Automorphisms of distance-regular graph with intersection array {25; 16; 1; 1; 8; 25} Текст научной статьи по специальности «Математика»

URAL MATHEMATICAL JOURNAL, Vol. 3, No. 1, 2017

AUTOMORPHISMS OF DISTANCE-REGULAR GRAPH WITH INTERSECTION ARRAY {25,16,1; 1,8, 25} 1

Konstantin S. Efimov

Ural Federal University, Ekaterinburg, Russia, Ural State University of Economics, Ekaterinburg, Russia [email protected]

Alexander A. Makhnev

N.N. Krasovskii Institute of Mathematics and Mechanics UB RAS, Ekaterinburg, Russia, Ural Federal University, Ekaterinburg, Russia [email protected]

Abstract: Makhnev and Samoilenko have found parameters of strongly regular graphs with no more than 1000 vertices, which may be neighborhoods of vertices in antipodal distance-regular graph of diameter 3 and with A = f. They proposed the program of investigation vertex-symmetric antipodal distance-regular graphs of diameter 3 with A = f, in which neighborhoods of vertices are strongly regular. In this paper we consider neighborhoods of vertices with parameters (25, 8, 3, 2).

Key words: Strongly regular graph, Distance-regular graph.

Introduction

We consider undirected graphs without loops and multiple edges. Given a vertex a in a graph r, we denote by r»(a) the subgraph induced by r on the set of all vertices, that are at the distance i from a. The subgraph [a] = r1 (a) is called the neighborhood of the vertex a. Let r(a) = r1(a), a± = {a} U r(a). If graph r is fixed, then instead of r(a) we write [a]. For the set of vertices X of graph r through X± denote %±.

Let r be an antipodal distance-regular graph of diameter 3 and A = ¡, in which neighborhoods of vertices are strongly-regular graphs. Then r has intersection array {k,i(r — 1), 1; 1,i,k}, and f f

spectrum , where f = (k + 1)(r — 1)/2. In the case r = 2 we obtain Taylor's

graph, in which k' = 2¡'. Conversely, for any strongly regular graph with parameters (v', 2, A', ¡') there exists a Taylor's graph, in which neighborhoods of vertices are strongly regular with relevant parameters.

In [1]there were chosen strongly-regular graphs with no more than 1000 vertices, which may be neighborhoods of vertices of antipodal distance-regular graph of diameter 3 and A = ¡. There is provided a research program of the study of vertex-symmetric antipodal distance-regular graphs of diameter 3 with A = ¡, in which neighborhoods of vertices are strongly regular with parameters from Proposition 1.

Proposition 1. Let A be a strongly-regular graph with parameters (v,k, A,i). If (r — 1)k = v — k — 1, v < 1000 and number (v + 1)(r — 1) is even, then either r = 2, or parameters (v, k, A, ¡, r) belong to the following list:

1This work is partially supported by RSF, project 14-11-00061-P (Theorem 1) and by the program of the government support of leading universities of Russian Federation, agreement 02.A03.21.0006 from 27.08.2013 (Corollary 1).

(1) (16, 5, 0, 2, 3), (81, 20,1, 6, 4), (l21, 30,11, 6, 4),

(169

(2) (225 (261 (289 (352

(3) (400 (484 (529 (540

(4) (625 (649 (729 (768

(5)

(837 (841 (848 (961

(25, 8, 3, 2, 3), (49,12, 5, 2, 4), (64, 21, 8, 6, 3), (81,16, 7, 2, 5), (85,14, 3, 2, 6), (99,14,1, 2, 7), (100, 33, 8,12, 3), (121, 20, 9, 2, 6), (121, 40,15,12, 3), (126, 25, 8, 4, 5), (133, 44,15,14, 3), (169, 24,11, 2, 7),

42, 5,12, 4), (169, 56,15, 20, 3), (176, 25, 0, 4, 7), (196, 39,14, 6, 5), (196, 65, 24, 20, 3);

28,13, 2, 8), (225, 56,19,12, 4), (243, 22,1, 2,11), (256, 51, 2,12, 5), (256, 85, 24, 30, 3), 52,11,10, 5), (288, 41, 4, 6, 7), (289, 32,15, 2, 9), (289, 48,17, 6, 6), (289, 72,11, 20, 4), 96, 35, 30, 3), (305, 76, 27,16, 4), (325, 54, 3,10, 6), (351, 50,13, 6, 7), (351, 70,13,14, 5), 39, 6, 4, 9), (361, 36,17, 2,10), (361, 72, 23,12, 5), (361, 90, 29, 20, 4), (361,120, 35, 42, 3);

57, 20, 6, 7), (400,133, 48, 42, 3), (441, 40,19, 2,11), (441, 88, 7, 20, 5), (441,110,19, 30, 4), 161, 48, 56, 3), (495, 38,1, 3,13), (505, 84, 3,16, 6), (507, 46, 5, 4,11), (512, 73,12,10, 7), 44, 21, 2,12), (529, 66, 23, 6, 8), (529, 88, 27,12, 6), (529,132, 41, 30, 4), (529,176, 63, 56, 3), 49, 8, 4,11), (576,115,18, 24, 5);

48, 23, 2,13), (625,156, 29, 42, 4), (625, 208, 63, 72, 3), (640, 71, 6, 8, 9), (649, 72,15, 7, 9),

216, 63, 76, 3), (676, 75, 26, 6, 9), 52, 25, 2,14), (729,104, 31,12, 7), 59,10, 4,13), (784, 261, 80, 90, 3);

(676,135,14,30,5), (729,182, 55, 42, 4),

(704, 37, 0, 2,19), (736,105,20,14, 7),

76, 15, 6, 11), 168, 47, 30, 5), 121, 24,16, 7), 160, 9, 30, 6),

(841, 56, 27, 2,15), (841, 210, 41, 56, 4), (901, 60, 3, 4, 15), (961,192, 23, 42, 5),

(841, 84, 29, 6,10), (841, 280, 99, 90, 3), (961, 60, 29, 2,16), (961,240,71,56,4),

(841,140,39,20,6), (847, 94, 21, 9, 9), (961,120,35,12, 8), (961, 320, 99, 100, 3),

(1000,111,14,12, 9).

Graphs with local subgraphs having parameters (64, 21,8,6), (81,16, 7,2), (85,14,3,2) and (99,14,1,2) were investigated in [2], [3], [4] and [5]. In this article we investigate parameters (25,8,3,2,3), i.e. this graph is locally 5 x 5-grid. In [6] it is proved that distance-regular locally 5 x 5-grid of diameter more then 2 is either isomorphic to the Johnson's graph J(10, 5) or has an intersection array {25,16,1; 1,8,25}.

Theorem 1. Let r be a distance-regular graph with intersection array {25,16,1; 1,8, 25}, G = Aut(r), g is an element of prime order p in G and Q = Fix(g) contains exactly s vertices in t antipodal classes. Then n(G) C {2, 3, 5,13} and one of the following assertions holds:

(1) Q is empty graph and p £ {2,3,13};

(2) p = 5, t = 1, 0.3(g) = 0, ai(g) = 50/ + 25 and 02(g) = 50 - 50/;

(3) p = 3, s = 3, t = 2, 5, 8, a:i(g) = 0, ai(g) = 30/ + 16 - 11t and 0.2(g) = 62 - 30/ + 8t;

(4) p = 2, and either s = 1, Q is t-clique, t = 2,4,6, a3(g) = 2t, a1(g) = 20/ — t + 6 and a2(g) = 72 — 20/ — 2t, or s = 3, t < 8, t is even, a3(g) = 0, a1(g) = 20/ — 11t + 6 and a2(g) = 72 — 20/ + 8t.

Corollary 1. Let r be a distance-regular graph with intersection array {25,16,1; 1,8,25} and a group G = Aut(r) acts transitively on the set of vertices of r. Then one of the following assertions holds:

(1) r is a Cayley graph, G is the a Frobenius group with the kernel of order 13 and with the complement of order 6;

(2) r is a arc-transitive Maton's graph and the socle of G is isomorphic to L2(25);

(3) G is an extension of a group Q of order 212 by the group T = L3(3), \Q : Q{F}\ = 2, T{F} is an extension of group E9 by SL2(3), T acts irreducibly on Q and for an element f of order 13 in G we have CQ(f) = 1.

1. Proof of the Theorem

Note that there is Delsarte boundary (proposition 4.4.6 from [7]) of maximum order of clique in distance-regular graph with intersection array {25,16,1; 1,8,25} and spectrum 251, 526, —125, —526 no more than 1 — k/dd = 1 + 25/5 = 6. If C is 6-clique in r, then each vertex not in C is adjacent to 0 or to b1/(9d + 1) + 1 — k/dd = 2 vertices in C.

Lemma 1. Let r be a distance-regular graph with intersection array {25,16,1; 1,8,25}, G = Aut(r) and g £ G. If ^ is the monomial representation of a group G in GL(78, C), %1 is the character of the representation ^ on subspace of eigenvectors of dimension 26, corresponding to the eigenvalue 5, %2 is the character of the representation ^ on subspace of dimension 25, then X1(g) = (10ao(g) + 2a1(g) — a2(g) — 5a3(g))/30, X2(g) = (ao(g) + 13(g))/3 — 1. If \g\ = p is prime, then x1(g) — 26 and X2(g) — 25 are divided by p.

1 1 1 1

26 26/5 —13/5 —13

25 —1 —1 25 "

26 —26/5 13/5 —13

Therefore x1(g) = (10a0(g) + 2a1(g) — a2(g) — 5a3(g))/30. Substituting a2(g) = 78 — a0(g) — a1(g) — a3(g), we obtain X1(g) = (11a0(g) + 3a1(g) — 4a3(g))/30 — 13/5.

Similarly, X2(g) = (25a0(g) — a1(g) — a2(g) + 25a3(g))/78. Substituting a1(g) + a2(g) = 78 — a0(g) — a3(g), we obtain X2(g) = (a0(g) + a3(g))/3 — 1.

The remaining assertions follow from Lemma 1 in [8]. The proof is complete. □

Let further in the paper r be a distance-regular graph with intersection array {25,16,1; 1,8,25}, G = Aut(r), g is an element of prime order p in G and Q = Fix(g).

Lemma 2. If Q is an empty graph, then either p = 13, a1(g) = 26 and a2(g) = 52, or p = 3, a3(g) = 9s + 6, s < 8, a1(g) = 54 + 12s — 30l and a2(g) = 18 — 21s + 30l, l < 5, or p = 2, (13(g) = 0, a1(g) = 20l + 6 and 12(g) = 72 — 20l, l < 3.

Proof. Let Q be an empty graph and ai(g') = pwi for i > 0. Since v = 78, we have p £ {2,3,13}.

Let p = 13. Then (3(g) = 0, 11(g) + 12(g) = 78 and X1(g) = (211(g) — (2(g))/30 = 13(w1 — 2)/10. This implies a1(g) = 26 and a2(g) = 52.

Let p = 3. Then X2(g)—25 = a3(g)/3—26 is divided by 3, a3(g) = 9s+6, s < 8 and a2(g) = 72— 9s — a1(g). Furthermore, the number X1(g) = (2a1 (g) — a2(g) — 45s — 30)/30 = (3w1 — 12s — 34)/10 is congruent to 2 modulo 3. This implies a1(g) = 54 + 12s — 30l and a2(g) = 18 — 21s + 30l, l < 5. In case s = 8 we have a3(g) = 78 and (g) acts regularly on each antipodal class. By lemma 4 in [9] 3 must divide k + 1 = 26, we have a contradiction.

Let p = 2. Then a3(g) = 0, a1(g) + a2(g) = 78, the number X1(g) = (a1(g) — 26)/10 is even, 11(g) = 20l + 6 and 12(g) = 72 — 20l, l < 3. □

In Lemmas 3-6 it is assumed that there are t antipodal classes intersecting the Q on s vertices. Then p divides 26 — t and 3 — s. Let F be an antipodal class, containing the vertex a £ Q, F n Q = {a, a2,..., as}, b £ Q(a). By F(x) we denote an antipodal class containing vertex x.

Proof. We have

Q =

Lemma 3. The following assertions hold:

(1) if t = 1, then p = 5, a3(g) = 0, a1(g) = 50/ + 25 and a2(g) = 50 — 50/;

(2) if p more than 3, then p = 5 and t = 1;

(3) if s = 1, then p = 2, t = 2,4,6, a3(g) = 2t, a1(g) = 20/ — t + 6 and a2(g) = 72 — 20/ — 2t.

Proof. If s = 3, then each vertex from r — Q is adjacent to t vertices in Q, so t < 8.

Let t = 1. As p divides 26 — t, then p = 5, s = 3, a2(g) = 75 — a1(g), the number x1(g) = (a1(g) — 15)/10 is congruent to 1 modulo 5. This implies a1(g) = 50/ + 25.

Let p > 3, a1(g) = pw1. Then s = 3, \Q\ = 3t, Q is a regular graph by degree t — 1 and p divides 26 — t.

If p > 7, then Q is a distance-regular graph with intersection array {t — 1,16,1; 1,8, t — 1}, we come to a contradiction.

Let p = 7. As p divides 26 — t, then t = 5, the subgraph Q(b) contains 2 vertices in a± and a vertex from [a2] and from [a3], so Q is a distance-regular graph with intersection array {4,1,1; 1,1,4}, it is a contradiction with the fact that r = 3.

Let p = 5. As p divides 26 — t, then t = 1,6. If t = 6, then the subgraph Q(b) contains a vertex in a±, 3 vertices from [a2] and 3 vertices from [a3], we come to a contradiction.

Let s = 1. Thenp = 2, t < 6, 0,3(9) = 2t, 0x2(9) = 78—a1(g)—3t, and %1(g) = (a1(g)+t—26)/10 is even. This implies that a1(g) =20/ — t + 6. □

Lemma 4. If p = 3, then s = 3, t = 2, 5, 8, a3(g) = 0, a1(g) = 30/ + 16 — 11t and a2(g) = 62 — 30/ + 8t.

Proof. Let p = 3. Then s = 3, t = 2, 5,8, a2(g) = 78 — a1(g) — 3t, and the number X1(g) = (11t + a1(g) — 26)/10 is congruent to 2 modulo 3. This implies that a1 (g) = 30/ + 16 — 11t. In the case t = 2 graph Q is a union of 3 isolated edges. □

Lemma 5. If p = 2, s = 3, then t is even, t < 8, a3(g) = 0, a1(g) = 20/ — 11t + 6 and a2(g) = 72 — 20/ + 8t.

Proof. Let p = 2, s = 3. Then t is even, t < 8, a3(g) = 0, a2(g) = 78 — 3t — a1(g).

The number X1(g) = (11t + a1(g) — 26)/10 is even, so a1(g) = 20/ — 11t + 6. □

Lemmas 2-5 imply the proof of the Theorem.

2. Proof of Corollary

Let the group G acts transitively on the set of vertices of the graph r. Then for a vertex a £ r subgroup H = Ga has index 78 in G. By Theorem we have {2, 3,13} C n(G) C {2,3, 5,13}.

Lemma 6. Let f be an element of order 13 in G. Then Fix(f) is an empty graph, a1(f) = 26 and the following assertions hold:

(1) if g is an element of prime order p = 13 in CG(f), then p = 2, Q is an empty graph, a1(g) = 26 and \Cc(f )\ is not divided by 4;

(2) either \G\ =78 or F(G) = 02(G);

(3) if G is nonsolvable group, then the socle T of the group G = G/F(G) is isomorphic to L2(25), L3(3), U3(4), L4(3) or 2F4(2)'.

Proof. By Lemma 2 Fix(f) is an empty graph and 11(f) = 26.

Suppose that g is an element of prime order p = 13 in CG(f). As f acts without fixed points on Q then by Theorem Q is an empty graph, p = 2 and 11(g) = 20l + 6 divided by 13. This implies that 11(g) = 26 and \CG(f)\ is not divided by 4.

Let Q = Op(G) = 1. If p = 13, then \G\ divides 26 • 12. In this case CG(f) = (f), otherwise for an involution g of Cg (f) we obtain a contradiction with the action of element of order 3 of G on {u \ d(u,ug) = 1}. Let the involution g inverts f, h is an element of order 3 in CG(g). From action h on {u \ d(u,ug) = 1} it follows that 11 (g) = 20l + 6 is divided by 3. In each case 11(g) is not divided by 4 and \G\ = 78.

If p = 3, then Q fixes some antipodal class. This implies that Q fixes each antipodal class. By Lemma 3 in [9] G does not contain subgroups of order 3, which are regular on each antipodal class, we come to a contradiction. So, if \G\ = 78 we have F(G) = O2(G).

Let T be the socle of the group G = G/F(G). Note that 13 divides \TT\ and by Theorem 1 in [10] group T is isomorphic to L2(25), L3(3), ^(4), L4(3), 2F4(2)'. □

Let us to prove the Corollary. As T contains a subgroup of index dividing 26, then the group T is isomorphic to L2(25) (and T{F} is the extension of a group of order 25 by group of order 12) or L3(3) (and T{F} is the extension of a group of order 9 by SL(2,3)).

In the first case F(G) fixes each antipodal class and F(G) = 1. This implies that r is the arc-transitiv Maton's graph.

In the second case for Q = F(G) we have \Q : Q{F^ = 2 and T acts irreducibly on Q. Further, for the element f of order 13 of G by Lemma 6 the number \Cq(f)\ divides 2. As Q is either 12-dimensional module over F2, or 16-dimensional module over F16, or 26-dimensional module over F2, then \Q\ = 212 and CQ(f) = 1. The Corollary is proved.

3. Conclusion

We found possible automorphisms of a distance regular graph with intersection array {25,16,1; 1, 8,25}. This completes the research program of vertex-symmetric antipodal distance-regular graphs of diameter 3 with A = ¡, in which neighborhoods of vertices are strongly regular with parameters from Proposition 1.

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