . , pm,m∏

i=1

p2(ei−1)i ci is

identical for the pair α(v) and α(w) and the pair v and w. Suppose |α(v)− α(w)| < pm+1. Thenα(v) 6≡ α(w) (mod pi) for all pm+1 ≤ i ≤ s. Hence, ci = p2

i − 3pi + 3 for all pm+1 ≤ i ≤ s. Since v ≡ w(mod pm+1), we have dm + 1 = p2

i − 3pi + 2. Thus, there are fewer paths of length 3 between v and w thanbetween α(v) and α(w). This gives a contradiction. Hence, |α(v)− α(w)| ≥ pm+1.

By the argument used in the odd case, this implies that α preserves congruence modulo pm + 1. Again,repeating this argument for each prime pi shows that α preserves congruence modulo pi for all i. Thiscompletes the proof.

�

11. Acknowledgements

We thank Dr. Reza Akhtar, our research director at SUMSRI, for his instruction, guidance, and supportthroughout the learning and research process this summer. Thanks to our wonderful graduate assistant,Amanda Phillips, for going above and beyond the call of duty, for staying up late helping us with this project,and for her unfailing confidence in us. Our thanks to Dr. Farmer for his instruction on mathematical writing,and his help revising the paper. Thanks also to the other participants and graduate assistants of SUMSRIfor their support and encouragement this summer. We are grateful to Dr. Davenport and Dr. Waikar,directores of SUMSRI, and all others who made this experience in SUMSRI possible for us. We would alsolike to thank the National Security Agency, the National Science Foundation, and Miami University for theirsupport of this program.

References

[1] Balakrishnan, R. and K. Ranganathan. A Textbook of Graph Theory. Springer-Verlag New York, Inc., New York, 2000.[2] Evans, Anthony B., Gerd H. Fricke, Carl C. Maneri, Terry A. McKee, and Manley Perkel. Representations of Graphs Modulo

n. Journal of Graph Theory, 18(8):801–815, 1994.[3] Foulds, L.R. Graph Theory Applications Springer-Verlag New York, Inc., New York, 1992.

[4] Gould, Ronald. Graph Theory. The Benjamin/Cummings Publishing Company, Inc, Menlo Park, California, 1988.

[5] West, Douglas B. Introduction to Graph Theory. 2nd ed. China Machine Press, 2004.

23