Siberian Mathematical Journal, Vol. 54, No. 6, pp. 1076–1087, 2013Original Russian Text Copyright c© 2013 Makhnev A.A., Paduchikh D.V., and Tsiovkina L.Yu.

EDGE-SYMMETRIC DISTANCE-REGULARCOVERINGS OF CLIQUES: THE AFFINE CASE

A. A. Makhnev, D. V. Paduchikh, and L. Yu. Tsiovkina UDC 519.17:512.54

Abstract: Let Γ be an edge-symmetric distance-regular covering of a clique. Then the group G =Aut(Γ) acts twice transitively on the set Σ of antipodal classes. We propose a classification for thegraphs based on the description of twice transitive permutation groups. This program is realizedfor a1 = c2. In this article we classify graphs in the case when the action of G on Σ is affine.

DOI: 10.1134/S0037446613060141

Keywords: distance-regular graph, edge-symmetric graph, automorphism group

We consider undirected graphs without loops and multiple edges. For a vertex a of a graph Γ, denoteby Γi(a) the i-neighborhood of a, i.e., the subgraph induced by Γ on the set of all vertices at distance ifrom a. Put [a] = Γ1(a) and a

⊥ = {a} ∪ [a].If vertices u and w are at distance i in Γ then denote by bi(u,w) (ci(u,w)) the number of vertices

in the intersection of Γi+1(u) (Γi−1(u)) with [w]. A graph Γ of diameter d is called distance-regularwith intersection array {b0, b1, . . . , bd−1; c1, . . . , cd} if bi(u,w) and ci(u,w) do not depend on the choice ofvertices u and w at distance i in Γ for each i = 0, . . . , d. Put ai = k − bi − ci. Note that, for a distance-regular graph, b0 is the degree of the graph, c1 = 1. Denote by p

lij(x, y) the number of vertices in the

subgraph Γi(x) ∩ Γj(y) for vertices x and y situated at distance l in Γ. In a distance-regular graph, thenumbers plij(x, y) are independent of the choice of the vertices x and y, are denoted by p

lij and called the

intersection numbers of Γ.A graph Γ of diameter d is called distance-transitive if the group G = Aut(Γ) acts transitively on

the vertex set and, for a vertex u in Γ, its stabilizer Gu acts transitively on Γi(u) for each i = 1, . . . , d.For a subset X of automorphism of a graph Γ, denote by Fix(X) the set of all vertices of Γ fixed by

all automorphisms in X.If H is a subgroup in the automorphism group of Γ then, on the set of H-orbits, we can define the

quotient of the graph Γ by assuming two orbits adjacent if there is a vertex in one of them adjacent in Γwith a vertex of the other.Denote the last term of the commutant series of G by G(∞).In [1] some description was obtained for the antipodal distance-transitive graphs of diameter 3. A more

general problem is that of describing edge-symmetric antipodal distance-regular graphs of diameter 3.A graph is called edge-symmetric if its group of automorphisms acts transitively on the set of its arcs

(ordered edges). In this case the group G = Aut(Γ) acts transitively on the vertex set and for a vertex uof Γ, its stabilizer Gu acts transitively on [u].A graph Γ of diameter d is called antipodal if the binary relation on the vertex set to coincide or be

located at distance d is an equivalence. The classes of this relation are called antipodal classes.An antipodal distance-regular graph Γ of diameter 3 has (see [2]) intersection array {k, μ(r − 1), 1;

1, μ, k}, v = r(k + 1) vertices, and spectrum k1, nf , (−1)k, (−m)g, where n and −m are the roots of theequation x2 − (λ− μ)x− k = 0 and f = m(r − 1)(k + 1)/(n+m), g = n(r − 1)(k + 1)/(n+m).

The authors were supported by the Russian Foundation for Basic Research (Grant 12–01–00012), the RFBR-NSFC (Grant 12–01–91155), the Program of the Division of Mathematical Sciences of the Russian Academy ofSciences (Grant 12–T–1–1003), and the Joint Program of the Ural and Siberian Divisions of the Russian Academyof Sciences (Grant 12–C–1–1018) with the National Academy of Sciences of Belarus (Grant 12–C–1–1009).

Ekaterinburg. Translated from Sibirskiı Matematicheskiı Zhurnal, Vol. 54, No. 6, pp. 1353–1367,

November–December, 2013. Original article submitted October, 29, 2012. Revision submitted February 20, 2013.

1076 0037-4466/13/5406–1076

c©

If μ �= λ then the eigenvalues of the graph are integer and the parameters of the graph are expressedvia r, n, and m: k = nm, μ = (m− 1)(n+1)/r, and λ = μ+ n−m. The integrality of the multiplicitiesof the eigenvalues give the divisibility condition: n+m divides (r − 1)m(m2 − 1).In this article we proceed with the program of the study of edge-symmetric antipodal distance-regular

graphs of diameter 3 (see [3]) based on the classification of twice transitive permutation groups (see [4]).Namely, the affine case is studied. In [3], we considered the cases of r = 2, r = k, and carried out the partof this program with λ = μ. Let Γ be an edge-symmetric distance-regular graph with intersection array{k, (r − 1)μ, 1; 1, μ, k}, G = Aut(Γ) and let Σ be the set of antipodal classes of Γ. By [3, Proposition 1],G acts twice transitively on Σ.The symbol GX stands for the permutation group of G on a set X. If Y ⊆ X then GY (G{Y }) is the

pointwise (global) stabilizer of Y in G.

Proposition 1 [1, Theorem 2.9]. Suppose that GX is a twice transitive permutation group of de-gree n, a ∈ X, H = Ga, and T is the socle of G. Then either(1) the almost simple case: G is an almost simple group and one of the following holds for (T, n):(i) alternating (An, n), n ≥ 5;(ii) linear (Lm(q), (q

m − 1)/(q − 1)), m ≥ 2, and (m, q) /∈ {(2, 2), (2, 3)};(iii) symplectic (Sp2m(2), 2

2m−1 ± 2m−1), m ≥ 3;(iv) unitary (U3(q), q

3 + 1), q ≥ 3;(v) Ree (2G2(q), q

3 + 1), q = 32a+1 ≥ 27;(vi) Suzuki (Sz(q), q2 + 1), q = 22a+1 ≥ 8;(vii) Mathieu (Mn, n), n ∈ {11, 12, 22, 23, 24};(viii) sporadic (L2(11), 11), (M11, 12), (A7, 15), (L2(8), 28), (HiS, 176), and (Co3, 276)

or(2) the affine case: G = TH, with T an elementary abelian group of order n = pe, and one of the

following holds:(i) linear e = cd, d ≥ 2, and SLd(pc) � H;(ii) symplectic e = ct, t is even, t ≥ 2, and Spt(pc) � H;(iii) G2-type e = 6c, p = 2, and G2(2

c)′ � H;(iv) one-dimensional H ≤ ΓL1(pe);(v) exceptional pe ∈ {92, 112, 192, 292, 592} and SL2(5) � H or pe = 24 and H ∈ {A6, S6, A7}, or

pe = 36 and SL2(13) � H;(vi) extraspecial pe ∈ {52, 72, 112, 232} and SL2(3) � H or pe = 34, R = D8 ◦ Q8 � H, H/R ≤ S5,

and 5 divides |H|.Given groups A and B having a unique central involution, denote by A ◦B the central product of A

and B in which |A ∩B| = 2.Theorem. Suppose that Γ is an edge-symmetric distance-regular graph with intersection array

{k, (r − 1)μ, 1; 1, μ, k} (2 < r < k), G = Aut(Γ), Σ is the set of antipodal classes in Γ, K is the kernel ofthe action of G on Σ, F ∈ Σ, a ∈ F , H = G{F}, and the socle T of G = G/K is an elementary abeliangroup of order pe. Then K = 1 and Γ is a graph with intersection array {15, 10, 1; 1, 2, 15} or |K| = rand one of the following holds:(1) p = 2, rμ = 2e, e = 2dc, d ≥ 1, r divides 2c, Γ has intersection array {22dc − 1, (r −

1)22dc/r, 1; 1, 22dc/r, 22dc − 1}, T is an elementary abelian group of order 22dcr not containing normalsubgroups of G of order 22dc, Sp2d(2

c) � Ha or d = 3 and G2(2c) � Ha;

(2) the extraspecial case (2vi) of Proposition 1 takes place and either(i) pe = 34, R = D8 ◦ Q8 � Ha, Ha/R ≤ S5, r = 3, T is an extraspecial group of order 3

5, Γ hasintersection array {80, 54, 1; 1, 27, 80}, or

(ii) pe = 52, SL2(3) ≤ Ha ≤ GL2(3), r = μ = 5, n = 4,m = 6, T is an extraspecial group oforder 53, and Γ has intersection array {24, 20, 1; 1, 5, 24}, or

1077

(iii) pe = 72, Ha = GL2(3), n = 6,m = 8, r = μ = 7, T is an extraspecial group of order 73, and Γ

has intersection array {48, 42, 1; 1, 7, 48};(3) the exceptional case (2v) of Proposition 1 takes place and either

(i) pe = 36, SL2(13) ≤ Ha, r = 3, Γ has intersection array {728, 486, 1; 1, 243, 728}, and T is anextraspecial group of order 37, or

(ii) pe = 92, |Ha : SL2(5)| = 2, r = 3, T is an extraspecial group of order 35, and Γ has intersectionarray {80, 54, 1; 1, 27, 80}, or

(iii) pe = 112, |Ha : SL2(5)| ≤ 2, r = μ = 11, T is an extraspecial group of order 113, and Γ hasintersection array {120, 110, 1; 1, 11, 120};(4) the one-dimensional case (2iv) of Proposition 1 takes place, e = 2, Ha is a cyclic group of order

p2−1, T is an extraspecial group of order p3, and Γ has intersection array {p2−1, (p−1)p, 1; 1, p, p2−1};(5) the symplectic case (2ii) of Proposition 1 takes place, e = 2dc, d ≥ 1, r divides pc, Sp2d(pc)�H, T

is a special group of order rp2dc, and Γ has intersection array {p2dc−1, (r−1)p2dc/r, 1; 1, p2dc/r, p2dc−1}.Corollary. Suppose that Γ is a vertex-symmetric distance-regular graph with intersection array

{k, (r − 1)μ, 1; 1, μ, k}, G = Aut(Γ), and, for a vertex a ∈ Γ, the group Ga acts transitively on [a]and Γ3(a). If the action of G on the set of antipodal classes of Γ is affine with kernel K, H is thestabilizer of the antipodal class F , a ∈ F , then either Γ is a distance-transitive graph in [1] or one of thefollowing holds:

(1) Γ has intersection array {8, 6, 1; 1, 3, 8}, |K| = r = 3, while T is an extraspecial group of order 33,H contains a subgroup of index 2 isomorphic to Z3 × SL2(3), and |H : CH(K)| = 2;(2) Γ has intersection array {15, 10, 1; 1, 2, 15}, |K| = 1, r = 6, and H ∈ {A6, S6} or H = GL2(4).Z2;(3) Γ has intersection array {80, 54, 1; 1, 27, 80}, |K| = r = 3, |H : CH(K)| = 2, while T is an

extraspecial group of order 35, R = D8 ◦Q8 � Ha, and Ha/R = S5;(4) Γ has intersection array {728, 486, 1; 1, 243, 728}, |K| = r = 3, SL2(13) � Ha, while T is an

extraspecial group of order 37, and |H : CH(K)| = 2.

§ 1. Auxiliary ResultsThis section contains the results that are needed in the proof of the theorem.

The proof of the theorem is based on Higman’s method of treating the automorphisms of a distance-regular graph as presenting in Cameron’s monograph [5, Chapter 3]. A graph Γ is considered as a sym-metric relation scheme (X,R) with d classes, where X is the vertex set of the graph, R0 is the equalityon X and, for i ≥ 1, the class Ri consists of the pairs (u,w) such that d(u,w) = i. Given u ∈ Γ, putki = |Γi(u)|, v = |X|. To a class Ri, there corresponds a graph Γi on the vertex set X in which vertices uand w are adjacent if (u,w) ∈ Ri. Let Ai be the adjacency matrix of Γi for i > 0 and let A0 = I be theidentity matrix. Then AiAj =

∑plijAl, where p

lij are the intersection numbers of the graph.

Let Pi be the matrix with plij as the (j, l) entry. Then the eigenvalues p1(0) = k, . . . , p1(d) of P1 are

eigenvalues of Γ of multiplicities m0 = 1, . . . ,md. The matrices P and Q having as (i, j) the entry pj(i)and qj(i) = mjpi(j)/ni respectively are called the first and second eigenvalue matrices of the graph andare related by the equalities PQ = QP = |X|I.Let uj and wj be the left and right eigenvalues of P1 with the eigenvalue p1(j) having the first

coordinate 1. Then the wj ’s are the columns of P and the mjuj ’s are the rows of Q. Let θ0 > · · · > θd+1be the eigenvalues of a distance-regular graph Γ of diameter d (the eigenvalues of the adjacency matrixA = A1 of Γ).

The permutation representation of G = Aut(Γ) on the vertices of Γ gives a matrix representation ψof G in GL(v,C) in a natural manner. The space Cv is the orthogonal direct sum of G-invarianteigensubspaces W0, . . . ,Wd of the adjacency matrix A of Γ. For every g ∈ G, the matrix ψ(g) commuteswith A, and hence the subspace Wi is ψ(G)-invariant. Let χi be the character of the representation ψWi .

1078

Then (see [5, § 3.7]) for g ∈ G we obtain

χi(g) = v−1

d∑

j=0

Qijαj(g),

where αj(g) is the number of the vertices x in X such that (x, xg) ∈ Rj . Note that the values of the

characters are algebraic integers, and if the right-hand side of the expression for χi(g) is a rational numberthen χi(g) is integer.

Lemma 1. Let Γ be a distance-regular graph with intersection array {k, μ(r − 1), 1; 1, μ, k}, v =r(k+1) and spectrum k1, nf , (−1)k, (−m)h, where f = m(r−1)(k+1)/(m+n), h = n(r−1)(k+1)/(m+n),and μ �= λ. If g ∈ Aut(Γ) and χ1 is the character of the projection of ψ to subspaces of dimension fcorresponding to an eigenvalue n, while χ2 is the character of the projection of ψ to the subspace ofdimension k corresponding to the eigenvalue −1, then αi(g) = αi(gl) for each natural l coprime with |g|,

χ1(g) = ((m(r − 1) + 1)α0(g) + (1−m)α3(g))/(r(m+ n)) + (α1(g)− (k + 1))/(m+ n),χ2(g) = (α0(g) + α3(g))/r − 1 = k − (α1(g) + α2(g))/r.

If |g| = p is a prime then χ1(g)− f and χ2(g)− k divide by p.Proof. We have

P1 =

⎛

⎜⎝

0 1 0 0k k − μ(r − 1)− 1 μ 00 μ(r − 1) k − μ− 1 k0 0 1 0

⎞

⎟⎠ .

Consider, for example, p1(1) = n. Then

P1 − nI =

⎛

⎜⎝

−n 1 0 0k k − μ(r − 1)− 1− n μ 00 μ(r − 1) k − μ− 1− n k0 0 1 −n

⎞

⎟⎠ .

If (1, x2, x3, x4) is a column vector in the kernel of the matrix P1−nI, then x2 = n/k, x3 = −1/(m(r−1)),and x4 = −1/(r − 1). Hence,

Q =

⎛

⎜⎝

1 1 1 1f f/m −(k + 1)/(m+ n) −f/(r − 1)k −1 −1 kh −h/n (k + 1)/(m+ n) −h/(r − 1)

⎞

⎟⎠ .

Therefore, χ1(g) = (m(r− 1)α0(g) + (r− 1)α1(g)−α2(g)−mα3(g))/r(m+ n). Since α2(g) = r(k+1)−α0(g)−α1(g)−α3(g), we have χ1(g) = ((m(r− 1)+ 1)α0(g)+ (1−m)α3(g))/(r(m+n))+ (α1(g)− (k+1))/(m+ n).Similarly, χ2(g) = (kα0(g)− α1(g)− α2(g) + kα3(g))/r(k + 1). Inserting α1(g) + α2(g) = r(k + 1)−

α0(g)− α3(g), we obtain χ2(g) = (α0(g) + α3(g))/r − 1 = k − (α1(g) + α2(g))/r.The remaining assertions of the lemma follow from [6, Lemma 1].

Lemma 2 [1, Theorem 2.5]. Suppose that Γ is a distance-regular nonbipartite graph with intersectionarray {k, μ(r − 1), 1; 1, μ, k}, K is an abelian subgroup in Aut(Γ) transitive on each antipodal class, andp is a prime divisor of r. Then p divides k + 1.

Assume that Γ is an edge-symmetric distance-regular graph with intersection array {k, (r− 1)μ, 1; 1,μ, k}, r > 2, G = Aut(Γ), g ∈ G, and Ω = Fix(g). Note that Γ contains k + 1 antipodal classes eachof which contains r vertices. Suppose that Σ is the set of antipodal classes of Γ, K is the kernel of theaction of G on Σ, F ∈ Σ, a ∈ F , and C is the kernel of the action of G{F} on F .If μ = 1 then, by Proposition 4 in [6], we see that k = 2e, L2(k)�G, and Γ is a graph with intersection

array {2e, 2e − 2, 1; 1, 1, 2e}. Henceforth, we assume that μ > 1.1079

Lemma 3. The following hold:(1) If Ω is the empty graph and |g| = p is a prime then either(i) p does not divide r, α3(g) = 0, α1(g) + α2(g) = v, and χ1(g) = (α1(g)− (k + 1))/(m+ n); or(ii) p divides r, α3(g) = tr, and χ1(g) = ((1 −m)t + α1(g) − (k + 1))/(m + n); if α3(g) = v then

χ1(g) = −m(k + 1)/(m+ n);(2) If Ω contains t > 0 antipodal classes then α3(g) = 0, α1(g) + α2(g) = r(k + 1 − t), χ2(g) =

α0(g)/r − 1, and χ1(g) = (m(r − 1) + 1)α0(g)/r(m+ n) + (α1(g)− (k + 1))/(m+ n);(3) If α2(g) = v then χ1(g) = −(k + 1)/(m+ n).Proof. By Lemma 1, we have

χ1(g) = ((m(r − 1) + 1)α0(g) + (1−m)α3(g))/r(m+ n) + (α1(g)− (k + 1))/(m+ n),χ2(g) = (α0(g) + α3(g))/r − 1 = k − (α1(g) + α2(g))/r.

Let Ω be the empty graph. If |g| does not divide r then α3(g) = 0, α1(g) + α2(g) = v, and χ1(g) =(α1(g)− (k + 1))/(m+ n).Suppose that |g| divides r. Then α3(g) = tr, χ2(g) = t − 1, and χ1(g) = ((1 −m)t + α1(g) − (k +

1))/(m+ n).If α3(g) = v then χ1(g) = −m(k + 1)/(m+ n). Item (1) is proved.Suppose that Ω contains t > 0 antipodal classes. Then α3(g) = 0, |g| divides k+1−t, α1(g)+α2(g) =

r(k+1− t), χ2(g) = α0(g)/r− 1, and χ1(g) = (m(r− 1)+1)α0(g)/r(m+n)+ (α1(g)− (k+1))/(m+n).Item (2) is proved.If α2(g) = v, then, by the equality χ1(g) = ((m(r−1)+1)α0(g)+(1−m)α3(g))/r(m+n)+(α1(g)−

(k + 1))/(m+ n), we have χ1(g) = −(k + 1)/(m+ n). The lemma is proved.§ 2. Proof of Proposition 2 and the Case of p = 2

Throughout the article, we assume that GΣ = TH, where T is a normal elementary abelian groupof order pe acting regularly on Σ and H is the stabilizer of the antipodal class F . As usual, T is the fullpreimage of T in G.

Lemma 4. If m+ n = ps, s < e, then p = 2, {m,n} = {2e/2 − 1, 2e/2 + 1}, rμ = 2e − 4 or rμ = 2e,and λ− μ = ±2.Proof. Suppose that m + n = ps, s < e. We have k = nm = pe − 1, rμ = (m − 1)(n + 1), and

n−m+ 2 = pe − rμ, e < 2s.Calculate the difference (m+ n)2 − rμ(rμ+ 4):

rμ(rμ+ 4) = (nm+m− n− 1)(nm+m− n+ 3)= (k + 1 +m− n− 2)(k + 1 +m− n+ 2) = (k + 1)2 + 2(k + 1)(m− n) +m2 + n2 − 2mn− 4;

therefore,(m+ n)2 − rμ(rμ+ 4) = −(k + 1)2 + 2(k + 1)(n−m+ 2) = p2e − 2perμ;

consequently, p2s − p2e = rμ(rμ+ 4− 2pe) and rμ+ 4 < 2pe.Next, (rμ+4−2pe, rμ) divides 2(2−pe). If p is odd then the p-part of the number rμ or rμ+4−2pe

is equal to p2s. If the p-part of rμ is equal to p2s then

rμ+ 4− 2pe = p2sh+ 4− 2pe = pe(p2s−eh− 2) + 4 > 0,where (h, p) = 1, h > 0; a contradiction. If the p-part of rμ + 4 − 2pe is equal to p2s then (rμ, p) = 1,rμ+4−2pe = p2sh, (h, p) = 1, h < 0, and rμ+4 = pe(2+p2s−eh); thus, the p-part of rμ+4 is equal to pe.But rμ+ 4 = pef < 2pe, (f, p) = 1, f > 0; therefore, f = 1 and 1 = 2 + p2s−eh, whence h = −1, 2s = e;a contradiction.Let p = 2. If rμ = pe then m − n = 2, p2s + p2e = rμ(rμ + 4) = 4pe + p2e, and p2s = 4pe; hence,

p2s−e = 4 and 2(s− 1) = e. Therefore, m = pe/2 + 1 and n = pe/2 − 1.Let rμ �= pe. The number (rμ+ 4− 2pe, rμ) divides 2(2− pe) = 4(1− 2e−1) and the p-part of rμ or

rμ+4− 2pe is at least p2s−2. If rμ = p2s−2x then rμ+4− 2pe = 22s−2x+4− 2e+1 = 2e+1(22s−2−e−1x−1) + 4 < 0 is possible only if s = e/2 + 1 and (m,n) = (2e/2 − 1, 2e/2 + 1), rμ = 2e − 4, e ≥ 4.1080

If rμ+4−2pe = p2s−2y then rμ+4 = 2e+1+22s−2y = 2e+1(1+22s−e−3y), y ≤ −1; hence, 2s−e−3 < 0and 2s < e+3. The case of 2s = e+1 is impossible; therefore, again 2s = e+2, (m,n) = (2e/2−1, 2e/2+1),and rμ = 2e − 4.Since λ− μ = n−m, we have λ− μ = ±2.Lemma 5. If m+ n = psx, x > 1, (x, p) = 1, s > 0, r > 2, then one of the following holds:

(1) p is odd and either (m,n) = (pe/2+1, pe/2−1) and rμ = pe or p does not divide rμ and the p-partof rμ+4−2pe is equal to p2s, e ≥ 2s; in particular, if e = 2s then rμ = pe−4, (m,n) = (pe/2−1, pe/2+1);(2) p = 2 and either (rμ, 8) = 4 and the 2-part of the number rμ + 4 − 2e+1 is equal to 22s−2 or

(m,n) = (2e/2 + 1, 2e/2 − 1) and rμ = 2e.Proof. Suppose that m+ n = psx, x > 1, (x, p) = 1, s > 0, and r > 2. We have k = nm = pe − 1,

rμ = (m− 1)(n+ 1), n−m+ 2 = pe − rμ. The condition r > 2 implies that m+ n = psx < pe.

As above, (m+ n)2− rμ(rμ+4) = p2e− 2perμ; so, p2sx2− p2e = rμ(rμ+4− 2pe) and rμ+4 < 2pe.Now, (rμ+ 4− 2pe, rμ) divides 2(2− pe). If p is odd then the p-part of rμ or rμ+ 4− 2pe is equal

to p2s. In the latter case −p2sz = rμ + 4 − 2pe, (z, p) = 1, z > 0, and e ≥ 2s; moreover, the equalitye = 2s is possible only for z = 1, rμ = pe − 4 (in particular, if e = 2), (m,n) = (pe/2 − 1, pe/2 + 1).Suppose that rμ = p2sy, (y, p) = 1. From the condition rμ < 2pe − 4 we obtain 2s ≤ e, and the equality2s = e is possible only for y = 1.

Put a = n + 1 and b = m − 1. Then ab = p2sy, a − b = pe − p2sy = p2s(pe−2s − y), whenceab = b2 + bp2s(pe−2s − y) = p2sy, ps divides b and a; therefore, a = b, e = 2s, y = 1, and (m,n) =

(pe/2 + 1, pe/2 − 1).Let p = 2. Since μ is even, 2 divides n+1 and m−1. Further, (rμ+4−2e+1, rμ) divides 4(1−2e−1),

and hence the 2-part of rμ or rμ+ 4− 2e+1 is equal to 22s−2, s ≥ 2.If rμ = 22s−2y, then rμ < 2e+1 − 4 yields 2s − 2 < e + 1. Put a = n + 1 and b = m − 1. Then

ab = 22s−2y, a−b = 22s−2(2e−2s+2−y), whence ab = b2+b22s−2(2e−2s+2−y) = 22s−2y and 2s−1 divides band a. Therefore, a = b, e = 2s−2, y = 1, (m,n) = (2e/2+1, 2e/2−1), and rμ = 2e. The lemma is proved.Proposition 2. One of the following holds:

(1) T does not contain normal subgroups in G of order pe (in particular, the action of G on Σ isnot exact);

(2) Γ has intersection array {15, 10, 1; 1, 2, 15}, G{F} acts twice transitively on F,K = 1, H ∈ {A6, S6}and the neighborhood of a vertex is a distance-regular graph with intersection array {4, 2, 1; 1, 1, 4} orH = GL2(4).Z2 and the neighborhood of a vertex is a union of three isolated cliques.

Proof. Suppose that T contains a normal subgroup N in G of order k + 1. Then each N -orbitcontains a unique element in each antipodal class. For a vertex a ∈ Γ, the group Ga acts transitivelyon [a]; therefore, each N -orbit consists of the vertices pairwise at distance 2 and there are t (t < r) orbitsin each of which a is adjacent to exactly α vertices, in particular, to k = tα. Given a nonunit elementg ∈ N , we obtain α2(g) = v, and by Lemma 3 we have χ1(g) = −(k + 1)/(m+ n). By Lemma 4, p = 2,{m,n} = {2e/2 − 1, 2e/2 + 1} and rμ = 2e − 4 or rμ = 2e.Further, Ga acts transitively on the set of nonunit elements in N , and the group NGa is a twice

transitive group of permutations of degree 2e.

A vertex in [a] is adjacent to α− 1 vertices in aN − {a}, between [a] and aN − {a} there are exactlyk(α− 1) edges; therefore, a vertex in aN −{a} is adjacent with α− 1 vertices in [a] “in average.” Hence,α − 1 = μ and k = t(μ + 1). Since b1 = (r − 1)μ, we have λ = k − b1 − 1 = t + μ − (r − t)μ − 1,t(μ+ 1) = rμ− μ+ 1 + λ, and t = r − (r − 1 + μ− λ)/(μ+ 1).If t = r − 1 then Ga acts transitively on Γ3(a) and H = G{F} acts twice transitively on F . Put

H = H/C and let S be the socle of H = H/C. The group K is isomorphic to a regular normal subgroup

in H. Therefore, K = 1 if H is an almost simple group and K is an elementary abelian 2-group of order r

if H is an affine group.

1081

We have k = (r− 1)(μ+1), λ = r− 2 and λ−μ = (r− 1)− (μ+1); therefore, n = r− 1, m = μ+1,and m + n = r + μ divides 2e. The inequality m ≤ n2 yields μ ≤ r2 − 2r. Now, m = μ + 1 = 2e/2 − 1,n = r − 1 = 2e/2 + 1, and rμ = 2e − 4 or m = μ+ 1 = 2e/2 + 1, n = r − 1 = 2e/2 − 1, and rμ = 2e.In the case of an almost simple action ofH on F , only the alternating, linear, symplectic, and G2-type

cases are possible. If S = Ar then r ∈ {6, 7} and pe = 24. With account taken of the equality 2e − 1 =(r − 1)(μ+ 1), we have r = 6, μ = 2. Further, |K| = 1 and Ha = A5. Computer calculations in GAP [7]show that, in this case, there exists a distance-regular graph with intersection array {15, 10, 1; 1, 2, 15}and the neighborhood of a vertex is a distance-regular graph with intersection array {4, 2, 1; 1, 1, 4}.If S = Ld(q) then r = (q

d − 1)/(q − 1) is even; so, d = 2, q = 5, and r = 6. Computer calculationsin GAP [7] show that there is a distance-regular graph with intersection array {15, 10, 1; 1, 2, 15}, H =GL2(4).Z2, and the neighborhood of a vertex is a union of three isolated 5-cliques.

If H = Sp2m(2) then r = 22m−1 ± 2m−1, m ≥ 3; a contradiction to the fact that pe = 22m and r − 1

does not divide 22m − 1.We may assume that the affine case is fulfilled for the action of H on F ; hence, S is a regular

elementary abelian group of order r = 2f , H = SHa, and one of the following is realized:

(i) linear f = cd, d ≥ 2, and SLd(2c) � Ha ≤ ΓLd(2c);(ii) symplectic f = cd, d is even, d ≥ 4, and Spd(2c) � Ha ≤ Z2c−1 ◦ ΓSpd(2c);(iii) G2-type f = 6c and G2(2

c)′ � Ha;(iv) one-dimensional Ha ≤ ΓL1(2f );(v) exceptional Ha ∈ {A6, S6, A7}.SinceK is a regular elementary abelian group of order 2f andN ≤ CG(K), the group CG(K) = K×N

is transitive on Σ and C = 1. Further, G = (K × N)Ha, G/N acts 2-transitively on F , and G/K acts2-transitively on Σ; moreover, |K| = 2e/2 and |N | = 2e; a contradiction.If t < r − 1 then Γ contains a 2(k + 1)-coclique and, in view of the Hoffman bound for cliques, we

have 2(k + 1) ≤ m(k + 1)r/(k +m); therefore, n + 1 ≤ r/2 and m − 1 ≥ 2μ. Since Ga acts transitivelyon the set Δ, which consists of N -orbits intersecting [a], each vertex in [a] is adjacent to s vertices in theorbits of Δ for some s < t− 1 (if s = t− 1 then the vertex set lying in aN and the vertices of the orbitsin Δ induce a connected subgraph in Γ).The number of the edges between the union of N -orbits not containing a and vertices adjacent to a

and [a] is equal to (r − t− 1)kμ. Therefore, a vertex in [a] is adjacent exactly to (r − t− 1)μ vertices inthis union, r− t−1 divides by μ+1, and (r− t−1)μ/(μ+1) = t− s−1. Hence, (r−2t+ s)μ = t− s−1,t − s − 1 = zμ, and r − t − 1 = z(μ + 1). Consequently, t = s + 1 + zμ and r = s + 2 + z(2μ + 1).Since s(μ+ 1) = λ+ (t− 1)μ, we have s = zμ2 + λ and t ≥ zμ2 + zμ+ 1. We infer t(μ+ 1) = k = mn,m ≥ 2μ+ 1; therefore, n ≤ t(μ+ 1)/(2μ+ 1).On the set N , the graph Ω = Γ is edge-regular with parameters (r, t, λΩ), where λΩ = s. A vertex in

Ω − a⊥ is adjacent with tμ/(μ + 1) vertices in Ω(a) “in average.” If Ω is a strongly regular graph thent = (μ + 1)y and s = yμ + (λ − μ)/(μ + 1); a contradiction to the fact that, by Lemma 4, λ − μ = ±2.Hence, Ω is not a strongly regular graph and Ga has at least two orbits on vertices in F−{a} not adjacentto vertices in aN .Assume that a vertex b in Ω − a⊥ is adjacent to zμ + z − 1 vertices in Ω − a⊥. Then b is adjacent

to s − z + 2 vertices in Ω(a). Therefore, s − z + 2 < tμ/(μ + 1) and zμ2 ≤ s < zμ2 + zμ + z − μ − 2;in particular, z > 1 and λ < zμ+ z − μ− 2.Note that, for a vertex a, the action of Ga on a

N −{a} is imprimitive with blocks of the form [b]∩aNfor t vertices b ∈ F adjacent to vertices in aN . Further, one of the following holds:(i) linear e = cd, d ≥ 3 and SLd(2c) � Ga ≤ ΓLd(2c);(ii) symplectic e = cd, d even, d ≥ 1, q = 2c, and Spd(q) � Ga ≤ Zq−1 ◦ ΓSpd(q);(iii) G2-type e = 6c, q = 2

c, and G2(q)′ � Ga ≤ Zq−1 ◦Aut(G2(q));

(iv) one-dimensional Ga ≤ ΓL1(2e);(v) exceptional pe = 24 and Ga ∈ {A6, S6, A7}.

1082

If pe = 24 then, by Lemma 4, either (m,n) = (3, 5), rμ = 12, t(μ + 1) = rμ + 3 = 15, and henceμ = 2, 4, r = 6, 3, but r > μ + 4; a contradiction; or (m,n) = (5, 3), rμ = 16, t(μ + 1) = rμ − 1 = 15,t = r− (r+1)/(μ+1), r > μ; hence, μ = 2, r = 8, but (r−μ) divides by (μ+1)2; a contradiction. Thus,e ≥ 6, and case (v) is impossible.In the one-dimensional case we obtain a contradiction to the action of a Singer cycle on nonzero

vectors. In the linear case with d ≥ 3 for E ∈ Σ− {F}, the subgroup Ga ∩G{E} acts transitively on E;a contradiction to the fact that [a] contains a single vertex of E.By [8], the minimal permutation degree of Spd(q) is equal to (q

d − 1)/(q − 1) with the exception ofthe cases of d = 2, q ∈ {5, 7, 9, 11}, and d = 4, q = 3. In the symplectic case with d > 2, the groupG(∞)a = Spd(q) has a permutation representation of odd degree dividing t. By [9], in this representation,the stabilizer of a point is contained in a parabolic maximal subgroup in Spd(q); therefore, the stabilizerof a point fixes a singular straight line in the corresponding symplectic space, μ+ 1 divides q − 1, and tdivides by (qd − 1)/(q − 1). Hence, G(∞)a acts trivially on the set of the vertices of F − {a} not adjacentto vertices in aN . A contradiction to the fact that then Ω(b) = Ω(a) for any vertex b ∈ Ω− a⊥. Hence,d = 2, 2e = q2, and zμ + z ≥ q + 1. Now, L2(2

c) � H/K ≤ GL2(2c).Zc, t(μ + 1) = 2

2c − 1, and byLemma 4 rμ is equal to 22c or 22c − 4. Note that the action of H(∞)a = Sp2(2

c) on one-dimensionalisotropic subspaces is primitive; therefore, μ+ 1 divides 2c − 1 and t divides by 2c + 1. Since the degreesucceeding the degree of the permutation representation of L2(2

c) is equal to 2c−1(2c− 1) or 2c/2(2c+1)(c is even), either zμ+ z divides by 2c + 1 or zμ+ z ≥ 2c−1(2c − 1) + (2c + 1) = 2c−1(2c + 1) + 1.Let rμ = 22c. Then μ = 2l, r = z(μ+1)2 + μ = 22c−l. Since 2c+1 does not divide 2l(22c−2l − 1), we

have r ≤ zμ+ z; a contradiction.Let rμ+4 = 22c. Then r = zμ2+2zμ+μ+z+4 and t = zμ2+zμ+μ+3. If zμ+z divides by 2c+1

then t is coprime with μ+ 1, (t, 2c − 1) = (μ+ 3, 2c + 1), and 2c + 1 divides μ+ 3. Thus, μ = 2c − 2 andr = 2c+2; a contradiction. Hence, r = mu(zμ+z)+(zμ+z)+μ+4 ≥ (μ+1)(2c−1(2c+1)+1)+μ+4 > 22c;a contradiction.By [10], the minimal permutation degree of a representation of G2(2

c) is equal to 50 for c = 1, 416,

for c = 2, and (q6 − 1)/(q − 1) for c > 2. Therefore, for the G2-type, the group H(∞)a = G2(2c)′ fixes

r − t − 1 = zμ + z vertices of F − {a} pointwise. For c > 2, we arrive at a contradiction as above. Forc = 1 we have 2e = 26 and zμ + z ≥ 50; a contradiction. For c = 2, we obtain 2e = 212 = 4096 andzμ + z ≥ 416. Hence, μ = 2 and t = 1365; a contradiction to the fact that μ + 1 does not divide t.Proposition 2 is proved.

In the sequel, we assume that Γ is not a graph with intersection array {15, 10, 1; 1, 2, 15}.Lemma 6. The following hold:(1) m+ n divides mpe;(2) if T contains a normal subgroup N in G of order dividing by pe then |N | divides by r;(3) |K| = r, K/K ′ is a p-group, (m,n) = (pe/2 + 1, pe/2 − 1), and rμ = pe.Proof. For a nonunit element g ∈ K, we have α3(g) = v and, by Lemma 4 , we obtain χ1(g) =

−m(k + 1)/(m+ n). Item (1) is proved.Suppose that T contains a normal subgroup N in G of order divisible by pe. If |N | does not divide r

then, on the set of K ∩N -orbits, the quotient Γ is edge-symmetric antipodal distance-regular graph andthe group T contains a normal subgroup in G of order pe. A contradiction to the fact that, for a graphwith intersection array {15, 10, 1; 1, 2, 15}, the maximal value of r is equal to 6. Item (2) is proved.It follows from item (2) that |K| = r. Lemma 2 implies that K/K ′ is a p-group. If r does not divide

by p then T contains a subgroup T0 of order pe. The action of T0 on the Sylow s-subgroup S of K implies

that T0 centralizes S/Φ(S); therefore, [S, T0] = 1 and T0 centralizes K. Now, T0 is a characteristicsubgroup in T ; a contradiction to item (2). Hence, r divides by p, and by Lemma 5 we infer that

(m,n) = (pe/2 + 1, pe/2 − 1) and rμ = pe.By Lemma 6, H is the semidirect product of K and Ga.

1083

Lemma 7. If p = 2 then the following hold:

(1) rμ = 2e, e is even, m = 2e/2 + 1, n = 2e/2 − 1, μ = 2l, Γ has intersection array {2e − 1, (2e−l −1)2l, 1; 1, 2l, 2e−1}, and if H contains an element of order 2e−1 then l = e/2−1 and α1(g) = (2e−1)2e/2;(2) the exceptional, one-dimensional, and linear cases are impossible;

(3) in the G2-type and symplectic cases, we have e = 2dc, d ≥ 1, r divides 2c, and Sp2d(2c) � Ha ord = 3 and G2(2

c)′ � Ha.Proof. Let p = 2. Apply Lemma 5. If rμ does not divide by 8 then the 2-part rμ = (m− 1)(n+1)

is equal to 4. In this case K contains a subgroup K0 of index 2, inverted by an involution in K −K0,K0 centralizes T , and T contains a normal subgroup in G of order 2

e; a contradiction to Lemma 6.Hence, rμ = 2e. Then e is even, m = 2e/2 + 1, n = 2e/2 − 1, μ = 2l, and Γ has intersection array{2e−1, (2e−l−1)2l, 1; 1, 2l, 2e−1} and spectrum (2e−1)1, (2e/2−1)2e/2−1(2e/2+1)(2e−l−1),−12e−1,−(2e/2+1)2

e/2−1(2e/2−1)(2e−l−1).If g is an element of order 2e−1, α1(g) = (2e−1)w, then χ1(g) = (m(r−1)+α1(g)−k)/(m+n) and

m(r− 1) +α1(g)− k = (2e/2 +1)(2e−l − 1) + (2e − 1)(w− 1) = (2e/2 +1)(2e−l − 1+ (2e/2 − 1)(w− 1)) =(2e/2 + 1)(2e−l + (2e/2 − 1)w− 2e/2)) divides by 2e/2+1; therefore, either w = 2e/2+1, l = e/2 or w = 2e/2,l = e/2−1. But in the first case we obtain a contradiction to the fact that α1(g) > kr. Item (1) is proved.

In the exceptional case pe = 24 and H ∈ {A6, S6, A7}. Then k = 15, n = 3, and m = 5. Further,CG(K) contains H

′; therefore, K ≤ Z(T ). If K ≤ T ′ then, for the subgroup K0 of index 2 in K, the

group T/K0 is extraspecial; a contradiction to the fact that the group O±4 (2) does not contain A6. For

the subgroup K0 of index 2 in K, the quotient Γ is a distance-transitive graph with r = 2 on the set ofK0-orbits. Therefore, the group Ha is isomorphic to A6 or S6.

If r = 4 then, by Lemma 6, the group T is the central product of Z4 and an extraspecial groupof order 32 or an elementary abelian group of order 64. In the latter case T does not contain normalsubgroups in G of order 32 and 64. Computer calculations in GAP [7] show that a desired module existsbut there is no graph with intersection array {15, 12, 1; 1, 4, 15}. In the former case computer calculationsin GAP [7] show that a graph with intersection array {15, 12, 1; 1, 4, 15} does not appear.If r = 8 then, for a subgroup K0 of order 2 in K normal in G, the graph Γ has r = 4 on the set of

K0-orbits; a contradiction to what was proved in the previous paragraph.

In the one-dimensional case H contains an element g of order 2e − 1, and, by item (1), Γ hasintersection array {2e − 1, (2e/2+1 − 1)2e/2−1, 1; 1, 2e/2−1, 2e −1}. Further, K = [K, g]CK(g) and |[K, g]|divides (2e − 1); therefore, K0 = CK(g) �= 1. Note that K0 centralizes T ; consequently, the groupG1 = CG(K0) acts twice transitively on Σ and, for a subgroup K1 of index 2 in CK(K0), the quotientof Γ is a distance-transitive graph with r = 2 on the set of K1-orbits; a contradiction.

In the linear and G2-type case K centralizes H(∞)a ; therefore, G1 = CG(K) acts twice transitively

on Σ and, for a subgroup K0 of index 2 in K, the quotient of Γ is a distance-transitive graph with r = 2on the set of K0-orbits. Hence, r divides 2

c and G2(2c) � Ha. Item (2) is proved.

In the symplectic case e = 2dc and Sp2d(2c) � Ha. As above, r divides 2

c. The lemma is proved.

§ 3. The Case p �= 2In the sequel, we assume p odd. Recall that by Lemma 6 (m,n) = (pe/2 + 1, pe/2 − 1) and rμ = pe.Lemma 8. In the extraspecial case one of the following holds:

(1) pe = 34, R = D8 ◦ Q8 � Ha, Ha/R ≤ S5, n = 8, m = 10, r = 3, T is an extraspecial group oforder 35, and Γ has intersection array {80, 54, 1; 1, 27, 80};(2) pe = 52, SL2(3) ≤ Ha ≤ GL2(3), r = μ = 5, n = 4, m = 6, T is an extraspecial group of order 53,

and Γ has intersection array {24, 20, 1; 1, 5, 24};(3) pe = 72, H = GL2(3), n = 6, m = 8, r = μ = 7, T is an extraspecial group of order 73, and Γ

has intersection array {48, 42, 1; 1, 7, 48}.1084

Proof. Suppose the extraspecial case. If pe = 34 then k = 80, R = D8 ◦ Q8 � Ha, Ha/R ≤ S5,5 divides |Ha|, rμ = 81, n = 8, m = 10, and Γ has intersection array {80, (34−l − 1)3l, 1; 1, 3l, 80}and spectrum 801, 845(3

4−l−1),−180,−1036(34−l−1), l ∈ {1, 2, 3}. In this case H is the extension of thedirect product K × R by Ha/R. If T is an abelian group then [T,Ra] is a characteristic subgroup in Gof order 81; a contradiction. Therefore, T is a nonabelian group and [R, T ] centralizes K. If T ′ doesnot contain K then, repeating the argument for the quotient graph on the set of T ′-orbits, we obtaina contradiction. Thus, T ′ = Z(T ) = K is an elementary abelian group and [Ra, T ] centralizes K. Themapping (g, h) �→ [g, h] for g, h ∈ T − Z(T ) defines an alternating bilinear form on the linear space Tover the field of |K| elements. If r > 3 then Ra is not included in Spd(|K|) (d = 34/|K|); a contradiction.If pe = 52 then k = 24, |Ha : R| ≤ 2, R ≤ SL2(3) ◦ Z4. Since rμ = 25, we have r = μ = 5, n = 4,

m = 6, and Γ has intersection array {24, 20, 1; 1, 5, 24} and spectrum 241, 460,−124,−640. In this case thesubgroup H is the extension of the direct product K ×R′ by Ha/R′. If T is an abelian group then [T,R]is a characteristic subgroup in G of order 25; a contradiction. Therefore, T is an extraspecial group andSL2(3) ≤ Ha ≤ GL2(3).If pe = 72 then k = 48, |Ha : R| ≤ 2, and R ≤ SL2(3) ◦ Z6. Next, rμ = 49, r = μ = 7, n = 6,

and m = 8, while Γ has intersection array {48, 42, 1; 1, 7, 48} and spectrum 481, 6168,−148,−8126. As forpe = 25, we infer that T is an extraspecial group and Ha = GL(2, 3).If pe = 112 then k = 120, |Ha : R| ≤ 2 and R ≤ SL2(3) ◦ Z10; hence, rμ = 121 r = μ = 11, n = 10,

m = 12, and Γ has intersection array {120, 110, 1; 1, 11, 120} and spectrum 1201, 10660,−1120,−12550.As for pe = 25, we see that T is an extraspecial group; a contradiction to the fact that SL2(3) ◦ Z10 isnot included in Sp2(11).If pe = 232 then k = 528, |Ha : R| ≤ 2, and R ≤ SL2(3) ◦ Z22. Further, rμ = 529, n = 22, m = 24,

r = μ = 23, and Γ has intersection array {528, 506, 1; 1, 23, 528} and spectrum 5281, 226072,−1528,−245566.As for pe = 25, we see that T is an extraspecial group; a contradiction to the fact that SL2(3) ◦ Z22 isnot included in Sp2(23).

Lemma 9. In the exceptional case one of the following holds:(1) pe = 36, Ha = SL2(13), k = 728, n = 26, m = 28, r = 3, T is an extraspecial group of order 3

7,and Γ has intersection array {728, 486, 1; 1, 243, 728};(2) pe = 92, k = 80, |Ha : R| ≤ 2, R ≤ SL2(5) ◦Z8, n = 8, m = 10, r = 3, T is an extraspecial group

of order 35, and Γ has intersection array {80, 54, 1; 1, 27, 80};(3) pe = 112, k = 120, |Ha : SL2(5)| ≤ 2, n = 10, m = 12, r = μ = 11, T is an extraspecial group of

order 113, and Γ has intersection array {120, 110, 1; 1, 11, 120}.Proof. If pe = 36, SL2(13) � H, then k = 728. Further, rμ = 729, n = 26, m = 28, and

Γ has intersection array {728, (36−l − 1)3l, 1; 1, 3l, 728}, 1 ≤ l ≤ 5, and spectrum 7281, 26378(36−l−1),−1728,−28351(36−l−1). In this case H = K ×Ha, SL2(13) � Ha, therefore, [Ha, T ] centralizes K. If T isan elementary abelian group of order 37 then T does not contain normal subgroups in G of order 36.Computer calculations show that there is a unique embedding of SL2(13) in GL7(3) (up to conjugationin GL7(3)). Hence, K ≤ T ′, T ′ = Z(T ) = K is an elementary abelian group. Since Spd(|K|) does notcontain SL2(13) for 3

d|K| = 36 and |K| > 3, we see that r = 3 and T is an extraspecial group of order 37.If pm = 92, then k = 80, |Ha : R| ≤ 2 and R ≤ SL2(5)◦Z8. By Lemma 6, r = 34−l, l = 1, 2, 3, n = 8,

m = 10, and Γ has intersection array {80, (34−l − 1)3l, 1; 1, 3l, 80} and spectrum 801, 845(34−l−1),−180,−1036(34−l−1). If T is an elementary abelian group of order 35 then T does not contain normal subgroupsin G of order 34. Computer calculations in GAP [7] show that this case is impossible. Thus, K ≤ T ′ andT ′ = Z(T ) = K is an elementary abelian group. Since Sp2(9) does not contain SL2(5) ◦ Z4, r = 3 andT is an extraspecial group of order 35. Hence, |Ha : SL2(5)| = 2 (Sp4(3) does not contain SL2(5) ◦ Z4).If pe = 112 then k = 120, |Ha : R| ≤ 2 and R ≤ SL2(5) ◦ Z10. Therefore, r = μ = 11, n = 10,

m = 12, while Γ has intersection array {120, 110, 1; 1, 11, 120} and spectrum 1201, 10110,−1120,−12132.If pe = 192 then k = 360, |Ha : R| ≤ 2, and R ≤ SL2(5) ◦ Z18. Further, rμ = 361, n = 18, m = 20,

μ = r = 19, while Γ has intersection array {360, 342, 1; 1, 19, 360} and spectrum 3601, 183420,−1360, 203078.1085

Since Ha contains a subgroup of index k then R contains SL2(5) ◦ Z6.If pe = 292 then k = 840, |Ha : R| ≤ 2 and R ≤ SL2(5)◦Z28. Further, rμ = 841, r = μ = 29, n = 28,

m = 30, while Γ has intersection array {840, 812, 1; 1, 29, 840} and spectrum 8401, 2812180,−1840,−3011368.Since Ha contains a subgroup of index r, then R contains SL2(5) ◦ Z14.If pm = 592, then k = 3480 and R ≤ SL2(5) ◦ Z58. Further, rμ = 3481, r = μ = 59, n = 58, m = 60,

while Γ has intersection array {3480, 3422, 1; 1, 59, 3480} and spectrum 34801, 58102660,−13480,−6099238.Since Ha contains a subgroup of index r, it follows that R contains SL2(5) ◦ Z29.In the case of p ≥ 11, T is an extraspecial group of order p3. If p ≥ 19 then we obtain a contradiction

to the fact that Sp2(p) does not contain R.

Lemma 10. In the one-dimensional case Ha ≤ GL1(pe).Ze, e = 2, n = −1 + p, m = 1 + p, while Γ

has intersection array {p2 − 1, (p− 1)p, 1; 1, p, p2 − 1}, and α1(g) = p(p2 − 1).Proof. Suppose that H ≤ GL1(p

e).Ze. Let R be a normal cyclic subgroup of order pe − 1 in Ha

generated by an element g, Ω = Fix(g). Then Ω = F , by Lemma 3, α3(g) = 0 and χ1(g) = (m(r − 1) +α1(g)− k)/(m+ n). Let y be a primitive element in R of prime order s and let R0 = 〈y〉.If K = K ′ then T = K× [T,R0] and T contains a normal subgroup in G of order pe; a contradiction.

Hence, K �= K ′, and, considering the quotient of Γ on the set of K ′-orbits, we may assume that K isan abelian p-group. If K does not lie in T ′ then, considering the quotient Γ on the set of T ′-orbits withautomorphism group G = G/T ′, we infer that T = K × [T , R0] and T contains a subgroup of order penormal in G; a contradiction to Lemma 5. Since K centralizes [T,R0], we have Z(T ) = T

′ = K, and theautomorphism group of T contains an element of order pe − 1. For the subgroup K1 of index p in K,the quotient group T/K1 is an extraspecial group; therefore, e = 2, n = −1 + p, m = 1 + p, and Γhas intersection array {p2 − 1, (p− 1)p, 1; 1, p, p2 − 1} and spectrum (p2 − 1)1, (−1 + p)p(p2−1)/2,−1p2−1,(−1− p)p(p−1)2/2.Since χ1(g) = (m(r − 1) + α1(g)− k)/(m+ n), it follows that m+ n = 2p divides (p2 − 1)α1(g) and

α1(g) = p(p2 − 1).

Lemma 11. In the symplectic case e = 2dc, d ≥ 2, Sp2d(pc) � Ha ≤ ΓSp2d(pc), k = p2dc − 1, m =pdc+1, n = pdc−1, rμ = p2dc, μ = pl, and Γ has intersection array {p2dc−1, (p2dc−l−1)pl, 1; 1, pl, p2dc−1}.Proof. Let Sp2d(p

c) � H. Further, CG(K) contains TH(∞), by Lemma 6 we have |Z(K)| = r, and

K is a subgroup of order pb in Z(T ). If K ≤ T ′ then K = T ′ is an elementary abelian p-group. If Kis not contained in T ′ then, considering the quotient of the graph Γ on the set of Φ(T )-orbits, we verifythat T/Φ(T ) is an indecomposable GF (p)Sp2d(p

c)-module; a contradiction to [11]. Thus, T is a specialp-group acting regularly on the vertex set of Γ; therefore, r divides pc.If rμ = p2dc then n = −1 + pdc and m = 1 + pdc, while Γ has intersection array {p2dc − 1,

(r − 1)p2dc/r, 1; 1, p2dc/r, p2dc − 1} and spectrum (pe − 1)1, (−1 + pe/2)pe/2(r−1)(1+pe/2)/2,−1pe−1, (−1 −pe/2)p

e/2(r−1)(−1+pe/2)/2.

Lemma 12. In the linear case SL2(pc) � Ha ≤ ΓL2(pc), r divides pc, and Γ has intersection array

{p2c − 1, (r − 1)p2c/r, 1; 1, p2c/r, p2c − 1}.Proof. In the linear case e = cd, d ≥ 2 and SLd(pc) � Ha ≤ ΓLd(pc). Further, CG(K) con-

tains TH(∞), by Lemma 6, we have |Z(K)| = r, and K is a subgroup in Z(T ). If K ≤ T ′ then K = T ′

is an elementary abelian p-group. If K is not contained in T ′ them considering the quotient of thegraph Γ on the set of Φ(T )-orbits, we verify that T/Φ(T ) is an indecomposable GF (p)Ld(p

c)-module;a contradiction to [6]. Thus, T is a special p-group acting regularly on the vertex set of Γ. For the sub-

group K0 of index p in K and G = G/K0, the group T is extraspecial. Consequently, SLd(pc) ⊆ Spd(pc),

d = 2, r divides pc, and Γ has intersection array {p2c − 1, (r − 1) p2c/r, 1; 1, p2c/r, p2c − 1} and spectrum(pe − 1)1, (−1+pe/2)pe/2(r−1)(1+pe/2)/2,−1pe−1, (−1 − pe/2)pe/2(r−1)(−1+pe/2)/2. The lemma is proved, andso is the theorem.

1086

References

1. Godsil C. D., Liebler R. A., and Praeger C. E., “Antipodal distance transitive covers of complete graphs,” European J.Combin., 19, No. 4, 455–478 (1998).

2. Brouwer A. E., Cohen A. M., and Neumaier A., Distance-Regular Graphs, Springer-Verlag, Berlin, Heidelberg, and NewYork (1989).

3. Makhnev A. A., Paduchikh D. V., and Tsiovkina L. Yu., “Edge-symmetric distance-regular coverings of cliques withλ = μ,” Dokl. Akad. Nauk, 448, No. 1, 22–27 (2013).

4. Cameron P. J., “Finite permutation groups and finite simple groups,” Bull. London Math. Soc., 13, No. 1, 1–22 (1981).5. Cameron P. J., Permutation Groups, Cambridge Univ. Press, Cambridge (1999) (London Math. Soc. Student Texts;V. 45).

6. Gavrilyuk A. L. and Makhnev A. A., “Geodesic graphs with homogeneity conditions,” Dokl. Math., 78, No. 2, 743–745(2008).

7. GAP—Groups, Algorithms and Programming. Version 4.4.12. 2008. (http://www.gap-system.org).8. Mazurov V. D., “Minimal permutation representations of finite simple classical groups. Special linear, symplectic, andunitary groups,” Algebra and Logic, 32, No. 3, 142–153 (1993).

9. Liebeck M. W. and Saxl J., “The primitive permutation groups of odd degree,” J. London Math. Soc., 31, No. 2,250–264 (1985).

10. Vasilyev V. A., “Minimal permutation representations of finite simple exceptional groups of types G2 and F4,” Algebraand Logic, 35, No. 6, 371–383 (1996).

11. Jones W. and Parshall B., “On the 1-cohomology of finite groups of Lie type,” in: Proc. Conf. Finite Groups/ eds.W. R. Scott, F. Gross, Academic Press, New York, 1976, pp. 313–327.

A. A. Makhnev; D. V. Paduchikh; L. Yu. Tsiovkina

Institute of Mathematics and Mechanics, Ekaterinburg, Russia

E-mail address: makhnev@imm.uran.ru; dpaduchikh@gmail.com; l.tsiovkina@gmail.com

1087