Apr 16, 1998 - as follows: Suppose that Professor X is blind and has a circular key ring with seemingly ... If G is a graph, then a symmetry or automo...

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Karen S. Potanka

Thesis submitted to the Faculty of the Virginia Polytechnic Institute and State University in partial fulfillment of the requirements for the degree of

Master of Science in Mathematics

Dr. Ezra Brown, Chair Dr. Joseph Ball Dr. Monte Boisen

April 16, 1998 Blacksburg, Virginia

Keywords: Petersen Graph, Symmetry-Breaking, Graph Theory Copyright 1998, Karen S. Potanka

Groups, Graphs, and Symmetry-Breaking

Karen S. Potanka

(ABSTRACT)

A labeling of a graph G is said to be r-distinguishing if no automorphism of G preserves all of the vertex labels. The smallest such number r for which there is an r-distinguishing labeling on G is called the distinguishing number of G. The distinguishing set of a group Γ, D(Γ), is the set of distinguishing numbers of graphs G in which Aut(G) ∼ = Γ. It is shown that D(Γ) is non-empty for any finite group Γ. In particular, D(Dn ) is found where Dn is the dihedral group with 2n elements. From there, the generalized Petersen graphs, GP (n, k), are defined and the automorphism groups and distinguishing numbers of such graphs are given.

Contents 1 Introduction

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2 Representation of Graphs

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3 Automorphism Groups and Group Actions

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3.1 Automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3.2 Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 Distinguishing Number of a Graph

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5 Groups and Graphs

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5.1 Cayley Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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5.2 Frucht’s Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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6 Distinguishing the Orbits of a Graph

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6.1 Distinguishing the Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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6.2 Size of the Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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6.3 Number of Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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7 Dihedral Groups

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7.1 Subgroups of Dn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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7.1.1

Types of Subgroups of Dn . . . . . . . . . . . . . . . . . . . . . . . .

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7.1.2

Properties of Subgroups of Dn . . . . . . . . . . . . . . . . . . . . . .

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7.2 Distinguishing number of graphs which realize Dn . . . . . . . . . . . . . . .

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7.2.1

Graphs with a stabilizer subgroup of Type 1 . . . . . . . . . . . . . .

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7.2.2

Graphs with a stabilizer subgroup of Type 2 or 3 . . . . . . . . . . .

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8 Generalized Petersen Graphs

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8.1 Automorphism Groups of the Generalized Petersen Graphs . . . . . . . . . .

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8.2 Distinguishing Numbers of Generalized Petersen Graphs . . . . . . . . . . .

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9 Conclusions and Conjectures

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A Tables of Congruences

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B Group Orders

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C 10-circuits in GP (26, 5)

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List of Figures 3.1 G1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4.1 K6 and W6

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4.2 G and G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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5.1 The Cayley color graph for C5 . . . . . . . . . . . . . . . . . . . . . . . . . .

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5.2 Cayley color graph for S3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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5.3 Cayley graph for S3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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5.4 The Frucht graph of S3 whose automorphism group is isomorphic to S3 . . .

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6.1 This graph has distinguishing number two while the minimum number of labels needed to distinguish each orbit separately is three . . . . . . . . . . .

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6.2 The graph on the left is labeled to distinguish each orbit separately while the graph on the right is labeled to distinguish the entire graph . . . . . . . . . .

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7.1 This graph illustrates the coloring in Lemma 10 . . . . . . . . . . . . . . . .

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7.2 The bipartite graph on the left is complete while the one on the right is empty 28 7.3 Vertex-transitive graphs with sizes 3,4 and 5 . . . . . . . . . . . . . . . . . .

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7.4 K3,2 and C5 ∨ K2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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8.1 The Petersen Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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8.2 GP(8,2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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8.3 Two cases for the action of α . . . . . . . . . . . . . . . . . . . . . . . . . .

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8.4 The action of α if an outer edge e is sent to a spoke . . . . . . . . . . . . . .

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8.5 For k = 1, there are only n 4-circuits in GP (n, 1) . . . . . . . . . . . . . . .

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8.6 GP (13, 5) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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8.7 A 3-distinguishing coloring of GP (4, 1) . . . . . . . . . . . . . . . . . . . . .

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Chapter 1 Introduction The concept of symmetry breaking in graphs made its first appearance as Problem 729 in the Journal of Recreational Mathematics, see Rudin [7]. The problem can be summed up as follows: Suppose that Professor X is blind and has a circular key ring with seemingly identical keys which open different doors. Someone suggests to Professor X that he use a variety of handle shapes to help in distinguishing the keys by touch. Assuming that a rotation of the key ring is undetectable from an examination of a key, what is the minimum number of shapes needed so that all keys would be distinguishable from one another? The question above can be made more general by allowing any particular shape for the key ring. In doing this, the keys become vertices, and two vertices are joined by an edge if the keys are adjacent on the key ring. We will be concerned with finding the minimum number of labels or shapes needed for the vertices, so that all vertices will be distinguishable. This minimum number will be called the distinguishing number of a graph. This paper begins with an introduction to the representation of graphs and some elementary terminology, followed by a discussion of the symmetry or automorphisms of a given graph. Upon knowing the automorphism group of a graph, the distinguishing number of a graph can be defined. We then discuss the relationship between groups and graphs; in particular, the Cayley graph of a group and the Frucht construction of a graph realizing a particular group. The paper will then continue with the theory developed in Collins [1] on distinguishing the orbits of a graph and how the size and number of orbits can help in distingushing the entire graph. Furthermore, we will present and prove the theorems given in [1] dealing with graphs which realize a dihedral group. In a number of cases, our proofs and discussion fill in many of the details omitted in [1]. Finally, we discuss the automorphism groups of the generalized Petersen graphs and their distinguishing numbers.

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Chapter 2 Representation of Graphs Definitions. A graph G consists of a nonempty vertex set V (G) together with an edge set E(G) (possibly empty), where each edge is an unordered pair of vertices. If both the vertex set and edge set are finite, then the graph G is said to be finite. We write uv for the edge {u, v}. If e = uv ∈ E(G), then we say that u and v are adjacent while u and e are incident, as are v and e, and we say that the edge e = uv joins u and v. A graph is called simple if it has no self-loops or multiple edges. In other words, e = uu ∈ / E(G) and there cannot be distinct e1 and e2 in E(G) with both e1 and e2 joining u and v. Unless otherwise specified, in what follows all of our graphs will be simple. The complement of a graph G, denoted G, consists of the vertex set V (G) = V (G), along with an edge set E(G) = {uv|uv ∈ / E(G)}. The degree of a vertex v in the graph G is the number of edges incident with v. A vertex with degree k is said to be k-valent. If all of the vertices have the same degree k, then G is called regular of degree k or simply a k-regular graph.

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Chapter 3 Automorphism Groups and Group Actions 3.1

Automorphisms

Definition. In what follows, we use both σx and σ(x) to denote the image of x under the map σ. If G is a graph, then a symmetry or automorphism of G is a permutation σ on the vertices of G so that for every edge xy of G, σ(xy) = σ(x)σ(y) is an edge of G. Thus, each automorphism of G is a one-to-one and onto mapping of the vertices of G which preserves adjacency. This implies that an automorphism maps any vertex onto a vertex of the same degree. Let G1 be the graph in Figure 3.1, let σ be the mapping (13)(24) and let τ be the mapping (123). To see that σ is an automorphism of G1 , notice that the images of all the edges are indeed edges of G1 : σ(12) = 34, σ(23) = 41, σ(34) = 12, σ(41) = 23, and σ(13) = 13. Now to see that τ is not an automorphism of G1 notice that τ (14) = 24 ∈ / E(G1 ). Lemma 1 Let Aut(G) be the collection of all automorphisms of G. Then Aut(G) is a group 1

2

4

3

Figure 3.1: G1 3

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Chapter 3. Automorphism Groups and Group Actions

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under composition. Proof: 1. (Identity) Clearly, e ∈ Aut(G). 2. (Closure) Suppose σ ∈ Aut(G) and τ ∈ Aut(G) and let xy be any edge of G, then στ (xy) = σ(τ (x)τ (y)) where τ (x)τ (y) is an edge, and σ(τ (x)τ (y)) = στ (x)στ (y) which is also an edge of G. 3. (Associativity) Associativity follows from the fact that if σ,τ , and γ are all in Aut(G), then they are all permutations of the vertices of G, and we know that composition of permutations is associative. 4. (Inverses) Suppose that σ ∈ Aut(G). Then σ is a permutation of the vertices and so is invertible. To show that σ −1 ∈ Aut(G), let xy ∈ E(G) and suppose that σw = x and σz = y (Since σ preserves adjacency, wz ∈ E(G)). Then σ −1(xy) = σ −1 (x)σ −1(y) = wz ∈ E(G). 2 Lemma 2 If G is a graph, then Aut(G) = Aut(G). Proof: Let σ ∈ Aut(G) and let xy ∈ E(G). By definition of the complement of a graph, we have that xy ∈ / E(G), so σ(x)σ(y) ∈ / E(G). (Otherwise, if σ(x)σ(y) ∈ E(G), then −1 −1 −1 σ (σ(x)σ(y)) = σ σ(x)σ σ(y) = xy ∈ E(G).) So Aut(G) ⊂ Aut(G). A similar argument shows that Aut(G) ⊂ Aut(G).

3.2

2

Group Actions

Definition. We can think of letting the automorphism group of a graph act on the set of vertices and edges of the graph. In doing this, we form orbits of both vertices and edges. We say that two vertices v1 and v2 of G are in the same orbit if there is an automorphism σ ∈ Aut(G) such that σv1 = v2. Similarly, we say that two edges e1 and e2 of G are in the same orbit if there is an automorphism τ ∈ Aut(G) such that τ e1 = e2. Definition. A graph G is said to be vertex-transitive if given any two vertices v1 and v2 of V (G), there is an automorphism σ ∈ Aut(G) such that σv1 = v2 . In other words, a graph is vertex-transitive if there is only one orbit of vertices under the action of the automorphism group on the set of vertices. Similarly, a graph G is said to be edge-transitive if given any two edges e1 and e2 of E(G), there is an automorphism τ ∈ Aut(G) such that τ e1 = e2. So, a graph is edge-transitive if there is only one orbit of edges under the action of the automorphism group on the set of edges.

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Chapter 3. Automorphism Groups and Group Actions

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Lemma 3 Suppose that the connected graph G is edge-transitive. Then either G is vertextransitive or there are only two orbits of vertices under the action of the automorphism group on the set of vertices. Proof: Suppose that there are three or more orbits of vertices. Then let u1 , u2 , and u3 be vertices from three different orbits. Now since G is connected, there must be at least one edge incident to each vertex u1 , u2, and u3. In particular there are edges u1 v1, u2v2 , and u3v3 ∈ E(G). Since G is edge-transitive, there is an automorphism σ ∈ Aut(G) such that σ(u1v1) = u2 v2. This can be done in only one way: σu1 = v2 and σv1 = u2, since σu1 6= u2 and σu2 6= u1 . Similarly, there exists τ ∈ Aut(G) with τ (u2v2 ) = u3v3 . Since u2 and u3 are in different orbits, we must have τ u2 = v3 and τ v2 = u3 . But then we would have τ (σ(u1)) = τ (v2) = u3 and this says that u1 and u3 are in the same orbit. 2 Definition. Suppose that V(G) is the vertex set of the graph G, and let Aut(G) be the automorphism group of the graph. For each v ∈ V (G) we define the stabilizer of v in Aut(G) to be the set of elements in Aut(G) which fix the vertex v or {σ ∈ Aut(G)|σv = v}. The stabilizer of the vertex v is denoted Stv , and it is straightforward to show that Stv is a subgroup of Aut(G). It is also important to note that the number of elements in Stv is equal to the order of the group divided by the number of elements in the orbit of v (provided that the orbit is finite): |Aut(G)| . |Stv | = |Ov | We now prove two results involving the stabilizer subgroup of a vertex which will be of use later in the paper: Lemma 4 Suppose that u and v are vertices in the same orbit under the action of the automorphism group of G on the set of vertices, and let Stu and Stv be their respective stabilizer subgroups. Then there is some element g ∈ Aut(G) with g(Stv )g −1 = Stu . Proof: Let Stv = {α1 , α2, . . . , αn } be the stabilizer subgroup of the vertex v, and let Stu = {β1, β2, . . . , βn } be the stabilizer subgroup of the vertex u, where u and v are in the same vertex orbit. Then there is some ϕ ∈ Aut(G) with ϕ(u) = v and ϕ−1 (v) = u. We want to find some g ∈ Aut(G) with g(Stv )g −1 = Stu . Let αi ∈ Stv , then ϕ−1 αi ϕ(u) = ϕ−1 αi (v) = ϕ−1 (v) = u. Thus, ϕ−1 αi ϕ ∈ Stu for each i, so ϕ−1 Stv ϕ ⊂ Stu .

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Chapter 3. Automorphism Groups and Group Actions

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Now let βj ∈ Stu . We want to express βj as ϕ−1 αi ϕ for any αi ∈ Stv . It is true that ϕβj ϕ−1 ∈ Stv for ϕβj ϕ−1 (v) = ϕβj (u) = ϕ(u) = v. So βj = ϕ−1 (ϕβj ϕ−1 )ϕ where ϕβj ϕ−1 ∈ Stv . Thus Stu ⊂ ϕ−1 (Stv )ϕ.

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Corollary 5 If O is an orbit under the action of the automorphism group of G on the set of vertices, and if v ∈ O with Stv not normal in Aut(G), then for any vertex u ∈ O, Stu is not normal in G. Proof: If Stu is the stabilizer subgroup of vertex u 6= v ∈ O, then by the previous lemma we have that there is some g ∈ Aut(G) with g(Stv )g −1 = Stu . If, for instance, Stu is a normal subgroup of Aut(G), then we have g(Stu )g −1 = Stu for every g ∈ Aut(G). Thus Stu = Stv . But this contradicts the fact that Stv is not normal. 2

Chapter 4 Distinguishing Number of a Graph Definition. If G is a graph then we can think of assigning to each vertex of the graph a color or label. With each labeling or coloring of a graph, we associate a function φ : V (G) → {1, 2, 3, . . . , r} where {1, 2, 3, . . . , r} are the sets of labels. We say that a labeling is r-distinguishing if no automorphism of G preserves all of the vertex labels. In other words, for every automorphism σ ∈ Aut(G) there is at least one vertex v ∈ V (G) with φ(v) 6= φ(σ(v)). Furthermore, if there is a labeling on graph G which is r-distinguishing, we say that G is r-distinguishable. The smallest such number r for which G is r-distinguishable is called the distinguishing number of G, and is denoted D(G). This idea is due to Albertson and Collins (see [1]). It should be clear that the automorphism group of any labeled graph is a subgroup of the automorphism group of the unlabeled graph. In some sense, the labels are “breaking the symmetry” of the graph. The goal will be to break all of the symmetries of a particular graph so that the automorphism group of the labeled graph contains only the identity automorphism. Using this terminology, we can now restate the key problem discussed in the introduction. Finding the minimum number of shapes for the keys so that each key can be distinguished from the others is the same as finding the distinguishing number of Cn , where Cn is the cycle with n vertices. It turns out that for n = 3, 4, 5, the distinguishing number for Cn is 3, while for n ≥ 6, the distinguishing number of Cn is 2. This can be seen in the following labeling ψ where the vertices of Cn are denoted v1 , v2, . . . , vn in order: ψ(v1) = 1, ψ(v2) = 2, ψ(v3) = 1, ψ(v4) = 1, ψ(vi) = 2 for 5 ≤ i ≤ n. It turns out that two graphs with the same automorphism group may have different distinguishing numbers. This idea is illustrated in the next example. Example. Suppose we consider the distinguishing numbers of the two graphs Kn , the 7

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Chapter 4. Distinguishing Number of a Graph

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complete graph with n vertices, and the graph Wn with 2n vertices obtained by attaching a single pendant vertex to each vertex in Kn .

Figure 4.1: K6 and W6 See Figure 4.1 for the graphs of K6 and W6 . It should be clear to the reader that Aut(Kn) ∼ = ∼ Aut(Wn) = Sn , the permutation group on n elements. It should also be clear that the distinguishing number of Kn is n. Now consider Wn . It turns out that there are n ordered pairs of vertices consisting of a vertex of Kn and its pendant neighbor ((vi , ui ), i = 1, 2, . . . , n, vi is a vertex of Kn , and ui is the pendant vertex adjacent to vi .) In order for Wn to be distinguished, we must have that each ordered pair be different. If we use k labels for the vertices in Wn , then there are k 2 possible ordered labels for the pairs (k choices for vi’s and k for ui .) We need for n ≤ k 2 . The√ distinguishing number will be the smallest such k satisfying the inequality. Thus D(Wn ) = d n e. Example. Because a graph G and its complement G have the same vertex set, and because each automorphism in Aut(G) has the same action on a given vertex in both G and G, D(G) = D(G). To illustrate this idea, Figure 4.2 contains a graph G and its complement both of which have distinguishing number 2. Definition. Suppose that Γ is a group. We say that the graph G realizes Γ if Aut(G) = Γ. Furthermore, we define the distinguishing set of a group by D(Γ) = {D(G) | G realizes Γ}. In other words, although two graphs with the same automorphism group Γ may have different distinguishing numbers, the possible distinguishing numbers are limited to the set D(Γ). A natural question to ask now is this: given an arbitrary group Γ, is D(Γ) 6= ∅? In other

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Chapter 4. Distinguishing Number of a Graph

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_ G

G

Figure 4.2: G and G words, given a group Γ, how do we know that there is even one graph G which realizes Γ? We discuss these questions in the following chapter.

Chapter 5 Groups and Graphs 5.1

Cayley Graphs

During the nineteenth century, a mathematician named Cayley invented the technique of representing a group with a graph, where vertices correspond to the elements of a group and the edges correspond to multiplication by group generators and their inverses. As a preliminary example, consider the the group C5 . Suppose that we let the elements of C5 = {e, a, a2, a3, a4} be vertices. Now a generator for the group is multiplication on the right by the element a. So two vertices vi and vj will be joined by a directed edge from vi to vj if vi a = vj . The resulting graph (Figure 5.1 ) is called the Cayley color graph for C5. Definition. Let Γ be a group and let Ψ = {ψ1 , ψ2, . . . , ψn } e

a4

a

a2

a3

Figure 5.1: The Cayley color graph for C5 10

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be a generating set for Γ. The Cayley color graph, denoted C(Γ, Ψ) has as its vertex set the group elements of Γ, and as its edge set the cartesian product Γ × Ψ. The edge (σ, ψi) has as its endpoints the group elements σ and σψi with direction from σ to σψi. To each generator we assign a color or other devices to distinguish one class of edges from another such as dashed lines, bold lines, etc. As an example of a Cayley color graph for a group with more than one generator, let us consider the group S3 = {(1), (123), (132), (12), (13), (23)}. The set of generators for this group are the elements (123) and (12). If multiplication on the right by (123) is represented by a solid line and (12) by a dashed line, then Figure 5.2 contains the Cayley color graph of this group. (132) (123) (23)

(12)

(12)

(13)

(1)

(123)

Figure 5.2: Cayley color graph for S3 The graphs for which we have defined an automorphism to this point have been only simple, nondirected graphs. In order to define an automorphism on a Cayley color graph, we would need to eliminate the directions on the edges and to collapse any multiple edges to one. The resulting graph is defined to be the Cayley graph. Note that multiple edges occur when we have a generator of order 2, called an involution, as we had in Figure 5.2 with the generator (12). See Figure 5.3 for the Cayley graph of S3 .

5.2

Frucht’s Construction

Frucht shows that every finite group is isomorphic to the automorphism group of some graph (See [3], page 70 or White [8] ). The graph which Frucht constructs to realize a given group Γ is the Cayley graph of Γ with modifications on each edge of the Cayley graph so that the automorphisms of the modified graph are forced to respect the direction and colors on

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Chapter 5. Groups and Graphs

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Figure 5.3: Cayley graph for S3

Figure 5.4: The Frucht graph of S3 whose automorphism group is isomorphic to S3 the edges of the Cayley color graph. These modifications are done on every edge except those edges resulting from an involution. The first modification to the Cayley graph is the insertion of two new vertices on each edge. For each edge corresponding to the generator ψi, attach to the new vertex near the initial end of the directed edge in the Cayley color graph a path of length 2i − 1 and attach to the other new vertex near the terminal end a path of length 2i. The resulting graph is called the Frucht graph of Γ, denoted F (Γ), and it turns out that the automorphism group of this resulting Frucht graph is isomorphic to the group Γ. See Figure 5.4 for the modified graph of S3 described above. In the case where none of the generators for a group is an involution, the graphs constructed above have h(n + 1)(2n + 1) vertices where h is the order of the group and n is the number of generators for the group. Frucht proves even further in [2] that given any finite group Γ of order h ≥ 3 which is generated by n of its elements, it is always possible to find a 3-regular graph with 2(n + 2)h vertices that has Γ as its automorphism group. Now we have that each

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Chapter 5. Groups and Graphs

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group Γ has at least one graph that realizes it, and in most cases there will be a 3-regular graph which realizes it. Hence we now know that D(Γ) will be a nonempty set. In fact, because there is a 3-regular graph that realizes a given group, we will see in the following theorem from [1] that 2 ∈ D(Γ) for any finite group Γ. Theorem 6 For any finite group Γ, 2 ∈ D(Γ). Proof: It is known that for any finite group Γ there is a connected 3-regular graph G which realizes Γ (see Frucht [2]). If G is a graph with n vertices, then construct a new graph from G by attaching to each vertex of G a path of length dlog2 ne. It should be clear that each automorphism of G is an automorphism of the new graph. Now to see that each automorphism of the new graph is also an automorphism of G, note that each vertex originally in G now has degree 4 and all the other vertices added to G have degree less than or equal to 2. Since automorphisms take vertices to vertices of the same degree. We have that a vertex v originally in G must go to another vertex u originally in G under this automorphism of the new graph. The path attached to this vertex v has only one place to go, namely the path attached to u. Now there are 2dlog2 ne ≥ n labelings of these paths using 2 labels. If we color each path differently, then we have distinguished the new graph using 2 colors. 2

Chapter 6 Distinguishing the Orbits of a Graph If Ov is the orbit containing the vertex v in a given graph G, then any automorphism must send v to another vertex in Ov . Thus, vertices in different orbits will be distinguished from one another. One approach to distinguish a graph would be to distinguish each orbit separately. In other words, one could say that an orbit is r-distinguishable if every automorphism that acts nontrivially on the orbit sends at least one vertex to another vertex with a different label. Though it may be easier to distinguish orbits separately than distinguishing the entire graph, the number of labels needed to distinguish the entire graph may be less than the number of labels needed to distinguish the orbits separately. For instance, consider the following example: Example. The automorphism group of the graph in Figure 6.1 is isomorphic to the cyclic group of order 3. It turns out that there are three orbits of vertices in this graph: 01 = {1, 2, 3}, 04 = {4, 6, 8}, and 05 = {5, 7, 9}. The distiguishing number for any of the orbits is three while the distiguishing number of the entire graph is just two. Figure 6.2 includes the labeling used to distinguish the orbits separately (note that this labeling distinguishes the entire graph as well) and also includes the labeling used to distinguish the entire graph with only two colors. As illustrated in the previous example, it is not necessary for a labeling to distinguish every orbit separately in order to distinguish the entire graph. It is worth pointing out that it might be beneficial to know the distinguishing number of a given orbit. We see in the next section that if a graph satisfies certain properties, then the distinguishing number of the graph is the distinguishing number of an orbit.

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Chapter 6. Distinguishing the Orbits of a Graph

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1

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4 Orbits: {1,2,3} {4,6,8} {5,7,9}

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5 2

3 7

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Figure 6.1: This graph has distinguishing number two while the minimum number of labels needed to distinguish each orbit separately is three

Figure 6.2: The graph on the left is labeled to distinguish each orbit separately while the graph on the right is labeled to distinguish the entire graph

Karen S. Potanka

6.1

Chapter 6. Distinguishing the Orbits of a Graph

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Distinguishing the Orbits

The following theorem as seen in [1] states a very nice case in which an orbit of a graph can be distinguished with two labels. Theorem 7 For a given graph G, if the stabilizer subgroup of a vertex v is normal in Aut(G), then the orbit of v can be distinguished with two colors. Proof: Label the vertex v red and the rest of the vertices in Ov blue. We will show that every nontrivial automorphism σ ∈ Aut(G) distinguishes Ov . Suppose there is a nontrivial automorphism h which does not distinguish Ov . In other words, h must preserve colors and so must send one blue vertex x ∈ V (G) to another blue vertex y ∈ V (G). In symbols, we have h(x) = y. It also must be true that h(v) = v since h preserves colors, and so h ∈ Stv . Now because x, y ∈ Ov , there must be elements g1 , g2 ∈ Aut(G) with g1 (v) = x and g2 (v) = y. So we have h(g1 (v)) = h(x) = y. Because Stv is normal in Aut(G), there must be some element h0 ∈ Stv with g1−1 hg1 = h0 which implies that hg1 = g1 h0. This would give x = g1 (v) = g1 (h0 (v)) = h(g1 (v)) = y. This contradicts the fact that h is nontrivial.

2

We define a hamiltonian group to be one in which each subgroup is normal. Thus, an immediate consequence of this theorem is that if the automorphism group of a graph is abelian or hamiltonian, then the stabilizer subgroup of each vertex in each orbit is normal and so each orbit can be distinguished with two colors. Hence the entire graph can be distinguished with two colors. Another interesting result given in [1] says that if we know the distinguishing number of a particular orbit is k, and we know that the intersection of the stabilizer subgroups of all of the vertices in that orbit is simply the identity, then the entire graph can be distinguished with k colors. Theorem 8 Let G be a graph with Aut(G) = Γ. If G has an orbit O = {v1 , v2, . . . , vs } that can be distinguished with k colors and ∩si=1 Stvi = {e}, then G can be distinguished with k colors. Proof: If we label G so that O is k-distinguishing, then any nontrivial automorphism will act nontrivially on the orbit O, since the only element in Aut(G) which stabilizes everything in O is the identity automorphism. 2

Karen S. Potanka

6.2

Chapter 6. Distinguishing the Orbits of a Graph

17

Size of the Orbit

In [1], we find that having an orbit of vertices large enough can force a graph to be 2distinguishable. Recall the Orbit-Stabilizer Theorem which states that the size of an orbit of a vertex v is equal to the order of the automorphism group divided by the size of the stabilizer subgroup of v. Thus, if a graph G has an orbit O the size of the automorphism group Aut(G), then every vertex in O has a stabilizer subgroup of size 1. In other words, the element {e} is the only automorphism to stabilize all of the vertices in O and so by the above theorem, the distinguishing number of the graph will be the distinguishing number of O. Now if we color one vertex of O red and the rest blue, then it follows that every nontrivial automorphism sends the red vertex to a blue vertex. Hence, G can be distinguished with two colors and we have the following corollary: Corollary 9 A graph which has an orbit the size of the automorphism group of the graph can be distinguished with two colors.

6.3

Number of Orbits

The following result from [1] says that having a large number of orbits can also force a graph to be distinguished with two colors, provided that a few conditions are met. Theorem 10 Suppose G is a graph with t orbits and the vertices v1, v2, . . ., vt are from the t different orbits with respective stabilizer subgroups Stv1 , Stv2 , . . ., Stvt . If ∩ti=1 Stvi = {e}, then D(G) = 2. Proof: Define the labeling ϕ as follows: (

φ(v) =

red, if v ∈ {v1 , v2, . . . , vt} blue, otherwise.

We want to show that for any nontrivial automorphism σ ∈ Aut(G), there is at least one vertex x ∈ V (G) such that φ(σ(x)) 6= φ(x). Let σ be a nontrivial automorphism of G. Since the intersection of the stabilizer subgroups of v1 , v2 , . . ., vt is the identity, σ must map at least one vi , 1 ≤ i ≤ t, to another vertex in Ovi , which by the definition of φ is blue. Thus, we have distinguished G with two colors. 2

Chapter 7 Dihedral Groups The dihedral groups arise as an important family of groups whose elements are the symmetries of geometric objects. The simplest of such objects are the regular n-gons which can be easily represented with a graph. We define Dn (n ≥ 3) to be the set of n rotations about the center of the n-gon along with the set of n reflections through the n lines of symmetry. More precisely, Dn =< σ, τ |σ n = τ 2 = e, τ σ = σ −1 τ >= {e, σ, σ 2, . . . , σ n−1 , τ, τ σ, τ σ 2, . . . , τ σ n−1 }. It is easy to see that the order of Dn is 2n. Furthermore, the set of involutions, or elements n of order 2, include only I = {τ σ i for 0 ≤ i ≤ n − 1} if n is odd, and I along with σ 2 if n is even.

7.1 7.1.1

Subgroups of Dn Types of Subgroups of Dn

It turns out that there are only three types of nontrivial subgroups of Dn : 1. A subgroup of < σ >, the cyclic half of Dn , 2. A subgroup isomorphic to Dm , where m|n, and 3. {e, τ σ i} where 0 ≤ i < n. It is obvious that sets of types 1 and 3 are subgroups of Dn . Below we will show that a set of type 2 is a subgroup of Dn . Since Dn is a finite group, to show that a set A of type 2 is a 18

Karen S. Potanka

Chapter 7. Dihedral Groups

19

subgroup, it is only necessary to show that for any two elements β, ρ ∈ A, βρ−1 ∈ A. Now if A ∼ = Dm where m|n, then A = {e, σ k , σ 2k , . . . , σ (m−1)k , τ σ i, τ σ k+1 , τ σ 2k+i, . . . , τ σ (m−1)k+i} where mk = n. We need to check several cases: 1. If β = σ kj and ρ = σ kl , then βρ−1 = σ kj (σ kl)−1 = σ kj σ −kl = σ k(j−l) ∈ A. 2. If β = σ kj and ρ = τ σ kl+i , then βρ−1 = σ kj (τ σ kl+i )−1 = σ kj σ −kl−i τ = σ k(j−l)−i τ = τ σ k(l−j)+i ∈ A. 3. If β = τ σ kl+i , ρ = σ kj , then βρ−1 = τ σ kl+i (σ kj )−1 = τ σ kl+i σ −kj = τ σ k(l−j)+i ∈ A. 4. If β = τ σ kl+i , ρ = τ σ kj+i , then βρ−1 = τ σ kl+i (τ σ kj+i )−1 = τ σ kl+i σ −kj−i τ = τ σ k(l−j) τ = σ k(j−l) ∈ A. To see why these are the only types of nontrivial subgroups, suppose that H is a nontrivial subgroup of Dn . We break the argument down into three cases depending on the number of elements of the type τ σ i . First suppose that H contains no elements of the type τ σ i. Then H must be a subgroup of {e, σ, σ 2, . . . , σ n−1 } =< σ >. Thus H =< σ d > for some d which divides n. Now suppose that H contains only one element of the type τ σ i . Then H must not contain σ j for any 0 < j < n. Otherwise, τ σ iσ j ∈ H and so τ σ i+j ∈ H. But this contradicts that τ σ i is the only element of its type in H. Therefore τ σ i and e are the only elements in H. Finally suppose that H contains two or more elements of the type τ σ i. Then there must be τ σ i ∈ H and τ σ j ∈ H with i 6= j and 0 ≤ i, j < n. Because H is a subgroup we must have that τ σ iτ σ j = τ 2 σ j−i = σ j−i ∈ H. (Note that σ j−i 6= e since i 6= j.) Thus H must contain at least one element σ p for some 0 ≤ p < n. Let d be the smallest positive integer with σ d ∈ H. If d = 1, then < σ >⊂ H which implies that H = Dn . So assume that d > 1; then d must divide n. Certainly < σ d > is a subgroup of H; if σ a ∈ H and a = qd + r with 0 ≤ r < d, then σ r = σ a−qd = σ a(σ d )−q ∈ H since both σ a ∈ H and σ d ∈ H. By the minimality of d it follows that r = 0 or a = qd so σ a = (σ d )q ∈< σ d >. Thus the only elements from the cyclic half contained in H are in the set < σ d >. So we must have that if τ σ i ∈ H and τ σ j ∈ H, then d must divide i − j or i = dk + j for some integer k with 0 ≤ k < nd . Thus τ σ i = τ σ dk+j . So H must contain nd elements of the form σ dk for 0 ≤ k < nd and nd elements of the form τ σ dk for 0 ≤ k < nd . So n n H = {e, σ d, σ 2d, . . . , σ ( d −1)d , τ σ i, τ σ i+d , τ σ i+2d, . . . , τ σ i+( d −1)d } ∼ = D nd .

Karen S. Potanka

7.1.2

Chapter 7. Dihedral Groups

20

Properties of Subgroups of Dn

Now for each type of subgroup of Dn , it will be useful to know the representation of the subgroup using generators, the subgroups conjugate to it, and the orbit of a vertex which has that subgroup as its stabilizer subgroup. This information is contained in the following table and is discussed below: Type of Subgroup < σk > where k|n < σk , τ σi > <τ >

Conjugate Subgroups < σk >

Intersection of Conjugate Subgroups

< σk , τ σ >, < σk , τ σ2 >, . . ., < σk , τ σk−1 > < τ σ >,< τ σ2 >, . . ., < τ σn−1 >

< σk >

Orbit of vertex v with stabilizer subgroup of this type {v, σ(v), σ2 (v), . . . , σk−1(v), τ (v), τ σ(v), τ σ2 (v), . . . , τ σk−1(v)} {v, σ(v), σ2 (v), . . . , σk−1(v)}

{e}

{v, σ(v), σ2 (v), . . . , σn−1 (v)}

< σk >

Type 1. Because < σ k > is a normal subgroup of Dn , we know that for any g ∈ Dn , g < σ k > g −1 =< σ k >. So < σ k > is its only conjugate and therefore the intersection of the conjugate subgroups is < σ k >. Now if a vertex v has < σ k > as its stabilizer subgroup, then clearly the orbit of v will be the set Ov = {v, σ(v), σ 2(v), . . . , σ k−1 (v), τ (v), τ σ(v), . . . , τ σ k−1(v)}. This is because σ p (v) = σ kl+j (v) = σ j σ kl(v) = σ j (v) ∈ Ov where p = kl + j and 0 ≤ j < k, and τ σ p(v) = τ σ kl+j (v) = τ σ j σ kl (v) = τ σ j (v) ∈ Ov where p = kl + j and 0 ≤ j < k. Type 2. Now consider a subgroup of the second type, < σ k , τ σ i > where km = n. To find the conjugate subgroups, choose an element g = σ p or g = τ σ p and compute g < σ k , τ σ i > g −1 . In the first case we have: σ p < σ k , τ σ i > σ −p = σ p{σ k , σ 2k , . . . , σ (m−1)k , τ σ i, τ σ k+i , τ σ 2k+i , . . . , τ σ (m−1)k+i }σ −p where p = kl + j and 0 ≤ j < k. But σ p σ ks σ −p = σ ks and σ pτ σ ks+i σ −p = τ σ ks−2p+i where 0 ≤ s < m. Thus in this case we have σ p < σ k , τ σ i > σ −p =< σ k , τ σ i−2p >. In the second case we have: τ σ p < σ k , τ σ i > σ −pτ = τ σ p{σ k , σ 2k , . . . , σ (m−1)k , τ σ i, τ σ k+i , τ σ 2k+i, . . . , τ σ (m−1)k+i}σ −p τ where p = kl + j and 0 ≤ j < k. But τ σ pσ ks σ −p τ = τ σ ks τ = σ −ks and τ σ pτ σ ks+i σ −p τ = σ ks−2p+i τ = τ σ 2p−ks+i where 0 ≤ s < m. Thus we have τ σ p < σ k , τ σ i > σ −p τ = < σ k , τ σ i+2p >.

Karen S. Potanka

Chapter 7. Dihedral Groups

21

Clearly the intersection of these conjugate subgroups is < σ k >. Now suppose that a vertex v has < σ k , τ σ i > as its stabilizer subgroup. Then the orbit of v will be the set Ov = {v, σ(v), σ 2(v), . . . , σ k−1 (v)}. This is because σ p(v) = σ kl+j (v) = σ j σ kl (v) = σ j (v) ∈ Ov where p = kl + j and 0 ≤ j < k, and τ σ p(v) = τ σ kl+j (v) = τ σ kl+(j−i)+i (v) = σ i−j τ σ kl+i (v) = σ i−j (v) ∈ Ov where p = kl + j and 0 ≤ j < k. Type 3. Finally, consider a subgroup of the last type, < τ >. To find the subgroups conjugate to < τ >, choose an element g = σ p or g = τ σ p and compute g < σ k , τ σ i > g −1 . In the first case we have: σ p < τ > σ −p = σ{e, τ }σ −p = {e, τ σ −2p} where p = kl + j and 0 ≤ j < k. In the second case we have: τ σ p < τ > σ −pτ = τ σ p{e, τ > σ −pτ = {e, τ σ pτ σ −p τ } = {e, σ −2pτ } = {e, τ σ 2p}. Clearly the intersection of the conjugate subgroups is just the identity. Now suppose that a vertex v has < τ > as its stabilizer subgroup. Then the orbit of v will be the set {v, σ(v), σ 2(v), . . . , σ n−1 (v)}. This is because τ σ p(v) = σ −p τ (v) = σ −p (v).

7.2

Distinguishing number of graphs which realize Dn

We are now ready to find the distinguishing number of graphs which realize Dn . We can do this by looking at the orbits of vertices and the stabilizer subgroups of the vertices. For instance, we will see from the results in [1] that if there is vertex in the graph with a stabilizing subgroup of Type 1, then the graph is automatically 2-distinguishable. If the graph has no vertex with stabilizing subgroup of Type 1, but does have an orbit with more than 6 elements, then we will see that the graph is 2-distinguishable. The case in which the graph satisfies neither of the above conditions will be handled by the last theorem in this section. In any case, a graph which realizes Dn will have a distinguishing number of either 2 or 3. Two lemmas are extremely useful in obtaining the results described above. One is Lemma 4 in Chapter 3, and the other is the following result given in [1]: Lemma 11 Let G be a graph which realizes Dn , and suppose that G has t orbits of vertices. If Stv1 , Stv2 , . . ., Stvt are the respective stabilizer subgroups of vertices v1, v2, . . ., vt from the t different orbits, then Stv1 ∩ Stv2 ∩ . . . ∩ Stvt ∩ < σ >= {e}.

Karen S. Potanka

Chapter 7. Dihedral Groups

22

Proof: Consider the conjugates of σ k ∈ Dn . First, σ lσ k σ −l = σ k for 0 ≤ l < n. Second, τ σ iσ k (τ σ i)−1 = τ σ i σ k σ −i τ = τ σ k τ = σ −k . So the conjugacy class of σ k is {σ k , σ −k }. Now if σ k is an element of any subgroup H, then so is σ −k . Let gHg −1 be any subgroup conjugate to H. If g = σ l for some 0 ≤ l < n, then gσ k g −1 = σ k ∈ gHg −1 . On the other hand, if g = τ σ l for some 0 ≤ l < n, then gσ k g −1 = τ σ lσ −k σ −l τ = τ σ −k τ = σ k ∈ gHg −1 . So if σ k ∈ H then σ k is an element of any subgroup conjugate to H. If σ k ∈ Stv1 ∩ Stv2 ∩ . . . ∩ Stvt , then σ k is in the conjugate of all of these subgroups. That would mean that σ k is in the stabilizer subgroup for every vertex of G by Lemma 4. So σ k would have to fix every element of G, and since G realizes Dn , σ k = e and k = n. Thus < σ > ∩Stv1 ∩ Stv2 ∩ . . . ∩ Stvt = {e} and so nothing in the cyclic half of Dn can stabilize every vertex of G. 2

7.2.1

Graphs with a stabilizer subgroup of Type 1

Theorem 12 Let G realize Dn . If G has a vertex whose stabilizer subgroup is of type 1, then G can be distinguished with 2 colors. Then Proof: Suppose that the vertex v has stabilizer subgroup Stv of type 1. j Stv =< σ > for some 0 ≤ j ≤ n where j|n. Now let v1, v2, . . . , vt be vertices from all the other different orbits of G where Stv1 , Stv2 , . . . , Stvt are their respective stabilizer subgroups. By lemma 11, we have that < σ > ∩Stv ∩ Stv1 ∩ . . . ∩ Stvt = {e}. But < σ > ∩ < σ j >=< σ j >, thus we have < σ j > ∩Stv1 ∩. . .∩Stvt = Stv ∩Stv1 ∩. . .∩Stvt = {e}. So by Theorem 10, D(G) = 2. 2

7.2.2

Graphs with a stabilizer subgroup of Type 2 or 3

Lemma 13 Let G realize Dn . If G has a vertex v whose stabilizer subgroup Stv is of type 2 or 3, and if the orbit of v, Ov , has more than six elements, the Ov can be distinguished with 2 colors. Proof: From the table above we see that Ov = {v, σ(v), σ 2(v), . . . , σ j−1 (v)} where 0 < j ≤ n and in the case that Stv =< τ σ i >, we have j = n. We may assume that j ≥ 6 since |Ov | ≥ 6. Let A = {v, σ 2(v), σ 3(v)}; color the vertices in A red and the rest of the vertices in Ov blue. Our goal is to show that this is a 2-distinguishing coloring of Ov . In other words, we need to show that every automorphism which acts nontrivially on Ov must send at least one vertex in A to vertex not in A. Suppose that the automorphism g ∈ Dn fixes A setwise. Then g must send v to v, σ 2(v), or σ 3(v). Now if g were to fix v, then g ∈ Stv . Similarly, if g(v) = σ 2 (v) then g ∈ σ 2Stv and if

Karen S. Potanka

Chapter 7. Dihedral Groups

23

g(v) = σ 3(v) then g ∈ σ 3Stv . Hence in order to show that g distinguishes Ov , we must show that g either acts trivially on A or does not fix A for all the cases of g listed above. So we need to look at the image of A under all of those automorphisms. To rule out the case that g acts trivially on Ov , note that if an automorphism acts trivially on Ov , then it is contained in the intersection of the stabilizer subgroups of each element in Ov . But by Lemma 4, we know that stabilizer subgroups of vertices in the same orbit are conjugate to one another. So the automorphisms which act trivially on A are in the intersection of such stabilizer subgroups. If Stv =< σ k , τ σ i >, then the intersection is just < σ k >, and if Stv =< τ σ i >, the intersection is simply {e}. First suppose Stv =< σ k , τ σ i >= {e, σ k , σ 2k , . . . , σ ( k −1)k , τ σ i, τ σ k+i , . . . , τ σ ( k −1)k+i }. Then if 0 ≤ d < nk we have: n

n

1. σ dk A = {σ dk (v), σ dk σ 2(v), σ dk σ 3(v)} = {v, σ 2(v), σ 3(v)} = A, 2. τ σ dk+i A = {τ σ dk+i (v), τ σ dk+iσ 2(v), τ σ dk+i σ 3(v)} = {v, σ n−2(v), σ n−3 (v)}, 3. σ 2σ dk A = {σ 2σ dk (v), σ 2σ dk σ 2(v), σ 2σ dk σ 3(v)} = {σ 2(v), σ 4(v), σ 5(v)}, 4. σ 2τ σ dk+i A = {σ 2τ σ dk+i (v), σ 2τ σ dk+i σ 2(v), σ 2τ σ dk+i σ 3(v)} = {σ 2(v), v, σ n−1(v)}, 5. σ 3σ dk A = {σ 3σ dk (v), σ 3σ dk σ 2(v), σ 3σ dk σ 3(v)} = {σ 3(v), σ 5(v), σ 6(v)}, 6. σ 3τ σ dk+i A = {σ 3τ σ dk+i (v), σ 3τ σ dk+i σ 2(v), σ 3τ σ dk+i σ 3(v)} = {σ 3(v), σ(v), v}. So since n ≥ 6, we see that only σ dk preserves A, but σ dk ∈< σ k > which acts trivially on Ov . Now suppose that Stv =< τ σ i >= {e, τ σ i}. Then the action of τ σ i,σ 2τ σ i, and σ 3τ σ i on A is as follows: 1. τ σ iA = {τ σ i(v), τ σ iσ 2(v), τ σ iσ 3 (v)} = {v, σ n−2(v), σ n−3(v)}, 2. σ 2τ σ i A = {σ 2τ σ i (v), σ 2τ σ iσ 2 (v), σ 2τ σ iσ 3(v)} = {σ 2(v), v, σ n−1(v)}, 3. σ 3τ σ i A = {σ 3τ σ i (v), σ 3τ σ iσ 2 (v), σ 3τ σ iσ 3(v)} = {σ 3(v), σ(v), v}. Again we see that because n ≥ 6, none of these preserves A.

2

Example. One motivation of this lemma is the original key problem with n keys on a circular ring. It was stated in Chapter 4 that for n ≥ 6, all keys could be distinguished using 2 colors. This is because there is only one orbit of vertices, O = {v1, v2, . . . , vn }, of

Karen S. Potanka

Chapter 7. Dihedral Groups

24

size ≥ 6 and the stabilizer subgroup of any vertex, vi , will be of type 3, τ σ j . Thus labeling the set A = {vi , vi+2, vi+3 } red and the rest of the vertices on the cycle blue distinguishes Ovi = V (G). Theorem 14 Let G realize Dn . If G has a vertex v whose stabilizer subgroup Stv is of type 2 or 3, and if the orbit of v, Ov , has more than six elements, the G can be distinguished with 2 colors. Proof: In the case that Stv =< τ σ i >, the intersection of the subgroups conjugate to Stv is just the identity. In other words, the identity is the only element in Dn which fixes every vertex in Ov . Furthermore, by the previous lemma, Ov can be distinguished with 2 colors. Thus by Theorem 8, we have that G can be distinguished with 2 colors. If Stv =< σ k , τ σ i >, then again by the previous lemma, Ov can be distinguished with 2 colors. With this coloring, every automorphism which acts nontrivially on Ov must send a red vertex to a blue vertex, distinguishing the graph. What about the automorphisms which act trivially on Ov ? We know that these are the automorphisms in the intersection of the stabilizer subgroups of the vertices in Ov . From the table above, we see that this intersection is just < σ k >. If we let < σ k > act on the vertices of G, then < σ k > makes vertex orbits V1 , V2 , . . . , Vs which are contained in the vertex orbits of G under the action of Dn . Since < σ k > stabilizes each vertex in Ov , Ov is broken into 1-orbits under the action of < σ k >. Consider the orbits Vi with |Vi | > 1. In each orbit of this type, choose a vertex vj ∈ Vi and color it red and color the rest of the vertices in Vi blue. Suppose that with this coloring, < σ k > fixes each vj which we colored red. We have n < σ k >= {e, σ k , σ 2k , . . . , σ ( k −1)k } and so if 0 ≤ d < nk , then σ dk (vj ) = vj . Now if ut is any other vertex in Vi , then there must be some number 0 ≤ r < nk with σ rk (vj ) = ut. So σ dk (ut) = σ dk (σ rk (vj )) = σ rk (σ dk (vj )) = σ rk (vj ) = ut . Thus σ dk fixes every element in Vi for 1 ≤ i ≤ s and so σ dk fixes all of G. This contradicts that G realizes Dn . Thus < σ k > must send at least one vertex to a blue vertex. So D(G) = 2. 2 Example. To illustrate this theorem, let G = GP (7, 2) be the graph in Figure 7.1. It turns out that this graph realizes D7 (see Chapter 8.) For any vertex in GP (7, 2), the stabilizer subgroup includes the identity automorphism and an involution (reflection through line through the vertex) and so is of type < τ σ i >. Furthermore, under the action of the automorphism group, the vertices are partitioned into two orbits, U = {u1, u2, . . . , u7} and V = {v1, v2, . . . , v7}. Thus GP (7, 2) satifies the conditions of Theorem 11. The coloring described in the proof of the theorem is shown in the figure.

Karen S. Potanka

Chapter 7. Dihedral Groups

25

u1 u7 v7

u2

v1

Ou

1

A

v2

v6 v3

u6

v5 u5

u3

v4 u4

Figure 7.1: This graph illustrates the coloring in Lemma 10 Lemma 15 If G realizes Dn , and pα is the largest power of p that divides n, then there is a vertex orbit O of size pα in G. Proof: Let p be a prime divisor of n, and suppose that pα is the largest power of p that divides n n. Certainly there is a subgroup Λ =< σ pα > which is cyclic. Note that there are only pα elements in Λ, so if v ∈ V (G) then by the Orbit-Stabilizer Theorem, |Ov ||Stv | = |Λ| = pα . Thus, the size of the orbit of any vertex must divide pα and so must be one of 1, p, p2 , . . . , pα . n

Let v ∈ V (G), and let d be the smallest positive integer such that σ d( pα )(v) = v. Then d is the size of the Λ-orbit that contains v. By the argument in the previous paragraph, d divides is an orbit O under the action of Λ of size pα . If not, then pα . It must be the case that there d( pnα ) n for each v ∈ V (G) we have σ (v) = v and d = pβ for some β < α. So d pnα = pβ pnα = p(α−β) (

n

)

n

and (α− β) ≥ 1. Thus,nσ p(α−β) (v) = v and so σ p (v) = v for each v ∈ V (G). In other words, we have found some σ p which fixes each vertex of G, contradicting the fact that G realizes Dn . So we conclude that there must be a Λ-orbit of size pα . Because the orbits under the action of Dn are at least as large as the Λ-orbit, we are guaranteed to have an orbit of size pα under the action of Dn . 2 This lemma is extremely useful in the proof of the following theorem, for it limits the cases needed to be considered. Theorem 16 D(Dn ) = {2} unless n = 3, 4, 5, 6, 10 in which case D(Dn ) = {2, 3}. Proof: By Theorem 12, if there is a vertex with stabilizer subgroup of type < σ k >, then the graph is 2-distinguishable. So we may assume that G does not have a vertex with stabilizer subgroup of type 1. Similarly, by Theorem 14, if a vertex has a stabilizer of type 2 or 3, and the orbit of that vertex has size greater than or equal to 6, then the graph is 2-distinguishable. So we may assume that for any v ∈ V (G), |Ov | < 6.

Karen S. Potanka

Chapter 7. Dihedral Groups

26

By the previous lemma, there must be an orbit of size pα where pα is the largest power of p which divides n. To have |Ov | < 6 for any vertex v ∈ V (G), we only need to handle now the cases in which pα < 6. So clearly, p ≤ 5 and n can have at most one factor of 5, one factor of 3, and two factors of 2. With this in mind, we need only consider the cases when n = 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. The cases when n ≥ 12 will be considered first. If the stabilizer subgroup of a vertex in G is of the type < τ σ i >, then the orbit of v contains 2n = n elements by the Orbit-Stabilizer 2 Theorem. For n ≥ 12, this would give an orbit of size greater than or equal to 12. So we may assume that the stabilizer subgroup of any vertex in G is of type < σ k , τ σ i >. Choose a vertex v ∈ V (G) and suppose that |Ov | = d (d = 1,2,3,4, or 5 and d divides n) and . Then Stv =< σ d , τ σ i > for some 1 ≤ i < d. Now the stabilizer subgroups of the |Stv | = 2n d vertices in Ov are conjugate to Stv and their intersection is simply < σ d > which means that 0 < σ d > fixes Ov . If v 0 is a vertex in another orbit Ov0 of size d2 with Stv0 =< σ d2 , τ σ i >, then the intersection of stabilizer subgroups of vertices in Ov0 is just < σ d2 >. Now if l = lcm(d, d2 ) then clearly < σ l > fixes both Ov and Ov0 . If we do this for each orbit O1 , O2 , . . ., Os of sizes d, d2 , . . ., ds respectively, then if L = lcm(d, d1 , . . . , ds ), then < σ L > will fix every vertex in the graph. Now the only automorphism in Dn that stabilizes the entire graph is e = σ n . So in order for G to realize Dn , G must have orbits with sizes whose least common multiple is n. The following table contains the values for n, and the possible sizes for the orbits so that their least common multiple is n. n 12 15 20 30 60

Possible size of orbits 2,3,4 3,5 2,4,5 2,3,5 2,3,4,5

We first show that when there are orbits of sizes which are not pairwise relatively prime, then the graph can be 2-distinguished. This can happen in the cases listed above when there are two orbits of the same size, or we have both a 2-orbit and a 4-orbit. To motivate this, consider the following example. Example. Consider a graph G which realizes D12 . Listed below are the possible stabilizer subgroups of a vertex with orbit size either 2,3 or 4:

Karen S. Potanka

Chapter 7. Dihedral Groups

Orbit Size

Possible Stabilizer Subgroup

List of elements

2

2

{e, σ , σ , σ , σ , σ 10, τ, τ σ 2, τ σ 4, τ σ 6, τ σ 8, τ σ 10} {e, σ 2, σ 4, σ 6, σ 8, σ 10, τ σ, τ σ 3, τ σ 5, τ σ 7, τ σ 9, τ σ 11} {e, σ 3, σ 6, σ 9, τ, τ σ 3, τ σ 6, τ σ 9} {e, σ 3, σ 6, σ 9, τ σ, τ σ 4, τ σ 7, τ σ 10} {e, σ 3, σ 6, σ 9, τ σ 2, τ σ 5, τ σ 8, τ σ 11} {e, σ 4, σ 8, τ, τ σ 4, τ σ 8} {e, σ 4, σ 8, τ σ, τ σ 5, τ σ 9} {e, σ 4, σ 8, τ σ 2, τ σ 6, τ σ 10} {e, σ 4, σ 8, τ σ 3, τ σ 7, τ σ 11}

3

4

< σ ,τ > < σ 2, τ σ > < σ 3, τ > < σ 3, τ σ > < σ 3, τ σ 2 > < σ 4, τ > < σ 4, τ σ > < σ 4, τ σ 2 > < σ 4, τ σ 3 >

2

4

6

27

8

So in order for G to realize D12 , G must have orbits with sizes whose least commmon multiple is 12. In other words, there must be an orbit of size 3 and 4. From the table above, we see that each of the possible stabilizer subgroups for a vertex in the 4-orbit intersects each of the possible stabilizer subgroups for a vertex in the 3-orbit at the identity and a non-cyclic automorphism τ σ h where 0 ≤ h ≤ 11. What happens if there is an orbit of size 2 as well? We see that we can choose a stabilizer subgroup from the 2-orbit, say < σ 2, τ σ >, and a stabilizer subgroup from the 4-orbit, say < σ 4, τ >, whose intersection is < σ 4 >. Thus < σ 4 > ∩{e, τ σ h} = {e}. So by Theorem 10, the graph can be distinguished with 2 colors. A similar argument can be made if we have two orbits of the same size. Resume proof of Theorem 16: As seen in the example, if we have two orbits O1 and O2 in which |O1 | = |O2 | or |O1 | = 2 and |O2| = 4, then the stabilizer subgroups from these orbits can be chosen so that their intersection is a subgroup of < σ >. If we throw in an orbit O3 with size relatively prime to |O1| or |O2|, then no matter which stabilizer subgroup St3 we choose for a vertex in O3 , the intersection of St3 with a subgroup of < σ > will be the identity. From Theorem 10, we have then that D(G) = 2. Now we must consider the cases when n ≥ 12 and the orbits of G have sizes which are pairwise relatively prime. Hence if n = 12, the orbits sizes are 3 and 4; n = 15, 3 and 5; n = 20, 4 and 5; n = 30, 2,3 and 5; n = 60, 3,4, and 5. The goal is to show that there cannot possibly be orbits of G with sizes which are pairwise relatively prime. This will be done in three steps: 1. Show that the bipartite graphs formed by vertices in two separate orbits and the edges between them is either complete or empty. 2. Show that Dn will be equal to the product of the automorphism groups of the orbits. 3. Consider the product of all vertex-transitive graphs of sizes 3,4 and 5. Step 1. We say that a graph G is a bipartite graph, if the set of vertices of G can be partitioned into 2 nonempty sets V1 and V2 such that any edge of G joins a vertex in V1 to

Karen S. Potanka

Chapter 7. Dihedral Groups Orbit 1 v1 1

Orbit 2 v2 1

v1 2 v1 3 v1 p

. . .

. . .

Orbit 1 v1 1

v2 2

v1 2

v2 3

v1 3

v2 q

v1 p

28

Orbit 2 v2 1 v2 2

. . .

. v2 3 . . v2 q

Figure 7.2: The bipartite graph on the left is complete while the one on the right is empty a vertex in V2 . Furthermore, we say that a bipartite graph is complete, if given any pair of vertices, one from V1 and the other from V2 , there is an edge joining them. We want to consider the bipartite graph formed by using vertices from 2 different orbis and the edges of G between them. We will show that this bipartite graph will either be complete or empty (see Figure 7.2). Form the bipartite graph described above where O1 = {v11 , . . . , v1p } and O2 = {v21 , . . . , v2q } are the two orbits and p and q are relatively prime. Suppose now that there is an edge in the bipartite graph. Without loss of generality, say an edge between v11 and v21 . Now choose any other vertex v1l and v2m from orbits 1 and 2 respectively. It must be shown that there is an automorphism that maps v11 to v1l and v21 to v2m . We see that |O1 | = p and |O2 | = q. Thus Stv11 =< σ p, τ σ i > for some 1 ≤ i < d1 and Stv21 =< σ q , τ σ j > for some 1 ≤ j < d2 . With this in mind, we can rewrite the orbits as follows: O1 = {v11 , σ(v11 ), σ 2(v11 ), . . . , σ p−1(v11 )} and O2 = {v21 , σ(v21 ), σ 2(v21 ), . . . , σ q−1 (v21 )}. Since v11 and v1l are in O1 , then there is an automorphism σ s such that σ s (v11 ) = v1l where 0 ≤ s < p. Similarly, there is some σ t with 0 ≤ t < q such that σ t(v21 ) = v2m . We need to find some σ r such that σ r (v11 ) = v1l and σ r (v21 ) = v2m . In other words we need to find the solution of r ≡ s (mod p) r ≡ t (mod q). By the Chinese Remainder Theorem, such a solution exists modulo pq. Because of the sizes of p and q, we have that pq ≤ n; thus there is some automorphism in Dn which takes v11 to v1l and v21 to v2m . So if there is an edge between v11 and v21 , then there is an edge between any other pair of vertices, v1i and v2j . Similarly, no edge between v11 and v21 means there is no edge between any other pair v1i and v2j .

Karen S. Potanka

Chapter 7. Dihedral Groups

29

6 G[O0 ] Step 2. We now show that if for every pair of orbits O and O0 in the graph with G[O] ∼ = the bipartite graph formed by the vertices in O and O0 and the edges between these orbits is either empty or complete, then Aut(G) = Aut(G[O]) × Aut(G[V − O]). Here we use the notation G[O] to mean the subgraph of G induced by O or the subgraph containing the vertices in O and all the edges in G which join two vertices in O. Suppose that v ∈ V (G) and h1 ∈ Aut(G[O]) and h2 ∈ Aut(G[V − O]). Define ω : Aut(G[O]) × Aut(G[V − O]) → Aut(G) by (

ω(h1 , h2 )(v) =

h1 (v) if v ∈ O h2 (v) if v ∈ V − O.

We want to show that ω is an automorphism of G. So we must show that ω preserves adjacency in G. It turns out that there are 3 kinds of edges in G: 1. edges in G[O], 2. edges in G[V − O], and 3. edges between O and V − O. If v1 v2 is an edge in G[O], then both v1 , v2 ∈ O. So ω(h1 , h2)(v1) = h1 (v1) and ω(h1 , h2 )(v2) = h1 (v2). Now h1 preserves adjacency in G[O], so ω(h1 , h2 )(v1)ω(h1 , h2)(v2 ) is an edge of G[O]. Similarly, we have that ω(h1 , h2 ) preserves adjacency in V − O. ω(h1, h2 ) also preserves adjacency between O and V − O since the graph formed by the vertices in O and V − O and the edges between them is either complete or empty. Thus Aut(G[O]) × Aut(G[V − O]) ⊂ Aut(G). Now conversely, if we have any automorphism of G, when restricted to O, is an automorphism of G[O] and when restricted to V − O is an automorphism of G[V − O]. Thus we have shown containment both ways. From the above discussion, we have that Dn = Aut(G[O]) × Aut(G[V − O]) where O is an orbit. In other words, Dn is the product of the automorphism group of each orbit considered as a subgroup of G. Step 3. Now, each orbit must form a vertex-transitive graph. So we must consider all orbits of sizes 3,4, and 5 which are vertex-transitive. The only graphs with 3 vertices which are vertex-transitive are C3 and C3 and both have the same automorphism group S3. The only

Karen S. Potanka

Chapter 7. Dihedral Groups

30

Orbit of size 3 C3

_ C3

Orbit of size 4 K4

_ K4

_ C4

C4

Orbit of size 5

K5

_ K5

C5

_ C5

Figure 7.3: Vertex-transitive graphs with sizes 3,4 and 5 graphs with 4 vertices which are vertex-transitive are K4 , K4 , C4 , and C4. The automorphism group of K4 and its complement is S4 , while the automorphism group of C4 and its complement is D4 . The only graphs with 5 vertices which are vertex-transitive are K5 , K5 , C5 and C5. The automorphism group of K5 and its complement is S5 and the automorphism group of C5 and its complement is D5 . These graphs are seen in Figure 7.3. 1. For n = 12, |D12 | = 24. Now |S3 | = 6, |S4| = 24 and |D4 | = 8, so the size of product of the automorphism groups for orbits of size 3 and 4 is at least 48. 2. For n = 15, |D15 | = 30. Now |S3| = 6, |S5 | = 120 and |D5 | = 10, so the size of product of the automorphism groups for orbits of size 3 and 5 is at least 60. 3. For n = 20, |D20 | = 40. The size of product of the automorphism groups for orbits of size 4 and 5 is at least 80. 4. For n = 30, |D30 | = 60. The size of product of the automorphism groups for orbits of size 2, 3 and 5 is at least 120. 5. For n = 60, |D60 | = 120. The size of product of the automorphism groups for orbits of size 3, 4 and 5 is at least 480.

Karen S. Potanka

Chapter 7. Dihedral Groups

K 3,2

31

C 5V K 2

Figure 7.4: K3,2 and C5 ∨ K2 Thus we see that we cannot have orbits of G with sizes which are relatively prime. Since we have exhausted all of the cases for n ≥ 12, we see that D(Dn ) = {2}. Now we will show that for the cases n = 3, 4, 5, 6, 10, D(Dn ) = {2, 3}. We know that 2 ∈ D(G) for any graph G. So we will find a graph in each case with distinghuishing number 3, and then we will show that every graph which realizes Dn can be distinguished with 3 colors (or less.) As observed in the original key problem when n=3,4,or 5, then it required 3 colors to disinguish the keys. Thus we have 3 ∈ D(D3 ), D(D4 ) and D(D5 ). It turns out that the complete bipartite graph K3,2 realizes D6 and requires 3 colors to be distinguished while the graph C5 ∨ K2 realizes D10 and also requires 3 colors to distinguish. See Figure 7.4 for the distinguishing coloring of each graph. It remains to be shown that any graph which realizes Dn for n = 3, 4, 5, 6, 10 can be distinguished with 3 colors. Suppose that v1 , v2, . . . , vt are vertices from the t different orbits of such a graph and Stv1 , Stv2 , . . . , Stvt are their respective stabilizer subgroups. By Lemma 11, we have < σ > ∩Stv1 ∩Stv2 ∩. . .∩Stvt = {e} which means that Stv1 ∩Stv2 ∩. . .∩Stvt = ∩tj=1 Stvj is of the third type. So we have ∩tj=1 Stvj =< τ σ i > for some 0 ≤ i < n. Color the vertices v1, v2, . . . , vt red. Choose one vertex v which is not fixed by τ σ i and color v green. Note that such a vertex exists, otherwise τ σ i would fix every vertex in G, and because G realizes Dn this cannot happen. Color the rest of the vertices in G blue. Thus every automorphism moves a red vertex to a vertex which is not red, except for the automorphism τ σ i which fixes all of the red vertices but moves the green vertex v to a blue vertex. Thus every graph can be distinguished with 3 colors. 2

Chapter 8 Generalized Petersen Graphs Having read Collins and Albertson’s paper, it became of interest to find the distinguishing numbers of the generalized Petersen graphs. Before attempting to solve this problem, it was necessary to find the automorphism groups of these graphs. In this chapter, I will discuss my attempts in identifying these automorphism groups as well as the results published by Frucht in 1971 on the groups of the generalized Petersen graphs (see [3]). Definition. The generalized Petersen graph GP (n, k) for positive integers n and k with 2 ≤ 2k < n is defined to have vertex set V (GP (n, k)) = {u0 , u2, . . . , un−1 , v0 , v1, v2, . . . , vn−1 } and edge set E(GP (n, k)) = {ui ui+1 , vivi+k , uivi | i = 0, 1, . . . , n − 1} where addition in the subscripts is modulo n. Notice that GP (n, k) is a regular 3-valent graph for any pair (n, k). We have already seen the generalized Petersen graph GP (7, 2) in Figure 7.1. The most familiar generalized Petersen graph is GP (5, 2), otherwise known as simply the Petersen graph. This graph can be found in Figure 8.1. As seen in the definition of the generalized Petersen graph, there are 3 types of edges: those edges between ui and ui+1 called outer edges, those between vi and vi+k called inner edges, and those between ui and vi called spokes. Since i = 0, 1, . . . , n − 1, we see that there are n edges of each type and the symbols Ω, I, and Σ will be used to denote the set of outer edges, inner edges, and spokes respectively. If we first join the n outer edges appropriately, we see that we form an n-circuit which will be called the outer rim. Similarly, if d = gcd(n, k), then by joining the n inner edges appropriately, we see that we form d pairwise-disjoint nd -circuits called inner rims. Notice that when n and k are relatively prime, then there will be only one inner rim of length n as seen in Figure 7.1 and Figure 8.1 with GP (7, 2) and GP (5, 2) 32

Karen S. Potanka

Chapter 8. Generalized Petersen Graphs

33

Figure 8.1: The Petersen Graph

Inner rims

Figure 8.2: GP(8,2) respectively. Figure 8.2 contains the graph of GP (8, 2). Notice that because 8 and 2 are not relatively prime, there are 2 inner rims of length 4.

8.1

Automorphism Groups of the Generalized Petersen Graphs

In finding the automorphism groups of the generalized Petersen graphs, it is important to notice that Dn ⊆ Aut(GP (n, k)) for all positive integers n and k. For some n and k, Dn is not the entire automorphism group of the graph. So finding the automorphism groups of the graphs now becomes a problem of seeing when there are automorphisms other than the basic rotations and reflections. The first question I wanted to answer was this: When is there an automorphism which switches the outer and inner rims? Clearly, if n and k are not relatively prime, then there would be no such automorphism since automorphisms must take n-circuits to n-circuits. So my attention turned to the cases when n and k were relatively prime. Lemma 17 Suppose that n and k are relatively prime. Then there is an automorphism α ∈ Aut(GP (n, k)) which switches the outer and inner rim if and only if k 2 ≡ ±1 (mod n).

Karen S. Potanka

Chapter 8. Generalized Petersen Graphs v0

vk

u0 v2k . .

u -k -1 u-2k-1 .

v0

v-k

v-k v-2k

u k-1 u 2k-1 .

OR

34 vk

u0 v-2k

. .

. .

u -k -1 u-2k-1

v2k

u -1 k

u 2k-1

.

.

. .

Figure 8.3: Two cases for the action of α Proof: Switching the outer rim and the inner rim if at all possible could be done in one of several ways. Without loss of generality, the only case needed to be considered is the one in which α(u0 ) = v0 and α(v0 ) = u0 . Again, we can assume without loss of generality that α(u1 ) = vk . Now the outer rim is the n-circuit u0u1 u2 . . . un−1 which must go to the inner rim if α is to be an automorphism. Since α(u1 ) = vk , we must have that α(ui ) = vik for i = 0, 1, . . . , n − 1 (Remember that subscripts are read modulo n.) In other words, what gets mapped to v1 , v2, . . . , vn−1 are the vertices uk−1 , u2k−1 , . . . , u−k−1 . Similarly, the inner rim is the n-circuit v0vk v2k . . . v−k which must get mapped to the outer rim. This can be done in one of two ways. Either α(vk ) = u1 or α(vk ) = u−1 . More generally, either α(vik ) = ui or α(vik ) = u−i . See Figure 8.3 for these two cases. Now α will be an automorphism if it preserves adjacency. We have already fixed the vertices so that the adjacency of vertices in the inner and outer rims were preserved, so now we need to check the spokes. In the first case α will be an automorphism if and only if lk and −lk−1 differ by n for l = 0, 1, . . . , n − 1; in other words, if and only if lk ≡ −lk −1 (mod n) or k ≡ −k −1 (mod n). In the second case α will be an automorphism if and only if lk and lk−1 differ by n; in other words, if and only if lk ≡ lk −1 (mod n) or k ≡ k −1 (mod n). Thus we see that α will be an automorphism if and only if k2 ≡ ±1 (mod n).

2

Corollary 18 If k 2 ≡ ±1 (mod n), then GP (n, k) is vertex-transitive. Proof: If k 2 ≡ ±1 (mod n), then α ∈ Aut(GP (n, k)), so vertices on the outer rim can be mapped to vertices on the inner rims and vice versa. Vertex-transitivity follows immediately. 2 Appendix A contains two tables, one of which contains values of n and k with n ≤ 150 and for which k 2 ≡ 1 (mod n), and the other of which contains values of n and k with n ≤ 150

Karen S. Potanka

Chapter 8. Generalized Petersen Graphs

35

and for which k 2 ≡ −1 (mod n). These would be the only cases in which there would be an automorphism which switches the outer and inner rims (given that n ≤ 150.) The next natural question to ask is this: Are there other symmetries of GP (n, k) other than those of Dn and other than α? My idea to answer this question was to look at possible images of the outer rim and then work from there. Before doing this, I used Mathematica and Groups and Graphs to generate a large number of these generalized Petersen graphs and to find the order of their automorphism groups. A table of the results I generated can be found in Appendix B. It should be clear to the reader that if the order of Aut(GP (n, k)) is greater than 2n, then there must be an automorphism in Aut(GP (n, k)) which is not in Dn . Thus, these automorphisms not in Dn must take a vertex from the outer rim to a vertex in the inner rim and vice versa. Looking at the table from Appendix B, I noticed that the only time the order of Aut(GP (n, k)) was greater than 2n was when k 2 ≡ ±1 (mod n). In other words, I saw that if the graph GP (n, k) were vertex-transitive, then α would be in Aut(GP (n, k)). This lead to the conjecture that if GP (n, k) is vertex-transitive, then k 2 ≡ ±1 (mod n) (hence the converse to the corollary above.) The proof of this is found in Frucht [3]. How Frucht proves that GP (n, k) is vertex-transitive if and only if k2 ≡ ±1 (mod n) is to first consider the subgroup B(n, k) of Aut(GP (n, k)) which fixes the set Σ set-wise. As we have already seen, the dihedral group Dn is a subgroup of Aut(GP (n, k)) but it is also contained in B(n, k). He then defines another mapping α of the vertices of GP (n, k) by: α(ui ) = vki , α(vi ) = uki for all i. Notice that this mapping exchanges Ω and I set-wise and is the same α which was defined above. Now we saw that α will be an automorphism of GP (n, k) if and only if k 2 ≡ ±1 (mod n). For a different approach to see why this is true, notice that α must map the spoke uivi onto the spoke vik uik , the outer edge uiui+1 onto the inner edge uki uki+k , and the inner edge vi vi+k onto the pair uki uki+k2 . Now this pair will be an edge if and only if their indices differ by ±1. Thus we have an edge and hence an automorphism if and only if k 2 ≡ ±1 (mod n) as we expected. Our first goal is to find B(n, k) for the three cases: k 2 6≡ ±1 (mod n), k 2 ≡ 1 (mod n), and k 2 ≡ −1 (mod n). We begin with the following two results found in [3]: Lemma 19 If λ ∈ Aut(GP (n, k)) fixes set-wise any of the sets Ω, I, or Σ, then either it fixes all three sets or fixes Σ set-wise and interchanges Ω and I. Proof: Suppose that an outer edge e gets mapped to a spoke. Then one of the two outer edges adjacent to e must get mapped to an outer edge while the other outer edge adjacent to e must go to an inner edge (see Figure 8.4). Thus if the automorphism does not preserve Σ

Karen S. Potanka

Chapter 8. Generalized Petersen Graphs e2

e e1

36

e2

α

e . . . e1 .

.

Figure 8.4: The action of α if an outer edge e is sent to a spoke set-wise, then it cannot preserve either of the other sets Ω or I. Since the set of outer edges and the set of inner edges are both separate components, any automorphism which fixes Σ set-wise must fix Ω and hence I as well, or map Ω onto I and hence I onto Ω. 2 Theorem 20

1. If k 2 6≡ ±1 (mod n) then B(n, k) =Dn with σ n = τ 2 = 1, τ στ = σ −1 .

2. If k 2 ≡ 1 (mod n), then B(n, k) =< σ, τ, α > with σ n = τ 2 = α2 = 1, τ στ = σ −1 , ατ = τ α, ασα = σ k . 3. If k 2 ≡ −1 (mod n), then B(n, k) =< σ, α > with σ n = α4 = 1, ασα−1 = σ k . In this case, τ = α2 . Hence, we can omit τ as a generator. Proof: Suppose that λ ∈ B(n, k). Then λ fixes Σ set-wise, and so by Lemma 19 λ either fixes Ω and I set-wise as well or λ interchanges Σ and I. For k 2 6≡ ±1 (mod n), we saw that there was no such automorphism which exchanged the sets Ω and I. So in this case, any automorphism must fix all three sets. Suppose λ fixes all three of the sets Ω, Σ, and I. The set of edges in Ω form an n-circuit whose automorphism group is Dn . It is not hard to see that any automorphism λ ∈ Aut(GP (n, k)) is uniquely determined by its action on Ω. Thus λ ∈ Dn =< σ, τ >. It is well known that the relations of the elements in the dihedral group are σ n = τ 2 = 1 and τ στ = σ −1 . For k 2 ≡ ±1 (mod n), we saw that there is an automorphism α which exchanges Ω and I. Now suppose that λ is another automorphism which fixes Σ but interchanges Ω and I. If we compose λ with an appropriate power of σ, then we can force u0 and v0 to be switched. Furthermore, by composing this automorphism with τ (if necessary), we can insist that u1 get mapped to vk . Thus since the outer rim is being mapped to the inner rim, the inner rim must also be an n-circuit. With u0 and u1 going to v0 and vk respectively, we must also

Karen S. Potanka

Chapter 8. Generalized Petersen Graphs

37

have ui going to vik for all i = 0, 1, . . . , n − 1. Now since the set of spokes is being preserved set-wise, we must have the spoke uivi going to uik vik . This forces vi to go to uik . Thus we have the the composition of τ , a power of σ, and λ is just the automorphism α. Thus we have that every element in B(n, k) which is not in Dn interchanges Ω and I and can be written as the composition of any number of σ’s and τ with the automorphism α. Since composing any two of these automorphism fixes each set Ω, Σ, and I, it is in Dn . Thus Dn has index 2 within B(n, k) and B(n, k) has order 4n. (Notice that this says that the order of Aut(GP (n, k)) is at least 4n. There could be other automorphisms which do not fix the set Σ set-wise. We will soon see when this is true.) It remains to be shown the defining relations for B(n, k) for k 2 ≡ ±1 (mod n). The easiest way to do this is to consider the action of B(n, k) on the set of spokes {s0 , s1, . . . , sn−1 } where si is the edge ui vi . Notice that σ(si) = si+1 , τ (si ) = s−i , and α(si ) = sik . First suppose that B(n, k) is not faithful on the set Σ. Then there must be a nontrivial automorphism λ ∈ B(n, k) which fixes each spoke. Certainly, the only automorphism fixing all sets Ω, Σ, and I while fixing each spoke si is the identity automorphism, so λ must interchange Ω and I. This can happen if and only if k = 1. In this case we have B(n, 1) ∼ = 2 −1 2 Dn × S2 . The remaining relations are α = ασασ = (τ α) = 1. Now assume that k > 1. We’ve already seen that B(n, k) acts faithfully on Σ. First suppose that k 2 ≡ 1 (mod n). For any si ∈ Σ, we have α2 (si ) = α(sik ) = sik2 = si . Thus α2 fixes every vertex and so α2 = 1. Similarly, ατ α(si ) = ατ (sik ) = α(s−ik ) = s−ik2 = s−i = τ (si ), ασα(si ) = ασ(sik ) = α(sik+1 ) = sik2 +k = si+k = σ k (si ). Thus we have proved the second part of the theorem. Now suppose that k 2 ≡ −1 (mod n). For any si ∈ Σ, we have α2 (si ) = α(sik ) = sik2 = s−i = τ (si ), ασα−1 (si ) = ασ(s−ik ) = α(s1−ik ) = sk−ik2 = sk+i = σ k (si ). Now because 1 = τ 2 = (α2 )2 = α4 , we may omit τ as a generator and thus we have proved the theorem. 2 To this point we can conclude that B(n, k) acts transitively on the vertices if and only if k 2 ≡ ±1 (mod n). From this we can deduce that if k 2 ≡ ±1 (mod n), then GP (n, k) is vertextransitive. In order to prove the converse, it will be necessary to find Aut(GP (n, k)) for each

Karen S. Potanka

Chapter 8. Generalized Petersen Graphs

38

pair (n, k). We do this by first enumerating the cases in which B(n, k) is a proper subgroup of the automorphism group of GP (n, k). We begin with the straight-forward result given in [3]. Lemma 21 The following are equivalent: 1. GP (n, k) is edge-transitive 2. There is an automorphism λ ∈ Aut(GP (n, k)) which maps some spoke onto an edge which is not a spoke. 3. B(n, k) is a proper subgroup of Aut(GP (n, k)). Proof: Suppose that GP (n, k) is edge-transitive; then clearly there must be some automorphism λ which maps some spoke onto an edge which is not a spoke. It is also clear by the definition of B(n, k) that λ is not contained in B(n, k) and so B(n, k) is a proper subgroup of Aut(GP (n, k)). Now suppose that GP (n, k) were not edge-transitive. Thus there must be at least two orbits of edges under the action of Aut(GP (n, k)) on the edges. But there are no more than three orbits of edges, since the set of outer edges are contained in one orbit, the set of spokes are contained in one, and the set of inner edges are contained in one orbit. Since B(n, k) ⊆ Aut(GP (n, k)), if we let B(n, k) act on the set of edges, then the edge-orbits we obtain will be subsets of the edge-orbits obtained by letting Aut(GP (n, k)) act on the edges. Now because B(n, k) has either 2 or 3 edge-orbits (B(n, k) fixes Σ set-wise and thus Ω and I are either in separate orbits or one together), Aut(GP (n, k)) and B(n, k) must have at least one edge-orbit in common. It turns out the Aut(GP (n, k)) must fix one of Ω, Σ, or I, but by Lemma 19, Aut(GP (n, k)) must fix Σ, and so Aut(GP (n, k)) = B(n, k) contrary to part 3 of the theorem. 2 Before proving the next theorem, we need some new notation. In particular if C is an arbitrary circuit in GP (n, k), let r(C) be the number of outer edges in C, s(C) be the number of spokes in C, and t(C) be the number of inner edges in C. Now if Cj is the set of j-circuits in GP (n, k), then let X Rj = r(C), C∈Cj

Sj =

X

s(C), and

C∈Cj

Tj =

X

t(C).

C∈Cj

The following lemma given in [3] is a very useful tool in showing that the entire automorphism group of GP (n, k) is equal to the subgroup B(n, k) for certain pairs of n and k. In particular, we have:

Karen S. Potanka

Chapter 8. Generalized Petersen Graphs

. . .

. .

.

.

. .

.

39

. .

Figure 8.5: For k = 1, there are only n 4-circuits in GP (n, 1) Lemma 22 If B(n, k) 6= Aut(GP (n, k)), then Rj = Sj = Tj for j = 3, 4, 5, . . .. Proof: Because σ ∈ Aut(GP (n, k)), then if any given outer edge e is in c j-circuits, then every outer edge is contained in c j-circuits. Thus we have Rj = nc. Similarly, there are constants c0 and c00 such that if s is a spoke contained in c0 different j-circuits, then Sj = nc0 and if f is an inner edge contained in c00 j-circuits, then Tj = nc00 . If B(n, k) 6= Aut(GP (n, k)), then GP (n, k) is edge-transitive by Lemma 21, and so if any edge in the graph is contained in c j-circuits, then they all are, so c = c0 = c00. Thus Rj = Sj = Tj . 2 Note that this lemma says that if we can find some j for which Rj 6= Sj , Rj 6= Tj , or Sj 6= Tj , then B(n, k) = Aut(GP (n, k)). The use of this lemma is illustrated in the following two lemmas from [3]. In particular, we will see that B(n, k) = Aut(GP (n, k)) except for the following pairs of n and k: (4, 1), (5, 2), (8, 3), (10, 2),(10, 3),(12, 5),(24, 5). We will begin with the case k = 1: Lemma 23 If n 6= 4, then B(n, 1) = Aut(GP (n, 1)) Proof: The method of proof will be that described above. In particular, we look at all of the possible 4-circuits in GP (n, 1). If n 6= 4, then the only possible 4-circuit will be of the form ui ui+1 vi+1 vi (See Figure 8.5), and there are exactly n of these. Note that in a circuit of this type, there is exactly one outer edge, one inner edge, and two spokes, so that R4 = n, S4 = 2n and T4 = n. Thus by the previous lemma, B(n, 1) = Aut(GP (n, 1)). 2 Lemma 24 If n 6= 5 or 10, then B(n, 2) = Aut(GP (n, 2)) Proof: If n 6= 5 or 10, then the outer rim would be an n-circuit and the inner rims would n n be circuits of length gcd(n,2) . Now because gcd(n, 2) = 1 or 2, the only time gcd(n,2) = 5 is when n = 5 or 10. Thus the only 5-circuits in GP (n, k) will be those with s(C) = 2 and either r(C) = 1 and t(C) = 2 or r(C) = 2 and t(C) = 1. The former case is impossible since s(C) = 2 forces the inner edges to be adjacent, yielding the circuit of type u0 v0v2v4 u4

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Chapter 8. Generalized Petersen Graphs

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which is not a circuit. Thus GP (n, 2) contains n 5-circuits of the form ui ui+1 ui+2 vi+2 vi. It follows that R5 = S5 = 2n and T5 = n. By Lemma 22 we have B(n, 2) = Aut(GP (n, 2)) for n 6= 5 or 10. 2 It turns out that by considering all of the possible 8-circuits in GP (n, k) for k > 2 and n > 5 most of the other cases fall out. We do this by looking at all possible cases for r(C), where C is an 8-circuit. First note that any circuit C in GP (n, k) must have an even number of spokes. Suppose that C is an 8-circuit: 1. Suppose r(C) = 8. Then there is only one possible way for this to happen: n = 8. 2. Suppose r(C) = 7. Then the other edge of the circuit would be either a spoke or an inner edge. Thus since there has to be an even number of spokes, the other edge would be an inner edge. The union of 7 outer edges with an inner edge can never be a circuit. 3. Suppose r(C) = 6, then we could have either s(C) = 2 and t(C) = 0 or s(C) = 0 and t(C) = 2. Note that in either case, the union of the 6 outer edges with the other two would not be a circuit. 4. Suppose r(C) = 5, then we must have s(C) = 2 and t(C) = 1. (The only other possible combination would be s(C) = 0 and t(C) = 3, but by the same reasoning used in (3), this would not be a circuit.) Now because s(C) = 2, the 5 outer edges are forced to be adjacent to obtain a circuit. Thus this type of 8-circuit would be u0u1u2 u3u4 u5v5 v0 and this is only possible if k = 5 or n − k = 5. 5. Suppose r(C) = 4, then we must have s(C) = 2 and t(C) = 2. Again because s(C) = 2, the 4 outer edges are forced to be adjacent and the 2 inner edges are forced to be adjacent. This type of 8-circuit would be u0u1 u2u3 u4v4v 1 (n+4) v0 and this is only 2 possible when k = 12 (n + 4) − 4 = n2 − 2 or 2k + 4 = n. 6. Suppose r(C) = 3, then either s(C) = 2 and t(C) = 3 or s(C) = 4 and t(C) = 1. Note that the latter case cannot happen because we have 4 endpoints of the 4 spokes to be joined by one inner edge. Thus since s(C) = 2 this forces the 3 outer edges to be adjacent and the 3 inner edges to be adjacent. This type of 8-circuit would be u0u1 u2u3 v3v3+k v3+2k v0 or u0u1 u2u3 v3v3−k v3−2k v0 and these are possible when n = 3+3k or 3 − 3k = −n respectively since 2k < n and n2 ≥ 3 ⇒ 3 + 3k < 3 + 32 n ⇒ 3 + 3k < 2n and 2k < n ⇒ 3 − 3k > 3 − 32 n ⇒ 3 − 3k ≥ −n. 7. Suppose r(C) = 2, then we can have s(C) = 2 and t(C) = 4 or s(C) = 4 and t(C) = 2. In the first case, because s(C) = 2, this forces the 2 outer edges to be adjacent and the 4 inner edges to be adjacent. This type of 8-circuit would be u0u1 u2v2v2+k v2+2k v2+3k v0 which occurs when n = 2 + 4k or 2n = 2 + 4k since 2k < n ⇒ 2 + 4k < 2 + 2n ⇒ 2 + 4k = n or 2 + 4k = 2n or it could be u0 u1u2 v2v2−k v2−2k v2−3k v0 which occurs when 2 − 4k = −n since 2k < n ⇒ −4k > 2n ⇒ 2 − 4k > 2 − 2n ⇒ 2 − 4k = −n

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Chapter 8. Generalized Petersen Graphs

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In the second case, because s(C) = 4, we cannot have the 2 outer edges adjacent, nor can we have the 2 inner edges adjacent. Each outer and inner edge must join two endpoints of a spoke, so that this type of 8-circuit is u0 u1v1v1+k u1+k u2+k v2+k v0 which occurs when 2 + 2k = n or it is u0u1 v1v1+k u1+k uk vk v0 which occurs for all n ≥ 4. Note that the case u0 u1v1 v1−k u1−k u2−k v2−k v0 can never happen and the case u0u1 v1v1−k u1−k u−k v−k v0 is analogous to the second circuit. Thus we have exhausted all cases. 8. Suppose r(C) = 1, then it must be true that s(C) = 2 which leaves t(C) = 5. This type of 8-circuit will be u0u1 v1v1+k v1+2k v1+3k v1+4k v0 which occurs when n = 1 + 5k or 2n = 1 + 5k since 2k < n and n2 ≥ 3 ⇒ 1 + 5k < 1 + 52 n ⇒ 1 + 5k < 3n. This type of 8-circuit could also be u0u1v1 v1−k v1−2k v1−3k v1−4k v0 which occurs when 1 − 5k = −n or 1 − 5k = −2n since 2k < n and n2 ≥ 3 ⇒ 1 − 5k > 1 − 52 n ⇒ 1 − 5k ≥ −2n. 9. Finally, suppose r(C) = 0, then we must also have s(C) = 0, which forces t(C) = 8. Thus we have an 8-circuit of type v0vk v2k v3k v4k v5k v6k v7k which occurs if n = 8k. Now since 2k < n, we have that 8k < 4n which tells us that this type of circuit could occur if 3n = 8k as well. The above information is summarized in the following table: Type 1 2 3 4 4’ 5 5’ 6 7 8 8’ 9 9’

Cycle representation u0 u1 u2 u3 u4 u5 u6 u7 u0 u1 u2 u3 u4 u5 v 5 v 0 u0 u1u2 u3u4 v4v 12 (n+4) v0 u0 u1u2 u3v3 v3+k v3+2k v0 u0 u1u2 u3v3 v3−k v3−2k v0 u0 u1u2 v2v2+k v2+2k v2+3k v0 u0 u1u2 v2v2−k v2−2k v2−3k v0 u0 u1v1 v1+k u1+k u2+k v2+k v0 u0 u1v1 v1+k u1+k uk vk v0 u0 u1v1 v1+k v1+2k v1+3k v1+4k v0 u0 u1v1 v1−k v1−2k v1−3k v1−4k v0 v0 vk v2k v3k v4k v5k v6k v7k v0 vk v2k v3k v4k v5k v6k v7k

Conditions n=8 k = 5 or n − k = 5 n = 2k + 4 n = 3k + 3 n = 3k − 3 n or 2n = 4k + 2 n = 4k − 2 n = 2k + 2 n≥4 n or 2n = 5k + 1 n or 2n = 5k − 1 n = 8k 3n = 8k

Number 1 n n n n n n 1 n 2 n n n k 1 n 8

r(C) 8 5 4 3 3 2 2 2 2 1 1 0 0

s(C) 0 2 2 2 2 2 2 4 4 2 2 0 0

t(C) 0 1 2 3 3 4 4 2 2 5 5 8 8

We have already seen that when n 6= 4 B(n, 1) = Aut(GP (n, 1)), and when n 6= 5 or 10 B(n, 2) = Aut(GP (n, 2)). We now use the 8-circuits in GP (n, k) for k > 2 to show that when (n, k) is not one of six pairs, we definitely have B(n, k) = Aut(GP (n, k)). Lemma 25 B(n, k) = Aut(GP (n, k)) if k > 2 and if (n, k) is not one of the pairs: (8, 3), (10, 3), (12, 5), (13, 5), (24, 5), or (26, 5)

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Chapter 8. Generalized Petersen Graphs

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Proof: Since (8, 1), (8, 2), and (8, 3) have already been excluded, we may assume that n 6= 8. We first consider the case when n = 8k and k > 2. The types of 8-circuits contained in GP (8k, k) are 7 and 9 as well as type 2 if k = 5. Thus there are two cases to consider. Suppose that k = 5, then R8 = 5n + 0 + 2n = 7n while S8 = 2n + 0 + 4n = 6n. Now suppose that k 6= 5, then similarly R8 = 0 + 2n = 2n and S8 = 0 + 4n = 4n. Thus in both cases we have B(8k, k) = Aut(GP (8k, k)) by Lemma 22. Now suppose that n 6= 8k. Note that because n cannot be equal to 3k + 3 and 3k − 3 simultaneously, there cannot be 8-circuits of types 4 and 4’ simultaneously. A similar statement can be made of 8-circuits of types 5 and 5’, 8 and 8’, and 9 and 9’. What we will do next is sum up the total number of outer edges in any 8-circuit, the total number of spokes in any 8-circuit, and the total number of inner edges in any 8-circuit. Before doing this, let xi = 1 if there are circuits of type i or i0 in GP (n, k) and let xi = 0 otherwise. Then we have: R8 = 5nx2 + 4nx3 + 3nx4 + 2nx5 + nx6 + 2n + nx8 , S8 = 2nx2 + 2nx3 + 2nx4 + 2nx5 + 2nx6 + 4n + 2nx8 , T8 = nx2 + 2nx3 + 3nx4 + 4nx5 + nx6 + 2n + 5nx8 + nx9 . We want to find the cases when B(n, k) = 6 Aut(GP (n, k)). Suppose that (R8−T8 ) B(n, k) 6= Aut(GP (n, k)). By Lemma 22 we would have = 0 or n 4x2 + 2x3 − 2x5 − x9 − 4x8 = 0 which gives 4x2 + 2x3 = 2x5 + x9 + 4x8 . Since the left side is always even, we must have that x9 = 0. Furthermore, it should be clear that x2 = x8 and x3 = x5. If x2 = x8 = 1, then we must have k = 5 and n = 12, 12, 24, or 26 all of which are excluded in the hypothesis. Now if x3 = x5, then intersecting the conditions gives 2k + 4 = 4k + 4 which cannot happen, 4k + 8 = 4k + 2 which cannot happen, and 2k + 4 = 4k − 2 ⇒ 2k = 6 ⇒ k = 3 ⇒ n = 10 which is also an excluded case. Thus we may assume that x2 = x3 = x5 = x8 = 0. We can similarly look at when or when

(R8 −S8 ) n

= 0. This occurs when 3x2 + 2x3 + x4 − x6 − 2 − x8 = 0

3x2 + 2x3 + x4 = x6 + 2 + x8. But since x2 = x3 = x5 = x8 = 0, we have x4 = x6 + 2. Because xi is either 0 or 1, the condition has no solution. 2 The next lemma gives two more cases in which B(n, k) = Aut(GP (n, k)). For an alternate proof of the second part of the lemma, please refer to [3], page 216. Lemma 26 B(13, 5) = Aut(GP (13, 5)) and B(26, 5) = Aut(GP (26, 5)).

Karen S. Potanka

Chapter 8. Generalized Petersen Graphs

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u0 u12

u1 v12

u11

v0 v1

v11

u2 v2

v10 u10

v3

u3

v9 v4 u9

v8

u4

v5 v7

u8 u7

v6

u5 u6

Figure 8.6: GP (13, 5) Proof: First consider GP (13, 5) as seen in Figure 8.7 and suppose that we look at all possible 7-circuits. Clearly there are no 7-circuits with 7,6, or 5 outer edges. All other cases are summarized in the following table: r(C) 4 3 3 2 2 1 1

s(C) 2 2 2 2 2 2 2

t(C) 1 2 2 3 3 4 4

Circuit representation u0 u1 u2 u3 u4 v 4 v 0 u0 u1 u2 u3 v 3 v 8 v 0 u0u1u2 u3v3 v11v0 u0u1u2 v2v7 v12v0 u0u1u2 v2v10v5v0 u0u1v1 v6v11v3v0 u0u1v1 v9v4v12v0

Circuit? No Yes No No Yes No No

It should be clear to the reader that the cases in which s(C) = 4 is impossible since it would leave r(C) = 2 and t(C) = 1 or r(C) = 1 and t(C) = 2. In either case, we would have to have one outer or inner edge joining the 4 endpoints of the spokes. From the table above, we see that there are only 2 types of 7-circuits in GP (13, 5) each type occuring n = 13 times. This leads to the calculations R7 = 5 × 13 = 65, S7 = 4 × 13 = 52, and so by Lemma 22, we have that B(13, 5) = Aut(GP (13, 5)). It turns out that in GP (26, 5) there are only two types of 10-circuits (See Appendix C for justification of this statement.) One type is u0u1 u2v2v7 u7u6 u5v5v0 while the other is

Karen S. Potanka

Chapter 8. Generalized Petersen Graphs

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u0u1 v1v6 v11u11 u10v10v5v0. Note that in the first type r(C) = 4, s(C) = 4, and t(C) = 2 and there are 26 circuits of this type. In the second type, r(C) = 2, s(C) = 4, and t(C) = 4 and there are also 26 circuits of this type so that R10 = 4 × 26 + 2 × 26 = 156 and S10 = 4 × 26 + 4 × 26 = 208. Thus by lemma 23 , B(26, 5) = Aut(GP (26, 5)). 2 We are now in a position to prove the converse of Corollary 18, namely: Theorem 27 If GP (n, k) is vertex-transitive, then k 2 ≡ ±1 (mod n) or the graph is GP (10, 2). Proof: As seen in the previous lemmas, there are only seven cases when B(n, k) 6= Aut(GP (n, k)), namely GP (4, 1), GP (5, 2), GP (8, 3), GP (10, 2), GP (10, 3), GP (12, 5), and GP (24, 5). We have also seen that B(n, k) acts transitively on the vertices if and only if k 2 ≡ ±1 (mod n) for graphs not in this list. Since Aut(GP (n, k)) = B(n, k) except for the seven cases, we have that if Aut(GP (n, k)) acts transitively on the vertices then k 2 ≡ ±1 (mod n). To complete the proof we need to show the statement is true for the seven exceptional cases. Consider first the graphs GP (4, 1), GP (8, 3), GP (12, 5), and GP (24, 5). It is shown in [3] that if we define λ on V (GP (n, k)) for these four cases by λ(u4i) = u4i , λ(v4i ) = u4i+1 , λ(u4i+1 ) = u4i−1 , λ(u4i−1 ) = v4i, λ(u4i+2 ) = v4i−1, λ(v4i−1) = v4i+5, λ(v4i+1 ) = u4i−2 , λ(v4i+2 ) = v4i−6 for all i then Aut(GP (n, k)) =< σ, τ, λ > with σ n = τ 2 = λ3 = 1, τ στ = σ −1 , τ λτ = λ−1 , λσλ = σ −1 , λσ 4 = σ 4λ. Thus α as defined earlier in the paper is given by α = λ−1 σλ−1 = τ λ−1 σλ−1 . So clearly in these four cases, the graph is vertex-transitive, and it can be verified that in each case k 2 ≡ 1 (mod n). Note that in each case |Aut(GP (n, k))| = 12n as seen in Appendix A.

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Chapter 8. Generalized Petersen Graphs

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Now consider the graphs GP (5, 2) and GP (10, 3). It has been shown that Aut(GP (5, 2)) is isomorphic to S5 while Aut(GP (10, 3)) is isomorphic to S5 × S2 . As given in [3], both graphs can be generated by σ and µ defined by (u2v1 )(u3v4)(u7 v6)(u8 v3)(v2v8)(v3 v7) for GP (10, 3) and (u2v1 )(u3v4)(v2 v3) for GP (5, 2). The defining relations for GP (10, 3) are σ 10 = µ2 = (µσ 2 )4 = (σ 2µσ −2 µ)3 = (σµσ −1 µ)2 = (σ 5µσ −5 µ) = 1 and those for GP (5, 2) are σ 5 = µ2 = (µσ 2)4 = (σ 2µσ −2 µ)3 = (σ 5µσ −5 µ) = 1. In either case we find that |Aut(GP (n, k))| = 24n and Aut(GP (n, k)) acts transitively on the vertices of GP (n, k). It can be verified that k 2 ≡ −1 (mod n). The last case to consider is the graph GP (10, 2) which is seen to be the graph of the regular dodecahedron. Its automorphism group is found to have order 120 and can be represented by Aut(GP (10, 2)) =< σ, ζ > where ζ is defined by ζ = (u0 v2v8)(u1 v4u8)(u2 v6u9 )(u3u6v9 )(u4u7 v1)(u5v7 v3). The defining relations for Aut(GP (10, 2)) are given by σ 10 = ζ 3 = (ζσ 2 )2 = σ 5ζσ −5 ζ −1 = 1. GP (10, 2) is seen to be vertex-transitive although k2 6≡ ±1 (mod n). Thus the theorem is proved. 2

8.2

Distinguishing Numbers of Generalized Petersen Graphs

Now that the automorphism groups of all of the generalized Petersen graphs have been defined, we can now find the distinguishing numbers for these graphs. Theorem 28 D(GP (n, k)) = 2 except for the pairs (4, 1) and (5, 2) in which case D(GP (n, k)) = 3.

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Chapter 8. Generalized Petersen Graphs

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Proof: Let’s first suppose that k 2 6≡ ±1 (mod n). In this case it must follow that n ≥ 6 and we know from the discussion above that Aut(GP (n, k) ∼ = Dn except for the pair (10, 2). Excluding the pair (10, 2), we must have that there are two orbits of vertices each of size n. Besides the identity automorphism, there is only one other automorphism, namely τ σ i for some 0 ≤ i < n, which stabilizes any given vertex in GP (n, k). Thus by Theorem 14 we have that GP (n, k) can be distinguished with two colors in these cases. We defined Aut(GP (10, 2)) =< σ, ζ > where ζ and the relations are given above. We can think of Aut(GP (10, 2)) as consisting of the rotations σ i, the reflections τ σ i , the order 3 rotations about the axis through antipodal vertices (i.e. u0 and u5 ), and the order 2 automorphism represented by ζσ 2 . If we color u0 , u2 and u3 red and the rest of the vertices in GP (10, 2) blue then clearly we have broken the symmetries including all of the rotations and reflections as well as all of the order 3 rotations about axes through opposite vertices. All that needs to be checked is that this coloring breaks the order 2 automorphisms described above. Using this coloring in Groups and Graphs, we can see that indeed all symmetries of GP (10, 2) are broken. Now consider the case when k2 ≡ ±1 (mod n). It was shown above that the automorphism groups of the graphs GP (n, k) included the rotations, reflections, and the automorphism α switching the outer rim and the inner rim except for the cases (4, 1), (5, 2), (8, 3), (10, 3), (12, 5), and (24, 5). Excluding these cases, if we color u0 , u2, and u3 red and the rest of the vertices blue, then clearly we see that all of the symmetries are broken. Using Groups and Graphs, it is possible to show that for the cases GP (8, 3), GP (10, 3), GP (12, 5), and GP (24, 5), D(GP (n, k)) = 2. In particular, if we color the vertices u0 , u2 and u3 red and the rest of the vertices in GP (n, k) blue as above then it turns out that this is a 2-distinguishing coloring for the cases GP (8, 3), GP (10, 3), GP (12, 5), and GP (24, 5). We can think of GP (4, 1) as the three dimensional cube. It is known that there are 48 automorphisms in Aut(GP (4, 1)). These can be described in terms of the cube so that there are 12 rotations about an axes through the center of a face (4 rotations in each of the 3 dimensions), 12 reflections about lines through two opposite vertices on a face or through the center of two opposite edges of a face (4 reflections in each of the three dimenstions), 12 rotations about axes through antipodal vertices in the cube (3 rotations for each of the 4 pairs of antipodal vertices), and 12 symmetries exchanging antipodal vertices and then rotating about axes through antipodal vertices (3 rotations for each of the 4 pairs of antipodal vertices.) It turns out that there is no 2-distinguishing coloring of GP (4, 1). Suppose that there were a 2-distinguishing coloring of GP (4, 1). Then we must find a set A of the 8 vertices to color red and a set B to color blue such that every nontrivial automorphism sends a red vertex to a blue vertex. Without loss of generality, there are only four cases to consider:

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Chapter 8. Generalized Petersen Graphs

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1. Suppose |A| = 1 and |B| = 7. Without loss of generality, suppose that u0 is the red vertex. Then every nontrivial automorphism not in Stv sends u0 to a blue vertex. There is at least one, for example a rotation about the axes through the center of any of the faces. 2. Suppose |A| = 2 and |B| = 6. Without loss of generality, suppose that u0 is one of the two red vertices. If the other red vertex is any vertex except the antipodal vertex, then there is a reflection exchanging the two, thus fixing the coloring. If the other vertex is the antipodal one, then there is a symmetry switching them. 3. Suppose |A| = 3 and |B| = 5. We first consider the case when two of the red vertices are antipodal, say for instance u0 and v2. Then the third red vertex will be adjacent to one of u0 or v2 and diagonal on a face from the other. Thus there is a (nontrivial) reflection about the line through that diagonal fixing each of the red vertices. Now suppose that none of the three are antipodal but two of them are adjacent, say u0 and u1. Then the third vertex cannot be v2 or v3 and so the three must lie on the same face. Thus there is a reflection about a line through one of them, switching the other two. Finally, suppose that none of the three are antipodal and none of them are adjacent. If we choose u0 and u2 to be red, then the other red vertex is either v1 ro v3. In either case, the three red vertices are adjacent to a common vertex (either u1 or u3), and there is a rotation about a line through that vertex rotating the three red vertices. 4. Suppose |A| = 4 and |B| = 4. First suppose that all three red vertices are on the same face. Then clearly any rotation about the axis through the center of that face is a symmetry fixing the red vertices. Now suppose that no more than three red vertices are on the same face. Without loss of generality, suppose that u0, u1 and u2 are red. Then the fourth red vertex could be v1 , but reflecting about the line through u1 and v1 fixes the red. The fourth could be v3, but reflecting through the plane containing u1, v1, u3, and v3 fixes the red. Finally the fourth could be v2 or v0 , but in either case the composition of two reflections fixes the red. Now suppose that no more than two red vertices lie on a face and suppose that u0 is red. Then the second red vertex could be u1 , in which case the other two red vertices are forced to be v2 and v3, but there is a reflection fixing all of these red vertices. (The case in which the second red vertex is u3 is similar.) If the second red vertex is u2, then the other two red vertices could be v0 and v2 or v1 and v3 . In the former case, there is a reflection fixing the red, and in the latter, there is a rotation fixing the red. Finally, the case in which a face contains no more than one red vertex is impossible. Thus we have exhausted all of the cases for a coloring using 2 colors for GP (4, 1) and we must conclude that the disinguishing number of GP (4, 1) must be greater than or equal to 3. Figure 8.7 contains a 3-distinguishing coloring of GP (4, 1) and so D(GP (4, 1)) = 3. It was already noted in [1] that D(GP (5, 2)) = 3, thus we have exhausted all of the cases. 2

Karen S. Potanka

Chapter 8. Generalized Petersen Graphs

Figure 8.7: A 3-distinguishing coloring of GP (4, 1)

48

Chapter 9 Conclusions and Conjectures As discovered in the results above, finding the distinguishing number of a given graph is not always a trivial problem. Nonintuitively, it turns out that in several cases the fewer the number of vertices the more colors needed to disinguish the graph. This was seen in the example of the cyclic graph Cn , other graphs which realize Dn , as well as the generalized Petersen graphs. An interesting idea to consider would be to let a proper nontrivial subgroup H of the automorphism group of a given graph G act on the vertices in G. Knowing that the distinguishing number of G under the action of H is k, then what can be said about the distinguishing number of G under the action of the entire automorphism group? Certainly there are cases in which an automorphism not in the subgroup fixes the coloring disintinguished under the action of the subgroup. Perhaps another coloring exists using the same number of colors such that the graph is disinguished under the action of the entire automorphism group. Perhaps more colors are needed to distinguish the graph under the action of the automorphism group. It would be interesting to see a pattern of such cases. For instance, is there an upper bound on the distinguishing number of a graph knowing the disinguishing number of a graph under the action of a subgroup? Another interesting idea to consider would be to see if there is a relation between the distinguishing number of two graphs G1 and G2 and the distinguishing number of G1 × G2 , G1 ∨ G2 or G1 ∧ G2 . Before analyzing this idea, it would be necessary to see if there is any relation between the automorphism groups of G1 and G2 and the automorphism groups of G1 × G2 , G1 ∨ G2 or G1 ∧ G2 . Finally, does the fact that a graph G is edge-transitive have anything to do with the distinguishing number of a graph? If so, does this give us an upper bound on the distinguishing number?

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Appendix A Tables of Congruences The first table contains values of n and k for which k2 ≡ 1 (mod n). The second contains values of n and k for which k 2 ≡ −1 (mod n). k 2 ≡ 1 (mod n) n 8 12 15 16 20 21 24 24 24 28 30 32 33 35 36 39 40 40 40 42 44

k 3 5 4 7 9 8 5 7 11 13 11 15 10 6 17 14 9 11 19 13 21

n 45 48 48 48 51 52 55 56 56 56 57 60 60 60 63 64 65 66 68 69 70

k 19 7 17 23 16 25 21 13 15 27 20 11 19 29 8 31 14 23 33 22 29

n 72 72 72 75 76 77 78 80 80 80 84 84 84 85 87 88 88 88 91 91 92

k 17 19 35 26 37 34 25 9 31 39 13 29 41 16 28 21 23 43 19 27 45

n 93 95 96 96 96 99 100 102 104 104 104 105 105 105 108 110 111 112 112 112 114

50

k 32 39 17 31 47 10 49 35 25 27 51 29 34 41 53 21 38 15 41 55 37

n 115 116 117 119 120 120 120 120 120 120 120 123 124 126 128 129 130 132 132 132 133

k 24 57 53 50 11 19 29 31 41 49 59 40 61 55 63 44 51 23 43 65 20

n 135 136 136 136 138 140 140 140 141 143 144 144 144 145 147 148 150

k 26 33 35 67 47 29 41 69 46 12 17 55 71 59 50 73 49

Karen S. Potanka

Appendix A. Tables of Congruences k 2 ≡ −1 (mod n) n 5 10 13 17 25 26 29 34 37 41 50 53

k 2 3 5 4 7 5 12 13 6 9 7 23

n 58 61 65 65 73 74 82 85 85 89 97 101

k 17 11 8 18 27 31 9 13 38 34 22 10

n 106 109 113 122 125 130 130 137 145 145 146 149

k 23 33 15 11 57 47 57 37 12 17 27 44

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Appendix B Group Orders The table on the following page contains the orders of GP (n, k) for n ≤ 27.

52

Karen S. Potanka Graph GP (5, 2) GP (7, 2) GP (7, 3) GP (8, 3) GP (9, 2) GP (9, 4) GP (10, 3) GP (11, 3) GP (11, 4) GP (11, 5) GP (12, 5) GP (13, 2) GP (13, 4) GP (13, 5) GP (13, 6) GP (14, 3) GP (14, 5) GP (15, 2) GP (15, 4) GP (15, 7) GP (16, 3) GP (16, 5) GP (16, 7) GP (17, 2) GP (17, 3) GP (17, 4) GP (17, 5) GP (17, 6) GP (17, 7) GP (17, 8) GP (18, 5) GP (18, 7) GP (19, 2)

Appendix B. Group Orders Order 120 = 24n 14 = 2n 14 = 2n 96 = 12n 18 = 2n 18 = 2n 240 = 24n 22 = 2n 22 = 2n 22 = 2n 144 = 12n 26 = 2n 26 = 2n 52 = 4n 26 = 2n 28 = 2n 28 = 2n 30 = 2n 60 = 4n 30 = 2n 32 = 2n 32 = 2n 64 = 4n 34 = 2n 34 = 2n 68 = 4n 34 = 2n 34 = 2n 34 = 2n 34 = 2n 36 = 2n 36 = 2n 38 = 2n

Graph GP (19, 3) GP (19, 4) GP (19, 5) GP (19, 6) GP (19, 7) GP (19, 8) GP (19, 9) GP (20, 3) GP (20, 7) GP (20, 9) GP (21, 2) GP (21, 4) GP (21, 5) GP (21, 8) GP (21, 10) GP (22, 3) GP (22, 5) GP (22, 7) GP (22, 9) GP (23, 2) GP (23, 3) GP (23, 4) GP (23, 5) GP (23, 6) GP (23, 7) GP (23, 8) GP (23, 9) GP (23, 10) GP (23, 11) GP (24, 5) GP (24, 7) GP (24, 11) GP (25, 2)

Order 38 = 2n 38 = 2n 38 = 2n 38 = 2n 38 = 2n 38 = 2n 38 = 2n 40 = 2n 40 = 2n 80 = 4n 42 = 2n 42 = 2n 42 = 2n 84 = 4n 42 = 2n 44 = 2n 44 = 2n 44 = 2n 44 = 2n 46 = 2n 46 = 2n 46 = 2n 46 = 2n 46 = 2n 46 = 2n 46 = 2n 46 = 2n 46 = 2n 46 = 2n 288 = 12n 96 = 4n 96 = 4n 50 = 2n

Graph GP (25, 3) GP (25, 4) GP (25, 6) GP (25, 7) GP (25, 8) GP (25, 9) GP (25, 11) GP (25, 12) GP (26, 3) GP (26, 5) GP (26, 7) GP (26, 9) GP (26, 11) GP (27, 2) GP (27, 4) GP (27, 5) GP (27, 7) GP (27, 8) GP (27, 10) GP (27, 11) GP (27, 13)

53 Order 50 = 2n 50 = 2n 50 = 2n 100 = 4n 50 = 2n 50 = 2n 50 = 2n 50 = 2n 52 = 2n 104 = 4n 52 = 2n 52 = 2n 52 = 2n 54 = 2n 54 = 2n 54 = 2n 54 = 2n 54 = 2n 54 = 2n 54 = 2n 54 = 2n

Appendix C 10-circuits in GP (26, 5) Type

1 1’

r(C) 7 6 6 5 4 4 4 4 4 4 4 4 4 4 3 3 3 3 3 3

s(C) 2 2 2 2 4 4 4 4 4 4 4 4 2 2 4 4 4 4 4 4

t(C) 1 2 2 2 2 2 2 2 2 2 2 2 4 4 3 3 3 3 3 3

Circuit representation u0 u1 u2 u3 u4 u5 u6 u7 v7 v0 u0 u1 u2 u3 u4 u5 u6 v6 v11 v0 u0 u1 u2 u3 u4 u5 u6 v6 v1 v0 u0 u1 u2 u3 u4 u5 v5 v10 v15 v0 u0 u1 u2 u3 v3 v8 u8 u9 v9 v0 u0 u1 u2 u3 v3 v8 u8 u7 v7 v0 u0 u1 u2 u3 v3 v24 u24 u23 v23 v0 u0 u1 u2 u3 v3 v24 u24 u25 v25 v0 u0 u1 u2 v2 v7 u7 u8 u9 v9 v0 u0 u1 u2 v2 v7 u7 u6 u5 v5 v0 u0 u1 u2 v2 v23 u23 u22 u21 v21 v0 u0 u1 u2 v2 v23 u23 u24 u25 v25 v0 u0 u1 u2 u3 u4 v4 v9 v14 v19 v0 u0 u1 u2 u3 u4 v4 v25 v20 v15 v0 u0 u1 u2 v2 v7 v12 u12 u13v13 v0 u0 u1 u2 v2 v7 v12 u12 u11v11 v0 u0 u1 u2 v2 v23 v18 u18 u17 v17 v0 u0 u1 u2 v2 v23 v18 u18 u19 v19 v0 u0 u1 u2 v2 v7 u7 u6 v6 v1 v0 u0 u1 u2 v2 v7 u7 u6 v6 v11 v0

54

Circuit?

No No No No No No No No No Yes Yes No No No No No No No No No

Appendix C. 10-circuits in GP (26, 5)

Karen S. Potanka Type

2

2’

r(C) 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1

s(C) 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 2 2 2 2

t(C) 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4

4 4 4 6 6 7 7

Circuit representation u0 u1 u2 v2 v7 u7 u8 v8 v3 v0 u0 u1 u2 v2 v7 u7 u8 v8 v13 v0 u0 u1 u2 v2 v23 u23 u22 v22 v17 v0 u0 u1 u2 v2 v23 u23 u22 v22 v1 v0 u0 u1 u2 v2 v23 u23 u24 v24 v3 v0 u0 u1 u2 v2 v23 u23 u24 v24 v19 v0 u0 u1 v1 v6 v11 v16 u16 u15 v15 v0 u0 u1 v1 v6 v11 v16 u16 u17 v17 v0 u0 u1 v1 v22 v17 v12 u12 u11 v11 v0 u0 u1 v1 v22 v17 v12 u12 u13 v13 v0 u0 u1 v1 v6 v11 u11 u10v10 v5 v0 u0 u1 v1 v6 v11 u11 u10v10 v15 v0 u0 u1 v1 v6 v11 u11 u12v12 v7 v0 u0 u1 v1 v6 v11 u11 u12v12 v17 v0 u0 u1 v1 v22 v17 u17 u16 v16 v11 v0 u0 u1 v1 v22 v17 u17 u16 v16 v21 v0 u0 u1 v1 v22 v17 u17 u18 v18 v13 v0 u0 u1 v1 v22 v17 u17 u18 v18 v23 v0 u0 u1 u2 v2 v7 v12 v17 v22 v1 v0 u0 u1 u2 v2 v23 v18 v13 v8 v3 v0 u0 u1 v1 v6 v11 v16 v21 v0 u0 u1 v1 v22 v17 v12 v7 v2 v23 v0

Circuit? No No No No No No No No No No Yes No No No No Yes No No No No No No

55

Bibliography [1] Michael O. Albertson, Karen L. Collins (1996). Symmetry Breaking in Graphs, The Electronic Journal of Combinatorics 3. [2] Robert Frucht. Graphs of Degree Three with a Given Abstract Group, Canadian Journal of Math. 1, (1949) 365-378. [3] Robert Frucht, J. E. Graver, M. Watkins. The groups of the generalized Petersen graphs, Proceedings of the Cambridge Philosophical Society 70, (1971), 211-218. [4] Jonathon L. Gross, Thomas W. Tucker. Topological Graph Theory. John Wiley & Sons, Inc.: New York, 1987. [5] Israel Grossman, Wilhelm Magnus. Groups and their Graphs. The Mathematical Association of America: Yale University, 1964. [6] Fred Roberts. Applied Combinatorics. Prentice-Hall, Inc.: Englewood Cliffs,NJ, 1984. [7] Frank Rudin, Problem 729 in Journal of Recreational Mathematics, volume 11, (solution in 12, 1980), p. 128, 1979. [8] Arthur T. White. Groups, Graphs and Surfaces. North Holland, New York, 1984, North Holland Math Studies 8.

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Vita Karen Sue Potanka was born in New Britain, Connecticut on February 18, 1975. Karen was raised in Cumberland, Maryland until she first attended Virginia Tech in August of 1993. As an undergraduate, she participated in two Research Experience for Undergraduates (REU) programs, one at the University of Oklahoma and the other at Northern Arizona University. Because of her experience in research, she was named a Barry M. Goldwater scholar in 1996 and was asked to present her research in graph theory at two national conferences in 1997. She graduated summa cum laude with an “in Honors” Bachelor of Science degree in Mathematics in May of 1997. She was named Outstanding Graduating Senior and received the Layman Prize for her research in graph theory. As a graduate student, Karen worked on another research project in graph theory. She received a Master of Science in Mathematics from Virginia Tech in 1998 as part of the Five Year Bachelor/Master’s program.

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