CUBIC SEMISYMMETRIC GRAPHS OF ORDER 2p3Aleksander Malni�c 1 Dragan Maru�si�c 1IMFM, Oddelek za matematiko IMFM, Oddelek za matematikoUniverza v Ljubljani Univerza v LjubljaniJadranska 19, 1111 Ljubljana Jadranska 19, 1111 LjubljanaSlovenija SlovenijaChangqun Wang2Department of System Science and MathematicsZhengzhou UniversityHenan Province, 450052People's Republic of China

1Supported in part by \Ministrstvo za znanost in tehnologijo Slovenije", proj. no.J1-0496-0101-99.2Supported in part by the grant from Zhengzhou University. Wang is grateful tothe Institute of Mathematics, Physics and Mechanics at the University of Ljubljana forhospitality and �nancial support during his visit that led to the completion of this work.1

SEMISYMMETRIC GRAPHS OF ORDER 2p3Dragan Maru�si�cIMFM, Oddelek za matematikoUniverza v LjubljaniJadranska 19, 1111 LjubljanaSlovenijaemail: dragan.marusic@uni-lj.si

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AbstractA regular graph is called semisymmetric if it is edge-transitive butnot vertex-transitive. It is proved that the Gray graph is the only cubicsemisymmetric graph of order 2p3, where p � 3 is a prime.

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1 IntroductionThroughout this paper graphs are assumed to be �nite, and, unless spec-i�ed otherwise, simple, undirected and connected. For the group-theoreticconcepts and notation not de�ned here we refer the reader to [3, 9, 21].Given a graph X we let V (X), E(X) and AutX be the vertex set, theedge set and the automorphism group of X, respectively. For two adjacentvertices u and v we write u � v, and use the symbol uv to denote eitherthe edge between u and v, or the arc from u to v. No ambiguity shouldarise for we always clearly state whether we refer to an edge or to an arc.If a subgroup G of AutX acts transitively on V (X) and E(X), we say thatX is G-vertex-transitive and G-edge-transitive, respectively. In the specialcase when G = AutX we say that X is vertex-transitive and edge-transitiverespectively. It can be shown that a G-edge- but not G-vertex-transitivegraph X is necessarily bipartite, where the two parts of the bipartition areorbits of G � AutX. Moreover, if X is regular these two parts have equalcardinality. A regular G-edge- but not G-vertex-transitive graph will bereferred to as a G-semisymmetric graph. In particular, if G = AutX thegraph is said to be semisymmetric.The study of semisymmetric graphs was initiated by Folkman [6] whoposed a number of problems which spurred the interest in this topic (see[1, 2, 4, 5, 11, 12, 13, 16]). Among other things he proved that there are nosemisymmetric graphs of order 2p or 2p2, for p a prime.This paper deals with (non)existence of cubic semisymmetric graphs oforder 2p3, where p is a prime. The �rst example of such a graph, the so calledGray graph, has order 54 and is descibed in [1]. Its discovery, according to[1], is due to Marion C. Gray in 1932, thus explaining its name. As shown in[15], it is the smallest cubic semisymmetric graph. Following [17], the Graygraph is a regular 9-fold cover of K3;3, with ZZ23 as the group of coveringtransformations (see Figure 1 below). For the purpose of this paper we takethis as the de�nition of the Gray graph. Alternative de�nitions can be foundin [17].Our object is to prove the following result.Theorem 1.1 Let X be a cubic semisymmetric graph of order 2p3, p � 3a prime. Then p = 3 and X is isomorphic to the Gray graph.The proof of Theorem 1.1 uses a combination of purely group-theoreticand combinatorial techniques, and is given in Section 3.4

Figure 1: The Gray graph as a cover of K3;3.2 PreliminariesAn epimorphism } : X ! Y of connected graphs is a regular coveringprojection if it arises essentially as a factorization X ! X=G �= Y , where theaction of G � AutX is semiregular on both vertices and edges of X. Notethat the graph Y may not be simple even if X is. The graph X is called thecovering graph and Y is the base graph. The preimage }�1(v), v 2 V (Y ),corresponds to an orbit of G on V (X) and is called the (vertex)-�bre over v.Similarly, edge-�bres correspond to orbits of G on E(X).It is well-known that a regular covering projection X ! Y �= X=G canbe reconstructed in terms of voltage assignments valued in G as follows(see [10]). First label arbitrarily a vertex in each �bre by 1 2 G, andthen label all other vertices by the right regular action of G � AutX oneach �bre. Consequently, given an arc uv in Y , the origins and termini ofarcs in }�1(uv) are labelled, respectively, by g and ag (g 2 G) for somea 2 G. This fact is recorded by assigning the voltage vol(uv) = a 2 G tothe corresponding arc uv. Clearly, inverse arcs carry inverse voltages. Theedges of X can thus be retrieved from Y by considering the left regularaction of G induced by the above labelling. A given voltage assignment can5

be modi�ed in such a way that the arcs of an arbitrarily prescribed spanningtree receive trivial voltages, and that the modi�ed assignment is associatedwith the same covering projection [10]. Moreover, the following propositionholds.Proposition 2.1 [18] Leaving the voltages of a spanning tree trivial andreplacing the voltage assignments on the cotree arcs by their images underan automorphism of the voltage group results in an equivalent covering pro-jection.Let } : X ! Y �= X=G be a regular covering projection. If ' 2 AutYand ~' 2 AutX satisfy ~'} = }' we call ~' the lift of ', and ' the projectionof ~'. (For the purpose of this paper, all functions are composed from leftto right.) Concepts such as the lift of a group of automorphisms and theprojection of a group of automorphisms are self-explanatory. The lifts andthe projections of groups are of course subgroups in AutX and AutY , re-spectively. In particular, CT(p) = G is the lift of the identity group and isknown as the group of covering transformations. Clearly, if G is normal inAutX then AutX does project (however, the projection need not be onto).The problem whether an automorphism ' lifts can be grasped in terms ofvoltages as follows. First observe that the voltage assignment on arcs ex-tends to an assignment on all walks in a natural way. De�ne the mapping'# : G! G, relative to a chosen base vertex v, by the rule(vol(C))'# = vol(C');where C ranges over all fundamental closed walks at v. Note that if Gis abelian, '# does not depend on the choice of the base vertex, and thefundamental closed walks at v can be substituted by the fundamental cyclesgenerating the cycle space of X.Proposition 2.2 [14] Let } : X ! Y �= X=G be a regular covering projec-tion. Then an automorphism ' of Y lifts along } if and only if '# extendsto an automorphism of G.Corollary 2.3 If X is a bipartite graph admitting an abelian subgroup Gof automorphisms acting regularly on each of the bipartition sets, then X isvertex-transitive.Proof. Consider the regular covering projection X ! X=G. The basegraph is a dipole dipd, that is, a graph with two vertices and d parallel6

edges, where d is the valency of X. Let � be the re ection of this dipolemapping each arc to its inverse. Then g�# = g�1. Since taking inverses inan abelian group is an automorphism, the re ection � lifts. Hence the graphX is vertex-transitive.Following [8], the coset graph �(G;H0;H1) of a group G with respect to�nite subgroups H0 and H1 is a bipartite graph with fH0g j g 2 Gg andfH1g j g 2 Gg as its bipartition sets of vertices, where H0g is adjacent toH1g0 whenever H0g \H1g0 6= ;. The following facts may be extracted from[8]:(i) �(G;H0;H1) is regular of valency d if and only if H0 \H1 has equalindex d in both H0 and H1;(ii) �(G;H0;H1) is connected if and only if G = hH0;H1i;(iii) G acts on �(G;H0;H1) by right multiplication. Moreover, this actionis faithful if and only if a normal subgroup of G which is contained inH0 \H1 coincides with the identity group;(iv) In the case when the action ofG is faithful, the coset graph �(G;H0;H1)is G-edge-transitive but not G-vertex-transitive.Proposition 2.4 Let X be a regular graph and G � AutX. If X is G-semisymmetric, then X is isomorphic to the coset graph �(G;Gu; Gv), whereu and v are adjacent vertices.Proof. De�ne a mapping ' : ug 7! Gug, vg 7! Gvg, g 2 G. Then ' iswell-de�ned. It is easy to see that ' is bijective. We verify that ' is a graphisomorphism of X onto �(G;Gu; Gv).Indeed, it is easy to see that X is bipartite. Moreover, let ug and vhbe two adjacent vertices in X, where g; h 2 G. Then u � vhg�1 . Since Gis edge-transitive, vhg�1 = vg0 for some g0 2 Gu. Hence there exists someh0 2 Gv such that hg�1g0�1 = h0. So h0�1h = g0g 2 Gug \ Gvh, thatis, Gug is adjacent to Gvh in �(G;Gu; Gv). Conversely, if Gug is adjacentto Gvh in �(G;Gu; Gv), then there exist some g0 2 Gu and h0 2 Gv suchthat g0g = h0h. Since u and v are adjacent and G � AutX, it followsthat ug = ug0g is adjacent to vh0h = vh. Therefore ' is the desired graphautomorphism.From this we can easily derive the following proposition.7

Proposition 2.5 The vertex stabilisers of a connected G-semisymmetriccubic graph have order 2r �3. Moreover, if u and v are two adjacent vertices,then G = hGu; Gvi, and the edge stabiliser Gu \ Gv is a common Sylow2-subgroup of Gu and Gv.The precise structure of the pair of admissible vertex stabilisers (Gu; Gv)was completely determined in [8]. In this article, however, a much weakerresult will be needed.Proposition 2.6 Let u and v be two adjacent vertices of a connected G-semisymmetric cubic graph. If Gu and Gv are both abelian, then Gu �= Gv �=ZZ3.Proof. If both Gu and Gv are abelian, then Gu \ Gv, the commonSylow 2-subgroup of Gu and Gv, is contained in Z(G) as G = hGu; Gvi. AsGu \Gv is the stabiliser of the edge uv in G and since G is edge-transitiveon X, any edge stabiliser of G is some G-conjugate of Gu \ Gv. It followsthat Gu \Gv = 1, which implies Gu �= Gv �= ZZ3.In general, the action of the full automorphism group of a semisymmetricgraph on its bipartition sets need not be faithful [4]. However, for cubicgraphs this action is always faithful.Proposition 2.7 Let X be a cubic semisymmetric graph. Then AutX actsfaithfully on each of the bipartition sets of X.Proof. Let V0 and V1 be the two bipartition sets of X, and let u 2 V0and v 2 V1 be adjacent.Assume that A = AutX is not faithful, say, on V0. Then the kernelCA(V0) of the action of A on V0 is nontrivial. By Proposition 2.5, eachvertex stabiliser of A is a f2; 3g-group. Since CA(V0) is the intersection ofvertex stabilisersAw for w 2 V0, it follows that CA(V0) is also a f2; 3g-group.If 3 divides jCA(V0)j, then there exists some element of order 3 in CA(V0).Since an element of order 3 in a vertex stabiliser of a semisymmetric cubicgraph is transitive on its neighbours, it follows that CA(V0) is transitive onthe neighbourhood X(w) of any vertex w 2 V0. By the connectivity of Xwe have X �= K3;3, a contradiction since K3;3 is vertex-transitive. HenceCA(V0) is a 2-group. It follows that CA(V0) � Au\Av = Auv. Since CA(V0)is normal in A (and A is faithful on E(X)), it follows that CA(V0) = 1,contrary to our assumption. 8

3 Proof of Theorem 1.1The following lemma which gives a detailed information on the local p-subgroups of the full automorphism group of cubic semisymmetric graphs oforder 2p3 is esential to the proof of our main theorem. (Note that the proofof this lemma depends on the classi�cation of �nite simple groups, appliedto groups of order 2r � 3 � p3.)Lemma 3.1 Let X be a cubic semisymmetric graph of order 2p3, p � 3 aprime. Then A = AutX contains a subgroup G isomorphic to ZZ2p, actingsemiregularly on vertices and edges of X. Moreover,(i) if p > 3 then G is normal in AutX; and(ii) if p = 3 then G is normal in a Sylow 3-subgroup of AutX, and thequotient graph X=G is isomorphic to K3;3.Proof. Denote the bipartition sets of X by V0 and V1 and let u 2 V0,and v 2 V1 be adjacent.We �rst consider the case p > 3. By Propositions 2.5 and 2.7, jAj =2r � 3 � p3 for some integer r. Let P be a Sylow p-subgroup of A. Then Pacts regularly on each Vi, and hence X is a regular P -cover of the dipoledip3 with three parallel edges. The covering projection X ! dip3 can bereconstructed in terms of the voltage group P , where the voltages on thethree arcs from V0 to V1 are 1, a and b. Since X is connected, we haveP = ha; bi. By Corollary 2.3, P is not abelian. It is well known thatthere are exactly two nonisomorphic nonabelian groups of order p3, givenby respective presentationsM(p) = ha; b j ap = bp = cp = 1; c = [a; b]; ac = ca; bc = cbiand M1(p) = hx; y j xp2 = yp = 1; [x; y] = xpi:We show that P = M1(p). Suppose, on the contrary, that P = M(p). Onecan show that in this case AutP acts transitively on the set of ordered pairsof generators of P .Therefore, without changing the covering projection we may assume thatthe generators of the voltage group P are the elements a and b as in theabove presentation of M(p). Let � 2 Aut dip3 be the re ection which maps9

each of the arcs to its inverse. Then a�# = a�1 and b�# = b�1. It is easy tosee that �# extends to a group automorphism of P . Hence � lifts, implyingthat X is vertex-transitive, a contradiction. Therefore P =M1(p).For a set � of prime divisors of the order of A, let O�(A) denote thelargest normal subgroup of A whose order is divisible only by primes in theset �. Furthermore, as usual, let �0 denote the set of all prime divisors ofthe order of A not in �. Now, Q = Op(A) is the maximal normal p-subgroupof A. Note that since Op0(A) is contained in every vertex stabiliser, we haveOp0(A) = 1.Suppose �rst that Q = 1. Let N be a minimal normal subgroup of A.In general, N is a direct product of isomorphic simple groups, which mustbe nonabelian, by our assumption. But 32 does not divide jAj and so Nis a nonabelian simple f2; 3; pg-group. By the classi�cation of �nite simplef2; 3; pg-groups, we have N �= PSL2(p), where p = 5 or 7. In particular, N 6=A and Out(N) = ZZ2. On the other hand, by [19, Theorem 6.11], we havethat A=CA(N) is isomorphic to a subgroup of AutN . Since CA(N)\N = 1,we have that CA(N)N = CA(N) � N , and so CA(N) � N is a subgroupof index at most 2 in A, implying that CA(N) is nontrivial. Moreover,since O3(A) (the smallest normal subgroup of A such that the order of thequotient A=O3(A) is a 3-group) is a subgroup of N , we have that CA(N) isa f2; pg-group. Thus, CA(N) is solvable, and so either Op(CA(N)) 6= 1 orOp0(CA(N)) 6= 1. Since these two groups are characteristic in CA(N), theyare both normal subgroups of A. Since Op0(A) = 1, this contradicts ourassumption that Op(A) = 1.Suppose now that Q �= ZZp. Then Q = Z(P ) for every Sylow p-subgroupP of A. Set C = CA(Q). Then P � C, and Op(C=Q) = 1. Let Op0(C=Q) =M=Q, where M � C. As M=Q is characteristic in C=Q and C=Q / A=Q,we have C=Q / A=Q. Hence M is normal in A. Since Q is a normal Sylowp-subgroup of M , it follows that Q has a p0-complement M1 in M . Conse-quently, M = QM1 = Q �M1 and so M1 = Op0(M) is normal in A. ThusM1 = 1 and Op0(C=Q) = 1. Hence C=Q is nonsolvable. Now we let N=Qbe a minimal normal subgroup of C=Q. By the same argument as in thepreceding paragraph (replacing A and N by C=Q and N=Q, respectively) asimilar contradiction is obtained.Suppose now that Q �= ZZp2 . Set C = CA(Q). Note that Q is a normalSylow p-subgroup of C. Let C1 be a p0-complement of Q in C. ThenC = QC1 = Q� C1, and so C1 = Op0(C). Since Op0(C) is normal in A andOp0(A) = 1, it follows that C1 = 1. Thus C = Q and so, by [19, Theorem10

6.11], we have that A=Q is isomorphic to a subgroup of AutQ �= ZZp(p�1).This implies that the vertex stabilisers Au and Av are both abelian. Inview of Proposition 2.6 we have Au �= Av �= ZZ3. Now jAj = 3p3, and bySylow's theorem it is easily seen that A has a normal Sylow p-subgroup,contradicting Q �= ZZp2 .To summarize, we have proved thus far that either Q is elementaryabelian of order p2 or Q = P is normal in A. In the �rst case we takethe group G to be Q. In the second case we have that 1(P ), the groupgenerated by all elements of order p in P , is a characteristic subgroup of P ,and hence normal in A. Since P = M1(p), we have 1(P ) �= ZZ2p, and wetake G = 1(P ) in this case.Moreover, since p > 3, we have that G �= ZZ2p is always semiregular onboth vertices and edges because the vertex stabilisers and edge stabilisers ofX are all p0-groups.Let us now consider the case p = 3. We have jP j = 34, jV0j = jV1j = 33,and jAj = 2r34. Let i 2 f0; 1g. Since A is transitive on Vi, and jVij = 33,it follows by [21, Theorem 3.4] that also the Sylow p-subgroup P of A istransitive on Vi. Further, P is nonabelian for its action on Vi is not regular.Note that each Sylow 2-subgroup of A is an edge stabiliser. Consequently,P acts edge-transitively and hence semisymmetrically on X.Let G be a normal subgroup of order 9 in P . We �rst show that Gmust be semiregular on each Vi. For if G is not semiregular, say on V0, thenG has nontrivial intersection with Pu �= ZZ3, for some u 2 V0, and hencePu � G. Choose an arbitrary w 2 V0. By the normality of G in P and thetransitivity of P on V0 it follows that Pw � G. Since Pw is transitive onthe neighbourhood N(w) of w, it follows that N(w) is contained in only oneorbit of G on V1. Moreover, for any other vertex w0 in the same orbit of Gon V0, the neighbourhoods N(w) and N(w0) are contained in the same orbitof G on V1. By the connectivity of X, the group G is transitive on V1, whichis impossible as jGj = 9 and jV1j = 27. Therefore G is semiregular on eachVi, and X ! X=G is a regular covering projection, where X=G is a cubicgraph of order 6. Since G is normal in P , the group P projects onto anedge-transitive subgroup of X=G. Consequently, the quotient graph X=G isisomorphic to K3;3.Note that the argument of the preceding paragraph implies that no nor-mal subgroup of order 9 of P contains a vertex stabiliser Pw, w 2 V (X).We now show that G �= ZZ23. Suppose not. Then G �= ZZ9. By [19,Theorem 6.11] the quotient P=CP (G) is isomorphic to a subgroup of AutG �=11

ZZ6. It follows that either CP (G) = P or CP (G) is abelian of order 27.If CP (G) = P , that is, G � Z(P ), then G = Z(P ) as P is not abelian.Recall that G and Pu have trivial intersection. It follows that hG;Pui �=ZZ9 � ZZ3. Now 1(hG;Pui) is characteristic in hG;Pui and hence normal inP . Note that 1(hG;Pui) contains Pu and is of order 9. This contradictsthe fact that P has no such subgroups. Therefore, CP (A) is abelian of order27. In other words, CP (G) is isomorphic either to ZZ9 � ZZ3 or to ZZ27.Suppose that CP (G) �= ZZ9 � ZZ3. By Corollary 2.3, CP (G) cannot beregular on both the V0 and V1. Consequently, it contains a vertex stabiliserPw for some w 2 V (X). The same argument as above now leads to acontradiction.It remains to exclude the case CP (G) cyclic. In fact, we claim thatP contains no cyclic subgroups of order 27. Assuming that P contains asubgroup R isomorphic to ZZ27, we have by Corollary 2.3 that R cannot betransitive on both V0 and V1. Suppose that R is intransitive on V0. ThenPu � R for some u 2 V0. Since R has index 3 in P , it is normal in P . By thetransitivity of P on V0, the group R contains Pw for all w 2 V0. As R is cyclicand Pu �= ZZ3, the groups Pw, w 2 V0, coincide and are therefore containedin the kernel of A on V0. This contradicts Proposition 2.7, completing theproof of Lemma 3.1.Proof of Theorem 1.1. We shall distinguish two cases.Case 1: p > 3.By Lemma 3.1, the automorphism group A = AutX has a normal sub-group G isomorphic to ZZ2p, acting semiregularly on vertices and edges of X.Consider the regular covering projection } : X ! X=G. Since G is normalin A, the group A projects to a subgroup of automorphisms of the graphY = X=G isomorphic to A=G, acting semisymmetrically on Y . But Y hasorder 2p and by the classical result of Folkman [6], we have that Y is abipartite, vertex-transitive and therefore also an arc-transitive graph. Thestructure of such graphs is well known. We may identify the vertex set ofY with two copies of ZZp, where the elements of the second copy are conve-niently dinstinguished from the elements of the �rst copy by the symbol 0.Furthermore there exists an element s 2 ZZ�p n f1g such that s2 + s+ 1 = 0.The edges of Y are then of the form y(y+1)0, y(y+s)0, y(y+s2)0, for y 2 ZZp.We reconstruct our covering by assigning the trivial voltage to the arcsof the natural hamiltonian path from 0 to 10. Furthermore, the voltage of12

the arc 010 is denoted by a, and the voltages of the chordal arcs from i(s�1)to (i(s� 1) + s2)0, i 2 ZZp, are denoted by ai (see Figure 2).

Figure 2: The voltage assignment in the case p > 3.Since A=G acts semisymmetrically on Y , at least two automorphisms ofY lift: the cyclic rotation �, mapping according to the rule y 7! y + s� 1,y0 7! (y + s� 1)0, for y 2 ZZp, and the element � of order 3 �xing 0 and 00,mapping according to the rule y 7! sy, y0 7! (sy)0. The requirement that �and � lift imposes strong restrictions to the cotree voltages. Recall that byProposition 2.2 an automorphism of Y lifts along } if and only if it inducesan automorphism of the voltage group G. Regarding G as a vector spaceover ZZp we may view AutG as a group of linear transformations of G.Considering the action of � and � on the fundamental cycles we �ndthe images of the voltages a and ai under �# and �#, thus restricting thepossibilities for the voltage assignments a and ai.13

In our analysis we shall distinguish two di�erent cases.Subcase 1.1: a 6= 0.Set bi = ( ai; i � s2ai + a; i > s2 :One can check that �# �xes a and maps bi 7! bi+1. We claim that bi =(i�; 1), which forces ai = ( (i�; 1); i � s2(i�� 1; 1); i > s2 : (1)Indeed, by the connectivity of the covering graph X, it follows thatG = ha; a0; a1; : : : ; ap�1i = ha; b0; b1; : : : ; bp�1i. So bi is nonzero for eachi 2 ZZp. Moreover, if b0 and a are linearly dependent, then hb0i = hai. As�# maps bi to bi+1, we have by induction that hbii = hai for all i 2 ZZp, acontradiction since G is not cyclic. We can therefore choose fa; b0g as a basisof G. There is an automorphism of G mapping a to (1; 0) and b0 to (0; 1).By Proposition 2.1 we may therefore assume a = (1; 0) and b0 = (0; 1).Let b1 = (�; �), where �; � 2 ZZp. By writing the cordinates as rowvectors, the matrix of �# with respect to the above basis isM� = � 1 0� � � :Clearly, (�#)p �xes a and each bi. So (�#)p = id, forcing �p = 1. Butsince � 6= 0 and p is a prime, we have � = �p and so � = 1. This impliesbi = (i�; 1), and (1) follows as claimed.We now investigate how �# imposes further restrictions to our voltageassignment. As we shall see, � = 0, implying that our covering graph X isuniquely determined. To this end we �rst describe the action of �# on thevoltages of the \small" base cycles, determined by the chords. Explicitly,the cycle Ci, i 2 ZZp, of length 2s + 2 with initial vertex i(s � 1) is of theform Ci = Ci;1C 0i;2Ci;3C 0i;4 : : : Ci;2s+1C 0i;2s+2Ci;1;where Ci;1 = i(s� 1); Ci;2 = i(s� 1) + s2;14

and for k = 1; 2; : : : ; sCi;2k+1 = (i+ s� k + 1)(s� 1); Ci;2k+2 = (i+ s� k + 1)(s� 1) + 1:It can be seen that � maps Ci to the cycleC�i = Di;1D0i;2Di;3D0i;4 : : : Di;2s+1D0i;2s+2Di;1;where Di;1 = i(s� 1)s; Di;2 = i(s� 1)s+ 1;and for k = 1; 2; : : : ; sDi;2k+1 = (i+ s� k + 1)(s� 1)s; Di;2k+2 = (i+ s� k + 1)(s� 1)s+ s:Note that vol(Ci) = bi for each i 2 ZZp. To calculate the voltage of themapped cycle C�i , observe that the chords of Y are distinguished from theedges on the natural hamiltonian cycle by the di�erence of their endvertices.The chordal arcs are of the form y(y + s2)0 for y 2 ZZp (see Figure 2).The cycle C�i has exactly s + 1 chordal arcs, namely, D0i;2kDi;2k+1 for k =1; 2; : : : ; s and D0i;2s+2Di;1. Since the only arc in C�i of the form y(y + 1)0 isDi;1D0i;2, it follows that among all base cycles Ci, i 2 ZZp, the cycle C0 is theonly one whose image under � contains 010. By computation,vol(C�0 ) = a� as2 � as(s�1) � as(s�2) � : : : � as � a0; (2)and, for i 6= 0,vol(C�i ) = �as(s+i) � as(s+i�1) � as(s+i�2) � : : :� as(i+1) � asi: (3)Similarly, the natural hamiltonian cycle C, which has voltage a, is mappedto the cycle C� withvol(C�) = �as(p�1) � as(p�2) � : : :� as � a0 = � p�1Xi=0 ai = s:Combining this with (2), we have that the matrix of �# with respect tothe basis fa; b0g is M� = � s 0s�=2 + 1 s2 � :15

Combining this with a direct computation using (3), it follows that �#maps b1 = a1 to (�; 1)M� = (3s�=2 + 1; s2) = ((1 + s=2)� + 1; s2). Sinces 6= 1 we get � = 0. Therefore for i 2 ZZpai = ( (0; 1); i � s2(�1; 1); i > s2 : (4)We now show that the unique covering graph obtained by this voltageassignment is not semisymmetric. To this end we show that the re ection �interchanging y with (1� y)0 lifts.Obviously, the natural hamiltonian cycle C is reversed by this re ection,and so vol(C� ) = �a. As for the cycles Ci, i 2 ZZp, observe that the chordalarc Ci;1C 0i;2 is mapped to the chordal arc C 0s2�i;2Cs2�i;1. Also, the arc 010appears only on cycles Ci with s2 < i < p. Moreover, the re ection �preserves the set of cycles Ci containing 010 and the set of cycles Ci notcontaining 010. This implies that vol(C�i ) = �bs2�i, for i 2 ZZp.Finally, using Proposition 2.2, one can check that � lifts, implying thatthe covering graph is not semisymmetric.Subcase 1.2: a = 0.We inherit the notation from Subcase 1.1. Since �# extends to an au-tomorphism of G, either all ai are zero or all ai are nonzero. Because ofthe connectivity of the covering graph X the second possibility must occur.Furthermore, the voltages ai, i 2 ZZp, are pairwise linearly independent.Suppose, on the contrary, that ai and aj are linearly dependent, for somei < j. Since �j�i induces a cyclic permutation of order p on the indices of ak,k 2 ZZp, all the voltages ak are linearly dependent. This contradicts the factthat G = ha0; a1; : : : ; ap�1i is not cyclic. It follows that all the subgroupshaii, i 2 ZZp, are pairwise distinct. But there are exactly p + 1 subspacesof dimension 1 in G, and so �# must �x the only remaining 1-subspacegenerated by some b 2 G.Now choose fb; a0g as a basis of G, and let the corresponding matrix of�# be M� = �� 0� � � ;for some � 2 ZZp and �; � 2 ZZ�p.Since (�#)p = id, we have Mp� = I. It follows that �p = �p = 1 and so� = � = 1. It is easily seen that ai = (i�; 1).16

As in the previous subcase, a short computation involving the actionof �# gives � = 0. Therefore ai = (0; 1) for i 2 ZZp, forcing G �= ZZp, acontradiction.We conclude that there is no cubic semisymmetric graph of order 2p3 forp > 3.Case 2: p = 3.By Lemma 3.1, a Sylow 3-subgroup P of A has a normal subgroup Gisomorphic to ZZ23 and acting semiregularly on vertices and edges of X. Aquotienting by the action of G results in a regular covering projection ofX onto Y = X=G �= K3;3. Let the bipartition sets of Y be f0; 2; 4g andf1; 3; 5g. Since G is normal in P and X is cubic, it follows that the group Pprojects to a subgroup of automorphisms of Y acting semisymmetrically onY . Hence the two automorphisms ' = (024) and = (135) do have a lift.Also, �1 = (01)(23)(45) and �2 = (01)(25)(34) do not lift, for otherwise thecovering graph would be vertex-transitive.In order to reconstruct this covering by voltages valued in G, let uschoose a tree with trivial voltages as shown in Figure 3, and let a, b, c and dbe the voltages of the remaining cotree arcs 32, 34, 25 and 45, respectively.

Figure 3: The voltage assignment in the case p = 3.The requirement that ' and lift imposes strong restrictions to thecotree voltages. In fact, we now show that such a covering projection mustgive rise to the Gray graph. The reader may verify, by checking the funda-mental cycles 03210, 03410, 01250 and 01450 of the graph Y , that '#, #,�#1 and �#2 map the voltages a, b, c and d as follows (we use the additive17

notation for the operation in the group G regarded as a vector space overZZ3. a b c d'# �a+ b �a �c+ d �c # �a� c �b� d a b�#1 �a c b �d�#2 d �b �c aTable 1: The mappings '#, #, �#1 and �#2 .By the connectivity of X we have G = ha; b; c; di. Observe from Table 1that none of a, b, c, d is trivial. We shall distinguish two cases.Suppose �rst that a and b are linearly dependent. It follows from the rowof '# in Table 1 that b = �a. Then from the row of # we have d = �c.Now G = ha; ci. There exists an automorphism of the voltage group G tak-ing a and c to (�1; 0) and (0;�1), respectively. By Proposition 2.1 we maytherefore assume a = (�1; 0) and c = (0;�1). Therefore the covering graphX, obtained from Y = K3;3 for the case when a and b are linearly depen-dent, is unique. We now show that it is indeed semisymmetric, and in factisomorphic to the Gray graph. By recalculating the cotree voltages relativeto the spanning tree with edges 01, 12, 03, 25, 14, we get an equivalent volt-age assignment where the arcs 23 and 34 receive voltage (0; 1), whereas thearcs 05 and 54 receive voltage (1; 0). It is known that such an assignmentgives rise to the Gray graph (see [17]).We now consider the case when a and b are linearly independent. Choosefa; bg as a basis of G. From Table 1 we get that the matrix for '#, withrespect to the basis fa; bg, isM' = ��1 1�1 0 � :Let us express c and d in terms of fa; bg. Regarding a, b, c and djust as symbols we have (c; d) = (a; b)M , where M is a 2 � 2 matrix overZZ3. Then (c; d)M' = (a; b)MM'. Since '# is a linear transformation,(c'# ; d'#) = (a'# ; b'#)M = (a; b)M'M . From the Table 1 we see that(c'# ; d'#) = (c; d)M', and hence M'M = MM'. It is easy to check thatM 2 f�I;�M';�M2'g. 18

We now consider how the automorphism # restricts the voltage assign-ments. Again from Table 1 we get that the matrixM for # with respect tothe basis fa; bg equals �M � I. We now have (c # ; d #) = (a # ; b #)M =(a; b)M M . On the other hand, Table 1 also implies (c # ; d #) = (a; b)I.Consequently, M M = I, that is, M2 + M + I = 0. This impliesM 2 fI;M';M2'g. One can check that if M =M', then �2 lifts, and that ifM = M2', then �1 lifts. Thus, in these two cases, the covering graph is notsemisymmetric. Therefore, M = I. Then a = c and b = d.There exists an automorphism of the voltage group G taking a and b to(1; 0) and (0;�1), respectively. By Proposition 2.1 we may assume a = (1; 0)and b = (0;�1). As above, recalculating the cotree voltages relative to thespanning tree with edges 01, 03, 05, 14 and 52, we get that the coveringgraph is again isomorphic to the Gray graph.We conclude that X is unique and isomorphic to the Gray graph, com-pleting the proof of Theorem 1.1.We remark that it is not hard to see that the above covering graph of Ycan be described in such a way that it makes its isomorphism with the Graygraph, as de�ned in [1], self-evident. Just substitute the vertices 0 and 2 bythree copies of K3;3, regard the vertices 1, 3 and 5 as inserted edge-vertices,and take the vertex 4 to be the nine vertices joining the edge-vertices.References[1] I. Z. Bouwer, An edge but not vertex transitive cubic graph, Bull. Can. Math. Soc.11 (1968), 533{535.[2] I. Z. Bouwer, On edge but not vertex transitive regular graphs, J. Combin. Theory,B 12 (1972), 32{40.[3] J. D. Dixon and B. Mortimer, \Permutation Groups", Springer-Verlag, New York,1996.[4] S. F. Du, Construction of Semisymmetric Graphs, Graph Theory Notes of New YorkXXIX (1995), 148{149.[5] S. F. Du and M. Y. Xu, A Classi�cation of Semisymmetric Graphs of Order 2pq,Comm. Algebra, to appear.[6] J. Folkman, Regular line-symmetric graphs, J. Combin. Theory 3 (1967), 215{232.[7] C. Godsil, On the full automorphism group of Cayley graphs, Combinatorica 1(1981), 143{156. 19

[8] D. Goldschmidt, Automorphisms of trivalent graphs, Ann. Math. 111 (1980), 377{406.[9] D. Gorenstein, \Finite Groups", Harper and Row, New York, 1968.[10] J. L. Gross and T. W. Tucker, \Topological Graph Theory", Wiley{Interscience,New York, 1987.[11] M. E. Io�nova and A. A. Ivanov, Biprimitive cubic graphs, Investigations in AlgebraicTheory of Combinatorial Objects, (Proceedings of the seminar, Institute for SystemStudies, Moscow, 1985) Kluwer Academic Publishers, London, 1994, pp 459{472.[12] A. V. Ivanov, On edge but not vertex transitive regular graphs, Ann. Discrete Math.34 (1987), 273{286.[13] M. H. Klin, On edge but not vertex transitive graphs, Coll. Math. Soc. J. Bolyai,(25. Algebraic Methods in Graph Theory, Szeged, 1978), Budapest, 1981, 399{403.[14] A. Malni�c, R. Nedela, M. �Skoviera, Lifting graph automorphisms by voltage assign-ments, Europ. J. Combin., in print.[15] A. Malni�c, D. Maru�si�c and C. Q. Wang, Semisymmetry of cyclic covers of K3;3, inpreparation.[16] D. Maru�si�c, Constructing cubic edge- but not vertex-transitive graphs, J. GraphTheory, to appear.[17] D. Maru�si�c and T. Pisanski, The Gray graph revisited, J. Graph Theory, to appear.[18] M. �Skoviera, A contribution to the theory of voltage graphs, Discrete Math. 61(1986), 281{292.[19] M. Suzuki, \Group Theory I", Springer-Verlag, New York, 1982.[20] W. T. Tutte, A family of cubical graphs, Proc. Cambridge Phil. Soc., 43(1948),459{474.[21] H. Wielandt, \Finite Permutation Groups", Academic Press, New York, 1964.

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