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9URAL MATHEMATICAL JOURNAL, Vol. 7, No. 2, 2021, pp. 136-158
DOI: 10.15826/umj.2021.2.010
ON A CLASS OF EDGE-TRANSITIVE DISTANCE-REGULAR ANTIPODAL COVERS OF COMPLETE GRAPHS1
Ludmila Yu. Tsiovkina
Krasovskii Institute of Mathematics and Mechanics,
Ural Branch of the Russian Academy of Sciences, 16 S. Kovalevskaya Str., Ekaterinburg, 620108, Russia
tsiovkina@imm.uran.ru
Abstract: The paper is devoted to the problem of classification of edge-transitive distance-regular antipodal covers of complete graphs. This extends the classification of those covers that are arc-transitive, which has been settled except for some tricky cases that remain to be considered, including the case of covers satisfying condition C2 = 1 (which means that every two vertices at distance 2 have exactly one common neighbour).
Here it is shown that an edge-transitive distance-regular antipodal cover of a complete graph with C2 = 1 is either the second neighbourhood of a vertex in a Moore graph of valency 3 or 7, or a Mathon graph, or a half-transitive graph whose automorphism group induces an affine 2-homogeneous group on the set of its fibres. Moreover, distance-regular antipodal covers of complete graphs with C2 = 1 that admit an automorphism group acting 2-homogeneously on the set of fibres (which turns out to be an approximation of the property of edge-transitivity of such cover) are described.
A well-known correspondence between distance-regular antipodal covers of complete graphs with C2 = 1 and geodetic graphs of diameter two that can be viewed as underlying graphs of certain Moore geometries, allows us to effectively restrict admissible automorphism groups of covers under consideration by combining Kantor's classification of involutory automorphisms of these geometries together with the classification of finite 2-homogeneous permutation groups.
Keywords: Distance-regular graph, Antipodal cover, Geodetic graph, Arc-transitive graph, Edge-transitive graph, 2-transitive group, 2-homogeneous group.
Introduction
A distance-regular antipodal cover of a complete graph can be defined as a connected graph whose vertex set admits a partition into n classes (called fibres) of the same size r > 2 such that each class induces a coclique, the union of any two distinct classes induces a perfect matching, and any two non-adjacent vertices from distinct classes have exactly c2 > 1 common neighbours. According to [8], such a graph will be referred to as an (n,r,c2)-cover. One can see that an (n, r,c2)-cover is indeed a cover (or a covering graph) of the complete graph Kn in the topological sense (see [8] or [7]), and that its diameter is 3.
To date, almost all arc-transitive (n, r, c2)-covers have been classified (see [14, 15, 19-22]), except for the following two tricky cases: when an arc-transitive automorphism group induces an affine permutation group on the set of fibres (see [22]) or c2 = 1 (see a discussion below in this section). Note that an arc- or, more generally, edge-transitive automorphism group of an (n, r, c2)-cover induces a 2-homogeneous action on its fibres. The purpose of this paper is to study the (n, r, 1)-covers whose automorphism group acts 2-homogeneously on the set of fibres, and to describe those that are edge-transitive.
1This work was supported by the Russian Science Foundation under grant no. 20-71-00122.
The afore-mentioned interplay between edge-transitivity and 2-homogeneity allows us to base our arguments on the classification of finite 2-homogeneous permutation groups, which follows from the classification of finite 2-transitive permutation groups and the Kantor's fundamental result [11]. To investigate admissible groups of automorphisms, we also exploit a remarkable correspondence between (n, r, 1)-covers and geodetic graphs of diameter two (see [2]) that are equivalent to certain Moore geometries. The classification of involutory automorphisms of these geometries that is due to Kantor [12] together with the Higman's technique for studying automorphisms of association schemes (e.g., see [4, Section 3.7]) turn out to be effective tools for their description.
Main results of this paper are presented by the following two theorems.
Theorem 1. Let A be a (k + 1, r, 1)-cover with s := k — r + 1 > 1, let £ be the set of fibres of A, and G = Aut(A). Denote by K and Gs the kernel and the image of the induced action of G on £, respectively. Then k = cs and r = cs — s + 1 for some c € Z, and the following statements hold:
(1) if Gs is a 2-homogeneous, but not 2-transitive group, then
Gs < ArLi(g), k + l = q = 3 (mod 4), -^cs + (s/2 - l)2 £ Z, and either K = 1, s = 2 and c = (q — 1)/2, or s is odd;
(2) if Gs is an almost simple 2-transitive group, then either K = 1, s = 2, c = 2e-1, Soc(G) ~ L2(2e) and A is a Mathon graph, or G acts intransitively on vertices of A;
(3) if Gs is an affine 2-transitive group, then G acts intransitively on arcs of A.
Theorem 2. Suppose A is an edge-transitive (k + 1, r, 1)-cover, let £ be the set of fibres of A, and G = Aut(A). Denote by K and Gs the kernel and the image of the induced action of G on £, respectively. Then either k = r € {2,6} and A is the second neighbourhood of a vertex in a Moore graph of valency k + 1, or k > r and one of the following statements holds:
(1) Gs is an almost simple 2-transitive group, K = 1, Soc(G) ~ L2(2e) and A is an arc-transitive Mathon graph of valency k = 2e;
(2) Gs is an affine 2-homogeneous group and A is a half-transitive graph.
Recall that the only Moore graphs of valency 3 or 7 are the Petersen graph or the Hoffman-Singleton graph, respectively (see [10]). Note that for each admissible k the resulting graph in Theorem 2 (1) is unique (up to isomorphism) and its construction is due to Mathon (e.g., see [3, Proposition 1.17.3]).
We also remark here that in [6, Proposition 4] it was claimed that each (k + 1, r, 1)-cover with s > 1 that possesses a group of automorphisms acting 2-homogeneously on the fibres necessarily has valency k = 2e and s = 2. Unfortunately, the proof of this result (see an exposition in [16]) is flawed; Theorem 1 shows that it holds under the additional assumption of arc-transitivity of the graphs under consideration. Thus, compared together with previous results (see [14, 15]), the classification of arc-transitive (n,r,c2)-covers in the almost simple case is complete.
The organization of the paper is as follows. In Section 1 we recall some basic definitions and facts on (n, r, 1)-covers. In Section 2 we obtain general results on automorphisms of such a graph. Section 3 is devoted to the proofs of Theorems 1 and 2.
1. Preliminaries
Throughout the paper we consider only finite undirected graphs without loops or multiple edges. By a subgraph of a graph r we mean a vertex-induced subgraph, and we also identify a subset X of vertices of r with the subgraph of r that is induced by X. The distance between vertices x and y of a graph r is denoted by dr(x, y), or simply d(x, y) if r is clear from the context. For a vertex a of a graph r, we denote by r(a) the i-th neighbourhood of a, that is the subgraph of r induced by the set {b € r | dr(a, b) = i}. The number of neighbours of a vertex a, i.e., the size of r1(a), is the valency of a in r. For a fixed graph r and any its vertex a, the subgraph r1(a) is also denoted by [a] if the graph r is clear from the context; we also put a± := {a} U [a]. A graph is said to be regular if all its vertices have the same valency; a graph is said to be biregular if it is not regular and every of its vertices has one of two possible valencies.
A graph is geodetic if every two of its vertices are joined by a unique shortest path. A biregular geodetic graph of diameter two that is not contained in a± for any its vertex a is referred to as a BRG -graph.
A connected graph r of diameter d is called distance-regular if there are integers c^a» and bj, for all i € {0,1,..., d}, such that for each pair of vertices x and y with dr(x, y) = i, the following equalities hold:
cj = |rj-1(x) n r1(y)|, a = |rt(x) n r1(y)| and bj = |rm(x) n r(y)|,
where bd = c0 = 0 by definition; in particular, |r1(x)| = b0 = cj+a»+bj holds for any i € {0,1,..., d}. The sequence {b0, b1,..., bd-1; c1,..., cd} is called the intersection array of r.
A distance-regular graph of diameter 2 is also called strongly regular. A graph is said to be edge regular if it is regular and there is a non-negative integer A such that every two adjacent vertices have exactly A common neighbours; a graph is said to be amply regular if it is edge-regular and there is a non-negative integer ^ such that every two vertices at distance 2 have exactly ^ common neighbours.
If the binary relation "to be at distance 0 or d" on the set of vertices of a connected graph r of diameter d is an equivalence relation, then the graph r is called antipodal; the classes of this relation are called antipodal classes or fibres of r. We will say that an antipodal graph r is an antipodal cover of a graph A, if r is not a complete graph and the following three conditions are satisfied: (i) every fibre of r induces a coclique, (ii) the union of any two distinct fibres of r induces a coclique or a perfect matching, and (iii) A is isomorphic to the graph r defined on the fibres of r, in which two vertices are adjacent if and only if the union of corresponding fibres forms a matching in r. By the Smith's theorem [3, Theorem 4.2.1], non-cyclic distance-regular graphs fall into families of primitive, bipartite or antipodal graphs. Every graph of diameter d from the latter family is a complete graph or a complete multipartite graph with parts of equal sizes if d = 1 or 2, and it is an antipodal cover of a distance-regular graph of diameter |_d/2j when d > 3 [5]. Hence distance-regular antipodal covers of complete graphs are precisely antipodal distance-regular graphs of diameter 3. They do not have a universal construction and form a large infinite class of graphs that is closely related to many interesting combinatorial objects, like projective planes or generalized quadrangles; we refer the reader to [3, 8, 16] for more background.
For a subset X of a group acting on a set Q, by Fixn(X) we denote the set of points in Q that are fixed by every element of X. When X = {g}, we write "Fixn(g)" instead of "Fixn({g})". We also write Fix(X) = Fixn (X) if Q is clear from the context. In what follows, for a graph r and a subset X C Aut(r), we identify the set Fix(X) with the subgraph of r that is induced by Fix(X).
A graph is called vertex-transitive or edge-transitive, if its automorphism group acts transitively on the set of its vertices or on the set of its edges, respectively. A graph is called arc-transitive, if
its automorphism group acts transitively on set of its arcs (ordered pairs of adjacent vertices). A graph is called half-transitive, if it is vertex- and edge-transitive, but not arc-transitive.
Our other terminology and notation are mostly standard and follow [1, 3].
Further in this section, we provide some auxiliary results that are used in the proofs of Theorems 1 and 2.
Throughout the rest of the paper, A is an (k + 1,r, 1)-cover, £ is the set of fibres of A and s := k — r + 1. By [8, Theorem 3.4] there is an integer c such that cs = k, the number cs + 1 is odd and s < c. Put
v = (cs + 1) (cs — s + 1) and D = cs — s + 1 + s2/4 = cs + (s/2 — 1)2. Then v is the number of vertices of A and its distinct eigenvalues are
d0 = cs, 01 = {s-2)/2 + VD, e2 =-1, 03 = {s - 2)/2 - Vd of respective multiplicities
(cs + 1)(cs — s) / s — 2\ (cs + 1)(cs — s) / s — 2\
mo = l, m1 =---(1-^j. ^ = "»3 =-j-+
Due to a result of Gardiner [5, Proposition 5.1] the eigenvalues of A are integral if s / 2. Hence for odd s the number 2\f~D is an odd integer (since D = cs + 1 — s + s2/4), while for even s > 2 already the number a/D is an integer.
Let us construct a graph A by adding a coclique A to A, whose vertices are identified with the fibres of A, together with a vertex b such that A(b) = A, and assuming that a vertex F € A is adjacent to just those vertices of A which belong to the fibre F € £. Note that each vertex from A has valency r + 1 in A.
It is easy to see that A is a geodetic graph of diameter two and hence by [3, Theorem 1.17.1] either A is a strongly regular graph and s = 1, or A is a BRG-graph with valencies r + 1 and k + 1, r < k, s > 2 and the following statements hold:
(1) if A and B denote the sets of vertices of A of valencies r + 1 and k + 1, respectively, then A is a coclique, for each vertex a € A the subgraph [a] is a coclique, and if x and y is a pair of adjacent vertices from B, then |[x] n [y]| = k — r = s — 1;
(2) |A| = (r + 1)(k + 1) + 1.
Moreover, each geodetic graph of diameter two that has no vertex adjacent to all others, can be viewed as the underlying graph of a Moore geometry, i.e. an incidence system of points and lines which satisfies the following axioms:
(i) there is at least one line, and each line has at least two points;
(ii) two points are on at most one line;
(iii) no point is collinear with all others;
(iv) two non-collinear points are both collinear with exactly one common point;
(v) a point not in a line is collinear with at most one point of the line;
(vi) there are no triangles or quadrangles of lines.
Also, by [5, Proposition 5.2], if s = 1, then A is a Moore graph (and c € {2,6, 56}). In what follows we assume that s > 2, so A is a BRG-graph, and its corresponding Moore geometry is said to have type (cs + 2, cs — s + 3, s + 1) in this case (see [12, p. 314]).
We say that A has type Da, if there is a projective plane (X, L) of order a = r + 1 with a polarity n such that A is isomorphic to the graph on X, in which two vertices x and y are adjacent if and only if x € yn (wherein k = r + 1 and A coincides with the set of absolute points of the polarity n).
Lemma 1. The following statements hold:
(1) c > 2, the number cs + 1 is odd, s < c, and the neighbourhood of each vertex in A is the disjoint union of c isolated cliques of size s;
(2) if c = 2, then A is a unique distance-regular graph with intersection array {4,2,1;1,1,4} (the line graph of the Petersen graph) and A has type D4;
(3) if 2 < c < 1000, then either s = 2 and A has type V2c, or the pair (s; D) is one of: (4; 25), (4; 49), (3; 169/4), (6; 100), (9; 625/4), (4; 81), (11; 1225/4), 4; 121), (18; 784), (4; 169), (35; 8649/4), (4; 225), (10; 676), (4; 289), (4; 361), (21; 7921/4),(46; 4900), (4; 441), (11; 5625/4), (4; 529), (4; 625), (26; 4356), (14; 2500), (4; 729), (4; 841), (4; 961), (4; 1089), (4; 1225), (8; 2601), (15; 20449/4), (4; 1369), (5; 7569/4), (4; 1521), (9; 14161/4), (4; 1681), (152; 70225), (4; 1849), (4; 2025), (20; 10201), (4; 2209), (4; 2401), (144; 93025), (4; 2601), (56; 38025), (44; 30625), (114; 81796), (4; 2809), (4; 3025), (4; 3249), (4; 3481), (7; 25281/4), (4; 3721), (4; 3969).
Proof. The first two statements follow by [3, Proposition 1.17.3] and [8, Theorem 3.4].
To prove the third statement, first observe that the number of cliques of size s + 1 of A equals (cs + 1)(cs — s + 1)c/(s + 1). Then, for 2 < s < c < 1000, the computer check in GAP (which uses integrality conditions for the eigenvalues of A and their multiplicities together with the condition of integrality of the number (cs + 1)(cs — s + 1)c/(s + 1)) gives just those feasible pairs (s; D) that are listed in (3). The lemma is proved. □
The above restrictions for parameters of A will be frequently used in following arguments, in particular, the list of feasible parameters from Lemma 1 (3) will be needed in Section 3 to rule out the existence of A in a series of special cases.
Lemma 2. Let $ be an amply regular graph with ^ = 1 and suppose there is an automorphism g of $ such that for a (g)-orbit ^ each vertex x € ^ is adjacent to xg. Then ^ is a cycle or a clique.
Proof. Suppose ^ is not a clique. Denote by i the least number in {2,..., — 1} such that the vertices x and xgi are not adjacent. Then {xg,xgi } C [x] n [xgi] and hence i = 2. Now let j denote the least number in {3,..., — 1} such that the vertices x and xgj are adjacent. Then {xg ,xgj} C [x] n [xgj ] and hence j = — 1. Thus, we conclude that ^ is a cycle.
The lemma is proved. □
Recall that (x, y) denotes the greatest common divisor of x and y.
Lemma 3 [16, Lemma 2.2.1]. The graph A has exactly cs + 1 fibres, each of size cs — s + 1, and the following statements hold:
(1) s + 1 divides c(c — 1)(c — 2) and each odd prime divisor of D divides (s — 2, c)(4c + 1, s — 4)(c — 1, s2 + 4);
(2) if c > 2, then there is a divisor d of cs such that s = d(d — 2)/(c — d) and y/D = (d + c(d — 2)/(c — d))/2, and if an odd prime p divides (D, s — 2), then the p-part of d is less than p-part of c;
(3) if cs = 2n, then s = 2;
(4) if s = 2, then A has type D2c.
P r o o f. Note that A has exactly cs + 1 fibres, each of size cs — s + 1.
(1) Since there are exactly c(cs + 1)(cs — s + 1)/(s + 1) cliques of size s + 1 in A, s + 1 divides c(c — 1)(c — 2). Let p be an odd prime divisor of D. Then p divides (cs + 1)(cs — s)(s — 2), (p, c) = (p, s — 2) and (p, cs + 1) = (p, s2/4 — s) = (p, 4c+1). So, we conclude (p, c — 1) = (p, 1+s2/4).
(2) Let c > 2. Put D = y2 and y — s/2 + 1 = d. Then
y2 — (s/2 — 1)2 = cs, y + s/2 — 1= cs/d.
Further,
y = (d + cs/d)/2, s/2 — 1 = (cs/d — d)/2,
hence
s = d(d — 2)/(c — d), y = (d + c(d — 2)/(c — d))/2. Suppose an odd prime p divides (D, s — 2). As
s — 2 = (d2 — 2c)/(c — d),
we get that p-part of d is less than p-part of c.
(3) Let cs = 2n. Suppose s > 2. Then
2n + (s/2 — 1)2 = D
is a square of a positive integer y. Hence
y — s/2 + 1 = 21, y + s/2 — 1 = 2n-1, y = 21-1 +2n-1-1, s = 2n- — 21 + 2.
Since s is a power of 2, we find l = n — l or l = 1. If l = 1, then s = 2n-1, which implies c = s = 2, while if l = n — l, then s = 2, a contradiction in both cases.
(4) If s = 2, then by [3, Proposition 1.17.2] A has type D2c (and thus it can be constructed on the points of a projective plane of order q = 2c = D with a polarity n, whose absolute points form a line A).
The lemma is proved. □
2. Automorphisms of (k + 1, r, 1)-covers
In this section, we prepare some technical results about automorphisms of A, which will be needed for the proof of Theorems 1 and 2.
The permutation representation of a group G < Aut(A) in its natural action on the vertex set of A gives rise to a matrix representation G ^ GLv (C). Recall that Cv is the orthogonal direct sum of the eigenspaces W0, ..., W3 of the adjacency matrix of A, where W corresponds to the eigenvalue As each Wj is a G-invariant subspace, it affords a character, say , of G. We can calculate values of this character using the theory of association schemes (see [4, Section 3.7]). Namely, let Q be the second eigenmatrix of A. (We assume that the first column of Q consists of the multiplicities mi's.) Then, for an element g € G, one has
1 3
Kü) = -^QijuAa), j=0
where aj(g) denotes the number of vertices x of A such that d(x,xg) = j. Recall that every character value must be an algebraic integer; in particular, if the value is rational, then it is an integer. The second eigenmatrix Q for A was determined in [16].
Lemma 4 [16, Lemma 2.2.2]. If g € Aut(A), then
Xi(g) = f(c2s2 - es2 + s/2 + VD- l)ao(g) + (s/2 + Vd - 1) x
(cs — s + 1)(2csvD) V
x(cs-s + l)cx\(g) - (cs - s/2 - Vd + l)a3(g)) - (cs + l)/(2/D),
( } = M9)+ as(g) _ XKyj cs-s + 1
Lemma 5. If a3(g) = v for an element g € Aut(A), then s > 2.
Proof. Suppose a3(g) = v for an element g € Aut(A). Then by Lemma 4 we have
~ — s/2 + l)(cs + 1)
xiig) = --
Now if s = 2, then Xi(g) = —(cs + 1)/2 € Z, but cs + 1 is odd, a contradiction. The lemma is proved. □
Lemma 6. If a2(g) = v for an element g € Aut(A), then s = 2.
Proof. Suppose a2(g) = v for an element g € Aut(A). Then by Lemma 4 we have
, , cs + 1 XWJ) =
If s is even and s > 4, then 2 divides cs + 1, which is an odd number by Lemma 1, a contradiction. Suppose s is odd. Then c is even and 4D = 4(cs + 1) + s(s — 4) divides (cs + 1)2, hence
(4(cs + 1) + s(s — 4))x = (cs + 1)2
for an integer x. Put y = (x, cs + 1). Then x = ay, cs + 1 = by, (s, by) = 1 and b > a. Thus
s(s — 4)a
4 ay + b = by,
and b divides s — 4. Hence c > s > a and
(s — 4)a
4 a + s^-—— = b, by
implying by divides (s — 4)a. But then s2 + 1 < cs + 1 < s(s — 4), a contradiction. The lemma is proved. □
In Lemmas 7-9 it is supposed that there is an element g € Aut(A) of prime order p and Q = Fix(g). For a vertex x € A, we put
Ri(x) = {y € [x] | d(y,yg) = i},
where i = 0,1,2,3, and by F(x) we denote the fibre of A containing x.
Lemma 7 (cf. [16, Lemma 2.2.3], [12, Theorem 4.10 (i)-(ii)]). Suppose Q = 0. Then
«3(g) = (cs — s + 1)t
with t = |Fixs(g)|, and s = 2 or the number cx\(g) + si/2 — cs — 1 is a multiple of \[~D; moreover, the following statements hold:
(1) if p = 2, then st is odd, a1(g) = (cs — s + 1)(cs + 1 — t), and if a3(g) < v, then t = s(c — 1) and the set {x € A | d(x, xg) = 1} is the disjoint union of cs — s + 1 isolated cliques of size s + 1;
(2) if p = 3 and (3, s + 1) = 1, then a1(g) = 0, the number cs — s + 1 is divisible by 3,
-(cs + 1 )(V~D - s/2 + 1) + 3l(VD - s/2)
Xiig) =
where I = (cs + 1 — t)/3, and (cs + l)(s/2 — 1) — 3ls/2 is a multiple of \[~D. Proof. First, note that (cs — s + 1, cs + 1) = 1.
If d(u, ) = 3 for a vertex u, then F(u) = F(ug) and hence p divides cs — s + 1 (the size of a fibre). In particular, if s is even, then p > 2.
By the integrality of x2(g), it follows that a3(g) = (cs — s + 1)t for a non-negative integer t. Further,
ai{y) ~ (VD — s/2)t — (cs + 1)
Xi(g) =
and if s > 2, then cx\(g) — (\/D — s/2)t — (cs + 1) is divisible by 2\[~D. (1) Let p = 2 and
$ = {x € A | d(x,xg) = 3}.
Note that d(x,xg) = 2 for any vertex x of A (as otherwise [x] n [xg] C Q, which contradicts our assumption), so that
A \ $ = {x € A | d(x,xg) = 1}
and a2(g) = 0. Since cs + 1 is odd, g fixes a fibre of A and st is odd. Also, we have a1(g) = (cs — s + 1)(cs + 1 — t) and
v , , (cs - s + l)(cs + 1 - t) - (VP - s/2)t - (cs + 1)
Xlig) =-WD- G
Suppose x € $. Then [x] = R1(x) U R3(x) and |R3(x)| = t — 1. Note that, for each edge {y,z} C [x], we have z = yg (otherwise yg € [x] n [xg], which is impossible by assumption).
If [x] contains an edge {y,z} C A \ $, then [xg] contains the edge {yg, zg} C A \ $, and, since {y,z} C R1(x), we get that {z,y,yg,zg} is a 4-cycle or a clique, a contradiction. Hence each (s + 1)-clique that contains an edge from A\ $, is contained in A\ $ itself. Since A\ $ is a regular graph of valency cs — t and a1 (g) = |A \ $|, we conclude that A \ $ is an edge regular graph with = s — 1 and the number of its edges equals (cs + 1 — s)(cs + 1 — t)(cs — t)/2. It follows that there are exactly (cs + 1 — s)(cs + 1 — t)(cs — t)/((s + 1)s) cliques of size s + 1 in A \ $.
Now suppose that x € A \ $. Then [x] = R1(x) U R3(x), |R3(x)| = t and, as it was proved above, [x] n [xg] C A \ $. Note that for each vertex y € R1(x) \ {xg} we have yg € [xg] n [y] and since {x, y, yg, xg} cannot be a 4-cycle, we get {y, yg} C [x] n [xg]. This implies |R1(x)| = s. On the other hand, |R1(x)| = cs — t, which, by the preceding equality, implies that s(c — 1) = t. Hence, there are exactly cs + 1 — s cliques of size s + 1 in A \ $ and a1(g) = (cs — s + 1)(s + 1), which implies that A \ $ is the disjoint union of cs — s + 1 isolated cliques of size s + 1.
(2) Let p = 3 and (3,s + 1) = 1. Then a1(g) = 0 (otherwise there is a (unique) (s + 1)-clique L that contains a 3-cycle {u, ug ,ug }, yielding L = Lg, which contradicts the assumption Q = 0). If 3 divides cs + 1, then a2(g) = v and by Lemma 6 we obtain s = 2, a contradiction. Assume that there are exactly 31 fibres that are not fixed by g. Then a2(g) = 31(cs — s + 1) and a3(g) = (cs + 1 — 31)(cs — s + 1). Hence
-(cs + 1)(Vd - s/2 + 1) + 3 1(Vd - s/2)
Xlig) =-wb-G z
and (cs + l)(s/2 — 1) — 3ls/2 is a multiple of \[~D. The lemma is proved. □
Remark 1. Note that Lemma 7 (1) specifies statements of [12, Theorem 4.10 (i), (ii)], and Lemma 7 (2) corrects [16, Lemma 2.2.3] (namely, the condition (3,s + 1) = 1 is missing there).
Lemma 8 (see [16, Lemma 2.2.4], [12, Theorem 4.10 (iv)-(vi)]). Suppose Q = 0 and p = 2. Then one of the following statements holds:
(1) Q is a fibre of A, and either s = 2, or cs'2 — s'2/2 + s — cx\(g) is a multiple of \/D;
(2) Q is an (s + 1) -clique and c = s = 2;
(3) Q is an (c's' + 1, c's' — s' + 1,1)-cover, and the parameters c, s, c' and s' satisfy the following equality:
( c'(s — s') \
(cs — s + l)(cs — c s ) = (c s + l)(c s — s + 1) ( cs — c s + —-—— J;
(s + 1)
moreover, (i) s' = 2 and 2c = (2c')2 if s = 2, and (ii) cs — s + 1 = (c's ' — s' + 1)2 if s > s' > 1.
Proof. Let p = 2 and a € Q. Note that, for each vertex e € Q n A3 (a), the valency of e in Q coincides with that of a. Indeed, for each vertex x € Q1(a), there is a unique vertex x' € [x] n [e], and x' € Q. Conversely, {x} = [x'] n [a].
(1) Suppose that all vertices in Q are at pairwise distance 3 in A and |Q| = w. Suppose further that for some vertex a € Q we have F(a) \ Q = 0 and let u € F(a) \ Q. Then d(u, ug) = 3 and for each vertex x € [u] we get x± C A \ Q, hence d(x,xg) = 2. If d(x,xg) = 3 for all x € [u], then by [8, Corollary 6.3] g = 1, a contradiction. It follows that there is a vertex x € [u] such that d(x, xg) = 1 and s = |[x] n [xg]| + 1 is odd. Hence w is even. Now if, for a vertex y € [u], d(y, yg) = 3, then g fixes a vertex in F(y) that has a neighbour in Q, which contradicts our assumption. It follows that R1(u) = [u]. Since s > 3, we may assume that [u] contains an edge {x,y}. Then [ug] contains the edge {xg,yg} and, as [ug] n [u] = 0, we get that {xg,x,y,yg} is a 4-cycle, a contradiction.
Hence, w = cs — s + 1, that is Q = F(a). Then a3(g) = 0, a2(g) is divisible by 2s, a1(g) = csw — a2(g) is divisible by 2s, and
Xi(9) = ^ ((cV " cs2 + s/'2 + ^ - 1) + № + VD- Dai(9)) ~
(cs + 1)/(2\f~D) = ((cs - s)(\/& - s/2 + 1) + ai(g) - cs)/(2\/Z>).
Thus, s = 2 or 2\[D divides s(c - 1 )(VD - s/2 + 1) + ai(g) - cs.
(2) Suppose Q is an w-clique. Then 1 < w < s + 1. Suppose further that there is a vertex x € A \ Q that has no neighbours in Q. Clearly, d(x,xg) = 2, and if d(x,xg) = 1, then, since |[a] \ Q| = cs — s is even, we get that s is even and [x] n [xg] contains a vertex from Q, a contradiction. Hence d(x,xg) = 3. Furthermore, each vertex of Q has exactly cs — w + 1 neighbours in A\Q, among which there are exactly cs — s vertices that do not belong to the maximal clique of A containing Q. Hence there are exactly s — w + 1 + w(cs — s) vertices in A \ Q that have a neighbour in Q. Thus, a1(g)+a2(g) = s+1—w+w(cs—s) and a3(g) = w(cs—s). Then v = w+s+1—w+w(cs—s)+w(cs—s), which implies c = s = 2 and w = 3.
(3) Suppose Q contains a pair of vertices a and b such that d(a, b) = 2. Put [a] n [b] = {c}. Then [a] contains a unique vertex e € A3(b) (which, obviously, belongs to Q) and Q1(b) contains a unique vertex f € A3(a). Further,
|Q1(a) n A2(b) \ cx| = |Q1(b) n A2(a) \ cx|.
Let X1,..., denote the fibres that intersect Q. Then a vertex in X1 nQ has a unique neighbour in each of the fibres X2, , hence Q is a regular graph of valency n — 1 and |Q| = n|Xj n Q|. Moreover, Q is a (|Q|/n)-cover of an n-clique, in which any two non-adjacent vertices from distinct fibres, say Q n X and Q n Xj, have exactly one common neighbour. It follows by [8, Lemma 3.1] that Q is an (c's' + 1, c's' — s' + 1,1)-cover, where c's' = n — 1 and, clearly,
s' — 1 = |Q1(x) n Q1(y)| = s — 1 (mod p).
Note that there are exactly n(n — s')(cs — c's') edges between Q and A\Q, and there are exactly c'n(n — s')/(s' +1) maximal cliques in Q. Hence we find that the number of vertices of A that have exactly s' + 1 neighbours in Q equals
7y+1 := c'n(n — s')(s — s')/(s' + 1), and the number of vertices of A that have exactly one neighbour in Q equals
Ti := — s')(cs — c's).
Clearly, if there is a vertex x € A \ Q that has no neighbour in Q, then F(a) C Q for all a € Q, and, as above, we obtain d(x,xg) = 2. Put
$ = {y € A | d(y,yg) = 1, [y] C A \ Q}.
First we prove that |Q| + n = s(c — 1) in the case $ = 0. Suppose x € $. Since g fixes the subgraph [x] n [xg] and [x] C A \ Q, it follows that s is odd. We have |R1(x)| = s, |R3(x)| = n (since d(w, W) = 3 if and only if g fixes a vertex in F(w)) and |R2(x)| = cs — |R1(x)| — n.
Let us compare the sizes of the sets R2(x) and R3(x). As Q contains no vertices from F(x) U F(xg), we get that Q contains a vertex b € A2(x) n A2(xg), and, since the number p22 = s(c — 1)(cs — 2) is even, the number of vertices in A2(x) n A2(xg) n Q is also even. For the vertex y € [b] n [x] we have yg € [b] n [xg] and d(y,yg) = 2 (otherwise {x,xg ,y,yg} is a clique and [b] n [x] contains y,yg, which is impossible). Pick a vertex w € [x]. If d(w,wg) = 2, then w,wg € [a] for a vertex a € Q, x has a unique neighbour u € F(a), xg has a unique neighbour ug € F(a) = F(u) and {a, w,x,xg, wg} is a 5-cycle. If d(w,wg) = 3, then w € F(wg) and, for each vertex a € Q such that a € A2(x) n A2(xg) n F(w), we get that {a, u, x,xg,ug} is a 5-cycle, where {u} = [a] n [x]. Since for each vertex w € R3 (x) there are exactly c's' — s' + 1 vertices in QnF(w)(nA2(x)), there are exactly c's' — s' +1 vertices y such that {y} = [x] n [a] C R2(x), where a € Q n F(w). Hence,
|R2(x)| = (c's' — s' + 1)|R (x)|,
which implies
s = cs — n(c's' — s ' + 2), and cs — s = n(c's ' — s' + 2),
that is
|Q| + n = s(c — 1).
Now consider the BRG-graph A and note that its corresponding Moore geometry G has type (cs + 2, cs — s + 3, s + 1) (in notation of [12, p. 314]). Since Q ^ zx for any vertex z € A, {b} U Q induces a subgeometry of G (recall, b denotes the vertex of A isolated in B), so by Lemma 7 and [12, Theorem 4.10] we obtain that one of the following three possibilities occurs: (i) s' = 1; (ii) s = s '; (iii) s > s' > 1 and cs — s + 1 = (c's' — s' + 1)2 (or equivalently, s(c — 1) = s'(c' — 1)(c's' — s' + 2)).
Hence $ = 0 and each vertex in A \ |JXj has exactly one or s' + 1 neighbours in Q and, for all
j
vertices x € A such that xx C A \ Q we have d(x, xg) = 3. Thus,
«3(g) = v — |Q| — T1 — 7V + 1,
and, on the other hand,
a3(g) = (c's' + 1)(cs — s + 1 — c's' + s' — 1),
which together give
c (s s )
(cs — s + 1) (cs — c's') = (c's' + 1) (c's' — s' + 1) ( cs — c's + —-—— J.
(s + 1)
In particular, if s = 2, then, since s' < s and
s' — 1 = |Q1(x) n Q1(y)| = s — 1 (mod 2),
we get s = 2 and
cs — s + 1 = (c s + 1)(c s — s + 1),
so that
2c — 1 = (2c' + 1)(2c' — 1) and 2c = (2c' )2.
The lemma is proved. □
Remark 2. Note that Lemma 8 specifies statements of [12, Theorem 4.10 (iv)-(vi)] and of [16, Lemma 2.2.4]. Also, the proof of Lemma 8 fills a gap in the proof of [16, Lemma 2.2.4 (3)], in which the case $ = 0 was not excluded properly.
Lemma 9 [16, Lemma 2.2.5]. If Q = 0 and p > 2, then one of the following statements holds:
(1) Q is contained in a fibre of A;
(2) Q is an w-clique and w < s + 1;
(3) Q is an (c's' + 1, c's' — s' + 1,1)-cover, where c's' + 1 is the number of fibres of A intersecting Q, and s' — 1 = |Q1(x) n Q1(y)| = s — 1 (mod p).
Proof. Let a € Q. Then for each vertex e € Q n A3(a) we have |Q1(a)| = |Q1(e)|.
Clearly, if Q consists of vertices that are at pairwise distance 3 in A, then the statement (1) is true, while if Q is a w-clique, then w < s + 1 and the statement (2) holds.
Now let Q contain two vertices a and b such that d(a, b) = 2. Then [a] contains a unique vertex that belongs to A3(b) (and, obviously, to Q) and Q1(b) contains a unique vertex that belongs to A3(a). Put [a] n [b] = {x}. Then |Q1(a) n A2(b) \ xx| = |Q1(b) n A2(a) \ xx|.
Let X1,...,Xn denote the fibres of A that intersect Q. Then each vertex in X1 n Q has a unique neighbour in each of the fibres X2, ...,Xn, hence Q is a regular graph of valency n — 1 and |Q| = n|Xj n Q|. Moreover, Q is a (|Q|/n)-cover of a n-clique, in which any two non-adjacent vertices from distinct fibres, say Q n Xj and Q n Xj, have exactly one common neighbour. Thus, the remaining claims of (3) follow by [8, Lemma 3.1].
The lemma is proved. □
3. Proofs of Theorems 1 and 2
In this section, we prove Theorems 1 and 2.
From now on we assume that there is a subgroup G < Aut(A) that induces a 2-homogeneous permutation group Gs on the set £ of fibres of A and we denote by K the kernel of the induced action of G, so that G/K ~ Gs. We also put m := |£| = cs + 1.
Note that by [8, Corollary 6.3] K is semiregular, in particular, a3(g) = v for each non-trivial element g € K. It also implies that K acts semiregularly both on the set of arcs of A and on the set of its cliques of size s + 1. For each subgroup X < K, we denote by AX the graph on the set of X-orbits, whose edges are the (unordered) pairs of X-orbits that are joined by an edge of A. In particular, if 1 < |X| < r, then by [8, Theorem 6.2, Corollary 6.3] AX is a non-bipartite (m, (cs — s + 1)/|X|, |X|)-cover.
First, in Lemma 10, we consider the case, when the group is 2-homogeneous, but not 2-transitive.
Lemma 10. If the group
is not 2-transitive, then Gs < ArL1(q), |£| = q = 3 (mod 4), Z, and either s = 2, c = (q — 1)/2 and A is a graph of type Dq-1, or s is odd.
Proof. Suppose that G induces 2-homogeneous, but not 2-transitive permutation group Gs on E. As m > 5 is odd, it follows by [11, Theorem 1] that G/K ~ Gs < ATLi(g) and |E| = q = 3 (mod 4). If \[D € Z, then cs = D — (s/2 — l)2 is an even difference of squares, which is impossible in this case. Hence \[~D qL Z and either s = 2 or c = (q — l)/2 is odd, or s is odd. Finally, if s = 2, then by Lemma 3 we obtain that A is a graph of type Dq-1. The lemma is proved. □
Further in Lemmas 11-14, we assume that the group is 2-transitive; in this case by a Burn-side's theorem, the group Gs is either almost simple, or affine, and we consider the corresponding cases in course, basing our argument on the classification of finite 2-transitive permutation groups (e.g. see [9, Theorem 2.9]).
Lemma 11. Suppose the group Gs is almost simple. Then either c = 2n-1, s = 2, A is a graph of type D2n, Soc(G) — L2(2n) and A is a Mathon graph, or the group G acts intransitively on the vertices of A.
Proof. Suppose that G induces an almost simple permutation group Gs on E. Then the socle H of Gs is a non-abelian simple group. In view of Lemma 1, we may assume that s = 2 (and the size of a fibre is m — 2) or |E| > 25. Fix F € E and a € F.
If H = Sp2n(2), then the number m € {2n-1(2n ± 1)} is even, a contradiction.
In the case H = 2G2(q) we have q = 32e+1 and m = q3 + 1 is even, a contradiction.
In the case H = U3(q) we have m = q3 + 1, hence q = 2e. By Lemma 3 we have s = 2 and A is a graph of type D23e. Then by Lemma 5 we have K = 1 and, since G{F} contains no subgroup of index q3 — 1, G cannot act transitively on the vertices of A.
In the case H = 2B2(q) we have q = 22e+1 and m = q2 + 1. By Lemma 3 we get s = 2 and A has type D22(2e+i). By Lemma 5 it follows that K = 1 and, since G{F} contains no subgroup of index q2 — 1, G cannot act transitively on the vertices of A.
If H is a Mathieu group Mm, then (since m is odd) m € {11,23} and, by Lemma 1 we have s = 2.
If the pair (H,m) is one of (L2(11), 11), (Mn, 12), (Altr, 15), (L2(8),28), (HiS, 176), or (Co3,276), then (since m is odd) m = 11, H = L2(11) or m = 15, H = Alt7 and, by Lemma 1 we have s = 2.
If m = 23, then by [8, Theorem 5.4] (—1)c-1 (2c — 1) = 21 = z2 (mod 11) for some z € Z and by the Euler's criterion 215 = 1 (mod 11), a contradiction.
Similarly, for m = 15 by [8, Theorem 5.4] we get (—1)c-1 (2c — 1) = 13 = z2 (mod 7) for some z € Z and by the Euler's criterion 133 = 1 (mod 7), a contradiction.
If m = 11, then by Lemma 5 we have Gs — G, and either H = L2(11) and H{f} — Alt5, or H = M11 and H{F} — Alt6 : Z2. But in both cases G contains no subgroup of index 99, hence G cannot act transitively on the vertices of A.
It remains to consider "alternating" and "linear" cases.
1. Let H = Alt
m. Then H contains an involution g that is a product of two independent transpositions and the number of the fibres that are fixed by g equals m — 4.
Note that for m = 5 we have c = s = 2 and by Lemma 5 we have L2(4) — Alt5 < Gs — G, and, moreover, if G is vertex-transitive, then it has a single orbit on arcs of A as well. So let m > 7.
1.1. First, suppose
K = 1. Then G ~ Gs and we may identify H with the socle of G.
Put Q = Fix(g). If Q = 0, then by Lemma 7 we obtain m — 4 = s(c — 1), that is s = 3, m > 13. In this case, H contains an involution g that is a product of four independent transpositions and the number of the fibres that are fixed by g equals m — 8, and by Lemma 7 we have Q = Fix(g ) = 0 (otherwise m — 8 = 3(c — 1), which is impossible). Hence Q' is distance-regular and its parameters
satisfy the equality given in Lemma 8(3), which, in view of Lemma 1, contradicts the restriction m > 13.
Hence by Lemma 8 we obtain that Q is an (c's ' + 1, c's ' — s' + 1,1)-cover,
( c' (s — s') \
4(m — s) = (m — 4)(m — s ) f m — 1 — c s + —-—— J
(s + 1)
and m = 7, that is c's' = 2 and Q is a 6-cycle. But s — 1 = s' — 1 (mod 2) and hence A has intersection array {6, 5,1; 1,1,6}, which contradicts the assumption s > 1.
1.2. Now let K > 1. If |K| is odd or coincides with the size of a fibre (so that G{F} = K : Ga), then there are involutions g € G \ K with |Fix^(g)| = m — 4 or m — 8, and we proceed as in the subcase 1.1.
Suppose that 1 < |K| < cs — s + 1 and G acts transitively on vertices of A. Then
Altm-1 < (G{f})s(^ G{f}/K) < Symm-i,
and the graph AK admits a vertex-transitive action of G/K, and the size of a fibre in A1 = AK is r' = (cs — s + 1)/|K|. If r' = 2, then G/K is a distance-transitive group of automorphisms of A1 with Altm-1 < GaK/K < (Aut(A1))x for some vertex x € A1, which implies that A1 is bipartite, a contradiction. Hence r' > 3. But the degree of a minimal permutation representation of Altm-1 is m — 1 unless m < 5, so we obtain either m = 5 and s = 1, or m > 7 and
Altm-1 < G„K/K < (Aut(A1))x
for some vertex x € A1 (and hence r' = |G{F}/K : GaK/K| < 2), a contradiction in both cases.
2. Next we assume H = Ld(q). Then £ can be regarded as the set of 1-dimensional subspaces of V = Fqd. Note that, since
(qd ~ 1)
m = ---
(9-1)
must be odd, q is even or d is odd.
Let d = 2. Then q = 2n,m = q + 1 and by Lemma 3 we have s = 2, which by Lemma 5 implies L2(q) <! G < PrL2(q). Note if the group G is vertex-transitive, then it has a single orbit on arcs of A, and moreover, its socle is also arc-transitive (otherwise A would be bipartite or disconnected, which is impossible), and hence A is a Mathon graph (see [3, Proposition 12.5.3]). Suppose further that d > 3 and fix a basis e1, e2,..., ed of V.
2.1. Assume first that K = 1. In the argument below, we identify H with the socle of G and consider various involutions g € H and subgraphs Q = Fix(g) of their fixed points.
2.1.1. Suppose q is odd. Then d is odd and there is an involution g € H such that its preimage in SL(V) fixes ed and, for all i € {1,2,..., d — 1}, it maps ej to —ej, so that
(qd-1 — 1)
If Q = 0, then s is odd and by Lemma 7 we have
(qd-1 — 1)
(q - 1)
But then
+ 1 = s(c - 1).
S = SC + 1-S(C-1)-1 = (9^1) _ (g^-1 - 1) _2= ä-i_2
(q - 1) (q - 1)
and
ld 1 1} +l = (/"1-2)(c-l),
(q — 1)
which contradicts the condition c > s > 2.
Hence by Lemma 8 we have that Q is an ((qd-1 — 1)/(q — 1) + 1, (qd-1 — 1)/(q — 1) — s', 1)-cover
and
- 1) - s) = (<fl=i> + l) ifc^ - ✓ + 0 (cs - c's + C'(S "
(q — 1) ) V (q — 1) A (q — 1) A (s' +1)
Hence (qd-1 — 1)/(q — 1) + 1 divides (q — 1)(s+q — 1), d = 3 and s = (q+11)/3. But (q+11, 3q(q+1)) divides 330, which implies that the corresponding equation has no solution in natural numbers, a contradiction.
2.1.2. Now let q be even.
2.1.2 (a). If d = 2/ + 1, then we assume that a preimage of g in SL(V) fixes ed and, for 1 < i < d — 1, interchanges ej with ed-1-i. Then
|FixE(9)| =
If Q = 0, then s is odd and by Lemma 7 we have
(qf+1 — 1)
(q - 1)
that is
= s(c - 1),
d-1 + ... + qf+1- 1
s = q" + ... + q
and
which contradicts the condition c > 2.
Hence, by Lemma 8 we have that Q is an ((qf+1 — 1)/(q — 1), (qf+1 — 1)/(q — 1) — s' — 1,1)-cover
and
(9-1) r (9-1)" (9-1) ^ (9-1) ACS CS+(S' + 1)J-
Hence
s = qf-1 + ... + q + 1, s + qf = sq + 1, c = q(qf + 1)
and
j2f+i (9-^-1) ^(9/+1~l) A f j c'{s-s')y Q (o-l) (o-l) {s, + l) )>
- 1) V (q - 1) A (s' + 1)
and, since qf > s > s', we get f = s = 1, a contradiction.
2.1.2 (b). For d = 2f, we assume that a preimage of g in SL(V) fixes both ed-1 and ed, and interchanges ej with ed-2-i, so that
(qf+1 -1)
|Fixs(g)| =
(q - 1)
If Q = 0, then s is odd and by Lemma 7 we have
(g/+1-D c(c u
that is
s = qd-1 + ... + q/+1 - 1
and
(g./+1"1} =(gd-1 + - + g/+1-l)(C~l), (q-1)
which contradicts the condition c > 2.
Hence, by Lemma 8, Q is an ((qf+1 - 1)/(q - 1), (qf+1 - 1)/(q - 1) - s', 1)-cover and
- s)^'1= {qr -i}. ({qi;l~i) - s>) (cs - ds+.
\ (q- 1) Jq (q-1) (q-1) \ (q - 1) A (S' +1) J
If
((Qf+1-1) (g^-ih = , I (q-1) ' (q-1) J '
-1) ' (q -1)
then f is even, qf + ... + q + 1 divides sq2 + q + 1, sq2 = x(qf + ... + q + 1) — q — 1, q2 divides x — 1 and
s > (q2 + 1)(qf-2 + ... + q + 1) + q + 1.
But s < c — 1 and cs = q2f-1 + ... + q2 + q, a contradiction. Hence, f is odd and
/(qW-1) (g/-i-lK
V (q-1) ' (q-1) ) q+ "
It follows that (qf+1 — 1)/(q2 — 1) divides sq2 + q + 1, sq2 = x(qf + ... + q + 1) — q — 1, q divides x — 1, q2 divides x — (q + 1) and x = zq2 + q + 1 for a positive integer z. But
(q^-1) 2 z^-Vf 2/ (q-1) >S > (q2 — l)2 ><? '
again a contradiction.
2.2. Now let K > 1. If |K| is odd or coincides with the size of a fibre, then there are involutions g € G\K with |Fixs(g)| as in the subcase 2.1, and by a similar argument we come to a contradiction. Suppose that 1 < |K| < cs — s + 1, s is odd and G acts transitively on vertices of A. Let H denote the full preimage of H(= Soc(Gs)) in G and put t = |H{F} : Ha|. We have
^ < PrLd(q), H{F}/K - (H{F- Eqd-i ■ SLd-i(q) ■ Z(,-i)M,-i) (e.g. see [18]), and (iHa)[a] is permutation isomorphic to (iHaK/K)s\{F}. So
" C ^ 1-|Gq, : HaI = \G : H\ = |GS :
t
divides (q — 1)e (where q = pe for a prime p), and |H{f} : HaK| = t/|K|. Hence Ha ~ HaK/K is isomorphic to a subgroup of Eqd-i ■ SLd-1(q) ■ Z(q-1)/(d,q-1) with index t/|K| dividing
Since H{f}/K contains a normal elementary abelian group R of size qd-1 that corresponds to a subgroup of SL(V), generated by transvections g with g(e1) = e1 and g(u) — u € (e1) for all u € V, we may assume R n HaK/K = 1 (otherwise qd-1 divides
t!m tf ~ 1) , /-i | ~ 1) s tm- (q-1) ~S-Q + (q- 1)
a contradiction).
Let q and d be odd. Then there is an element h € Ha of order p such that
(qd-1 — 1)
I^wi - Vtr
and, since the valency of Q = Fix(h) is odd, by Lemma 9 Q is a clique of size
(qd-1 — 1)
- 1)
< s + 1,
a contradiction.
Let q be even. Then there is an involution h € Ha such that
(qd-1 _ 1)
and, by Lemma 8 0 = Fix(h) is an ((qd-1 — 1)/(q — 1), (qd-1 — 1)/(q — 1) — s', 1)-cover. Finally, it is easy to check that the equality given by Lemma 8(3) is not satisfied in this case, a contradiction. The lemma is proved. □
Lemma 12. If the group is affine and the group G acts transitively on arcs of A, then s + 1 divides c(pe, c — 1), s > 2, |K| = cs — s + 1 and |K| is divisible by 1 + Ip, where cs + 1 = pe, p is a prime and l is a positive integer.
Proof. Let cs + 1 = pe for a prime p, and denote by T the full preimage of Soc(Gs) in G. Since K acts semiregularly on each fibre, (|K|, cs + 1) = 1 and hence each element g € T of order p has no fixed points. Besides, K has a complement To in T that is an elementary abelian group of order pe, and K = Op(T). Hence, by [1, 37.7], NT(To) = CT(To).
Suppose that G acts transitively on arcs of A. Pick F € E. Then for each vertex a € F the group Ga acts transitively on [a]. We have T{F} = K and |T : Ta| = (cs + 1)|K|. Hence T acts transitively on the vertices of A if and only if K acts transitively on F.
Suppose that K acts intransitively on F. Since T-orbits comprise an imprimitivity system of G, each T-orbit is a coclique (otherwise, a T-orbit containing an edge induces a subgraph of valency cs in A, which is impossible by assumption). Hence a2(g) = v for each element g € T of order p, and by Lemmas 6 and 5 we have s = 2 and K = 1.
Then |F | = 2c — 1, T is a subgroup of order pe, normal in G, and each T-orbit contains a unique vertex from every fibre of A. Note that T acts semiregularly on the set of 3-cliques of A. Hence the number of 3-cliques of A is divisible by 2c + 1, which implies 3 divides c(c — 2), and p > 3.
Further, there are exactly t (t < cs — s + 1) T-orbits that intersect [a], so that cs = tj for some positive integer j. Let us show that t = c. Indeed, Ga acts transitively on the set of non-trivial elements of T and the group TGa induces 2-transitive permutation group of degree pe on aT. Now, since each vertex of [a] is adjacent to exactly j — 1 vertices from aT \ {a} and there are exactly
cs(j — 1) edges between [a] and aT \ {a}, each vertex from aT \ {a} is adjacent "on the average" to j — 1 vertices from [a]. Hence j — 1 = 1 and cs = 2t, that is t = c.
Denote by a1,...,ac the vertices of F that have a neighbour in aT \ {a}. Then the set a = {[a^] n aT}f=1 comprises an imprimitivity system of Ga on aT\{a}, and the set n = {[a]naiT}c=1 comprises an imprimitivity system of Ga on [a].
For a block {x,y} € n, there is a 2-element h € Ga that interchanges the vertices x and y, so x = yh and h2 € Ga)X;y, and hence, Ga contains an involution z that fixes {x,y}.
Suppose c > 2. Put Q = Fix(z). Then F \ {a, a1,..., ac} = 0 and by Lemma 8 we have either Q = F, or Q is an (2c' + 1,2c' — 1,1)-cover and 2c = (2c')2. But in the second case by Lemma 4 we obtain x1(g) = —(2c + 1)/(4c') € Z for an element g € T of order p, a contradiction. Hence (z) < Gf < G{f} and Ga>x (= GaXyy) is a 2'-group, that is h2 = 1 and we may assume that h = z. It implies that h also fixes the block {x',y'} € n with {x'} = [a] n [x] and {y'} = [a] n [y], and (|G|)2 divides 2c.
Thus, for each edge {x,x'} C [a], there is a (unique) edge {y,y'} C [a] such that {x,y} € n and {x',y'} € n. On the other hand, for a block {x,y} € n, there is an element h € Ga such that d(x,xh) = 1, and, since [a] is the disjoint union of edges, h2 € Ga)X>y. Without loss of generality, we may assume that h is a 2-element. But Ga,x is a 2'-group, that is h is an involution of Ga such that Q = Fix(h) = F (since h interchanges distinct orbits xT and (x')T). By Lemma 8 it follows that Q is an 2c + 1, 2c — 1, 1 -cover and 2c = (2c )2, which together with Lemma 4 imply X1(g) = —(2c + 1)/(4c') € Z for an element g € T of order p, a contradiction.
Hence c = s = 2 and again by Lemma 4 we obtain xi(g) = -(2c+l)/(2\/2c) Z for an element g € T of order 5, a contradiction.
Now suppose that K acts transitively on F. Then G{F} = K : Ga and, by Lemma 5 we have s > 2. Since K acts semiregularly on the set of cliques of size s + 1 of A, the number of cliques of size s + 1 of A is divisible by cs — s + 1 and hence s + 1 divides c(pe, c — 1).
Further, in view of Lemmas 6 and 2, for each element g € T of order p there is a (g)-orbit that is a cycle or a clique. Besides, | Sylp(T)| divides |K|. If | Sylp(T)| = 1, then To = Op(T) and, hence, T0 < G. But in this case, each T0-orbit contains an edge and hence induces a subgraph of valency cs of A, a contradiction. Therefore, |K| is divisible by 1 + Ip for some positive integer l. The lemma is proved. □
Further for a finite group X we denote by Xthe last term of the commutator series of X.
Lemma 13. Suppose that the group Gs is affine. Then the group G acts intransitively on arcs of A or Gs < ATL1(q).
Proof. Note that Gs can be identified with a subgroup of ArLd(q), where q is a power of an odd prime p. Thus, the socle T of Gs is regarded as the additive group of a linear space V with dimension d over Fq, cs = qd — 1, and the stabiliser G0 of the zero vector in Gs acts transitively on the set of non-zero vectors of V. Fix a basis e1,..., ed of V.
First, the case G0(œ) = G2(q), as well as the cases G0 € {Alte, Alt7,U3(3)} are immediately ruled out, since m = |£| must be odd.
Suppose further that G acts transitively on arcs of A. Take F € £. Then by Lemma 5 we have |K| = cs — s + 1, s > 2 and G0 ~ G{F}/K ~ Ga for each vertex a € F, thus we may identify the groups Ga[a] and G0V\{0} in what follows.
Since K acts regularly on each fibre, (|K|, cs + 1) = 1. Hence the full preimage T of Soc(Gs) in G contains an element y of order p that has no fixed points, and by Lemma 2 we obtain that each (y)-orbit containing an edge is a cycle or a clique, while by Lemma 6 we have a2(y) < v.
Note if some (y)-orbit is a clique, then y fixes an (s + 1)-clique containing it, implying p divides (c - 1,s + 1).
1. Suppose G0(œ) = SLd(q) or Spd(q), where d > 2, and take an involution g € G0(œ) that maps ej to -ej for 1 < i < 2 and fixes ej when i > 3 (in case G0(œ) = Spd(q) we assume that (e1, e2) is a hyperbolic pair). Then the number of fibres that are fixed by g equals qd-2. Put Q = Fix(g). By Lemma 8 we obtain that Q is an (c's' + 1, c's' - s' - 1,1) -cover, c's' = qd-2 - 1 and
(<T2 - 1 )(qd -S) = (qd~2 - s>) (cs - c's + •
Thus, qd-2 - s' divides (q2 - 1)(q2s' - s) and 3 = d < 8.
For 5 < d < 8, there is an involution g' € G0(œ) that maps ej to -ej if 1 < i < 4 and fixes ej if i > 5 (in case G0(œ) = Spd(q) we assume that (e^e2) and (e3,e4) are hyperbolic pairs). The number of fibres that are fixed by g' equals qd-4, and, again by Lemma 8, Q' = Fix(g') is an (c'V + 1, c''s'' - s' - 1, 1)-cover, c''s'' = qd-4 - 1 and
(94 " 1X9* -s) = (qd~' - s") (cs - c"s + ,
which contradicts the assumption 5 < d < 8.
If d = 4, then there is an involution g' € that fixes both e3 and e4, and, for 1 < i < 2,
maps ej to -ej, so that there are exactly q2 fibres that intersect Q' = Fix(g'), and, by Lemma 8, Q' is an (q2, q2 - 1 - s', 1)-cover, q2 - 1 = c''s'' and
(q4 - s) (q2 - 1) = (q2 - s») (V - 1 - c" s + .
If s'' = 1, then c'' = q2 - 1 and q4 - s = q4 - 1 - (q2 - 1)s + (q2 - 1)(s - 1)/2, a contradiction. So s > 1, and
(q4 - s)(q2 - 1) > (q2 - s") (q4 - 1 - c"S + ,
again a contradiction.
Hence, d = 2, Z(G0œ)) contains a unique involution g that, for 1 < i < 2, maps ej to -ej, A is an (q2, q2 - s, 1)-cover and, by Lemma 8, Q = Fix(g) is a fibre. This implies a0(g) = cs - s + 1 and, by Lemma 4 we have
xi{g) = (cs-s)(^D-s/2)-s + al(g) ^ ^
Suppose s is odd. Then ot\(g) = 0 (otherwise g fixes a vertex in [a,] for some a, € Q, which is impossible) and c is even. In this case the odd number 2 \[~D divides
VDs + s2(c - l)/2 + s = s{VD + s{c - l)/2 + 1).
Since (s,2y/D) = 1, 2\fD divides cs — s + 2 = q2 — s + 1. Put = (cs + 1, 2\fD). By Lemma 3 we have that x equals 1 or is a power of 3, and, since (cs — s + 2, cs — s) = 1 and (x, s — 2) = 1, 2\[~D divides x(c, s — 2). If x = 1, then s — 2 < 2\[~D < (c,s — 2), a contradiction. Hence q2 = 3e and there is an element of order 3 in G \ K that has no fixed points. But (3, s + 1) = 1 and, by Lemma 7 we obtain that 3 divides cs - s + 1 = q2 - s, which contradicts the fact (|Kcs + 1) = 1.
Thus s is even and | K| is odd. It follows that K is solvable and K is a normal subgroup of G that is properly contained in K. Hence the graph AK admits an arc-transitive action of G/K',
and the abelian group K/K ' can be considered as the group of all automorphisms of AK' fixing its fibres. But the size of a fibre in coincides with |K/K '| and (|K/K'|,cs + 1) = 1, a contradiction to [9, Theorem 2.5].
2. Let either m = p2 and p € {5, 7,11,23,19,29, 59}, or m = 36, or m = 34. As it was shown above, s > 2.
If m = p2, SL2(5) < G0 and p € {11,19,29, 59}, then we may assume by Lemma 1 that the triple (c; s; D) is one of (30;4;121), (90;4;361), (210;4;841) or (870;4;3481).
If m = 36 = 729 and SL2(13) < Go, then by Lemma 1 (c; s; D) = (182; 4; 729).
If Gs is solvable, m = p2, SL2(3) <1 G0 and p € {5, 7,11,23}, then by Lemma 1 we may assume that the triple (c; s; D) is one of (6;4;25), (12;4;49), (30;4;121), or (132;4;529).
If m = 34 = 81 and G0 contains a normal extraspecial subgroup H of order 32, then (c; s; D) = (20; 4; 81).
Since in all these cases s = 4, there is an (y)-orbit that is a cycle or p = 5 and 5 divides c — 1. Then |K| = m — 4, 5 divides c(c — 1,pe) and | Sylp(T)| = 1 + Ip divides |K|. It implies m = p2 and p € {5,11,19, 29, 59} or m = 34.
Let m = p2. We have p2 — 4 = t(1 + Ip) and (t; p) = (1; 5) (otherwise, t = t'p — 4 > 1 and p = t'(1 + Ip) — 4l, which is impossible). Hence |K| = 21. If K is cyclic, then by [8, Theorem 9.2], 21 divides m, a contradiction. Hence, the subgroup K' ~ Z7 of K is normal in G. Then the graph AK admits an arc-transitive action of G/K', and K/K' can be considered as a group of all automorphisms of AK fixing each its fibre. But K/K' ~ Z3 and hence, by [8, Theorem 9.2], 3 divides m, a contradiction.
Let p = 3. Then K is a cyclic group of order 77 and, by [8, Theorem 9.2], 77 divides m, a contradiction.
Thus, the only remaining possibility is Gs < ArL1(q). The lemma is proved. □
Lemma 14. Suppose Gs < ArL1(q), where q = pe for a prime p. Let H1 be the stabiliser of a fibre F in Gs, H = H1 n AGL1(q), f and g be two elements of H, whose orders are 2'-part and 2-part of |H|, respectively, and let z be an involution in (g). Denote by z, f and g some representatives of the preimages of the elements z, f and g in G, respectively. Then Fix(z) is a fibre,
a0(z)=q-s, a3(z) = 0, xi(z) = («1(2) + (■y/D - s/2 + l)(cs - s) - cs)/(2/D) and the following statements hold.
(1) If K = 1, then s is even, s = 2 or cs is divisible by 4, (\/D,s) < 2 and \fg\ divides a\(z), and, in particular,
(i) if s = 4, then c = d(d + 2)/4, \f~D = d+1 for some even integer d, a\(z) = 2(pe/2/ + 2), where l is an even integer, and a1(z) is divisible by (q — 1)/(e, q — 1);
(ii) if p = 3, then s + 1 is divisible by 3.
(2) G acts intransitively on arcs of A.
Proof. First we show that g = 1. On the contrary, suppose that g = 1. Then |H| is odd,
H1/H < Ze and since H1 is transitive on £ \ {F}, e is even and (e)2 > (q — 1)2 > 4. If p — 1 is
divisible by 4, then (pe — 1)2 = (e)2(p — 1)2 > (e)2, a contradiction. Hence, (p — 1)2 = 2 and the
number p2 — 1 is divisible by 4 and divides pe — 1, and again (pe — 1)2 = (e/2)2(p2 — 1)2 > (e)2, a contradiction.
(1) Let K = 1. Let z, f and g denote some representatives of the preimages of z, f and g in G, respectively. Then the involution z € G{f} does not fix any fibre from E \ {F}. If Fix(z) = 0, then a2(z) = 0 and a1(z) = cs(cs + 1 — s), that is z fixes an (s + 1)-clique and by Lemma 7 we have cs = s + 1 = 2, a contradiction. It follows by Lemma 8 that Fix(z) is a fibre, that is Fix(z) = F, a0(z) = q — s and a3(z) = 0.
Since for each nontrivial element h € (fg) we have a0(h) + a3(h) = cs — s + 1,
, (VD - s/2 + l)(a0(h) - 1) + ai(/?) - cs Xlih) =--'
hence
. (a/D - s/2)(cs - s) + ai(z) - s
X1(Z) =-WD-*
Suppose s is odd. Then a\(z) = 0 and c is even. In this case, Xi(z) € Z and the odd number
2 \[~D divides
s/Ds + s2(c - l)/2 + s = s(V~D + s(c - l)/2 + 1).
As (s, 2a/D) = 1, we also get that 2\f~D divides cs — s + 2 = q — s + 1. By repeating the argument from Lemma 13, we obtain that 3 divides (s — 1,2\/~D), q = 3e and there is an element of order
3 in G that has no fixed points. But (3,s + 1) = 1 and, by Lemma 7 we conclude that 3 divides cs — s + 1 = q — s,a contradiction.
Thus, s is even.
If cs is not divisible by 4, then c is odd. But then
D = cs + (s/2 — 1)2 = 2 (mod 4),
which yields s = 2. If s is divisible by 4, then D is odd. Hence (■>/D, s) < 2.
As the element fg does not fix any vertex u such that uz € [u] and centralizes z, we get that |fg| divides a1(z).
(1i) Let s = 4. Then
c = d(d + 2)/4, q = {d+ l)2 and VD = d + 1. Since x1(z) € Z, 2(d + 1) divides a1(z) — 4 and hence a1(z) = 2(pe/2l + 2) is divisible by (q —
1)/(e, q — 1).
Suppose that a1(z) is not divisible by 4. As c > 2, we get a1(z) > 0 and 2(q — 1, e)2 > (q — 1)2. If p — 1 is divisible by 4, then (pe — 1)2 = (e)2(p — 1)2 > 2(e)2, a contradiction. Hence, (p — 1)2 = 2 and the number p2 — 1 is divisible by 4 and divides pe — 1, and
(pe — 1)2 = (e/2)2(p2 — 1)2 > 2(e)2,
a contradiction. Thus l is even.
(1m) If y is an element of order p of the socle of G, then a.i(y) = 0 for i = 0,3 and Xi(u) = (a\(y) — q)/(2\f~D). In the case p = 3 in view of Lemma 7 we conclude that 3 divides s + 1.
(2) Suppose that G acts transitively on arcs of A. Then by Lemma 12 we have |K| = cs — s +1 and Ga ~ G{F}/K (a € F). Let z, f and g denote some representatives of the preimages of the elements z, f and g in Ga. Then we may assume that z is an involution and it does not fix any fibre from E \ {F}. Hence by Lemma 8, Fix(z) is a fibre, a0(z) = q — s and a3(z) = 0.
Since for each nontrivial element h € (fg) we have a0(h) + a3(h) = cs — s + 1,
/ _ {y/D - s/2)(cs - s) + ai(z) - s
Xl{Z) WD '
Suppose s is odd. Then a\(z) = 0 and c is even. In this case, Xi(z) € Z, and the odd number 2\f~D divides
VDs + s2(c - l)/2 + s = s(VD + s(c - l)/2 + 1).
Again, by repeating the argument from Lemma 13, we obtain a contradiction to Lemma 7.
Thus s is even and |K| is odd. It follows that K is solvable and K' < K. Hence the graph AK is a non-bipartite antipodal distance-regular graph of diameter 3 that admits an arc-transitive action of G/K', and the abelian group K/K' can be considered as the group of all automorphisms of AK fixing its fibres. But the size of a fibre in coincides with | K/K | and (|K/K'|,cs + 1) = 1, a contradiction to [9, Theorem 2.5]. The lemma is proved. □
Proof of Theorem 1 follows immediately from Lemmas 10, 11, 13 and 14. □
Proof of Theorem 2. Assume A is not a 6-cycle. First note that each edge-transitive group of automorphisms of A induces a 2-homogeneous permutation group on £. It is also clear that if there is an edge-transitive group of automorphisms of A, then it acts transitively on its vertices as well, because A is a non-bipartite graph whenever (c; s) = (2; 1). The case (c; s) = (56; 1) cannot occur, as otherwise the order of G would be divisible by 57 ■ 56, which is impossible by [17] (see also [13]). Thus, it remains to apply Theorem 1. □
4. Open Questions
We conclude with few open questions.
1. Is there a half-transitive (n,r, 1)-cover?
2. Is there any (n,r, 1)-cover with n — r > 1 that possesses a group of automorphisms acting 2-homogeneously on the fibres when n — 1 is not a power of 2 or n — r > 2?
Acknowledgements
The author thanks A.A. Makhnev and A.L. Gavrilyuk for discussions and comments on earlier versions of the manuscript.
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