On the fixing number of graphs with abelian ...

On the fixing number of graphs with abelian automorphism group J. Bradley Escuela de Matem´atica Universidad de Costa Rica San Jos´e, Costa Rica P. Dankelmann, D. Erwin, B.G. Rodrigues School of Mathematical Sciences University of KwaZulu-Natal Durban, South Africa October 1, 2008

Abstract The fixing number of a graph G is the cardinality of a smallest set of vertices of G that is not fixed by any non-trivial automorphism of G. In this paper we investigate the fixing number of finite graphs with abelian automorphism group.

1

Introduction

For notation not explained here, we refer the reader to the books by Chartrand and Lesniak [5] and Godsil and Royle [8]. Let G be a finite graph with vertex set V (G), automorphism group Γ(G), and let S ⊆ V (G). The pointwise stabilizer of S is the subgroup ΓS (G) = {α ∈ Γ(G) : for every v ∈ S, v α = v}. The set S is said to fix G if it has trivial pointwise stabilizer, i.e., ΓS (G) = {1}. So a set S fixes G if every nontrivial automorphism of G moves at least one member of S. The fixing number of a graph G, fix(G), is the cardinality of a minimum fixing set, i.e., fix(G) = min{|S| : S fixes G}. The fixing number of a graph was introduced by Harary [9, 10] and studied further by Erwin and Harary [6], who determined a formula for the fixing number of a tree, and pointed out that, in general, graphs with isomorphic automorphism groups may have different fixing 1

numbers. Amongst other applications the fixing number provides a lower bound for wellstudied invariants such as the metric dimension [11, 13]. The fixing number has also been studied by Boutin [2], who investigated fixing numbers of Kneser Graphs, and by Gibbons and Laison [7]. A similar approach, in which we color a graph in such a way that there are no nontrivial color-preserving automorphisms, has been studied by Albertson and Collins [1] and others. We are interested in two questions. Given a graph G with automorphism group isomorphic to some group Γ, what does Γ tell us about the fixing number of G? And, given a group Γ and a number k, is it possible to construct a graph with automorphism group Γ and fixing number k? In this paper we consider the case where Γ is abelian.

2

Definitions and preliminary results

¯ and hence fix(G) = fix(G). ¯ Furthermore, if We note that for every graph G, Γ(G) = Γ(G) ¯ G is disconnected, then G is connected. From this we can see that whenever there exists a graph with automorphism group Γ and fixing number k, then there exists a connected graph with automorphism group Γ and fixing number k. In this paper, most of the graphs we construct are disconnected but, in light of this observation, all of our results hold for connected graphs. If Γ is a group, then a chain of subgroups is a sequence of subgroups Γ = Γ0 > Γ1 > · · · > Γn = {1}, where by > we mean that Γi is a proper subgroup of Γi−1 . A discussion of chains can be found in [3]. The number n is called the length of the chain. Let S = {v1 , v2 , . . . , vfix(G) } be a minimum fixing set for a graph G with automorphism group Γ. Let Γ0 = Γ and Γi = Γ{v1 ,...,vi } . Then Γ = Γ0 > Γ1 > · · · > Γfix(G) = {1} is a chain of subgroups of Γ. Hence the fixing number of G is at most l(Γ), the length of the longest chain of subgroups of Γ. For a natural number n, we denote by π(n) the number of prime factors of n, including repetitions. Thus, for example, π(12) = 3. Clearly, l(Γ) ≤ π(|Γ|), so π(|Γ|) is an upper bound for fix(G). For finite abelian groups we have l(Γ) = π(|Γ|) although we show that this upper bound is not always attained. We can see that {v} is a fixing set of cardinality 1 for G if and only if Γv is trivial, i.e., by the Orbit Stabilizer Theorem (see, for example, [8]), v is in an orbit of length |Γ|. This observation prompted Sabidussi [12] to point out the following: Theorem 1. For every group Γ, there exists a graph G with automorphism group Γ and fixing number 1. Proof. For the group Γ with generating set Γ \ {1}, consider the construction given by Frucht for a graph G with automorphism group Γ (the construction is well known and can be found in many textbooks, e.g., [5]). Every vertex of G that corresponds to an element 2

of Γ is similar to every other such vertex; consequently, these vertices comprise an orbit of cardinality |Γ| and the result follows from the Orbit Stabilizer Theorem. Before we proceed further, we shall need some standard results about finite abelian groups (see, for example, [4]). Let Γ be a finite abelian group. The group Γ is isomorphic to a group of the form (Zpn1,1 × · · · × Zpn1,k1 ) × · · · × (Zpnt,1 × · · · × Zpnt,kt ), where p1 , . . . , pt 1

t

1

t

are distinct primes and ni,j ≥ ni,j+1 ; moreover, this decomposition is unique. We call the number of cyclic groups of prime power order in this decomposition the rank of Γ. What we are defining as the rank of a finite abelian group is not the same as the torsion free rank of an abelian group, which is 0 for every finite abelian group. We shall make use of the fact that every subgroup of Γ is isomorphic to a group of the form (Zpm1,1 × · · · × Zpm1 ,l1 ) × 1

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· · · × (Zpmt,1 × · · · × Zpmt,lt ) with mi,j ≥ mi,j+1 , li ≤ ki and mi,j ≤ ni,j . By counting the t t elements of order pi for each i, we obtain the following result: Theorem 2. If Γ is a finite abelian group of the form described previously, then Γ has a unique subgroup isomorphic to (Zp1 × · · · × Zp1 ) × · · · × (Zpt × · · · × Zpt ), where Zpi appears ki times.

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Cyclic groups

For a natural number n, we denote by π ? (n) the number of distinct prime factors of n (e.g., π ? (12) = 2). Proposition 1. Let G be a finite graph with non-trivial cyclic automorphism group Γ. Then 1 ≤ fix(G) ≤ π ? (|Γ|). Proof. Let |Γ| = pe11 pe22 · · · pekk , where p1 , p2 , . . . , pk are distinct primes, e1 , e2 , . . . , ek are positive integers, and k = π ? (|Γ|). We claim that for each i there is an orbit Oi whose length is a multiple of pei i . Let gi be an element of Γ of order pei i . We can view this element as a permutation of the vertices of G. As Γ is the automorphism group of G, its action is faithful and the order of an element is the lowest common multiple of its cycle lengths. All cycles of gi have order a power of pi and there is at least one cycle of length pei i . If we let xi be a vertex in this cycle, then it is clear that the length of the orbit containing xi is a multiple of piei . We now claim that the set S = {x1 , x2 , . . . , xk } is a fixing set for G. Let Γxi be the stabilizer in Γ of xi . Since piei divides the length of Oi and |Γ| = |Oxi | · |Γxi |, we can see that pi does not divide |Γxi |. Now ΓS , the pointwise stabilizer of S, is the intersection of all the Γxi . If g ∈ ΓS , then, for all i, g ∈ Γxi and hence pi does not divide the order of g. It follows that ΓS is trivial and S is a fixing set of cardinality π ? (|Γ|) for G.

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The following is a converse to Proposition 1: Proposition 2. Let Γ be a cyclic group. For all l ∈ {1, . . . , π ? (|Γ|)} there exists a graph G with automorphism group Γ and fixing number l. Proof. Let |Γ| = pe11 pe22 · · · pekk , where p1 , p2 , . . . , pk are distinct primes, e1 , e2 , . . . , ek are positive integers, and k = π ? (|Γ|). By Theorem 1, there exists a graph G1 with automorphism group Zpel × Zpe2 × . . . × Zpek−l+1 and fixing number 1. Again by Theorem 1, for each l

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k−l+1

i ∈ {2, . . . , l} there is a graph Gi with automorphism group Zpek−l+i and fixing number 1. k−l+i S Let G = li=1 Gi . We can assume that the connected components of G are the graphs Gi , and that no connectedQcomponent is isomorphic to another. Then the automorphism group of G is isomorphic to li=1 Γ(Gi ), which is in turn isomorphic to Γ. A fixing set for G must consist of the union of fixing sets of each of its connected components. Since there are l connected components, each having fixing number 1, the fixing number of G must be l. We can combine these two propositions to obtain the following theorem: Theorem 3. Given a finite cyclic group Γ, there exists a graph with fixing number l and automorphism group Γ if and only if 1 ≤ l ≤ π ? (|Γ|). In the sequel we prove a generalization of this result for abelian groups.

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Direct products

We begin by proving a result about direct products of groups that we shall need later. Proposition 3. If G1 and G2 are graphs with automorphism groups Γ1 and Γ2 and fixing numbers k1 and k2 , respectively, then there exists a graph H with automorphism group Γ1 × Γ2 and fixing number k1 + k2 . Proof. If G1 is trivial, then we let H = G2 and the result follows. Hence we can assume that neither graph is trivial. By the remarks in Section 1 we can also assume that both graphs are connected. If G1 and G2 are not isomorphic, let H = G1 ∪ G2 . The graph H has automorphism group Γ1 × Γ2 and fixing number k1 + k2 . If G1 ∼ = G2 , then for each vertex v ∈ G1 we add a vertex xv and join v and xv to form the graph G01 . Since G1 is non-trivial and connected, every vertex in G01 that was originally in G1 has degree at least 2 in the new graph G01 , but the vertices xv that were added to G1 to form G01 have degree 1. Hence any automorphism of G01 must map vertices originally in G1 to other vertices originally in G1 and therefore the automorphism group of G01 is Γ1 . Also it 4

is clear that G1 and G01 have the same fixing number. Since G01 and G2 are not isomorphic we are in the previous case and the result holds. By induction we have the following corollary: Corollary 1. If G1 , . . . , Gn are graphs with automorphism groups Γ1 , . . . , Γn and fixing numbers k1 , . . . , kn , respectively, then there exists a graph H with automorphism group Γ1 × · · · × Γn and fixing number k1 + · · · + kn . We can now generalize to obtain the following theorem about direct products of groups. Theorem 4. Let G1 , . . . , Gn be graphs with automorphism groups Γ1 , . . . , Γn and fixing numbers k1 , . . . , kn , respectively. If (a) k = 1, or, (b) k = k1 + · · · + kn , or, P (c) k = 1 + i∈I ki , where I is any nonempty proper subset of {1, . . . , n}, then there exists a graph with automorphism group Γ1 × · · · × Γn and fixing number k. Proof. The cases k = 1 and k = k1 + · · · + kn are dealt with by Theorem 1 and Corollary 1, respectively. For the remaining case, we assume without loss of generality that I = {1, . . . , l}. We can find a graph G0 with automorphism group Γl+1 × · · · × Γn and fixing number 1 by Theorem 1. We now apply Corollary 1 to the graphs G1 , . . . , Gl , G0 .

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Abelian groups

We are now ready to state our main result: Theorem 5. Let Γ be a finite abelian group of rank r. Then there exists a graph with automorphism group Γ and fixing number k if and only if 1 ≤ k ≤ r. Proof. Let k ∈ {1, . . . , r}. Since Γ is the direct product of r groups, and since for each group there exists a graph with fixing number 1, it follows by Theorem 4 that there is a graph with automorphism group Γ and fixing number k. We now show that every graph with automorphism group Γ has a fixing set of cardinality at most r. Let Γ ∼ = (Zpn1,1 × · · · × Zpn1,k1 ) × · · · × (Zpnt,1 × · · · × Zpnt,kt ), with ni,j ≥ ni,j+1 1

t

1

5

t

and r = k1 + · · · + kt , be the automorphism group of a graph G. Let α be an element of Γ of maximal prime power order, say pn say. As we have seen in the proof of Proposition 1, when we view α as a permutation of the vertices of G it must have a cycle of length pn . Let v be a vertex on this cycle. By construction Γv has trivial intersection with hαi. As Γv is a subgroup of Γ, it is isomorphic to a group of the form (Zpm1,1 ×· · ·×Zpm1,l1 )×· · ·×(Zpmt,1 ×· · ·×Zpmt,lt ) 1

1

t

t

with mi,j ≥ mi,j+1 , li ≤ ki and mi,j ≤ ni,j (see, for example, [4], Chapter 4, Item 25). Hence the rank of Γv is at most r. If the rank of Γv is r then it has a subgroup isomorphic to (Zp1 × · · · × Zp1 ) × · · · × (Zpt × · · · × Zpt ), with Zpi appearing ki times in the product. But, by Theorem 2, Γ has only one subgroup of this isomorphism type and this has non-trivial intersection with hαi and so cannot be contained in Γv . Hence the rank of Γv is strictly less than that of Γ. We can continue this process by selecting an element of maximal prime power order in Γv and finding a vertex u such that Γ{v,u} has rank strictly smaller than the rank of Γv . Since Γ has finite rank r, this process will terminate in a set of cardinality at most r whose pointwise stabilizer has rank 0, i.e., is trivial. Hence we have found a fixing set of cardinality at most the rank of Γ. The following is a simple corollary, but it is worth stating explicitly. Corollary 2. If G is a finite graph with abelian automorphism group Γ, then the fixing number of G is less than or equal to the rank of Γ.

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[9] F. Harary, Survey of methods of automorphism destruction in graphs, Invited address, 8th Quadrennial International Conference on Graph Theory, Combinatorics, Algorithms and Applications. Kalamazoo, Michigan, 1996. [10]

, Methods of destroying the symmetries of a graph, Bull. Malays. Math. Sci. Soc. (2) 24 (2001), no. 2, 183–191 (2002).

[11] F. Harary and R.A. Melter, On the metric dimension of a graph, Ars Combin. 2 (1976), 191–195. [12] G. Sabidussi, private communication, 2004. [13] P.J. Slater, Leaves of trees, Congr. Numer. 14 (1975), 549–559.

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