Shifts in Cayley graphs

Discrete Mathematics 309 (2009) 3748–3756

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Discrete Mathematics

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Shifts in Cayley graphsI

Gabriel VerretInstitute of Mathematics, Physics and Mechanics, Jadranska 19, 1000 Ljubljana, Slovenia

a r t i c l e i n f o

Article history:Received 22 July 2007Received in revised form 5 October 2008Accepted 7 October 2008Available online 2 December 2008

Keywords:Cayley graphShiftMobility

a b s t r a c t

An automorphism of an undirected simple graph is called a shift if it maps every vertex toan adjacent one. For all finite groups G, we determine the minimum nonzero valency of aCayley graph on G that does not admit a shift. We also get a classification of groups withall involutions central and such that for every pair a, b of elements of the group, one ofab = ba, aba−1 = b−1, bab−1 = a−1 or a2 = b±2 holds.

© 2008 Elsevier B.V. All rights reserved.

1. Introduction

An automorphism of a simple undirected graph is called a shift if it maps every vertex to an adjacent one. Shifts and otherrelated notions have been studied by several authors: Larose, Laviolette and Tardiff [7] studied shifts in Cartesian productsof graphs. It is known [3] that compact regular graphs admit shifts. Shifts are an example of shunts, that is, automorphismsthat map an arc to an adjacent arc. Shunts play a central role in the study of arc-transitive graphs [2,10]. Shifts are also aspecial case of adjacency automorphisms [8,11], that is, automorphisms that map each vertex to itself or a neighbor.Further, shifts are closely related to a notion ofmobility in graphs. Themobility of an automorphism σ ∈ Aut(Γ ), denoted

bymob(σ ), was defined in [9] to bemindist(v, σ (v)) : v ∈ V(Γ ). (Ridder and Bodlaender [4] use the term k-distance lowerbounded for an automorphism ofmobility k.) An automorphism σ ofΓ hasmobility greater than 1 if and only if itmaps everyvertex of Γ to a nonadjacent vertex. In other words, if and only if it is a shift of Γ c .This paper deals with the question of existence of shifts in Cayley graphs. A nonempty, inverse-closed and identity-free

subset S of a finite groupG is called a connection set. The Cayley graphΓ = Cay(G, S) onGwith connection set S has V(Γ ) = Gand u∼Γ v if and only if v−1u ∈ S. Note that, because of the restrictions imposed above on connection sets, all Cayley graphsin this paper will be nonedgeless, undirected and loopless, but not necessarily connected.A connection set S of a group G such that Cay(G, S) admits a shift will also be said to admit a shift. Note that if σ is a

shift of Cay(H, S) and H is a subgroup of G, then Cay(G, S) is a disjoint union of copies of Cay(H, S) and σ is also a shift ofCay(G, S), hence there is no ambiguity.Theminimal shiftless valency of a group G is the minimum size of a connection set for G that does not admit a shift. It will

be denoted by msv(G). We write msv(G) = ∞ if all connection sets of G admit shifts. Note that a Cayley graph of valency 1or 2 always admits a shift, and hence for any group G, msv(G) ≥ 3. We now discuss some groups that are needed to statethe main result of this paper, which is the classification of finite groups according to minimal shiftless valency.The notation Cn and Dn will be used for the cyclic and dihedral group of order n, respectively, while G32 = 〈a, b, c|a4 =

b4 = c4 = 1, b2 = a2c2, ab = ba, cac−1 = a−1, bcb−1 = c−1〉 denotes the unique nonmetacyclic group of order 32 withthe property that all of its proper subgroups are metacyclic. There are two more groups that will be needed to state some

I Some of this work was part of the author’s Master’s Thesis written under the supervision of Mateja Šajna at the University of Ottawa.E-mail address: [email protected].

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results: G64,1 denotes the Sylow 2-subgroup of SU(3, 4)which has order 64, while G64,2 = 〈a, b, c, d|a4 = b4 = c4 = d4 =1, a2b2 = c2 = d2, aba−1 = b−1, cdc−1 = d−1, ac = ca, ad = da, bc = cb, bd = db〉 has order 64 and is a central productof Q and C4 o C4.A generalized dicyclic group is an extension by C2 of an even-order abelian group A. Here is how it is constructed: let y be

an element of order 2 in A. Wewill denote by Dic(A, y) the group generated by A and an element x (not in A) such that x2 = yand xax−1 = a−1 for all a ∈ A. Clearly A has index 2 in Dic(A, y). Another easily proved fact that will be used is that thedirect product of a generalized dicyclic group with C2 is also a generalized dicyclic group. In the special case that A is cyclicof order 2n, there is a unique choice for y and the resulting group is said to be dicyclic and denoted by Dic4n. For example,the quaternion group Q = Dic8 is dicyclic.The main result of this paper is the following theorem.

Theorem 1. If G is a finite group, thenmsv(G) ∈ 3, 4, 6,∞. In fact,(1) msv(G) = ∞ if and only if G is abelian or isomorphic to one of D6, D8, D10, Dic12, Dic16, Dic20, Q× C2n or (C4 o C4)× C2nfor some n ≥ 0.

(2) msv(G) = 6 if and only if G is generalized dicyclic but not listed above or is isomorphic to H × C2n for some n ≥ 0, where His isomorphic to one of Q× C3, Q× C4, G32, Q2, G64,1 or G64,2.

(3) msv(G) = 4 if and only if G is not in one of the above families and all involutions of G are central.(4) msv(G) = 3 if and only if G has a noncentral involution and G is not isomorphic to one of D6, D8 or D10.As part of the proof of this result, we also get a classification of a family of groups which we now define.

Definition 2. A groupGwith all involutions central will be said to be of type 1 if, for all a, b ∈ G, one of ab = ba, aba−1 = b−1,bab−1 = a−1 or a2 = b±2 holds.

Theorem 3. A group is of type 1 if and only if it is abelian, generalized dicyclic or isomorphic to H × C2n for some n ≥ 0, whereH is isomorphic to one of Q× C3, Q× C4, G32, Q2, G64,1, or G64,2.The paper is divided into twomain parts. In Section 2, we prove Theorem 1 by assuming Theorem 3which, being a purely

group theoretical result, will be proven in Section 3.

2. Proof of Theorem 1

In Section 2.1, we consider some necessary conditions for a graph to admit a shift and apply them to Cayley graphs.In Sections 2.2 and 2.3 we use these conditions to characterize groups with minimal shiftless valency equal to 3 and 4,respectively. Theorem 3 will allow us to deal with the remaining groups, that is, groups with minimal shiftless valencystrictly greater than 4.

2.1. Conditions for existence of shifts

Here is an easy sufficient condition for a Cayley graph to admit a shift.

Lemma 4. Let Γ = Cay(G, S) be a Cayley graph on a group G. For each g ∈ G, let σg : G → G be the automorphism of Γdefined as σg(v) = gv. Then σg is a shift if and only if S contains the entire conjugacy class of g. In particular, if S contains anelement that is central in G then Γ admits a shift.The proof is straightforward. It turns out that a graph admitting a shift must have certain 4-cycles.

Lemma 5. Let Γ be a graph admitting a shift σ , let v ∈ V(Γ ) be a vertex of degree d, let Ev be the set of edges of Γ incident to vand let E ′v be the subset of Ev consisting of the edges vσ(v) and vσ

−1(v). Then, any two edges e and e′ such that e ∈ Ev \ E ′v ande′ ∈ E ′v belong to a common 4-cycle. In particular, there is an edge incident to v that is contained in at least d−2 distinct 4-cycles.Moreover, if d ≥ 3, then every edge incident with v is an edge of some 4-cycle.Proof. See Fig. 1.Let N′(v) be the subset consisting of neighbors u of v such that uv is an edge of Ev \ E ′v . For any u ∈ N

′(v), let Cu be the4-cycle (u, v, σ (v), σ (u)) and let C ′u be the 4-cycle (u, v, σ

−1(v), σ−1(u)). These both contain the edge e = uv and one ofthe two edges of E ′v , thereby proving our first claim. As |N

′(v)| ≥ d − 2, and Cu1 6= Cu2 for distinct u1 and u2 in N′(v), the

edge vσ(v) lies in at least d− 2 different 4-cycles. The last claim follows easily.

Part of this result will be used many times; hence the following definition will be useful.

Definition 6. A graph having the property that for each vertex v of degree 3 ormore, every edge incidentwith v is containedin a 4-cycle, and at least one of these edges is contained in at least deg(v) − 2 distinct 4-cycles, will be said to be 4-cyclerich. Furthermore, a subset S of a group G such that S is a connection set and Cay(G, S) is 4-cycle rich will also be said to be4-cycle rich.Note that, by Lemma 5, any Cayley graph Cay(G, S) of valency |S| ≥ 3 admitting a shift is 4-cycle rich. Note also that if H

is a subgroup of G containing S, then Cay(H, S) is 4-cycle rich if and only if Cay(G, S) is 4-cycle rich, hence there is no needto mention the full group G in the above definition of a 4-cycle rich connection set S.

3750 G. Verret / Discrete Mathematics 309 (2009) 3748–3756

Fig. 1. The proof of Lemma 5.

Fig. 2. The 4-cycle C from the proof of Lemma 7.

Fig. 3. The proof of Lemma 8.

Lemma 7. Let S be a 4-cycle rich subset of a group G such that |S| ≥ 3. Then the following hold:(1) For each g ∈ S, there exists a triple (h, k, l) ∈ S3 such that gh = kl 6= 1 with g 6= k and h 6= l.(2) For some g ∈ S, there exist |S| − 2 distinct triples satisfying the above conditions.

Proof. See Fig. 2.Let g ∈ S. By definition, the edge 1g is contained in a 4-cycle C , say 1gxk. In particular, x 6= 1 and, by definition, there

exist h, l ∈ S such that gh = x = kl, and since g 6= kwe have h 6= l. The second part of the lemma follows easily.

2.2. 4-cycle rich 3-subsets

In this subsection, we apply the last result to 3-subsets and use this to characterize groupswithminimal shiftless valencyequal to 3.

Lemma 8. Let G be a group and let S be a 4-cycle rich 3-subset of G. Then S contains an involution that either commutes withthe other two elements of S or conjugates one of them to the other.

Proof. See Fig. 3.By Lemma 7, for each g ∈ S, there exists a triple (h, k, l) ∈ S3 such that gh = kl 6= 1, g 6= k and h 6= l.If S = a, a−1, b consists of one involution b and an inverse pair a, a−1 of non-involutions, then setting g = b yields

h 6= b 6= k, so h, k ∈ 〈a〉. If l 6= b then b ∈ 〈a〉 hence a and b commute. If l = b, then bab = a±1.


If S = a, b, c consists of 3 distinct involutions, then suppose that bab 6= c , aba 6= c and cac 6= b. For g = a, the onlypossibilities left are (h, k, l) = (b, b, a) or (h, k, l) = (c, c, a) which imply ab = ba or ac = ca. Similarly, g = b impliesba = ab or bc = cb, and g = c implies ca = ac or cb = bc. Combining these cases shows that at least one element of Scommutes with the other two unless aba = c , bab = c or cac = b.

This result allows us to determine the structure of subgroups generated by involutions in groups with 4-cycle rich 3-subsets.

Corollary 9. Let G be a group all of whose 3-subsets are 4-cycle rich. Then any subgroup H generated by involutions in G isisomorphic to one of D6, D8, D10 or C2n for some n ≥ 1.Proof. Let S be a minimal set such that S generates H and S contains only involutions. Let |S| = d. If d = 1, then H ∼= C2.If d = 2, then H is generated by two involutions σ and ρ. Let m = |σρ|. We have that H ∼= Dm is dihedral of order 2m (weallow the degenerate case D2 ∼= C22). Lemma 8 shows that σ , ρ, ρ(σρ)2 is not a 4-cycle rich subset if m ≥ 6, and hencewe must havem < 6.If d = 3, let S = a, b, c. Byminimality, no two of these involutions can generate a subgroup containing the third, hence,

by Lemma 8, one of the elements (say c) must commute with the other two. Since c 6∈ 〈a, b〉, we have H = 〈a, b〉 × 〈c〉. Bythe above, 〈a, b〉 is isomorphic to one of D6, D8, D10 or C22, and hence H is isomorphic to one of D6 × C2 ∼= D12, D8 × C2,D10 × C2 ∼= D20 or C23. We have just seen that dihedral groups of order at least 12 contain 3-subsets which are not 4-cyclerich. Similarly, if H ∼= D8 × C2 with (ab)4 = 1, then, by Lemma 8, the subset b, a, ac is not 4-cycle rich. We conclude thatH ∼= C23.If d > 3, let S ′ = a, b, c ⊆ S be a 3-subset of S. The group 〈S ′〉 cannot be generated by a set S ′′ consisting of two or fewer

involutions, or else (S \ S ′)∪ S ′′ would generate H , contradictingminimality. This means we can apply the case d = 3 to 〈S ′〉,hence 〈S ′〉 ∼= C23 and a, b, c commute. Since the choice of a, b, c was arbitrary, this shows that all elements of S commute,and H ∼= C2n for some n.

We can now fully characterize groups in which all 3-subsets are 4-cycle rich.

Theorem 10. Let G be a group. Then the following are equivalent:(1) All involutions of G are central or G is isomorphic to one of D6, D8 or D10.(2) msv(G) > 3, that is, each cubic Cayley graph on G admits a shift.(3) All 3-subsets of G are 4-cycle rich.Proof. We first show that if all involutions of G are central or G is isomorphic to one of D6, D8 or D10, then each cubic Cayleygraph on G admits a shift. If G is isomorphic to one of D6, D8 or D10, then this can be checked by hand, while if all involutionsof G are central, then Lemma 4 allows us to reach the same conclusion since any connection set of size 3 must contain aninvolution.Next, by Lemma 5, msv(G) > 3 implies that all 3-subsets of G are 4-cycle rich.Finally, we assume that all 3-subsets of G are 4-cycle rich, and prove that either all involutions of G are central or G is

isomorphic to one of D6, D8 or D10. Consider the subgroup H of G generated by all involutions of G. By Corollary 9, H isisomorphic to one of D6, D8, D10 or C2n for some n ≥ 1. Now we show that for all b ∈ H and a 6∈ H we have ab = ba.Let S = a, a−1, b. By Lemma 8, either ab = ba or bab = a−1, but the latter implies (ba)2 = 1, hence ba ∈ H whichis a contradiction. We conclude that a commutes with all involutions in H and hence commutes with all elements of H . IfH ∼= C2n, then, again by the above remark, H is in the center and all involutions are central. If H is dihedral, suppose G 6= Hand let a ∈ G \ H . Take two noncommuting involutions b and c in H . By the above remark, b commutes with both a andabc , and hence commutes with a−1abc = bc and hence with c , which is a contradiction. Therefore, we have G = H and G isisomorphic to one of D6, D8 or D10.

2.3. 4-subsets

Theorem10 shows that, apart from small exceptions, msv(G) = 3 for every finite groupG having a noncentral involution.We now turn our attention to groups where all involutions are central and determine when these have the property that alltheir 4-subsets are 4-cycle rich. A 4-subset of G containing a central involution is 4-cycle rich, hence we need only consider4-subsets that do not contain involutions.

Lemma 11. Let G be a groupwith all involutions central and let a, a−1, b, b−1 be a 4-cycle rich subset of non-involutions. Then,one of the following holds: a2 = b±2, ab = ba, aba−1 = b−1 or bab−1 = a−1.Proof. Let S = a, a−1, b, b−1 and suppose that a2 6= b±2, ab 6= ba, aba−1 6= b−1 and bab−1 6= a−1. Without loss ofgenerality, we assume that the element g ∈ S guaranteed by part (2) of Lemma 7 is a, hence there exist 2 distinct triples(h1, k1, l1), (h2, k2, l2) ∈ S3 such that ahi = kili 6= 1, a 6= ki and hi 6= li for i ∈ 1, 2. All but one of these possibilities canbe eliminated. For example, if ab = b−1a−1, then ab is an involution and is central, which implies that a and b commute.Similarly, ab−1 = ba−1 also implies ab = ba. The only possibility left is (h1, k1, l1) = (h2, k2, l2) = (a, a−1, a−1), but thenthe triples are not distinct, which is a contradiction.

We now define two standard notions that will be needed. The lexicographic product Γ ∆ of graphs Γ and∆ is definedon V(Γ ∆) = V(Γ ) × V(∆) with vertices (u, v) and (u′, v′) adjacent if either uu′ ∈ E(Γ ) or u = u′ and vv′ ∈ E(∆).


Clearly, if Γ admits a shift, then Γ ∆ also does. Let Γ = Cay(G, S) and let H be a normal subgroup of a group G such thatS ∩ H = ∅. The quotient Cayley graph Γ /H is defined as Γ /H = Cay(G/H, S/H), where S/H = Hs : s ∈ S ⊂ G/H .

Lemma 12. Let S be a connection set of a group G, and let g ∈ G be such that s2 = g for all s ∈ S and 〈g〉 is normal in G. ThenΓ = Cay(G, S) admits a shift if and only if Γ /〈g〉 does.

Proof. Note that for any s ∈ S, s2 = s−2 = g , hence s4 = 1 and g2 = 1. If g ∈ S, then g2 = g = 1, contradicting the fact thatS is identity-free. It is easy to check that Γ ∼= Γ /〈g〉 K2 and hence, if Γ /〈g〉 admits a shift, then Γ also does. Now supposeΓ admits a shift σ . Define a relation ≡ on V(Γ ): v ≡ u if and only if N(u) = N(v), that is, u and v have exactly the sameneighbors. Note that ≡ is an equivalence relation and is invariant under Aut(Γ ). Define a graph Γ ′ with the equivalenceclasses of ≡ as vertices and U ∼Γ ′ V if and only if u∼Γ v for some u ∈ U, v ∈ V . It is easy to see that this is well-definedand, since N(u) = N(v) if and only if N(σ (u)) = N(σ (v)), σ induces a shift on Γ ′. Note that, since Γ is loopless, u ≡ vimplies that uΓ v. But x ≡ xg for all x hence Γ /〈g〉 ∼= Γ ′ Km for somem and therefore Γ ′ admits a shift.

Using this, we can characterize groups with minimal shiftless valency more than 4.

Theorem 13. Let G be a group with all involutions central. Then the following are equivalent:(1) msv(G) > 4.(2) All 4-subsets of G are 4-cycle rich.(3) G is of type 1.

Proof. If msv(G) > 4, then each 4-regular Cayley graph on G admits a shift and hence, by Lemma 5, all 4-subsets of G are4-cycle rich.If all 4-subsets of G are 4-cycle rich, then, by Lemma 11 and Definition 2, G is of type 1.Finally, suppose that G is of type 1. We must show that msv(G) > 4, that is each cubic and each quartic Cayley graph

on G admits a shift. Let Γ = Cay(G, S), with |S| ∈ 3, 4. If S contains an involution z (in particular if |S| = 3), then z iscentral and, by Lemma 4, Γ admits a shift. Otherwise, |S| = 4, S = a, a−1, b, b−1 and one of the following holds: a2 = b±2,ab = ba, aba−1 = b−1 or bab−1 = a−1. In the last three cases, Lemma 4 again implies that Γ admits a shift. If a2 = b±2 = g ,then Γ /〈g〉 is 2-regular hence admits a shift. By Lemma 12, Γ also does.

Theorem 13 leads us to the question of classifying groups of type 1, or more specifically, to proving Theorem 3. We willdelay this proof to the next section but assume the result in the meantime. Under this assumption, we now determine theminimal shiftless valency of all remaining groups.

Theorem 14. Let G be a group such that msv(G) > 4. Thenmsv(G) ∈ 6,∞, and(1) msv(G) = ∞ if and only if G is abelian or isomorphic to one of D6, D8, D10, Dic12, Dic16, Dic20, Q× C2n or (C4 o C4)× C2nfor some n ≥ 0.

(2) msv(G) = 6 if and only if G is generalized dicyclic but not listed above or is isomorphic to H × C2n for some n ≥ 0, where His isomorphic to one of Q× C3, Q× C4, G32, Q2, G64,1 or G64,2.

Proof. If G is abelian, then msv(G) = ∞ by Lemma 4. By Theorem 10, either G is isomorphic to one of D6, D8 or D10, orotherwise all involutions of G are central. If G is isomorphic to one of D6, D8 or D10, then it is easy to check that all Cayleygraphs on G are isomorphic to Cayley graphs on abelian groups, hence msv(G) = ∞. If all involutions of G are central, thenLemma 4 implies that each 5-regular Cayley graph on G admits a shift, hence msv(G) > 5. Moreover, Theorem 13 impliesthat G is of type 1, and hence, by Theorem 3 (which is proved in Section 3), G is abelian, generalized dicyclic or is isomorphicto H × C2n for some n ≥ 0, where H is isomorphic to one of Q × C3, Q × C4, G32, Q2, G64,1 or G64,2. It is easy to check thatG64,1 and G64,2 both have a subgroup isomorphic to G32, while Q2 clearly has a subgroup isomorphic to Q× C4. Therefore, Gis either abelian or generalized dicyclic or has a subgroup isomorphic to one of Q× C3, Q× C4, G32.IfGhas a subgroupH isomorphic toQ×C3, with presentation 〈i, j, a|i4 = j4 = a3 = 1, i2 = j2, iji = j, ai = ia, aj = ja〉, let

k = ij, let S = ia, (ia)−1, ja, (ja)−1, ka, (ka)−1 and let Γ = Cay(H, S). It is not hard to check that each edge incident withthe identity vertex lies in at most one 4-cycle. By Lemma 5, this implies that Γ does not admit a shift, hence msv(G) = 6.IfGhas a subgroupH isomorphic toQ×C4, with presentation 〈i, j, a|i4 = j4 = a4 = 1, i2 = j2, iji = j, ai = ia, aj = ja〉, let

k = ij and let S = ia, (ia)−1, ja, (ja)−1, ka, (ka)−1. By Lemma 12,Γ = Cay(H, S) admits a shift if and only ifΓ ′ = Γ /〈i2a2〉does, but a simple calculation shows that Γ ′ has no 4-cycles. Since Γ ′ is 3-regular, Lemma 5 implies that Γ ′ does not admita shift and hence neither does Γ . Therefore Γ is a 6-regular Cayley graph on Gwithout a shift, hence msv(G) = 6.If G has a subgroup H isomorphic to G32, with presentation 〈a, b, c|a4 = b4 = c4 = 1, b2 = a2c2, ab = ba, cac−1 =

a−1, bcb−1 = c−1〉, let S = b, b−1, bc, (bc)−1, abc, (abc)−1 and apply the above argument to Γ = Cay(H, S) to concludethat msv(G) = 6.Finally, let G = Dic(A, y) be a generalized dicyclic group, and let Γ = Cay(G, S) a Cayley graph on G that does not admit

a shift. By Lemma 4, we have A ∩ S = ∅. In particular, s2 = y 6∈ S for all s ∈ S hence, by Lemma 12, Γ admits a shift if andonly if Γ /〈y〉 does. This argument shows that msv(G) = 2 · msv(G/〈y〉), where 2 · ∞ = ∞. There are now three cases toconsider.(1) If A/〈y〉 ∼= C2n for some n ≥ 1, then one of the following happens:(a) A ∼= C2n+1 and G is abelian,


(b) A ∼= C4 × C2n−1 and G ∼= Q× C2n−1, or(c) A ∼= C4 × C2n−1 and G ∼= (C4 o C4)× C2n−2.In each of these three sub-cases, G/〈y〉 is abelian, hence msv(G) = msv(G/〈y〉) = ∞.

(2) If A/〈y〉 ∼= Cm withm ∈ 3, 4, 5, then either(a) A ∼= C2m and G ∼= Dic4m,(b) A ∼= C4 × C2 and G ∼= Q× C2, or(c) A ∼= C4 × C2 and C4 o C4.In this case, G/〈y〉 is dihedral of order at most 10, hence msv(G) = msv(G/〈y〉) = ∞.

(3) Finally, if A/〈y〉 C2n, C3, C4, C5, then G/〈y〉 contains a noncentral involution and is not dihedral of order at most 10,hence msv(G) = 2 ·msv(G/〈y〉) = 6 by Theorem 10.

Theorem 14, together with Theorem 10, concludes the proof of Theorem 1. In the next section, we prove Theorem 3,which we skipped earlier.

3. Proof of Theorem 3

Recall that a group of type 1 is a group whose involutions are all central and such that, for all its elements a, b one of thefollowing holds: ab = ba, aba−1 = b−1, bab−1 = a−1, a2 = b2 or a2 = b−2. We now classify these groups. This proof isdivided into three main steps. The first step is to prove a theorem (Corollary 17), which tells us that groups of type 1, exceptfor one exceptional case, must follow even stricter conditions. These conditions motivate the definition of a smaller class ofgroups (of type 1A), which we investigate in Sections 3.1 and 3.2.It is possible to slightly refine the type 1 equations, as the next result shows.

Lemma 15. Let G be a group of type 1. Then, for all a, b ∈ G, one of the following holds:(1) ab = ba,(2) aba−1 = b−1,(3) bab−1 = a−1,(4) a2 = b2 = (ab)−2,(5) a2 = b−2 = (ab−1)−2,(6) a2 = b2 and a4 = b4 = 1.

Proof. Let a, b ∈ G, and suppose that none of (1), (2), (3) holds. Applying the type 1 conditions to a and b implies thata2 = b±2. If a2 6= b2, then a2 = b−2 and applying the type 1 conditions to a and ab−1 implies a2 = (ab−1)−2. If a2 6= b−2,then a2 = b2 and applying the type 1 conditions to a and ab implies a2 = (ab)−2. Finally, if a2 = b2 and a2 = b−2, then (6)occurs.

Apart from a special case, we can eliminate some of the equations, hence we introduce the following definition.

Definition 16. A group G with all involutions central will be said to be of type 1A if, for all a, b ∈ G, one of the followingholds:(1) ab = ba,(2) aba−1 = b−1,(3) bab−1 = a−1,(4) a2 = b2 and a4 = b4 = 1.

Corollary 17. Let G be a group of type 1. Then G is either of type 1A or isomorphic to Q× C3 × C2n for some n ≥ 0.Proof. Suppose that there exist a, b ∈ G such that none of (1), (2), (3) and (6) of Lemma 15 holds. It follows that one of(4) or (5) holds. Moreover, by exchanging the role of b and b−1, we may assume that (4) holds, that is a2 = b2 = (ab)−2.We will show that G ∼= Q × C3 × C2n in this case. First, we show that H = 〈a, b〉 ∼= Q × C3. Write g = ab. We have thatb2 = a2 = g−2 and aga−1 = a2ba−1 = a4g−1. Since g2 = a−2 commutes with a, this implies a−2 = g2 = ag2a−1 =(aga−1)2 = (a4g−1)2 = a8g−2 = a10 and we get a12 = 1. Note that a6 = 1 implies aga−1 = a−2g−1 = g , while a4 = 1implies aga−1 = g−1, which both contradict our assumptions, hence |a| = 12 and |H| = 2|a| = 24.Now, let i = a9, j = g3 and z = a4, which have orders 4, 4 and 3 respectively, and generate H . Also, note that i2 = j2,

iji = j, and that z commutes with both i and j. Since |H| = 24, we conclude that H = 〈i, j, z〉 ∼= Q× C3.We will now show that G ∼= Q×C3×C2n for some n. If G = H , this is trivial, otherwise take c ∈ G \H . We will first show

that c2 ∈ H = 〈a, b〉. We apply the type 1 conditions to a and c and to b and c.If c2 = a±2 = b±2 (recall that a2 = b2), then clearly c2 ∈ H . If bcb−1 = c−1 or aca−1 = c−1, applying the type 1

conditions to b and cb2 or a and ca2, respectively, implies c2 ∈ H . Finally, if cac−1 = ax and cbc−1 = by with x, y ∈ −1, 1,then applying the type 1 conditions to b and ac implies c2 ∈ H .We now show that H is normal in H ′ = 〈a, b, c〉. Note that aca−1 = c−1 implies cac−1 = c2awhich is in H since c2 ∈ H .

If a2 = c2 = (ac)−2, recall that a2 = (ab)−2 and hence (ac)2 = (ab)2 which implies cac−1 ∈ H . Thus, in all cases, cac−1 ∈ H .An analogous reasoning shows that cbc−1 ∈ H and hence H is normal in H ′. In particular, H ′ = H ∪ cH and H ′ has order 48.All that remains to show is that, in each case, there exists h ∈ H such that (ch)2 = 1. Since involutions are central, this

will suffice to show thatH ′ = H×〈ch〉. The necessary calculations are very tedious.We therefore propose an alternative.We


know that H ′ has order 48, contains a subgroup isomorphic to Q×C3, all involutions of H ′ are central and all 4-subsets of H ′are 4-cycle rich. An exhaustive search, easily performed by computer, reveals that the only possibility is thatH ′ ∼= Q×C3×C2and that H ′ = 〈H, c ′〉where c ′ is an involution, as needed.Write G = 〈H, c1, . . . , cn〉. We can repeat the argument above to show that G = 〈H, c ′1, . . . , c

′n〉where c

′

i is a involution.Since involutions are central in G, this implies G ∼= Q× C3 × C2n for some n ≥ 0.

Our next goal is to characterize groups of type 1A.

3.1. Groups of type 1A

We now prove two lemmas which allow us to prove that a group of type 1A and of exponent different than 4 is abelianor generalized dicyclic.

Lemma 18. Let G be a group of type 1A such that there exists a ∈ G with a4 6= 1. Then, for every b ∈ G, we have either ab = ba,or b4 = 1 and bab−1 = a−1.Proof. We first remark that for c, d ∈ G, dcd−1 = c−1 6= c implies d4 = 1 by applying the type 1A conditions to d and cd2.Now, suppose that ab 6= ba. Since involutions are central, it follows that b2 6= 1, and the type 1A conditions imply eitherbab−1 = a−1 6= a or aba−1 = b−1 6= b. In the first case, the previous remark ensures that b4 = 1, while in the second caseit ensures that a4 = 1, which is a contradiction.

Lemma 19. Let G be a group of type 1A such that there exist a, b, c ∈ G with b4 = 1 and a4 6= 1 6= c4. Then ac = ca, and eitherab = ba and bc = cb, or bab−1 = a−1 and bcb−1 = c−1.Proof. By Lemma 18, we know that ac = ca, and one of ab = ba or bab−1 = a−1 holds. For the same reason, one of bc = cbor bcb−1 = c−1 also holds. Assuming ab = ba and bcb−1 = c−1 and applying the type 1A conditions to b and ac yield acontradiction. Exchanging the role of a and c and repeating the argument again yield a contradiction. Hence, either ab = baand bc = cb, or bab−1 = a−1 and bcb−1 = c−1.

Theorem 20. Let G be a group of type 1A such that there exists a ∈ G with a4 6= 1. Then, G is abelian or generalized dicyclic.Proof. Let A be the centralizer of a in G and let b, c ∈ A. If b4 6= 1 6= c4, then bc = cb by Lemma 18. If b4 = 1 6= c4 orb4 6= 1 = c4, then bc = cb by Lemma 19. Finally, if b4 = 1 = c4, then note that (ac)4 = a4c4 = a4 6= 1 and we canconclude, by Lemma 19, that b(ac) = (ac)b. This implies again cb = bc and A is abelian in all cases.If G = A, then G is abelian. If G 6= A, let x ∈ G \ A. By Lemma 18, we know that xax−1 = a−1 and x4 = 1. In fact every

element of G either centralizes a or conjugates a to its inverse, so |G : A| = |G : CG(a)| = 2 and x2 ∈ A. Now, let c ∈ A. Ifc4 6= 1, then xcx−1 = c−1 by Lemma 19. If c4 = 1, then (ac)4 = a4c4 = a4 6= 1. We can then apply Lemma 19 to a, b and acto conclude that x(ac)x−1 = (ac)−1 which implies xcx−1 = c−1. Hence, we have G = 〈A, x〉 = Dic(A, x2).

This leaves us with the case where G has exponent 4.

3.2. Groups of type 1A and exponent 4

Every finite group of exponent 4 (indeed every finite 2-group) has a normal subgroup of index 2. The main idea is toprove that whenever G is a group of type 1A and exponent 4, if this subgroup is abelian or generalized dicyclic, then, apartfrom some exceptions which can be dealt with separately, the group itself must also be abelian or generalized dicyclic. First,we need a few intermediate results.

Lemma 21. Let G be a group of type 1A and exponent 4.(1) If g ∈ G and H ≤ G, then |〈H, g〉| ≤ 4|H|.(2) Let H be a subgroup of index 2 in G. If H = H ′ × E for some H ′ and E ∼= C2n for n ≥ 1, then there exist G′ and E ′ ∼= C2n−1such that G = G′ × E ′.

(3) All subgroups of G of order at most 256 are either abelian, generalized dicyclic or isomorphic to H × C2n for some n ≥ 0,where H is isomorphic to one of Q× C4, G32, Q2, G64,1 or G64,2.

(4) If A ∼= C43 is a subgroup of G and x ∈ G, then either x centralizes A or xax−1 = a−1 for all a ∈ A.Proof. (1) The type 1A conditions imply that g2 is a central involution and that 〈H, g2〉 is normal in 〈H, g〉 hence |〈H, g〉| =

|〈H,g2〉||g||〈H,g2〉∩〈g〉|

≤2|H|||g|2 ≤ 4|H|.

(2) Let x ∈ G \ H . We have G = 〈H, x〉. Since E is elementary abelian, there exists E ′ ≤ E such that E = E ′ × 〈e〉, E ′ ∼= C2n−1and E ′ ∩ 〈x〉 = 1. Let G′ = 〈H ′, e, x〉. It is easy to check that G = G′ × E ′.

(3) A computer algebra system with a library of small groups like GAP [5] can be used to construct all the 2-groups of type1A and exponent 4 of order at most 256 and verify that they are either abelian, generalized dicyclic or isomorphic to oneof the groups listed.

(4) Let B = 〈x, A〉. By part (1), |B| ≤ 256. It is clear that none of Q×C4, G32, Q2, G64,1 or G64,2 contains a subgroup isomorphicto C43 hence, by part (3), B is either abelian or generalized dicyclic. If B is abelian, then clearly x centralizes A. If B isgeneralized dicyclic, it is not hard to see that xax−1 = a−1 for all a ∈ A.


In the next two results, we use part (4) of Lemma 21 to study groups of type 1A and exponent 4 with a subgroupisomorphic to C43.

Lemma 22. Let G be a group of type 1A and exponent 4 with an abelian subgroup A such that |G : A| = 2 and A contains asubgroup isomorphic to C43. Then G is abelian or G = Dic(A, x2) for some x ∈ G \ A.

Proof. Let x ∈ G \ A. If x is an involution, then G is abelian (since all involutions are central), hence we assume |x| = 4.Let A = B × E = 〈a1〉 × · · · × 〈am〉 × E where m ≥ 3, |ai| = 4 and E is elementary abelian. For i ∈ 1, . . . ,m − 2, letAi = 〈ai, ai+1, ai+2〉 ∼= C43. By part (4) of Lemma21, x either centralizesAi or xax−1 = a−1 for all a ∈ Ai. For j ∈ 1, . . . ,m−3,we have aj+1 ∈ Aj ∩ Aj+1 hence, if x centralizes Aj, then it must centralize Aj+1. By induction, if x centralizes A1, it must alsocentralize Ai for all i ∈ 1, . . . ,m − 2. In this case, x centralizes B and it follows that G is abelian. An analogous reasoningshows that if xax−1 = a−1 for all a ∈ A1, then xbx−1 = b−1 for all b ∈ B in which case G = Dic(A, x2).

Corollary 23. Let G be a group of type 1A and exponent 4 with a generalized dicyclic subgroup H = Dic(A, h2) of index 2 in G.Suppose that A contains a subgroup isomorphic to C43. Then G contains an abelian subgroup A′ with A ≤ A′ and |G : A′| ≤ 2.

Proof. Let h ∈ H \ A, let g ∈ G \ H and let H ′ = 〈A, g〉. Since H is generalized dicyclic, all elements of H \ A have order 4and hence g2 6∈ H \ A. But g2 ∈ H hence g2 ∈ A and it follows that A≤2 H ′≤2 G. Note that A has index 2 in both H and H ′hence H ∩ H ′ = A. Since A is abelian, we can apply Lemma 22 to H ′.If H ′ is abelian, then take A′ = H ′. Otherwise, we have H ′ = Dic(A, g2). If gh ∈ A ⊆ H , then g ∈ H , since h ∈ H . This

contradicts the fact that g ∈ H ′ \ A therefore we conclude that gh 6∈ A. Notice that gh centralizes A from which it followsthat A′ = 〈A, gh〉 is an abelian subgroup of index at most 2 in G.

This allows us to complete our classification of groups of type 1A.

Theorem 24. A group of exponent 4 is of type 1A if and only if it is abelian, generalized dicyclic or isomorphic to H×C2n for somen ≥ 0, where H is isomorphic to one of Q× C4, G32, Q2, G64,1 or G64,2.

Proof. Let G be a group of exponent 4 and type 1A. Let E be the largest elementary abelian subgroup of G such that thereexists H ≤ G with G = H × E. We must show that H is either abelian, generalized dicyclic or isomorphic to one of Q× C4,G32, Q2, G64,1 or G64,2.We prove this by induction on |H|. If |H| ≤ 256, this follows from part (3) of Lemma 21. We now assume that |H| ≥ 512

and consider a subgroup K of index 2 in H . By induction, K = K ′ × E ′ where E ′ ∼= C2n and K ′ is either abelian, generalizeddicyclic or isomorphic to one of Q×C4, G32, Q2, G64,1 or G64,2. If K ′ is not abelian or generalized dicyclic, note that |K | ≥ 256and hence n ≥ 2. But, by part (2) of Lemma 21, H = H ′× E ′′ for some H ′ and E ′′ ∼= C2n−1, contradicting the hypothesis on H .Hence, wemay assume that K ′ is abelian or generalized dicyclic and in particular, K is also abelian or generalized dicyclic.

We deduce that there exists an abelian subgroup A of K of index 2 and hence |A| ≥ 128. Write A ∼= C4m × C2n. If m ≤ 2,then n ≥ 3. By part (2) of Lemma 21 applied twice, there exists H ′ such that H = H ′ × E ′′ with E ′ ∼= C2n−2, contradictingminimality of E. If m ≥ 3, Corollary 23 allows us to conclude that H contains an abelian subgroup A′ with A ≤ A′ and|H : A′| ≤ 2. By Lemma 22, H is abelian or generalized dicyclic.The fact that generalized dicyclic groups, Q × C4, G32, Q2, G64,1 and G64,2 are of exponent 4 and type 1A can be checked

by simple computation. Combined with the observation that type 1A groups are closed under direct product with C2, thiscompletes the proof.

Theorem 24, together with Corollary 17 and Theorem 20, concludes the proof of Theorem 3.

Remark. A surprising connection between our classification of groups of type 1A and a recent unpublished result of Jankocame to our attention after this paper was already submitted.A p-groupH isminimal nonabelian if and only ifH = 〈a, b〉 and |H ′| = p [1, Lemma 3.1 and 3.2]. A J-group is a nonabelian

2-group all of whose minimal nonabelian subgroups are isomorphic to Q or C4 o C4. It is not hard to see that groups of type1A are J-groups. More surprisingly perhaps, it turns out that the converse is true in the exponent 4 case. In other words,J-groups of exponent 4 are precisely nonabelian type 1A groups of exponent 4. We give a brief proof of this fact.

Let G be a J-group of exponent 4. We first show thatΩ1(G) ≤ Z(G). Since D8 is not a subgroup of G,Ω1(G) is elementaryabelian. Let a, b ∈ G such that b2 = 1. Note that [a, b]b = [a, b]−1. Since D8 is not a subgroup of G, [a, b]2 = 1 and[a, b, b] = 1. Also, 1 = [a2, b] = [a, b][a, b, a][a, b] hence [a, b, a]=1. If [a, b] 6= 1, then 〈a, b〉 is minimal nonabelian, andhence isomorphic to Q or C4 o C4. This is a contradiction since all involutions in Q and C4 o C4 are central. Now, let a, b beany elements in G. Since G has exponent 4 andΩ1(G) ≤ Z(G), we have [a, b] = a2b2(ab)2 and hence [a, b]2 = 1. This showsthat 〈a, b〉 is minimal nonabelian, and hence isomorphic to Q or C4 o C4. Both of these groups are of type 1A, and hence sois 〈a, b〉. This shows that G is of type 1A.Janko has recently classified J-groups [6, Theorem 1.3]. His proof of the exponent 4 case yields an independent proof of

our Theorem 24. His proof avoids the use of computer algebra systems; hence it is longer, having to deal with a lot of smallcases by hand.


Acknowledgement

The author would also like to thank Primož Potočnik for many helpful comments.

References

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[4] H.N. de Ridder, H.L. Bodlaender, Graph automorphismswithmaximal projection distances, in: Fundamentals of Computation Theory, in: LectureNotesin Comput. Sci., vol. 1684, 1999, pp. 204–214.

[5] The GAP Group, GAP—Groups, Algorithms, and Programming, Version 4.46; Aachen, Braunschweig, Fort Collins and St Andrews, 2006.http://www.gap-system.org/.

[6] Z. Janko, On minimal nonabelian subgroups in finite p-groups, 2008 (submitted for publication).[7] B. Larose, F. Laviolette, C. Tardiff, On normal Cayley graphs and hom-idempotent graphs, European J. Combin. 19 (1998) 867–881.[8] T. Pisanski, T.W. Tucker, B. Zgrablić, Strongly adjacency-transitive graphs and uniquely shift-transitive graphs, Discrete Math. 244 (2002) 389–398.[9] P. Potočnik, M. Šajna, G. Verret, Mobility of vertex-transitive graphs, Discrete Math. 307 (2007) 579–591.[10] W.T. Tutte, A family of cubical graphs, Proc. Cambridge Philos. Soc. 43 (1947) 459–474.[11] B. Zgrablić, On adjacency-transitive graphs, Discrete Math. 182 (1998) 321–332.

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