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6URAL MATHEMATICAL JOURNAL, Vol. 4, No. 2, 2018, pp. 69-78
DOI: 10.15826/umj.2018.2.008
AUTOMORPHISMS OF A DISTANCE-REGULAR GRAPH WITH INTERSECTION ARRAY {39, 36,4;1,1,36}1.
Konstantin S. Efimov
Ural State University of Economics, 62 March 8th Str., Ekaterinburg, Russia, 620144
konstantin.s.efimov@gmail.com
Alexander A. Makhnev
Krasovskii Institute of Mathematics and Mechanics,
Ural Branch of the Russian Academy of Sciences, 16 S. Kovalevskaya Str., Ekaterinburg, Russia, 620990 makhnev@imm.uran.ru
Abstract: Makhnev and Nirova have found intersection arrays of distance-regular graphs with no more than 4096 vertices, in which A = 2 and ^ =1. They proposed the program of investigation of distance-regular graphs with A = 2 and ^ =1. In this paper the automorphisms of a distance-regular graph with intersection array {39, 36,4; 1, 1, 36} are studied.
Keywords: Strongly regular graph, Distance-regular graph.
Introduction
We consider undirected graphs without loops and multiple edges. Our terminology and notation are mostly standard and could be found in [1]. Given a vertex a in a graph r, we denote by r(a) the subgraph induced by r on the set of all the vertices of r, that are at the distance i from a. The subgraph [a] = r1(a) is called the neighbourhood of a vertex a. Let r(a) = r1(a), a± = {a} U r(a). If graph r is fixed, then we write [a] instead of r(a).
The incidence system with the set of points P and the set of lines L is called a-partial geometry of order (s,t) if each line contains exactly s + 1 points, each point lies exactly on t + 1 lines, any two points lie on no more than one line, and for any antiflag (a, l) € (P, L) there are exactly a lines passing through a and intersecting l. This geometry is denoted by pGa(s,t).
In the case a = 1, the geometry pGa(s,t) is called a generalized quadrangle and is denoted by GQ(s,t). A point graph of this geometry is defined on the set of points P and two points are adjacent if they lie on a line. The point graph of a geometry pGa(s,t) is strongly regular with parameters v = (s + 1)(1 + st/a), k = s(t + 1), A = s — 1 + t(a — 1), ^ = a(t + 1). A strongly regular graph with such parameters for some natural numbers a, s, t is called a pseudo-geometric graph for pGa (s, t).
If vertices u, w are at distance i in r, then by bi(u,w) (respectively, c(u,w)) we denote the number of vertices in Ti+1(u) n [w] (respectively, ri-1(u) n [w]). A graph r of diameter d is called distance-regular with intersection array {b0, b1,..., bd-1; c1,..., cd} if the values bi(u, w) and ci(u, w) do not depend on the choice of vertices u, w at distance i in r for each i = 0,..., d. Note that, for a
1This work is partially supported by RSF, project 14-11-00061P
distance-regular graph, b0 is the degree of the graph and c\ = 1. For a subset X of automorphisms of a graph r, Fix(X) denotes the set of all vertices of r, fixed with respect to any automorphism of X. Further, by pj(x, y) we denote the number of vertices in a subgraph r^(x) n r (y) for vertices x,y at distance l in r.
A graph is said to be vertex-symmetric if its automorphism group acts transitively on the set of its vertices.
In [2], intersection arrays of distance-regular graphs with A = 2, i = 1 and with the number of vertices at most 4096 were found. A.A. Makhnev and M.S. Nirova proposed an investigation program of automorphisms of distance-regular graphs from the obtained list.
Proposition 1. [2] Let r be a distance-regular graph with A = 2, ¡i = 1, which has at most 4096 vertices. Then r has one of the following intersection arrays:
(1) {21,18; 1,1}(v = 400);
(2) {6,3,3,3; 1,1,1,2} (r is a generalized octagon of order (3,1), v = 160), {6,3,3;1,1,2} (r is a generalized hexagon of order (3,1), v = 52), {12,9,9; 1,1,4} (r is a generalized hexagon of order (3,3), v = 364), {6,3,3,3,3,3; 1,1,1,1,1,2} (r is a generalized dodecagon of order (3,1), v = 1456);
(3) {18,15,9; 1,1,10}(v = 1 + 18 + 270 + 243 = 532, r is a strongly regular graph); {33, 30, 8; 1, 1, 30}, {39, 36, 4; 1, 1, 36}, {21, 18, 12, 4; 1, 1, 6, 21}.
In this paper we study automorphisms of a hypothetical distance-regular graph r with intersection array {39,36,4; 1,1,36}. The maximal order of a clique C in r is not more than 4. A graph with intersection array {39,36,4; 1,1,36} has v = 1 + 39 + 1404 + 156 = 1600 vertices and the spectrum 391, 7675, -1156, -6768.
Theorem 1. Let r be a distance-regular graph with intersection array {39,36,4;1,1,36}, 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} and one of the following statements holds:
(1) Q is an empty graph and either p = 2, a1(g) = 10r + 26m + 12 and aa(g) = 80r or p = 5, ai(g) = 65n + 10l + 10 and aa(g) = 200l;
(2) Q is an n-clique and one of the following statements holds:
(i) n = 1, p = 3, ai(g) = 15l + 24 + 39m and aa(g) = 120l + 36,
(ii) n = 2, p = 2, ai(g) = 10l + 26m and aa(g) = 80l - 8,
(in) n = 4, p = 2, a1(g) = 10l + 26m + 14 and aa(g) = 80l — 16 or p = 3, a1(g) = 10l + 39m + 1, l is congruent to —1 modulo 3 and aa(g) = 120l + 24;
(3) Q consists of n vertices pairwise at distance 3 in r, p = 3, n € {4,7,..., 40}, aa(g) = 120l + 40 — 4n and 01(g) = 15l + 30 + 39m — 6n;
(4) Q contains an edge and is a union of isolated cliques, any two vertices of different cliques are at distance 3 in r, and either p = 3 and the orders of these cliques are 1 or 4, or p = 2 and the orders of these cliques are 2 or 4;
(5) Q contains vertices that are at distance 2 in r and p < 3.
If r is a distance-regular graph with the intersection array {39,36,4; 1,1,36} then ra is a pseudo-geometric for pGa(39,3).
Theorem 2. Let r be a strongly regular graph with parameters (1600,156,44,12), G = Aut(r), g is an element of prime order p in G and A = Fix(g). Then p < 43 and the following statements hold:
(1) if A is an empty graph, then p = 2 and a1(g) = 80s or p = 5 and a1(g) = 200t;
(2) if A is an n-clique, then one of the following statements holds:
(i) n = 1, p = 2 and a1(g) = 80s — 4, or p = 3 and a1(g) = 120t + 36, or p = 13 and a1(g) = 5201 + 156,
(ii) n € {4, 7,10,..., 40}, p = 3 and a1(g) = 120t + 40 — 4n,
(iii) n = 9, p = 37 and a1(g) = 444;
(3) if A is an m-coclique, where m > 1, then either p = 2, m € {4,6,8,..., 40} and a1(g) = 80s — 4m or p = 3, m € {4, 7,10,..., 40} and a1(g) = 120t + 40 — 4m;
(4) if A contains an edge and is an union of isolated cliques, then p = 3;
(5) if A contains a geodesic 2-path, then p < 43.
Corollary 1. Let r be a distance-regular graph with intersection array {39,36,4;1,1,36} and nonsolvable group G = Aut(r) acts transitively on the set of vertices of r. If a is a vertex of r, T is the socle of the group G = G/O5'(G), then T = L x M, and each of subgroups L, M is isomorphic to one of the following groups: Z5,A5,A6 or PSp(4,3).
If |T : T0| = 402, then O5'(G) = 1 and this case is realized if one of the following statements holds:
(1) L ^ M = PSp(4, 3), |L : Lo| = |M : Mo| = 40,
(2) L = PSp(4, 3), |L : Lo| =40, M ^ A3 and |Mo| = 9,
(3) L ^ M = A6 and |Lo| = |Mo| = 9.
1. Proof of Theorem 2
First we give auxiliary results.
Lemma 1. [2, Theorem 3.2] Let r be a strongly regular graph with parameters (v,k, A,^) and with the second eigenvalue r. If g is an automorphism of r and A = Fix(g), then
|A| < v ■ max{A, ^}/(k — r).
By Lemma 1, for a strongly regular graph with parameters (1600,156,44,12) we have |A| < 1600 ■ max{44,12}/(156 — 36), |A| < 586.
Lemma 2. Let r be a distance regular graph with intersection array {39,36,4; 1,1,36}. Then for intersection numbers of r the following statements hold:
(1) p111 = 2, p112 = 36, p122 = 1224, p213 = 144, p313 = 12;
(2) p121 = 1, p212 = 34, p213 = 4, p222 = 1229, p223 = 140, p323 = 12;
(3) p32 = 36, p33 = 3, p22 = 1260, p33 = 108, p33 = 44.
Proof. This follows from [1, Lemma 4.1.7]. □
The proofs of Theorems 1 and 2 are based on Higman's method of working with automorphisms of a distance-regular graph, presented in the third chapter of Cameron's book [4].
Let r be a distance-regular graph of diameter d with v vertices. Then we have a symmetric association scheme (X, R) with d classes, where X is the set of vertices of r and Ri = {(u, w) € X2| d(u, w) = i}. For a vertex u € X we set ki = |ri(u)|. Let Ai be an adjacency matrix of graph ri. Then AiAj = pijAi for some integer numbers pj > 0, which are called the intersection numbers. Note that pj = |ri(u) n Tj(w)| for any vertices u, w with d(u, w) = 1.
Let Pi be a matrix in which in the (j, 1)-th entry is pj. Then eigenvalues k = p1(0), ...,p1(d) of the matrix P1 are eigenvalues of r with multiplicities m0 = 1,..., md, respectively. The matrices P
and Q with Pj = pj (i) and Qji = mjPi(j)/ki are called the first and the second eigenmatrices of r, respectively, and PQ = QP = vI, where I is an identity matrix of order d +1.
The permutation representation of the group G = Aut(r) on the vertex set of r naturally gives the monomial matrix representation ^ of a group G in GL(v, C). The space Cv is an orthogonal direct sum of the eigenspaces W0, Wi,..., Wd of the adjacent matrix A = A1 of r. For every g € G, we have ^(g)A = A^(g), so each subspace Wi is ^(G)-invariant. Let xi be the character of a
representation . Then for g € G we obtain Xj(g) = v number of vertices x of X such that d(x,xg) = j.
Qj aj (g), where aj (g) is the
j=0 ^ ij"j
Lemma 3. Let r be a strongly regular graph with parameters (1600,156,44,12) and with the spectrum 1561,36156, —4144a, G = Aut(r). If g € G, %1 is the character of ^Wl, where dim(W1) = 156, then a^g) = ai(gl) for any natural number l, coprime to |g|, X1(g) = (4ao(g) + a1(g))/40 — 4. Moreover, if |g| = p is a prime, then x1(g) — 156 is divisible by p.
Proof. We have
1
1
1
Q = | 156 36 —4 1443 —37 3
So, X1(g) = (39ao(g) + 9a1 (g) — a2(g))/400. Note that 02(g) = 1600 — ao(g) — 01 (g), so X1(g) = (4a0(g) + a1(g))/40 — 4. The remaining statements of the lemma follow from Lemma 2 [5]. □
Lemma 4. Let r be a distance-regular graph with intersection array {39,36,4;1,1,36}, G = Aut(r). If g € G, x1 is the character of ^Wl, where dim(W1) = 675, x2 is the character of , where dim(W2) = 156, then ai(g) = ai(g1) for any natural number l coprime to |g|, X1(g) = (44ao(g) + 801(g) — aa(g))/104 — 25/13 and X2(g) = (4ao(g) + aa(g))/40 — 4. Moreover, if |g| = p is a prime, then x1(g) — 675 and x2(g) — 156 are divisible by p.
P r o o f. We have
Q =
1 1 1 1 675 1575/13 -25/13 -225/13
156
-4
-4
36
V 768 -1536/13 64/13 -256/13 J
This means x1(g) = (351ao(g) + 63a1(g) — a2(g) — 9aa(g))/832. Note that a2(g) = 1600 — ao(g) — 01(g) — aa(g), so x1(g) = (44ao(g) + 801(g) — aa(g))/104 — 25/13.
Similarly, x2(g) = (390o(g) — 01(g) — 02(g) + 90a(g))/400. Note that 01(g) + 02(g) = 1600 — 0o(g) — 0a(g), so x2(g) = (40o(g) + 0a(g))/40 — 4.
The remaining statements of this lemma follow from Lemma 2 of [5].
In Lemmas 5-7 we suppose that r is a strongly regular graph with parameters (1600,156,44,12), G = Aut(r), g is an element of prime order p from G, ai(g) = pwi for i > 0 and A = Fix(g). By Delsarts's boundary the maximal order of a clique K in r is not greater than 1 — k/^d, so |K| < 40. Due to Hoffman's boundary the maximum order of a coclique C in r is not greater than —v0d/(fc — 0d), so |C| < 40.
Lemma 5. The following statements hold:
(1) if A is an empty graph, then either p = 2 and 01(g) = 80s or p = 5 and 01(g) = 200t;
(2) if A is an n-clique, then one of the following statements holds:
(i) n = 1, p = 2 and a1(g) = 80s — 4, or p = 3 and a1(g) = 120t + 36, or p = 13 and a1(g) = 5201 + 156,
(ii) n € {4, 7,10,..., 40}, p = 3 and a1(g) = 120t + 40 — 4n,
(iii) n = 9, p = 37 and a1(g) = 444;
(3) if A is an m-coclique, where m > 1, then p = 2, m € {4,6,8,..., 40} and a1 (g) = 80s — 4m or p = 3, m € {4, 7,10,..., 40} and a1(g) = 120t + 40 — 4m;
(4) if A contains an edge and is an union of isolated cliques, then p = 3.
Proof. Let A be an empty graph. As v = 26 ■ 25, then p is equal to 2 or 5.
In the case p = 2 we have x1(g) = a1(g)/40 — 4 and a1(g) = 80s.
In the case p = 5 we have x1(g) = a1(g)/40 — 4 and a1(g) = 200t.
Let A be an n-clique. If n = 1, then p divides 156 and 1443, therefore p € {2,3,13}. In the case p = 2 we have x1(g) = (4 + a1(g))/40 — 4 and a1(g) = 80s — 4.
In the case p = 3 we have x1(g) = (4 + a1(g))/40 — 4 and the number (4 + 3w1)/40 is congruent to 1 modulo 3. Hence, 4 + 3w1 = 120t + 40 and a1(g) = 120t + 36.
In the case p = 13 we have x1(g) = (4 + 13w1 )/40 — 4 and the number (4 + 13w1 )/40 is congruent to 4 modulo 13. Hence, 4 + 13w1 = 5201 + 160 and a1(g) = 5201 + 156.
If n > 1, then for any two vertices a, b € A the element g acts without fixed points on [a]n[b] — A, on [a] — b± and on r — (ax U bx). Hence, p divides 46 — n, 111 and 1332, therefore p € {3,37}.
In the case p = 3 we have n € {4, 7,10,..., 40}. Further, x1(g) = (4n + a1(g))/40 — 4 and the number (4n + a1(g))/40 is congruent to 1 modulo 3. Hence, 4n + 3w1 = 120t + 40 and a1(g) = 120t + 40 — 4n.
In the case p = 37 we have n = 9. Further, x1(g) = (36 + a1(g))/40 — 4 and the number (36 + a1(g))/40 is congruent to 12 modulo 37. Hence, a1(g) = 444.
Let A be an m-coclique, where m > 1. Then for any two vertices a, b € A the element g acts without fixed points on [a] n [b], on [a] — b± and on r — (ax U b± U A). Hence, p divides 12, 144 and 1300 — m, therefore p € {2,3}.
In the case p = 2 we have m € {4,6,8,..., 40}. Further, x1(g) = (4m + a1(g))/40 — 4 and the number (4m + a1(g))/40 is even. Hence, a1(g) = 80s — 4m.
In the case p = 3 we have m € {4, 7,10,..., 40}. Further, x1(g) = (4m + a1(g))/40 — 4 and the number (4m + a1(g))/40 is congruent to 1 modulo 3. Hence, a1(g) = 120t + 40 — 4m.
Let A contains an edge and is a union of isolated cliques. Then p divides 12 and 111, therefore p = 3. □
Lemma 6. If [a] C A for some vertex a, then for any vertex u € r2(a) — A the orbit of u^ is a clique or a coclique, and one of the following statements holds:
(1) if on r — A there are no coclique orbits, then a1(g) = 1600 — a0(g), a0(g) = 401 and either
(i) 1 = 4, p = 2,3 or
(ii) 1 = 5, p = 5, 7, or
(iii) 1 = 6, p = 2, or
(iv) 1 = 7, p = 3, 11, or
(v) 1 = 8, p = 2, or
(vi) 1 = 10, p = 2, 3, 5, or
(vii) 1 = 12, p = 2, 7, or
(viii) 1 = 13, p = 3, or
(ix) 1 = 14, p = 2;
(2) if on r — A there is a coclique orbit, then p < 3, and if a± = A, then p = 3 and a1(g) = 1201 + 12.
Proof. Let [a] C A for some vertex a. Then for any vertex u € r2(a) — A the orbit doesn't contain a geodesic 2-pathes and is a clique or a coclique.
In the case p > 13 a subgraph [a] n [u] is a 12-clique and for two vertices b, c € [a] n [u] a subgraph [b] n [c] contains a, 10 vertices from [a] n [u] and p vertices from u^, so 11 + p < 44, therefore p < 31.
If on r — A there are no coclique orbits, then ai(g) = v — |A| and for a vertex u' € — {u} a subgraph [u] n [u'] contains p — 2 vertices from u^ and 12 vertices from A. Further, x1(g) = (3a0(g) + 1600)/40 — 4, x1(g) — 156 is divisible by p and p divides 3a0(g)/40 — 120. We denote ao(g) = 401. Then 4 < 1 < 14, p divides 40(40 — 1) and 3(40 — 1). Thus, either 1 = 4, p = 2,3, or 1 = 5, p = 5, 7, or 1 = 6, p = 2,17, or 1 = 7, p = 3,11, or 1 = 8, p = 2, or 1 = 9, p = 31, or 1 = 10, p = 2, 3, 5, or 1 = 11, p = 29, or 1 = 12, p = 2, 7, or 1 = 13, p = 3, or 1 = 14, p = 2,13. In the case p > 13 a subgraph [a] n [u] is a 12-clique and p < 23.
Let p = 17 and b € A — ax. Then |A(b) — ax| < 82 and |[b] — A| > 68. For w € [b] — A we have
[a] n [w] = [a] n [b] (otherwise w^ is contained in [b] n [c] for c € [a] n [w] — [b]). A contradiction with a fact that for two vertices c, d € [a] n [w] a subgraph [c] n [d] contains 68 vertices from [b] — A.
Let p = 13. Then |A — ax| = 403. If b € A — a± and |[b] — A| = 13, then for any w € [b] — A we have [a] n [w] = [a] n [b] (otherwise w^ is contained in [b] n [c] for a vertex c € [a] n [w] — [b]). Further,
[b] n [w] contains 12 vertices from w^ and 32 vertices from A(b). Hence, for w' € w^ — {w} a subgraph [w] n [w'] contains b, 32-clique from A(b) and 11 vertices from w^. A contradiction with a fact that the order of a clique in r is not greater than 40.
If b € A —ax and |[b] — A| = 26, then [b]—A = Uw<«>. As above, [a]n[u] = [a]n[w] = [a]n[b], therefore a subgraph {b} U ([a] n [b]) U U is a 39-clique. If e € [u] n A(b) — [w], then [e] n [w] contains 13 vertices from a contradiction. So, {b} U ([u] n A(b)) U u^ U is a 46-coclique, a contradiction. If b € A — a± and |[b] — A| > 39, then for any two vertices c, d € [a] n [b] a subgraph
[c] n [d] contains a, b and 39 vertices from [b] — A, a contradiction. Statement (1) is proved.
Let on r — A there is a coclique orbit u^. Then [ugi] n [ugj ] does not intersect r — A for distinct vertices ugi,ugj, so 145p < |r — A| < 1443, therefore p < 7.
Let us show that p < 3.
Let c € [a] n [u] and [c] n [u] contains exactly 7 vertices from [a] n [u]. Then [c] n [u] contains 44 — 7 vertices outside of A (lying in distinct (g)-orbits) and p(44 — 7) < |[c] — A| < 156 — 45 = 111. Hence, 32p < 111.
If a± = A, then a0(g) = 157, p divides 1443 and p = 3. Further, x1(g) = (628 + a1(g))/40 — 4, (628 + a1(g))/40 is congruent to 1 modulo 3 and a1(g) = 1201 + 12. □
Lemma 7. The following statements hold :
(1) r does not contain proper strongly regular subgraphs with parameters (v',k', 44,12);
(2) p < 43.
Proof. Assume that r contains proper strongly regular subgraph S with parameters (v',k', 44,12). Then 4(k' —12) + 322 = n2, therefore n = 21, k' = 12 — 244, 1 > 16, S has nonprincipal eigenvalues 16 +1,16 — 1 and multiplicity of 16 +1 is equal to (1 — 17)(12 — 244)(12 +1 — 260)/241. If 1 is odd, then 8 divides (1 — 17)(12 +1 — 20), 1 divides 17 ■ 61 ■ 65 and 1 € {5,13}. If 1 is even, then 3 divides (1 — 2)(12 — 1)(12 +1 — 2) and 1 = 16. In all cases we have contradictions.
If p > 47, then A is a strongly regular graph with parameters (v',k', 44,12), so A = r, a contradiction. □
Theorem 2 follows from Lemmas 5-7.
2. Proof of Theorem 1
In Lemmas 8-9 it is assumed that r is a distance-regular graph with intersection array {39,36,4; 1,1,36}, G = Aut(r), g is an element of prime order p from G, 0i(g) = pwi for i > 0 and Q = Fix(g).
Lemma 8. The following statements hold:
(1) if Q is an empty graph, then either p = 2, 01(g) = 10r + 26m + 12 and 0a(g) = 80r = 1600 — 01(g) or p = 5, 01(g) = 65n + 10l + 10 and 0a(g) = 200l;
(2) if Q is an n-clique, then one of the following statements holds:
(i) n = 1, p = 3, 01(g) = 15l + 24 + 39m and 0a(g) = 120l + 36,
(ii) n = 2, p = 2, 01(g) = 10l + 26m and 0a(g) = 80l — 8,
(iii) or n = 4, p = 2, 01(g) = 10l + 26m + 14 and 0a(g) = 80l — 16 or p = 3, 01(g) = 10l + 39m + 1, l is congruent to —1 modulo 3 and 0a(g) = 120l + 24;
(3) if Q consists of n vertices at distance 3 in r, then p = 3, n € {4,7,10,..., 40}, 0a(g) = 120l + 40 — 4n and 01(g) = 15l + 30 + 39m — 6n;
(4) if Q contains an edge and doesn't contain vertices at distance 2 in r, then Q is an union of isolated cliques and any two vertices from different cliques are at distance 3 in r, either p = 3 and the orders of these cliques are equal to 1 or 4, or p = 2 and the orders of these cliques are equal to 2 or 4.
Proof. Let Q be an empty graph and (g) = pwi for i > 1. As v = 1600, then p is equal to 2 or 5.
Let p = 2. Then w1 + w2 + wa = 800 and x2(g) = wa/20 — 4. Hence, wa = 40r. Further, the number x1(g) = (2w1 — 10r — 25)/13 is odd, therefore w1 = 13m + 6 + 5r. Finally, 02(g) = 0 (if d(u,ug) = 2, then the only vertex from [u] n [ug] belongs to Q, a contradiction). Therefore 01(g) = 10r + 26m + 12 = 1600 — 80r.
Let p = 5. Then w1 + w2 + wa = 320 and x2(g) = wa/8 — 4. Hence, wa = 40l. Finally, x1(g) = (5w1 — 25l — 25)/13, therefore w1 = 13n + 5l + 5. Statement (1) is proved.
Let Q be an n-clique. If n = 1, then p divides 39 and 315, therefore p = 3. We have x1(g) = (801(g) — 0a(g) — 156)/104, x2(g) = (4 + 0a(g))/40 — 4. Therefore the number (4 + 0a(g))/40 is congruent to 1 modulo 3, 0a(g) = 120l + 36 and the number x1(g) = (01(g) — 15l — 24)/13 is divisible by 3. Hence, 01(g) = 15l + 24 + 39m.
If n > 1, then p divides 4 — n and 36, therefore either n = 2, p = 2, or n = 4, p = 2, 3. In the first case the number x2(g) = (8 + 0a(g))/40 — 4 is even and 0a(g) = 80l — 8. Further, the number x1(g) = (01(g) — 10l)/13 — 1 is odd and 01(g) = 10l + 26m. In the second case x2(g) = (16 + 0a(g))/40 — 4 and either p = 2, 0a(g) = 80l — 16, or p = 3 and 0a(g) = 120l + 24. Further, x1(g) = (176 + 801(g) — 0a(g))/104 — 25/13 and either p = 2, 01(g) = 10l + 26m + 14, or p = 3 and 01(g) = 10l + 39m + 1, l is congruent to —1 modulo 3.
Let Q consists of n vertices at distance 3. As paa = 3, paa = 44, then p divides 3 and 46 — n. Hence, p = 3 and n € {4, 7,10,..., 40}. We have x2(g) = (4n + 0a(g))/40 — 4 and the number (4n + 0a(g))/40 is congruent to 1 modulo 3, therefore 0a(g) = 120l + 40 — 4n. Further, the number x1(g) = (6n + 01(g) — 15l — 30)/13 is divisible by 3 and 01(g) = 15l + 30 + 39m — 6n.
Let Q contains an edge and does not contain vertices at distance 2 in r. Then Q is an union of isolated cliques, any two vertices from distinct cliques are at distance 3 in r. As orders of these cliques are at most 4, then p < 3. If p = 3, then the orders of these cliques are equal to 1 or 4. If p = 2, then the orders of these cliques are equal to 2 or 4. □
Lemma 9. If Q contains vertices a, b at distance 2 in r, then p < 3.
Proof. Let Q contains vertices a, b at distance 2 in r and Qo is a connected component of Q containing a, b.
Assume that the diameter of graph Qo is equal to 2. Then by [1, 1.17.1] one of the following statements holds:
(i) Qo C a± and Qo(a) is an union of isolated cliques;
(ii) Qo ia a strongly regular graph;
(iii) Qo is a biregular graph with degrees of vertices 0, 5, where 0 < 5, and if A and B are sets of vertices from Qo with degrees 0 and 5, then A is a coclique, the lines between A and B have order 2, the lines from B have order l = 5 — 0 + 2 > 2, and |Qo| = + 1.
Last case is impossible because c2 = 1 in r.
In the case (i) we have p € {2,3} because of p1a = 12.
In the case (ii) either p = 2 and Qo is the pentagon, Petersen graph or Hoffman-Singletone graph, or p > 2 and Qo is a strongly regular graph with parameters (v', k', 2,1).
Let p > 2. Then Q(a) consists of e isolated triangles and either e = 1, p = 3, or e = 2, p = 3,11, or e = 3, p = 3, 5, or e = 4, p = 3, or e = 5, p = 3, or e = 6, p = 3, 7, or e = 7, p = 3, or e = 8, p = 3, 5, or e > 9, p = 3.
In case p = 11 graph Q is a regular graph of degree 6, |Q n r2(a)| = 18, |Q n ra(a)| = 24 and |ra(a) — Q| is not divisible by 11.
In case p = 7 graph Q is a regular graph of degree 18, |Qnr2(a)| = 270, |Qnra(a)| = 270-4/15 = 64 and |ra(a) — Q| is not divisible by 7.
In case p = 5 graph Q contains vertices of degrees 9 and 24. Assume that |Q(a)| = 24, Q(a) contains 5 vertices of degree 24 in Q and Qa(a) contains 7 vertices of degree 24 in Q. Then the number 215 + 6(24 — 5) = |Q n r2(a)| is congruent to 4 modulo 5 and 4|Q n r2(a)| = 21y + 6(|Q n ra(a)| — 7). Hence, ^(a) n Q| = (144 + 155) and 576 + 605 = 15y + 6|Q n ra(a)|, a contradiction with the fact that |Q n ra(a)| is divisible by 5.
So, Q is an amply regular graph with parameters (v', 9,2,1), 54 = |Qnr2(a)| and |Q nra(a)| = 36. Again we have a contradiction with the fact that |Q n ra(a)| is divisible by 5.
The lemma is proved. □
Theorem 1 follows from Lemmas 8-9.
3. Proof of Corollary 1
Until the end of the paper we will assume that r is a distance-regular graph with intersection array {39,36,4;1,1,36} and the nonsolvable group G = Aut(r) acts transitively on the set of vertices of this graph. For the vertex a € r we get |G : Ga| = 1600. In view of Theorem 1 we have p € {2,3, 5}. Let T be the socle of the group G = G/O5'(G).
Lemma 10. If f is an element of order 5 of G, g is an element of order p < 5 of CG(f) and Q = Fix(g), then one of the following statements holds:
(1) Q is an empty graph, p = 2, 0a(g) = 80r, r < 19, 01(g) = 10r + 26m + 12 = 1600 — 80r, and m € {—7, —2, 3, 8,..., 58};
(2) Q consists of n vertices at distance 3 in r, p = 3, n € {10,25,40}, 0a(g) = 120l + 40 — 4n, 01(g) + 0a(g) = 135l — 10n + 39m + 70 < 1600 and m is divisible by 5;
(3) p = 3, 0a(g) = 120s, 0o(g) = 30t + 10, 01(g) = 39l — 165t + 15s — 30 or 0a(g) = 120s + 60, 0o(g) = 30t — 5, 01(g) = 195l — 165t + 15s + 60;
(4) p = 2, 0a(g) = 80s — 40o(g) and 01(g) = 10s + 26l + 38 — 60o(g), l is congruent to 2 modulo 5.
Proof. In view of Theorem 1 Fix(f) is empty graph, a1(f) = 65n+101 + 10 and a3(f) = 2001.
If Q is an empty graph, then p = 2, a3(g) = 80r and a1(g) = 10r + 26m + 12 = 1600 — 80r is divisible by 5. Hence, 13m + 6 is divisible by 5 and m € {—7, —2,3,8,..., 58}. Finally, 26m + 12 = 1600 — 90r, therefore m is congruent to 2 modulo 3 and m € {—7,8,23,38, 53}.
If Q is an n-clique, then n is divisible by 5, we have got a contradiction.
If Q consists of n vertices at distance 3 in r, then p = 3, n € {10,25,40}, the numbers a3(g) = 1201 + 40 — 4n and a1(g) = 151 + 30 + 39m — 6n are divisible by 5. Hence, m is divisible by 5, a1(g) + a3(g) = 1351 — 10n + 39m + 70 < 1600.
If p = 3, then x2(g) = (4a0(g)+ a3(g))/40 — 4 and the number (4a0(g)+ a3(g))/40 is congruent to 1 modulo 3. Further, the number x1(g) = (44a0(g) + 8a1(g) — a3(g))/104 — 25/13 is divisible by 3, a3(g) is divisible by 60. If a3(g) = 120s, then a0(g) = 30t + 10, a1(g) = 391 — 165t + 15s — 30. If a3(g) = 120s + 60, then a0(g) = 30t — 5, a1(g) = 1951 — 165t + 15s + 60.
If p = 2, then x2(g) = (4a0(g) + a3(g))/40 — 4, 4a0(g) + a3(g) = 80s. Further, a1(g) = —6a0(g) + 10s + 261 + 38 and 131 + 19 is divisible by 5, therefore 1 € {2, 7,...}. Finally, 1600 — 5a0(g) + 80s = —6a0(g) + 10s + 261 + 38, 1600 = —70s — a0(g) + 261 + 38. □
Lemma 11. The following statements hold:
(1) T = L x M, and each of subgroups L, M is isomorphic to one of the following groups Z5,A5,A6 or PSp(4,3);
(2) in case |T : To| = 402 we have O5' (G) = 1 and this case is realized if one of the following statements holds:
(i) L ^ M ^ PSp(4,3), or
(ii) L = PSp(4, 3), |L : Lo| =40, M = A6 and |Mo| = 9, or
(iii) L ^ M = A6 and |Lo| = |Mo| = 9.
Proof. Recall that a nonabelian simple {2,3, 5}-group is isomorphic to A5, A6 or PSp(4,3) (see, [6, Table 1]). Hence, in view of Theorem 1 we have T = L x M, each of subgroups L,M is isomorphic to one of the following groups A5, A6 or PSp(4,3).
If T = PSp(4,3), then the group To has an index 40 in T and is isomorphic to Eg .SL2(3) or
E27.S4.
If T = A6, then the group To has an index in T, divisible by 10, and dividing 40.
If T = A5, then the group To has an index in T, divisible by 10, and dividing 20.
In case |T : To| = 402 we have O5'(G) = 1 and this case is realized if one of the following statements holds: either L = M = PSp(4,3), or L = PSp(4,3), M = A6 |Mo| = 9, or L = M = A6 and |Lo| = |Mo| =9. □
Corollary is proved.
4. Conclusion
We found possible automorphisms of a distance-regular graph with intersection array {39, 36, 4; 1, 1, 36}. In particular this graph is not arc-transitive.
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