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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
Alexander A. Makhnev
Krasovskii Institute of Mathematics and Mechanics,
Ural Branch of the Russian Academy of Sciences, 16 S. Kovalevskaya Str., Ekaterinburg, Russia, 620990 [email protected]
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.
REFERENCES
1. Brouwer A. E., Cohen A.M., Neumaier A. Distance-Regular Graphs. New York: Springer-Verlag, 1989.
495 p. DOI: 10.1007/978-3-642-74341-2
2. Makhnev A. A., Nirova M.S. On distance-regular graphs with A = 2. J. Sib. Fed. Univ. Math. Phys.,
2014. Vol. 7, No. 2. P. 204-210.
3. Behbahani M., Lam C. Strongly regular graphs with nontrivial automorphisms. Discrete Math., 2011. Vol. 311, No. 2-3. P. 132-144. DOI: 10.1016/j.disc.2010.10.005
4. Cameron P. J. Permutation Groups. London Math. Soc. Student Texts, No. 45. Cambridge: Cambridge Univ. Press, 1999.
5. Gavrilyuk A. L., Makhnev A. A. On automorphisms of distance-regular graph with the intersection array {56,45,1; 1,9,56}. Doklady Mathematics, 2010. Vol. 81, No. 3. P. 439-442. DOI: 10.1134/S1064562410030282
6. Zavarnitsine A. V. Finite simple groups with narrow prime spectrum. Sib. Electron. Math. Izv., 2009. Vol. 6. P. S1-S12.
