PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 98(112) (2015), 251–263 DOI: 10.2298/PIM150304024K

SIMPLE GROUPS WITH THE SAME PRIME GRAPH AS ^{2}D_{n}(q)

Behrooz Khosravi and A. Babai

Abstract. In 2006, Vasil’ev posed the problem: Does there exist a positive integer k such that there are no k pairwise nonisomorphic nonabelian finite simple groups with the same graphs of primes? Conjecture: k = 5. In 2013, Zvezdina, confirmed the conjecture for the case when one of the groups is alternating. We continue this work and determine all nonabelian simple groups having the same prime graphs as the nonabelian simple group ^{2}D_{n} (q).

1. Introduction

If n is an integer, then we denote by π(n) the set of all prime divisors of n. If G is a finite group, then π(|G|) is denoted by π(G). The spectrum of a finite group G which is denoted by ω(G) is the set of its element orders. We construct the prime graph of G, denoted by Γ(G), as follows: the vertex set is π(G) and two distinct primes p and q are joined by an edge (we write p ∼ q) if and only if G contains an element of order pq. Let s(G) be the number of connected components of Γ(G) and let π_{i}(G), i = 1, …, s(G), be the connected components of Γ(G). If 2 ∈ π(G) we always suppose that 2 ∈ π_{1}(G). The connected components of the prime graph of nonabelian simple groups with disconnected prime graph are listed in [13]. In graph theory a subset of vertices of a graph is called an independent set if its vertices are pairwise non-adjacent. Denote by t(G) the maximal number of primes in π(G) pairwise non-adjacent in Γ(G). In other words, if ρ(G) is an independent set with the maximal number of vertices in Γ(G), then t(G) = |ρ(G)|. Similarly if p ∈ π(G), then let ρ(p, G) be an independent set with the maximal number of vertices in Γ(G) containing p and t(p, G) = |ρ(p, G)|. In [11, Tables 2-9], independent sets also independence numbers for all simple groups are listed.

Hagie [6] determined finite groups G satisfying Γ(G) = Γ(S), where S is a sporadic simple group. The same problem is considered for some finite simple groups (see [1, 2, 3, 8, 14]).

2010 Mathematics Subject Classification: 20D05; 20D60.

Key words and phrases: prime graph, simple group, Vasil’ev conjecture.

The first author was supported in part by a grant from IPM (93200043).

Communicated by Zoran Petrović.

Vasil’ev formulated the following problem in [9]:

Problem 16.26. Does there exist a positive integer k such that there are no k pairwise nonisomorphic nonabelian finite simple groups with the same graphs of primes? Conjecture: k = 5.

In [16], the problem was solved when one of the two groups is an alternating group. The conjecture is true in this case.

Here we continue this work and determine all nonabelian simple groups, with the same prime graph as ^{2}D_{n}(q).

Throughout the paper, we use the classification of finite simple groups, all groups are finite and by simple groups we mean nonabelian simple groups. All further unexplained notations are standard and refer to [4]. Also for a natural number n and a prime number p, we denote by (n)_{p}, the p-part of n, i.e., (n)_{p} = p^{α}, such that p^{α} | n, but p^{α+1} ∤ n.

2. Preliminary results

In this section, we will quote some useful facts which will be used during the proof of the main theorem.

Remark 2.1. [10] Let p be a prime number and (q, p) = 1. Let k ⩾ 1 be the smallest positive integer such that q^{k} ≡ 1 (mod p). Then k is called the order of q with respect to p and we denote it by ord_{p}(q). Obviously by Fermat’s little theorem it follows that ord_{p}(q)|(p − 1). Also if q^{n} ≡ 1 (mod p), then ord_{p}(q)|n. Similarly if m > 1 is an integer and (q, m) = 1, we can define ord_{m}(q). If a is odd, then ord_{a}(q) is denoted by e(a, q), too. If q is odd, let e(2, q) = 1 if q ≡ 1 (mod 4) and e(2, q) = 2 if q ≡ −1 (mod 4).

Lemma 2.1. [5, Remark 1] The equation p^{m} −q^{n} = 1, where p and q are primes and m, n > 1 has only one solution, namely 3^{2} − 2^{3} = 1.

Lemma 2.2. [5, 7] Except the relations (239)^{2}−2(13)^{4} = −1 and (3)^{5}− 2(11)^{2} =1 every solution of the equation

p^{m} − 2q^{n} = ±1; p, q prime; m, n > 1,

has exponents m = n = 2; i.e., it comes from a unit p − q2^{1/2} of the quadratic field Q(2^{1/2} ) for which the coefficients p, q are primes.

Lemma 2.3. (Zsigmondy Theorem) [15] Let p be a prime and n a positive integer. Then one of the following holds:

(i) there is a primitive prime p′ for p^{n} −1, that is, p′ | (p^{n} −1) but p′ ∤ (p^{m} −1), for every 1 ⩽ m < n, (usually p′ is denoted by r_{n})

(ii) p = 2, n = 1 or 6,

(iii) p is a Mersenne prime and n = 2.

We denote by D_{n}^{+} (q) the simple group D_{n}(q), and by D_{n}^{−} (q) the simple group 2D_{n}(q).

SIMPLE GROUPS WITH THE SAME PRIME GRAPH AS ^{2} D_{n} (q) 253

Table 1. 2-independence numbers for group ^{2}D_{n}(q)

G  conditions  t(2, G)  ρ(2, G)
2D_{n}(q)  n ≡ 0 (mod 2), n ⩾ 4  2  {2, r_{2n} }
n ≡ 1 (mod 2), n > 4, q ≡ 1 (mod 4)  2  {2, r_{2n} }
n ≡ 1 (mod 2), n > 4, q ≡ 7 (mod 8)  2  {2, r_{2(n−1)} }
n ≡ 1 (mod 2), n > 4, q ≡ 3 (mod 8)  3  {2, r_{2(n−1)} , r_{2n} }

Lemma 2.4. [12, Proposition 2.5] Let G = D_{n}^{ε}(q) be a finite simple group of Lie type over a field of characteristic p. Define

η(m) = \{ \begin{matrix}m & if m is odd, \\ m/2 & otherwise. \end{matrix}

Let r and s be odd primes and r, s ∈ π(G) ∖ {p}. Put k = e(r, q) and l = e(s, q), and 1 ⩽ η(k) ⩽ η(l). Then r and s are non-adjacent if and only if 2η(k) + 2η(l) > 2n − (1 − ε(−1)^{k+l} ), and l/k is not an odd natural number, and if ε = +, then the chain of equalities n = l = 2η(l) = 2η(k) = 2k is not true.

Lemma 2.5. [12, Proposition 2.4] Let G be a simple group of Lie type, B_{n}(q) or C_{n}(q) over a field of characteristic p. Let r, s be odd primes with r, s ∈ π(G) ∖ {p}. Put k = e(r, q) and l = e(s, q), and suppose that 1 ⩽ η(k) ⩽ η(l). Then r and s are non-adjacent if and only if η(k) + η(l) > n, and l/k is not an odd natural number.

Lemma 2.6. [11, Proposition 2.1] Let G = A_{n−1} (q) be a finite simple group of Lie type over a field of characteristic p. Let r and s be odd primes and r, s ∈ π(G) ∖ {p}. Put k = e(r, q) and l = e(s, q), and suppose that 2 ⩽ k ⩽ l. Then r and s are non-adjacent if and only if k + l > n, and k does not divide l.

Lemma 2.7. [11, Proposition 2.2] Let G = ^{2}A_{n′−1} (q) be a finite simple group of Lie type over a field of characteristic p. Define

ν(m) = \{ \begin{matrix}m & if m ≡ 0 (mod 4); \\ m/2 & if m ≡ 2 (mod 4); \\ 2m & if m ≡ 1 (mod 4). \end{matrix}

Let r and s be odd primes and r, s ∈ π(G) ∖ {p}. Put k = e(r, q) and l = e(s, q), and suppose that 2 ⩽ ν(k) ⩽ ν(l). Then r and s are non-adjacent if and only if ν(k) + ν(l) > n, and ν(k) does not divide ν(l).

3. Prime graph of simple classical Lie type groups

In this section, we denote by r_{i}, a primitive prime divisor of q^{i} − 1 and we consider R_{i}(q) as the set of all primitive prime divisors of q^{i} − 1.

Remark 3.1. Let G =^{2} D_{n}(q), where q = p^{α} and n ⩾ 4. Using [11, Table 6], we give the 2-independence number for the simple group ^{2}D_{n}(q) in Table 1. Let n be odd. By [11, Proposition 3.1], we have ρ(p, G) = {p, r_{2(n−1)} , r_{2n} }. By Lemma 2.4, we know that ρ(r_{1}, G) = {r_{1}, r_{2n} } and also ρ(r_{2}, G) = {r_{2}, r_{2(n−1)} }.

Let r_{k} ≁ r_{i}, where k ⩾ 3 is a fixed odd number. Hence 2k + 2η(i) > 2n − (1 + (−1)^{i+k} ) . Therefore, i ∈ A ∪ B, where A = {2n, 2(n − 1), …, 2(n − k + 1)} and B = {n − 2, n − 4, …, n − k + 1}. Since k ⩾ 3, so r_{k} is not adjacent to r_{2n}, r_{2(n−1)} and r_{2(n−2)} . Moreover, {r_{2(n−2)} , r_{2(n−1)} , r_{2n} } is an independent set. So {r_{k}, r_{2(n−2)} , r_{2(n−1)} , r_{2n} } ⊆ ρ(r_{k}, G). Therefore, t(r_{k}, G) ⩾ 4. Let r_{k} ≁ r_{i}, where k ⩾ 4 is a fixed even number. Hence k + 2η(i) > 2n − (1 + (−1)^{i+k} ). Define A = {2n, 2(n − 1), …, 2(n − k/2)} and B = {n − 2, n − 4, …, n − k/2 + a}, where if k ≡ 0 (mod 4), then a = 2 and otherwise, a = 1. Therefore, i ∈ A ∪ B. Let k = 4. If n ≡ 1 (mod 4), then ρ(r_{4}, G) = {r_{4}, r_{2(n−2)} , r_{2(n−1)} , r_{2n} }, otherwise ρ(r_{4}, G) = {r_{4}, r_{2(n−2)} , r_{2n} }. If k ⩾ 6, then t(r_{k}, G) ⩾ 5.

Let n be even. We know that ρ(p, G) = {p, r_{n−1} , r_{2(n−1)} , r_{2n} }, by [11, Proposition 3.1]. By Lemma 2.4, we know that ρ(r_{1}, G) = {r_{1}, r_{2n} } and ρ(r_{2}, G) = {r_{2}, r_{2n} }.

Let r_{k} ≁ r_{i}, where k ⩾ 3 is a fixed odd number. Hence 2k + 2η(i) > 2n − (1 + (−1)^{i+k} ). Suppose A = {2n, 2(n − 1), …, 2(n − k + 1)} and B = {n − 1 , n − 3 , …, n − k}. Therefore, i ∈ A ∪ B. Since k ⩾ 3, so t(r_{k}, G) ⩾ 5. Let r_{k} ≁ r_{i}, where k ⩾ 4 is a fixed even number. Hence k + 2η(i) > 2n − (1 + (−1)^{i+k} ). Define A = {2n, 2(n − 1), …, 2(n − k/2)} and B = {n − 1, n − 3, …, n − k/2 + a}, where if k ≡ 0 (mod 4), then a = 1, otherwise a = 2. Therefore, i ∈ A ∪ B. Let k = 4. If n ≡ 0 (mod 4), then ρ(r_{4}, G) = {r_{4}, r_{n−1} , r_{2(n−1)} , r_{2n} }, otherwise ρ(r_{4}, G) = {r_{4}, r_{n−1} , r_{2(n−2)} , r_{2(n−1)} }. Also if k ⩾ 6, then t(r_{k}, G) ⩾ 5.

Remark 3.2. Let G = D_{n}(q), where q = p^{α} and n > 4. Let n be odd. By [11, Proposition 3.1], we have ρ(p, G) = {p, r_{n}, r_{2(n−1)} }. By Lemma 2.4, we know that ρ(r_{1}, G) = {r_{1}, r_{2(n−1)} } and ρ(r_{2}, G) = {r_{2}, r_{n}}. Let r_{k} ≁ r_{i}, where k ⩾ 3 is a fixed odd number. Hence 2k + 2η(i) > 2n − (1 − (−1)^{i+k} ). Suppose A = {2(n − 1) , 2(n − 2) , …, 2(n − k)} and B = {n, n − 2, …, n − k + 1}. Therefore, i ∈ A ∪ B. Since k ⩾ 3, so t(r_{k}, G) ⩾ 4. Let r_{k} ≁ r_{i}, where k ⩾ 4 is a fixed even number. Hence k + 2η(i) > 2n − (1 − (−1)^{i+k} ). Define A = {2(n − 1) , 2(n − 2) , …, 2(n − k/2 + 1) } and B = {n, n − 2, …, n − k/2 + a}, where if k ≡ 0 (mod 4), then a = 0, otherwise a = 1. Therefore, i ∈ A ∪ B. Let k = 4, if n ≡ 1 (mod 4), then ρ(r_{4}, G) = {r_{4}, r_{n−2} , r_{n}, r_{2(n−1)} }, otherwise ρ(r_{4}, G) = {r_{4}, r_{n−2} , r_{n}}. If k ⩾ 6, then t(r_{k}, G) ⩾ 4.

Let n be even. By [11, Proposition 3.1], we have ρ(p, G) = {p, r_{n−1} , r_{2(n−1)} }. By Lemma 2.4, we know that ρ(r_{1}, G) = {r_{1}, r_{2(n−1)} } and ρ(r_{2}, G) = {r_{2}, r_{n−1} }. Let r_{k} ≁ r_{i}, where k ⩾ 3 is a fixed odd number. Hence 2k + 2η(i) > 2n − (1 − (−1)^{i+k} ) . Therefore, i ∈ A∪B, where A = {2(n−1), 2(n−2), …, 2(n−k)} and B = {n−1, n−3, …, n−k+2}. Since k ⩾ 3, if n ≠ 4, then t(r_{k}, G) ⩾ 4. Let r_{k} ≁ r_{i}, where k ⩾ 4 is a fixed even number. Hence k + 2η(i) > 2n − (1 − (−1)^{i+k} ). Define A = {2(n − 1), 2(n − 2), …, 2(n − k/2 + 1)} and B = {n − 1, n − 3, …, n − k/2 + a}, where if k ≡ 0 (mod 4), then a = 1, otherwise a = 0. Therefore, i ∈ A ∪ B. Similarly to the above, if k = 4, then ρ(r_{4}, G) = {r_{4}, r_{n−1} , r_{2(n−1)} } otherwise, t(r_{k}, G) ⩾ 4.

Remark 3.3. Let G = C_{n}(q) or G = B_{n}(q), where q = p^{α} and n > 3. Let n be odd. By [11, Proposition 3.1], we have ρ(p, G) = {p, r_{n}, r_{2n} }. By Lemma 2.5, we know that ρ(r_{1}, G) = {r_{1}, r_{2n} } and ρ(r_{2}, G) = {r_{2}, r_{n}}. Let r_{k} ≁ r_{i}, where k ⩾ 3 is a fixed odd number. Hence k+η(i) > n. Suppose A = {2n, 2(n−1), …, 2(n−k+1)} and B = {n, n−2, …, n−k+1}. Therefore, i ∈ A∪B. Since k ⩾ 3, so t(r_{k}, G) ⩾ 4. Let r_{k} ≁ r_{i}, where k ⩾ 4 is a fixed even number. Hence k/2 + η(i) > n. Define A = {2n, 2(n − 1), 2(n − 2), …, 2(n − k/2 + 1)} and B = {n, n − 2, …, n − k/2 + a}, where if k ≡ 0 (mod 4), then a = 2, otherwise a = 1. Therefore, i ∈ A ∪ B. If k ≠ 4, then t(r_{k}, G) ⩾ 4. Let k = 4, if n ≡ 1 (mod 4), then t(r_{4}, G) = 4, otherwise ρ(r_{4}, G) = {r_{4}, r_{n}, r_{2n} }.

Let n be even. By [11, Proposition 3.1], we have ρ(p, G) = {p, r_{2n} }. By Lemma 2.5, we know that ρ(r_{1}, G) = {r_{1}, r_{2n} } and ρ(r_{2}, G) = {r_{2}, r_{2n} }. Let r_{k} ≁ r_{i}, where k ⩾ 3 is a fixed odd number. Hence k + η(i) > n. Suppose A = {2n, 2(n−1), …, 2(n−k+1)} and B = {n−1, n−3, …, n−k+2}. Therefore, i ∈ A ∪ B. Since k ⩾ 3, so t(r_{k}, G) ⩾ 4. Let r_{k} ≁ r_{i}, where k ⩾ 4 is a fixed even number. Hence k/2 + η(i) > n. Define A = {2n, 2(n − 1), …, 2(n − k/2 + 1)} and B = {n − 1, n − 3, …, n − k/2 + a}, where if k ≡ 0 (mod 4), then a = 1, otherwise, a = 2. Therefore, i ∈ A ∪ B. Let k = 4, if n ≡ 0 (mod 4), then ρ(r_{4}, G) = {r_{4}, r_{n−1} , r_{2(n−1)} , r_{2n} }, otherwise ρ(r_{4}, G) = {r_{4}, r_{n−1} , r_{2(n−1)} }. Let k ⩾ 6, so t(r_{k}, G) ⩾ 4.

Remark 3.4. Let G = A_{n−1} (q). By [11, Proposition 3.1], we have t(p, G) = 3 and also by [11, Proposition 4.1], we know that 2 ⩽ t(r_{1}, G) ⩽ 3. Let r_{k} ≁ r_{i}, where k ≠ 1 is a fixed number, hence i ∈ {n, n − 1, …, n − k + 1}. Therefore, by Lemma 2.6, we have t(r_{2}, G) = 2 and t(r_{3}, G) = 3. Let k ⩾ 4, so t(r_{k}, G) ⩾ 4.

Remark 3.5. Let G = ^{2}A_{n−1} (q). By [11, Proposition 3.1], we have t(p, G) = 3 and also by [11, Proposition 4.2], we know that 2 ⩽ t(r_{2}, G) ⩽ 3. Let r_{k} ≁ r_{i}, where k ≠ 2 is a fixed number, hence i ∈ {n, n − 1, …, n − k + 1}. Therefore, by Lemma 2.7, we have t(r_{1}, G) = 2 and t(r_{3}, G) = 3. Let k ⩾ 4, so t(r_{k}, G) ⩾ 4.

4. Main results

In the sequel, we denote by r_{i} and u_{i}, a primitive prime divisor of q^{i} − 1 and q^{′i} − 1, respectively. Also we consider R_{i}(q) and U_{i}(q′) as the set of all primitive prime divisors of q^{i} − 1 and q^{′i} − 1, respectively.

Theorem 4.1. Let G = ^{2}D_{n}(q), where n ⩾ 4 and q = p^{α}, and also S be a classical simple group of Lie type over the field GF(q′), where q′ = p^{′β}. Then Γ(S) = Γ(G) if and only if one of the following holds:

(1) S = G.

(2) S = ^{2}D_{n}(q′), where q′ = p^{′β}, 4 | n, p′ ≠ p, p′ ≡ 1 (mod 4), p ≡ 1 (mod 4), π(q^{2} − 1) = π(q^{′2} − 1), R_{2n} (q) = U_{2n} (q′) and {p} ∪ R_{4}(q) =

{p′} ∪ U_{4}(q′).

(3) S = B_{n}(q′) or S = C_{n}(q′), where q′ = p^{′β}, 4 | n, p′ ≠ p, R_{1}(q) ∪ R_{2}(q) = {p′} ∪ U_{1}(q′) ∪ U_{2}(q′), R_{2n} (q) = U_{2n} (q′) and either {p} ∪ R_{4}(q) = U_{3}(q′) ∪

U_{4}(q′) ∪ U_{6}(q′) or {p} ∪ R_{4}(q) = U_{4}(q′).

Proof. We know that Γ(S) = Γ(G), therefore t(S) = t(G), t(2, S) = t(2, G) and for every r ∈ π(G), we have t(r, G) = t(r, S). We know that t(p, G) ⩾ 3 and t(r_{1}, G) = t(r_{2}, G) = 2 and for every r_{i} ∈ π(G), where i ∉ {1, 2}, we have t(r_{i}, G) > 2, by Remark 3.1. Now we consider each possibility for S by [13, Tables 1a–1c]. Case 1. Let S = ^{2}D_{n′}(q′), where q′ = p^{′β}. We have t(S) = t(G) so [(3n′+ 4) /4] = [(3n + 4)/4]. Therefore, n = n′, n + 1 = n′ or n′ + 1 = n. Also t(p, S) ⩾ 3, t(u_{1}, S) = t(u_{2}, S) = 2 and for every u_{i} ∈ π(S), where i ∉ {1, 2}, we have t(u_{i}, S) > 2 , by Remark 3.1. Therefore, R_{1}(q) ∪ R_{2}(q) = U_{1}(q′) ∪ U_{2}(q′). Now we consider the following subcases: 1.1. Let n be odd. 1.1.1 Let n + 1 = n′. It is clear that p′ ≠ 2 otherwise, t(2, S) = 4, which is a contradiction, since t(2, G) ⩽ 3, by [11, Tables 4, 6]. Therefore, ρ(2, S) = {2, u_{2n′} }, hence t(2, G) = 2. Consequently, p ≠ 2 otherwise, t(2, G) = 3, which is a contradiction. Now we consider the following two cases: 1.1.1.1. Let ρ(2, G) = {2, r_{2n} }. Therefore, R_{2n} (q) = U_{2n′} (q′). We know that r_{1} ≁ r_{2n} ∼ r_{2}, and u_{1} ≁ u_{2n′} ≁ u_{2}, which is a contradiction. 1.1.1.2. Let ρ(2, G) = {2, r_{2(n−1)} }. Therefore, R_{2(n−1)} (q) = U_{2n′} (q′). We know that r_{1} ∼ r_{2(n−1)} ≁ r_{2}, and u_{1} ≁ u_{2n′} ≁ u_{2}, which is a contradiction. Similarly, if n′+ 1 = n, then we get a contradiction. 1.1.2. Let n = n′, now we consider the following subcases: 1.1.2.1. Let q ≡ 1 (mod 4), hence ρ(2, G) = {2, r_{2n} }. Therefore, t(2, S) = 2, and so q′ ≢ 3 (mod 8). If q′ ≡ 1 (mod 4), then ρ(2, S) = {2, u_{2n} } and so R_{2n} (q) = U_{2n} (q′). We know that r_{1} ≁ r_{2n} ∼ r_{2}, and u_{1} ≁ u_{2n} ∼ u_{2}. Consequently, R_{1}(q) = U_{1}(q′) and R_{2}(q) = U_{2}(q′). Moreover, we know that u_{2} is adjacent to all vertices except u_{2(n−1)} and also r_{2} is adjacent to all vertices except r_{2(n−1)} , which implies that R_{2(n−1)} (q) = U_{2(n−1)} (q′). Consequently, R_{2(n−1)} (q) ∪ R_{2n} (q) = U_{2(n−1)} (q′) ∪ U_{2n} (q′). Therefore, every r_{2n} and r_{2(n−1)} can be regarded as u_{2n} and u_{2(n−1)} . For convenience in the sequel we write {r_{2n} , r_{2(n−1)} } ≈ {u_{2n} , u_{2(n−1)} } to illustrate the above statement. By Remark 3.1, we know that p is the only vertex in Γ(G), which is adjacent to all vertices except r_{2(n−1)} and r_{2n}, and similarly p′ is the only vertex in Γ(S), which is adjacent to all vertices except u_{2(n−1)} and u_{2n}. Consequently, p = p′. Since π(S) = π(G), so α = β, by Lemma 2.3, which implies that S = G. If q′ ≡ 7 (mod 8), then R_{2n} (q) = U_{2(n−1)} (q′). Similarly to the above, by the above notation {r_{2(n−1)} , r_{2n} } ≈ {u_{2(n−1)} , u_{2n} }, and by Remark 3.1, p = p′ and so q = q′, which is a contradiction, since q ≡ 1 (mod 4). 1.1.2.2. Let q ≡ 7 (mod 8), hence ρ(2, G) = {2, r_{2(n−1)} }, completely similar to the above case we get that S = G. 1.1.2.3. Let q ≡ 3 (mod 8), hence ρ(2, G) = {2, r_{2(n−1)} , r_{2n} } so t(2, S) = 3. It follows that ρ(2, S) = {2, u_{2(n−1)} , u_{2n} }, by [11, Tables 4, 6]. Therefore, R_{2(n−1)} (q)∪ R_{2n} (q) = U_{2(n−1)} (q′) ∪ U_{2n} (q′). Therefore, {r_{2(n−1)} , r_{2n} } ≈ {u_{2(n−1)} , u_{2n} }. We know that p and 2 are the only vertices which are adjacent to all vertices in Γ(G) except r_{2n} and r_{2(n−1)} . Also we know that p′ and 2 are the only vertices which are adjacent to all vertices in Γ(S) except u_{2n} and u_{2(n−1)} . Consequently, {2, p} = {2, p′}. Since q ≡ 3 (mod 8), so p ≠ 2. Therefore, p = p′. Since π(S) = π(G), so α = β, by Lemma 2.3, which implies that S = G. 1.1.2.4. Let q = 2^{α}, hence ρ(2, G) = {2, r_{2(n−1)} , r_{2n} } so t(2, S) = 3. Consequently, either q′ ≡ 3 (mod 8) or q′ = 2^{β}, so ρ(2, S) = {2, u_{2(n−1)} , u_{2n} }. Therefore, similarly to the above {r_{2(n−1)} , r_{2n} } ≈ {u_{2(n−1)} , u_{2n} }. If q′ ≡ 3 (mod 8), then p′ and 2 are the only vertices, which are adjacent to all vertices except u_{2n} and u_{2(n−1)} , by [11, Tables 4, 6] and Remark 3.1. On the other hand, p is the only vertex, which is adjacent to all vertices except r_{2n} and r_{2(n−1)} . Consequently, {p} = {p′, 2}, which is a contradiction. It follows that p′ = 2 = p. Since π(G) = π(S), so α = β, by Lemma 2.3. Therefore, S = G. 1.2 Let n be even. 1.2.1. Let n + 1 = n′. It is clear that p ≠ 2, since t(2, S) ≠ t(2, G). Therefore, ρ(2, G) = {2, r_{2n} } and we know that t(2, S) = 2. Now we consider the following two cases: 1.2.1.1. Let ρ(2, S) = {2, u_{2n′} }. Therefore, R_{2n} (q) = U_{2n′} (q′). We know that r_{1} ≁ r_{2n} ≁ r_{2}, and u_{1} ≁ u_{2n′} ∼ u_{2}, which is a contradiction. 1.2.1.2 Let ρ(2, S) = {2, u_{2(n′−1)} }. Therefore, R_{2n} (q) = U_{2(n′−1)} (q′). We know that r_{1} ≁ r_{2n} ≁ r_{2}, and u_{1} ∼ u_{2(n′−1)} ≁ u_{2}, which is a contradiction. Similarly, if n′+ 1 = n, then we get a contradiction. 1.2.2. Let n = n′. If p = 2, then t(2, G) = 4. It follows that p′ = 2, by [11, Tables 4, 6]. Consequently, p = p′ and so similarly to the above we have G = S. If p ≠ 2, then t(2, G) = 2. Hence p′ ≠ 2, since otherwise, t(2, S) = 4. Consequently, p ≠ 2 and p′ ≠ 2. Since n is even, so ρ(2, G) = {2, r_{2n} } and ρ(2, S) = {2, u_{2n} }. Therefore, R_{2n} (q) = U_{2n} (q′). By Remark 3.1, p and r_{4} are the only vertices in Γ(G) such that their independence numbers are 4. Also p′ and u_{4} are the only vertices in Γ(S) such that their independence numbers are equal to 4. Therefore, R_{4}(q) ∪ {p} = U_{4}(q′) ∪ {p′}. Now we consider the following two cases: 1.2.2.1. Let n ≡ 2 (mod 4). If p = p′, then similarly to the above we have S = G. Otherwise, there exists u_{4} such that p = u_{4}. We know that {p, r_{n−1} , r_{2(n−1)} , r_{2n} } is the unique maximal independent set in Γ(G) which contains p. Also we know that {u_{4}, u_{n−1} , u_{2(n−2)} , u_{2(n−1)} } is the unique maximal independent set in Γ(S) which contains u_{4}. So R_{n−1} (q)∪R_{2(n−1)} (q)∪R_{2n} (q) = U_{n−1} (q′)∪U_{2(n−2)} (q′)∪U_{2(n−1)} (q′), which is a contradiction, since R_{2n} (q) = U_{2n} (q′). 1.2.2.2. Let n ≡ 0 (mod 4). Thus {r_{n−1} , r_{2(n−1)} , r_{2n} } is equal to ρ(r_{4}, G) ∖ {r_{4}} and ρ(p, G) ∖ {p}. Similarly we can consider ρ(u_{4}, S) ∖ {u_{4}} = ρ(p′, S) ∖ {p′}. If p = p′, then similarly to the above we have S = G. Otherwise, there exist r_{4} and u_{4} such that p = u_{4} and p′ = r_{4}. Consequently, S = ^{2}D_{n}(r_{4}^{β}).

Case 2. Let S = D_{n′}(q′), where q′ = p^{′β}.

If n′ = 4, then t(S) = 3, and so t(G) = 3. Therefore, n = 3, which is a contradiction. Consequently n′ > 4. We know that t(p, S) = 3 and t(u_{1}, S) = t(u_{2}, S) = 2 and for every u_{i} ∈ π(S), where i > 2, we have t(u_{i}, S) > 2, by Remark 3.2. Therefore, R_{1}(q) ∪ R_{2}(q) = U_{1}(q′) ∪ U_{2}(q′).

Let n be even. By Remark 3.1, there is no vertex in Γ(G), whose independence number is 3, while p′ ∈ π(S) and t(p′, S) = 3, which is a contradiction.

Therefore n is odd. We know that t(S) = t(G).

2.1. If t(S) = [(3n′+ 1)/4], then n′ = n, n′ = n + 1 or n′ = n + 2. 2.1.1. Let n = n′. We know that t(2, S) = 2 or 3. We consider the following two cases: 2.1.1.1. Let t(2, S) = 3, hence ρ(2, S) = {2, u_{n}, u_{2(n−1)} }, by [11, Tables 4, 6]. So t(2, G) = 3 and ρ(2, G) = {2, r_{2(n−1)} , r_{2n} }, by [11, Tables 4, 6]. Therefore, similarly to the above, R_{2n} (q) ∪ R_{2(n−1)} (q) = U_{n}(q′) ∪ U_{2(n−1)} (q′) and consequently {r_{2(n−1)} , r_{2n} } ≈ {u_{n}, u_{2(n−1)} }. By Remarks 3.1 and 3.2, {p, 2} = {p′, 2}. Therefore, p = p′. Since π(S) = π(G), so 2nα = 2(n − 1)β, by Lemma 2.3, so (α)_{2} > (β)_{2}. Let x be a primitive prime divisor of p^{2nα} − 1, so x is a primitive prime divisor of q^{2n} − 1. By assumption, x ∈ U_{n}(q′) ∪ U_{2(n−1)} (q′). If x ∈ U_{n}(q′), then x | (p^{nβ} − 1), which implies that 2nα ⩽ nβ, and this is a contradiction. Therefore, R_{2n} (q) = U_{2(n−1)} (q′) and R_{2(n−1)} (q) = U_{n}(q′). Let x be a primitive prime divisor of p^{2(n−1)α} − 1. Then similarly to the above 2(n − 1)α ⩽ nβ. If y is a primitive prime divisor of p^{nβ} − 1, then similarly we have nβ ⩽ 2(n − 1)α. Therefore, nβ = 2(n − 1)α. Since nα = (n − 1)β, so (n − 2)α = β, which is a contradiction, since n is odd and (α)_{2} > (β)_{2}. 2.2.1.2. Let t(2, S) = 2 and so p′ ≠ 2. Let q′ ≡ 3 (mod 4) and so ρ(2, S) = {2, u_{n}}. We know that t(2, G) = t(2, S) = 2. Now we consider the following two cases: 2.2.1.2.1. Let q ≡ 1 (mod 4), so R_{2n} (q) = U_{n}(q′). We know that r_{1} ≁ r_{2n} ∼ r_{2} and u_{1} ∼ u_{n} ≁ u_{2}. Therefore, R_{2}(q) = U_{1}(q′) and so R_{2(n−1)} (q) = U_{2(n−1)} (q′). 2.2.1.2.2. Let q ≡ 7 (mod 8), so R_{2(n−1)} (q) = U_{n}(q′). We know that r_{1} ∼ r_{2(n−1)} ≁ r_{2} and u_{1} ∼ u_{n} ≁ u_{2}. Therefore, R_{1}(q) = U_{1}(q′) and so R_{2n} (q) = U_{2(n−1)} (q′). Consequently, R_{2n} (q)∪R_{2(n−1)} (q) = U_{n}(q′)∪U_{2(n−1)} (q′) and similarly to the above {r_{2(n−1)} , r_{2n} } ≈ {u_{n}, u_{2(n−1)} }. By Remarks 3.1 and 3.2, p = p′. Similarly to the above we have 2nα = 2(n − 1)β and so (α)_{2} > (β)_{2}. If R_{2n} (q) = U_{2(n−1)} (q′), then similarly to the above we get a contradiction. Therefore, R_{2n} (q) = U_{n}(q′) and similarly 2nα = nβ, which is a contradiction. Let q′ ≡ 1 (mod 4), then similarly to the above we get a contradiction. 2.1.2. Let n = n′ + 1. If p′ = 2, then t(2, S) = t(2, G) = 3. We know that ρ(2, S) = {2, u_{n′−1} , u_{2(n′−1)} } and ρ(2, G) = {2, r_{2(n−1)} , r_{2n} }, by [11, Tables 4,6]. Therefore, R_{2(n−1)} (q) ∪ R_{2n} (q) = U_{n′−1} (q′) ∪ U_{2(n′−1)} (q′) and so {r_{2n} , r_{2(n−1)} } ≈ {u_{n′−1} , u_{2(n′−1)} }. By Remarks 3.1 and 3.2, we have {2, p} = U_{4}(q′)∪{p′}. If p = 2, then we get a contradiction. It follows that U_{4}(q′) has one member and p = u_{4}, and so there exists a natural number m such that p^{m} = q^{′2}+ 1. By Lemma 2.1, we have m = 1. On the other hand, we know that R_{1}(q) ∪ R_{2}(q) = U_{1}(q′) ∪ U_{2}(q′) or in other words π(q^{2} − 1) = π(q^{′2}− 1). Consequently, π((q^{′2}+ 1)^{2α} − 1) = π(q^{′2}− 1), which is a contradiction. Therefore, p′ ≠ 2 and since n′ is even, so t(2, S) = 2. If q′ ≡ 3 (mod 4), so ρ(2, S) = {2, u_{n′−1} }. We know that t(2, G) = t(2, S) = 2. Now we consider the following two cases: 2.1.2.1. Let q ≡ 1 (mod 4), so R_{2n} (q) = U_{n′−1} (q′). We know that r_{1} ≁ r_{2n} ∼ r_{2} and u_{1} ∼ u_{n′−1} ≁ u_{2}. Therefore, R_{2}(q) = U_{1}(q′) and so R_{2(n−1)} (q) = U_{2(n′−1)} (q′). 2.1.2.2. Let q ≡ 7 (mod 8), so R_{2(n−1)} (q) = U_{n′−1} (q′). We know that r_{1} ∼ r_{2(n−1)} ≁ r_{2} and u_{1} ∼ u_{n′−1} ≁ u_{2}. Therefore, R_{1}(q) = U_{1}(q′) and so R_{2n} (q) = U_{2(n′−1)} (q′). Thus, similarly to the above we have {r_{2(n−1)} , r_{2n} } ≈ {u_{n′−1} , u_{2(n′−1)} }. By Remarks 3.1 and 3.2, we have {p} = {p′} ∪ U_{4}(q′), which is a contradiction. If q′ ≡ 1 (mod 4), then ρ(2, S) = {2, u_{2(n′−1)} } and similarly to the above we get a contradiction. 2.1.3. Let n′ = n + 2, so n′ is odd. It is clear, either n ≡ 1 (mod 4) or n′ ≡ 1 (mod 4). If n ≡ 1 (mod 4), then p is the only vertex in Γ(G), whose independence number is 3. Also independence number of p′ and u_{4} are 3 in Γ(S). Therefore, {p} = {p′} ∪ U_{4}(q′), which is a contradiction. Similarly, if n′ ≡ 1 (mod 4), then we get a contradiction. 2.2. If t(S) = (3n′ + 3)/4, then n = n′ and similarly to the above we get a contradiction.

Case 3. Let S = C_{n′}(q′) or S = B_{n′}(q′), where q′ = p^{′β}.

If n′ ⩽ 3, then t(S) ⩽ 3, and so t(G) ⩽ 3. Therefore, n ⩽ 3, which is a contradiction. Consequently, n′ > 3. We have t(S) = t(G) so [(3n′ + 5)/4] = [(3n + 4)/4]. Therefore, n = n′+ 1 or n = n′. 3.1. Let n be odd. 3.1.1. Let n = n′. We know that t(p, S) = 3, t(u_{1}, S) = t(u_{2}, S) = 2 and for every u_{i} ∈ π(S), where i > 2, we have t(u_{i}, S) > 2, by Remark 3.3. Therefore, R_{1}(q)∪R_{2}(q) = U_{1}(q′)∪U_{2}(q′). Let p′ = 2. Then p = 2, since π(q^{2}−1) = π(q^{′2}−1). Since π(G) = π(S), so α = β, by Lemma 2.3. Therefore, S = C_{n}(q) or S = B_{n}(q). We know that r_{n} ∈ π(S) ∖ π(G), which is a contradiction. Consequently, p′ ≠ 2 and so t(2, S) = t(2, G) = 2. Therefore ρ(2, S) = {2, u_{2n} } or ρ(2, S) = {2, u_{n}}. Since the proofs are similar, for convenience we give a proof for ρ(2, S) = {2, u_{2n} } and the proof of the other case is similar. Let ρ(2, S) = {2, u_{2n} }. 3.1.1.1. If q ≡ 1 (mod 4), then R_{2n} (q) = U_{2n} (q′). We know that r_{1} ≁ r_{2n} ∼ r_{2} and u_{1} ≁ u_{2n} ∼ u_{2}. Consequently, R_{2}(q) = U_{2}(q′) and so R_{2(n−1)} (q) = U_{n}(q′). Similarly to the above {r_{2(n−1)} , r_{2n} } ≈ {u_{n}, u_{2n} }. By Remarks 3.1 and 3.3, if n ≡ 1 (mod 4), then p = p′. Similarly to the above we can see that S = C_{n}(q) or S = B_{n}(q), which is a contradiction. Otherwise, {p} = {p′} ∪ U_{4}(q′), which is a contradiction. 3.1.1.2. If q ≡ 7 (mod 8), then R_{2(n−1)} (q) = U_{2n} (q′) and similarly to the above we get a contradiction. 3.1.2. Let n = n′ + 1, so n′ is even. Hence ρ(2, S) = {2, u_{2n′} } and t(p′, S) = 2 , by [11, Tables 4, 6]. By Remarks 3.1 and 3.3, and similarly to the above, R_{1}(q) ∪ R_{2}(q) = U_{1}(q′) ∪ U_{2}(q′) ∪ {p′} and u_{2n′} is not adjacent to u_{1}, u_{2} and p′. If q ≡ 1 (mod 4), then R_{2n} (q) = U_{2n′} (q′). On the other hand, we know that r_{2n} ∼ r_{2}, which is a contradiction. Similarly, if q ≡ 7 (mod 8), then R_{2(n−1)} (q) = U_{2n′} (q′), while r_{2(n−1)} ∼ r_{1}, which is a contradiction. 3.2 Let n be even. It is clear that p ≠ 2, otherwise, t(2, S) ≠ t(2, G), which is a contradiction. Therefore, ρ(2, G) = {2, r_{2n} }. 3.2.1. Let n = n′, so ρ(2, S) = {2, u_{2n} }. Therefore, R_{2n} (q) = U_{2n} (q′). By Remarks 3.1 and 3.3, R_{1}(q) ∪ R_{2}(q) = U_{1}(q′) ∪ U_{2}(q′) ∪ {p′}. Let n ≢ 0 (mod 4), so t(u_{4}, S) = 3. But we know that t(x, G) ≠ 3, for every x ∈ π(G), which is a contradiction. Consequently, n ≡ 0 (mod 4). By Remark 3.1, we have p and r_{4} are the only vertices in Γ(G) such that their independence numbers are equal to 4. If 3 | (n − 1), then u_{3}, u_{4} and u_{6} are the only vertices in Γ(S) such that their independence numbers are equal to 4. In this case, we have {p} ∪ R_{4}(q) = U_{3}(q′) ∪ U_{4}(q′) ∪ U_{6}(q′). Otherwise, 3 ∤ (n − 1) and so {p} ∪ R_{4}(q) = U_{4}(q′). Similarly, we can find some relations between other vertices. 3.2.2. Let n = n′+ 1. Since n is even, so by Remark 3.1, there is not any vertex in Γ(G) such that its independence number is 3. On the other hand, n′ is odd, so t(p′, S) = 3, which is a contradiction.

Case 4. Let S = A_{n′−1} (q′) or ^{2}A_{n′−1} (q′), where q′ = p^{′β}.

Since the proofs for these groups are similar, we state the details of the proof for one of them, say A_{n′−1} (q′). So in the sequel let S = A_{n′−1} (q′), where q′ = p^{′β}.

Let n be even. By Remark 3.1, there is not any vertex in Γ(G), whose independence number is 3, while t(p′, S) = 3, which is a contradiction. Therefore, n is odd. We know that by [11], t(u_{1}, S) is equal to 2 or 3. 4.1. Let t(u_{1}, S) = 3, so by Remarks 3.1 and 3.4, R_{1}(q) ∪ R_{2}(q) = U_{2}(q′). Now we consider the following cases: 4.1.1. Let n′_{2} < (q′ − 1)_{2}, so ρ(2, S) = {2, u_{n′}}. Hence t(2, G) = 2. If q ≡ 1 (mod 4), then ρ(2, G) = {2, r_{2n} }. Therefore, R_{2n} (q) = U_{n′}(q′). We know that r_{1} ≁ r_{2n} ∼ r_{2}, which is a contradiction, since R_{1}(q) ∪ R_{2}(q) = U_{2}(q′). Consequently, q ≡ 7 (mod 8), and so R_{2(n−1)} (q) = U_{n′}(q′). We have r_{1} ∼ r_{2(n−1)} ≁ r_{2}, which is a contradiction. 4.1.2. Let n′_{2} > (q′ − 1)_{2} or n′_{2} = (q′ − 1)_{2} = 2, so ρ(2, S) = {2, u_{n′−1} }. If q ≡ 1 (mod 4), then R_{2n} (q) = U_{n′−1} (q′). We know that r_{1} ≁ r_{2n} ∼ r_{2}, which is a contradiction, since R_{1}(q) ∪ R_{2}(q) = U_{2}(q′). Similarly, if q ≡ 7 (mod 8), then R_{2(n−1)} (q) = U_{n′−1} (q′), while we know that r_{1} ∼ r_{2(n−1)} ≁ r_{2}, which is a contradiction. 4.1.3. Let 2 < n′_{2} = (q′ − 1)_{2}, so ρ(2, S) = {2, u_{n′−1} , u_{n′}}. Therefore, q ≡ 3 (mod 8), hence similarly to the above we get that {r_{2(n−1)} , r_{2n} } ≈ {u_{n′−1} , u_{n′}}. It follows that R_{2n} (q) = U_{n′}(q′) or R_{2n} (q) = U_{n′−1} (q′) and similarly to the above we get a contradiction. 4.2. Let t(u_{1}, S) = 2, hence R_{1}(q) ∪ R_{2}(q) = U_{1}(q′) ∪ U_{2}(q′). Let n ≡ 1 (mod 4). Let t(2, S) = t(2, G) = 2. By Remark 3.4, we know that p′ and u_{3} are the only vertices in Γ(S) such that their independence number is 3. On the other hand, p is the only vertex in Γ(G) such that its independence number is 3 so {p} = {p′} ∪ U_{3}(q′), which is a contradiction. So t(2, S) = t(2, G) = 3. Similarly to the above we have, {2, p} = {2, p′}∪U_{3}(q′). Also p′ = 2 if and only if p = 2, since π(q^{2}−1) = π(q^{′2}−1). It follows that u_{3} = 2, which is a contradiction. Therefore, p = p′ ≠ 2 and so {2} = {2} ∪ U_{3}(q′), which is a contradiction. Hence, n ≢ 1 (mod 4).

Now we claim that t(2, S) = t(2, G) = 3. Otherwise, t(2, S) = t(2, G) = 2 and {p} ∪ R_{4}(q) = {p′} ∪ U_{3}(q′). If p′ = p, then {r_{2(n−1)} , r_{2n} } ≈ {u_{n′−1} , u_{n′}}.

Let R_{2n} (q) = U_{n′}(q′) and R_{2(n−1)} (q) = U_{n′−1} (q′). If x is a primitive prime divisor of p^{2nα} − 1, then x is a primitive prime divisor of q^{2n} − 1. Therefore, x | (p^{n′β} − 1), which implies that 2nα ⩽ n′β. Let y be a primitive prime divisor of p^{n′β} −1, so y | (p^{2nα} −1), which implies that n′β ⩽ 2nα. Consequently, n′β = 2nα, similarly we have (n′ − 1)β = 2(n − 1)α. It follows that 2α = β. On the other hand, t(S) = t(G) so n′ ∈ {(3n − 1)/2, (3n + 1)/2, (3n + 3)/2}. Consequently, S = A_{n′−1} (p^{2α} ), which is a contradiction, since π(S) ≠ π(G).

Let R_{2n} (q) = U_{n′−1} (q′) and R_{2(n−1)} (q) = U_{n′}(q′). Similarly to the above, we have 2nα = (n′− 1)β and 2(n − 1)α = n′β, which is a contradiction.

Therefore, p ≠ p′ and so p is a primitive prime divisor of q^{′3} − 1 and p′ is a primitive prime divisor of q^{4} − 1. Hence we can consider p′ = r_{4} and p = u_{3}.

Since p′ = r_{4}, so ρ(r_{4}, G) ∖ {r_{4}} = ρ(p′, S) ∖ {p′}, hence R_{2(n−2)} (q) ∪ R_{2n} (q) = U_{n′−1} (q′) ∪ U_{n′}(q′). On the other hand, u_{3} is not adjacent to two elements of {u_{n′−2} , u_{n′−1} , u_{n′}} in Γ(S) and ρ(p, G) = {p, r_{2(n−1)} , r_{2n} }. Consequently, 3∤ (n′− 2) and u_{3} ≁ u_{n′−2} , since p = u_{3}. Therefore, either 3 | (n′− 1) and so R_{2(n−1)} (q) ∪ R_{2n} (q) = U_{n′−2} (q′) ∪ U_{n′}(q′) or 3 | n′ and R_{2(n−1)} (q) ∪ R_{2n} (q) = U_{n′−2} (q′) ∪ U_{n′−1} (q′). Moreover, we know that t(S) = t(G), which implies 2n′∈{3n−1, 3n+1, 3n + 3}.

Let 3n + 1 = 2n′. It is clear that neither 3 | (n′ − 1) nor 3 | n′, which is a contradiction.

By Remark 3.1, if 3 ∤ (n − 2), then for every x ∈ π(G), we have t(x, G) ≠ 4, while t(u_{4}, S) = 4, which is a contradiction. So 3 | (n − 2) and t(r_{3}, G) = 4. Since u_{4} is the only vertex in Γ(S), whose independence number is equal to 4, so r_{3} = u_{4}. Therefore, ρ(r_{3}, G) ∖ {r_{3}} = ρ(u_{4}, S) ∖ {u_{4}}. It follows that R_{2(n−2)} (q) ∪ R_{2(n−1)} (q) ∪ R_{2n} (q) = ρ(u_{4}, S) ∖ {u_{4}}. By the above discussion, we get that R_{2(n−2)} (q)∪R_{2(n−1)} (q)∪R_{2n} (q) = U_{n′−2} (q′)∪U_{n′−1} (q′)∪U_{n′}(q′). Hence 4 | (n′−3), by Lemma 2.6, and so n′ is odd.

If 3n + 3 = 2n′, then 3(n + 1) = 2n′, which is a contradiction, since n ≢ 1 (mod 4). If 3n − 1 = 2n′, then 3(n − 1) = 2(n′− 1), we get a contradiction.

Hence, t(2, S) = t(2, G) = 3, so {2, p} ∪ R_{4}(q) = {2, p′} ∪ U_{3}(q′). If p = 2, then p′ = 2, since π(q^{2} − 1) = π(q^{′2} − 1). Therefore, p = p′ and similarly to the above we get a contradiction. Consequently, since r_{4} ≠ 2 and u_{3} ≠ 2 so {p} ∪ R_{4}(q) = {p′} ∪ U_{3}(q′). Now completely similar to the above we get a contradiction. □

Theorem 4.2. Let G = ^{2}D_{n}(q), where q = p^{α} and n ⩾ 4, and also S be an exceptional group of Lie type. Then Γ(S) and Γ(G) are not equal.

Proof. We consider the following cases:

(1) Let S = E_{8}(q′). Since s(E_{8}(q′)) ⩾ 4 and s(G) ⩽ 3, by [13, Tables 1a–1c], so we get a contradiction. Similarly S ≠ ^{2}B_{2}(q′), where q′ = 2^{2m+1} .

(2) Let S = G_{2}(q′). We know that t(S) = t(G). Therefore, [(3n + 4)/4] = 3, so n = 3, which is a contradiction. Similarly S ≠ ^{3}D_{4}(q′) and ^{2}F_{4}(2′).

(3) Let S = E_{6}(q′). Since t(S) = t(G), so [(3n + 4)/4] = 5, hence n = 6. We know that Γ(S) has two components so s(G) = 2, which is a contradiction, by [13, Tables 1a–1c]. Similarly S is not isomorphic to ^{2}E_{6}(q′), ^{2}G_{2}(q′), where q′ = 3^{2m+1} , 2F_{4}(q′), where q′ = 2^{2m+1} and F_{4}(q′), where q′ > 2.

(4) Let S = E_{7}(q′). So [(3n + 4)/4] = 8, hence n = 10. Therefore, t(2, G) = 2 or 4, while t(2, S) = 3, which is a contradiction. Similarly S ≠ F_{4}(2). □

Theorem 4.3. Let G = ^{2}D_{n}(q), where q = p^{α} and n ⩾ 4, and also S be an alternating or sporadic group. Then Γ(S) and Γ(G) are not equal.

Proof. We consider the following cases:

(1) Let S = M_{22}. Since s(M_{22}) = 4 and s(G) ⩽ 3, by [13, Tables 1a–1c], so we get a contradiction. Similarly S ≠ J_{1}, J_{4}, ON, Ly, F′_{24} and F_{1}.

(2) Let S = M_{11}. We know that t(S) = t(G). Therefore, [(3n+4)/4] = 3, so n = 3, which is a contradiction. Similarly S is not isomorphic to M_{12}, J_{2}, J_{3}, He, McL, HN and HiS.

(3) Let S = F_{3}. Since t(S) = t(G), so [(3n + 4)/4] = 5, hence n = 6. We know that Γ(S) has three components so s(G) = 3, which is a contradiction, by [13, Tables 1a–1c]. Similarly S ≠ F i_{23} and F_{2}.

(4) Let S = M_{23}. So [(3n + 4)/4] = 4, hence n = 4 or 5. By [13, Tables 1a–1c], s(G) = 1 while s(S) > 1, which is a contradiction. Similarly for the other sporadic groups, we get a contradiction.

By [16], it is clear that S cannot be equal to an alternating group. □

Corollary 4.1. (i) If n is a natural number such that 4 ∤ n, then the Vasil’ev Conjecture is true for the nonabelian simple group ^{2}D_{n}(p^{α}).

(ii) Let 4 | n and q = p^{α}. If S is a nonabelian simple group such that Γ(S) = Γ(^{2}D_{n}(q)), then one of the following holds:

(1) S = G.

(2) S = ^{2}D_{n}(q′), where q′ = p^{′β}, p′ ≠ p, p′ ≡ 1 (mod 4), p ≡ 1 (mod 4), π(q^{2} − 1) = π(q^{′2}− 1), R_{2n} (q) = U_{2n} (q′) and {p} ∪ R_{4}(q) = {p′} ∪ U_{4}(q′).

(3) S = B_{n}(q′) or S = C_{n}(q′), where q′ = p^{′β}, p′ ≠ p, R_{1}(q) ∪ R_{2}(q) = {p′} ∪ U_{1}(q′) ∪ U_{2}(q′), R_{2n} (q) = U_{2n} (q′) and either {p} ∪ R_{4}(q) = U_{3}(q′) ∪

U_{4}(q′) ∪ U_{6}(q′) or {p} ∪ R_{4}(q) = U_{4}(q′).

Finally we state

Conjecture. Cases (2) and (3) in above corollary can not occur.

It is clear that if the conjecture is true, then Vasil’ev’s conjecture will be true for ^{2}D_{n}(q), for each n and prime power q.

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Dept. of Pure Math., Faculty of Math. and Computer Sci., (Received 23 04 2014) Amirkabir University of Technology (Tehran Polytechnic), Tehran, Iran and School of Mathematics, Institute for Research in Fundamental sciences (IPM), Tehran, Iran.

khosravibbb@yahoo.com

Department of Mathematics, University of Qom, Qom, Iran

a_babai@aut.ac.ir