PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 95 (109) (2014), 255–266 DOI: 10.2298/PIM1409255K

ON RECOGNITION BY PRIME GRAPH OF THE PROJECTIVE SPECIAL LINEAR GROUP OVER GF(3)

Bahman Khosravi, Behnam Khosravi, and Hamid Reza Dalili Oskouei

Communicated by Žarko Mijajlović

Abstract. Let G be a finite group. The prime graph of G is denoted by Γ(G). We prove that the simple group PSL_{n} (3), where n ⩾ 9, is quasirecognizable by prime graph; i.e., if G is a finite group such that Γ(G) = Γ(PSL_{n} (3)), then G has a unique nonabelian composition factor isomorphic to PSL_{n} (3). Darafsheh proved in 2010 that if p > 3 is a prime number, then the projective special linear group PSL_{p}(3) is at most 2-recognizable by spectrum. As a consequence of our result we prove that if n ⩾ 9, then PSL_{n} (3) is at most 2-recognizable by spectrum.

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 which is 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). In graph theory a subset of vertices of a graph is called an independent set if its vertices are pairwise nonadjacent. Denote by t(G) the maximal number of primes in π(G) pairwise nonadjacent in Γ(G). In other words, if ρ(G) is some 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 some independent set with the maximal number of vertices in Γ(G) containing p and let t(p, G) = |ρ(p, G)|.

2010 Mathematics Subject Classification: Primary 20D05, 20D60; Secondary 20D08.

Key words and phrases: prime graph, simple group, recognition, quasirecognition.

A finite group G is called recognizable by prime graph whenever if for a finite group H, we have Γ(H) = Γ(G), then H ≅ G. A nonabelian simple group P is called quasirecognizable by prime graph if every finite group whose prime graph is Γ(P) has a unique nonabelian composition factor which is isomorphic to P (see [16]). Obviously recognition (quasirecognition) by prime graph implies recognition (quasirecognition) by spectrum, but the converse is not true in general. Also some methods of recognition by spectrum cannot be used for recognition by prime graph. If Ω is a nonempty subset of the set of natural numbers, we denote by h(Ω) the number of nonisomorphic groups G with ω(G) = Ω. If G is a finite group, then h(ω(G)) is denoted by h(G). If h(G) = ∞, then G is called nonrecognizable by spectrum. If h(G) = r, then G is called r-recognizable by spectrum.

Hagie in [12], determined finite groups G satisfying Γ(G) = Γ(S), where S is a sporadic simple group. It is proved that if q = 3^{2n+1} (n > 0), then the simple group ^{2}G_{2}(q) is uniquely determined by its prime graph [16, 35]. A group G is called a CIT group if G is of even order and the centralizer in G of any involution is a 2-group. In [18], finite groups with the same prime graph as a CIT simple group are determined. Also in [19], it is proved that if p > 11 is a prime number and p ≢ 1 (mod 12), then PSL_{2}(p) is recognizable by prime graph. In [17, 23], finite groups with the same prime graph as PSL_{2}(q), where q is not prime, are determined. It is proved that the simple group F_{4}(q), where q = 2^{n} > 2 (see [15]) and ^{2}F_{4}(q) (see [1]) are quasirecognizable by prime graph. In [14], it is proved that if p is a prime number which is not a Mersenne or Fermat prime and p ≠ 11, 13, 19 and Γ(G) = Γ(PGL_{2}(p)), then G has a unique nonabelian composition factor which is isomorphic to PSL_{2}(p) and if p = 13, then G has a unique nonabelian composition factor which is isomorphic to PSL_{2}(13) or PSL_{2}(27). Then it is proved that if p and k > 1 are odd and q = p^{k} is a prime power, then PGL_{2}(q) is uniquely determined by its prime graph [2]. In [20,21,22,24,27,28] finite groups with the same prime graph as PSL_{n}(2), U_{n}(2), D_{n}(2), B_{n}(3) and ^{2}D_{n}(2) are obtained. In [3, 4], it is proved that ^{2}D_{2^{m} +1} (3) is recognizable by prime graph.

The projective special linear groups defined over a finite field of order 3, called the ternary field, are denoted by PSL_{n}(3), PSL(n, 3), L_{n}(3) or A_{n−1} (3) as a finite group of Lie type. In this paper as the main result we prove that the simple group PSL_{n}(3), where n ⩾ 9, is quasirecognizable by prime graph; i.e., if G is a finite group such that Γ(G) = Γ(PSL_{n}(3)), then G has a unique nonabelian composition factor isomorphic to PSL_{n}(3). In [8], it is proved that the projective special linear group PSL_{p}(3), where p > 3 is a prime number, is at most 2-recognizable by spectrum, i.e., if G is a finite group such that ω(G) = ω(PSL_{p}(3)), where p > 3 is an odd prime, then G is isomorphic to PSL_{p}(3) or PSL_{p}(3) · 2, the extension of PSL_{p}(3) by the graph automorphism. As a consequence of our result we prove that if n ⩾ 9, then PSL_{n}(3) is at most 2-recognizable by spectrum, i.e., if G is a finite group such that ω(G) = ω(PSL_{n}(3)), then G is isomorphic to PSL_{n}(3) or PSL_{n}(3) · 2, the extension of PSL_{n}(3) by the graph automorphism.

In this paper, all groups are finite and by simple groups we mean nonabelian simple groups. All further unexplained notations are standard and refer to [7]. Throughout the proof we use the classification of finite simple groups. In [31, Tables 2–9], independent sets also independent numbers for all simple groups are listed and we use these results in the proof of the main theorem of this paper.

2. Preliminary results

Lemma 2.1. [33, Theorem 1] Let G be a finite group with t(G) ⩾ 3 and t(2, G) ⩾ 2. Then the following hold:

(1) there exists a finite nonabelian simple group S such that S ⩽ Ḡ = G/K ⩽ Aut(S) for the maximal normal soluble subgroup K of G;

(2) for every independent subset ρ of π(G) with |ρ| ⩾ 3 at most one prime in ρ divides the product |K||Ḡ/S|. In particular, t(S) ⩾ t(G) − 1;

(3) one of the following holds:

(a) every prime r ∈ π(G) nonadjacent to 2 in Γ(G) does not divide the product |K||Ḡ/S|; in particular, t(2, S) ⩾ t(2, G);

(b) there exists a prime r ∈ π(K) nonadjacent to 2 in Γ(G); in which case t(G) = 3, t(2, G) = 2, and S ≅ Alt_{7} or PSL_{2}(q) for some odd q.

Remark 2.1. In Lemma 2.1, for every odd prime p ∈ π(S), we have t(p, S) ⩾ t(p, G) − 1.

Lemma 2.2. [26] Let N be a normal subgroup of G. Assume that G/N is a Frobenius group with Frobenius kernel F and cyclic Frobenius complement C. If (|N|, |F |) = 1, and F is not contained in NC_{G}(N)/N, then p|C| ∈ ω(G), where p is a prime divisor of |N|.

Lemma 2.3 (Zsigmondy’s Theorem). [36] Let p be a prime and let n be 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.

Lemma 2.4. [13] Let G be a finite simple group.

(1) If G = C_{n}(q), then G contains a Frobenius subgroup with kernel of order q^{n} and cyclic complement of order (q^{n} − 1)/(2, q − 1).

(2) If G = ^{2}D_{n}(q), and there exists a primitive prime divisor r of q^{2n−2} − 1, then G contains a Frobenius subgroup with kernel of order q^{2n−2} and cyclic complement of order r.

(3) If G = B_{n}(q) or D_{n}(q), and there exists a primitive prime divisor r_{m} of q^{m} − 1 where m = n or n − 1 such that m is odd, then G contains a Frobenius subgroup with kernel of order q^{m(m−1)/2} and cyclic complement of order r_{m}.

Remark 2.2. [30] 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 the 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 an odd prime, then ord_{a}(q) is denoted by e(a, q), too. If q is odd, then e(2, q) = 1 for q ≡ 1 (mod 4) and e(2, q) = 2 for q ≡ −1 (mod 4).

Lemma 2.5. [32, 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. Define

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

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 nonadjacent if and only if η(k) + η(l) > n, and l/k is not an odd natural number.

Lemma 2.6. [31, 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 nonadjacent if and only if k + l > n, and k does not divide l.

Lemma 2.7. [31, 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 2). \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 nonadjacent if and only if ν(k) + ν(l) > n, and ν(k) does not divide ν(l).

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

Lemma 2.8. [32, Proposition 2.5] Let G = D_{n}^{ε}(q) be a finite simple group of Lie type over a field of characteristic p and let function η(m) be defined as in Lemma 2.5. 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 nonadjacent if and only if 2η(k) + 2η(l) > 2n − (1 − ε(−1)^{k+l} ), l/k is not an odd natural number, and if ε = +, then the equality chain n = l = 2η(l) = 2η(k) = 2k is not true.

Lemma 2.9. [5, Lemma 3.1] Let G be a finite group satisfying the conditions of Lemma 2.1, and let the groups K and S be as in the claim of Lemma 2.1. Let there exist p ∈ π(K) and p′ ∈ π(S) such that p ≁ p′ in Γ(G), and S contains a Frobenius subgroup with kernel F and cyclic complement C such that (|F |, |K|) = 1. Then p|C| ∈ ω(G).

Lemma 2.10. [34, Theorem 1] Let L = PSL_{n}(q), where n ⩾ 5 and q = p^{α}. If L acts on a vector space W over a field of characteristic p, then ω(L) ≠ ω(W ⋋ L).

3. Main Results

Theorem 3.1. The simple group PSL_{n}(3), where n ⩾ 9, is quasirecognizable by prime graph; i.e., if G is a finite group such that Γ(G) = Γ(PSL_{n}(3)), then G has a unique nonabelian composition factor which is isomorphic to PSL_{n}(3).

Proof. Let D = PSL_{n}(3), where n ⩾ 9, and G be a finite group such that Γ(G) = Γ(D). Using [32, Tables 4-8], we conclude that t(D) = [\frac{n+1}{2} ] ⩾ 5 and t(2, D) = 2. Therefore t(G) ⩾ 5 and t(2, G) = 2. Also ρ(D) = \{r_{i} | [\frac{n}{2} ] < i ⩽ n\}, where r_{i} is a primitive prime divisor of 3^{i} − 1. Also using [32, Table 6], it follows that ρ(2, PSL_{n}(3)) = {2, r_{n−1} } if n is even and ρ(2, PSL_{n}(3)) = {2, r_{n}} if n is odd. Using Lemma 2.1, we conclude that there exists a finite nonabelian simple group S such that S ⩽ Ḡ = G/N ⩽ Aut(S), where N is the maximal normal soluble subgroup of G. Also t(S) ⩾ t(G) − 1 ⩾ 4 and t(2, S) ⩾ t(2, G) ⩾ 2, by Lemma 2.1. Now we consider each possibility for S, by the tables in [32]. Step 1. Let S ≅ A_{m}, where m ⩾ 5.

We know that t(S) ⩾ 4. Thus m ⩾ 17. So 17 | |A_{m}|. Since e(17, 3) = 16, we conclude that n ⩾ 16. Therefore t(G) = [(n + 1)/2] ⩾ 8, and so m > 19. If p is a prime number such that p ⩽ m−17, then p ∼ 17 in Γ(S). Let A := [m−17, m]∩ℤ. Then 12 = [18/2] + [18/3] − [18/6] elements of A are divisible by 2 or 3. Therefore t(17, S) ⩽ 7.

On the other hand, we know that e(17, 3) = 16. Let k = e(p, 3). Then p is not adjacent to 17 in Γ(G) if and only if 16 + k > n and 16 ∤ k if 16 ⩽ k, and k ∤ 16 if k ⩽ 16. There are 16 consecutive numbers in (n−16, n]∩ℤ. So 16 can divide exactly one of them. Also at most 5 of them divide 16. Hence t(17, G) ⩾ 16 − 1 − 5 = 10. Therefore 7 ⩾ t(17, S) ⩾ t(17, G) − 1 ⩾ 10 − 1 = 9, which is a contradiction. Step 2. In this step, we prove that the simple group S is not isomorphic to a simple group of Lie type over GF(p^{α}), where p ≠ 3. Using Table 8 in [32], we consider the independent set B = {r_{i} | n − 4 ⩽ i ⩽ n} in Γ(G), since n ⩾ 9 and so [\frac{n}{2} ] < n − 4. By Lemma 2.1, |B ∩ π(S)| ⩾ 4. Case 1. Let S ≅ PSL_{m}(q), where q = p^{α}, p ≠ 3.

We know that t(S) ⩾ t(G) − 1. Therefore [\frac{m+1}{2} ] ⩾ t(S) ⩾ [\frac{n+1}{2} ] − 1 ⩾ 4, which implies that m ⩾ 7 and m ⩾ n − 3. Also t(p, S) ⩽ 3 by Table 4 in [32]. Thus t(p, G) ⩽ 4. So we conclude that p ∉ B. Therefore p is joined to at least two elements of B in Γ(G). In the sequel we consider one case and other cases are similar to it. We assume that p is joined to r_{n−4} and r_{n−3} in Γ(G). Let t = e(p, 3) and note that e(r_{n−4} , 3) = n − 4 and e(r_{n−3} , 3) = n − 3. By Lemma 2.6, one of the following subcases occurs:

(1) t + n − 3 ⩽ n and t + n − 4 ⩽ n; (3) t + n − 4 ⩽ n and t | (n − 3);

(2) t + n − 3 ⩽ n and t | (n − 4); (4) t|(n − 4) and t | (n − 3).

Therefore in each case we conclude that t ⩽ 4. Therefore p ∈ {2, 5, 13}.

• If p = 5, then S ≅ PSL_{m}(5^{α}). We note that e(71, 5) = 5 and e(71, 3) = 35. We know that m ⩾ 6 and e(71, 5^{α}) divides e(71, 5) = 5. Therefore 71 ∈ π(S) ⊆ π(G). Now by Lemma 2.6, if e(x, 5^{α}) ⩽ m − 5, then x is joined 71 in Γ(S). Therefore t(71, S) ⩽ 6. On the other hand, by Lemma 2.6, we conclude that if e(y, 3) ⩽ n−35, then y ∼ 71 in Γ(G). Therefore ρ(71, G) ⊆ {r_{i} | n − 35 < i ⩽ n}. Also we know that r_{i} ≁ 71 if and only if n − 35 < i ⩽ n and i/35, 35/i are not integers. Let C = {n − 34, …, n}. Thus there is only one i ∈ C such that i/35 is an integer. Also since 35 = 5 × 7, there are at most 4 elements i ∈ C such that 35/i is an integer. Thus t(71, G) ⩾ 35 − 5 = 30, which is a contradiction since 29 ⩽ t(71, G) − 1 ⩽ t(71, S) ⩽ 6.

• If p = 2, then S ≅ PSL_{m}(2^{α}). Since e(31, 2) = 5 and e(31, 3) = 30, similarly we get a contradiction.

• If p = 13, then S ≅ PSL_{m}(13^{α}). We know that e(30941, 13) = 5 and e(30941, 3) = 30940. Now similarly to the above we get a contradiction. Case 2. Let S ≅ U_{m}(q), where q = p^{α}, p ≠ 3. Since t(S) = [(m + 1)/2], similarly to Case 1 we have m ⩾ 7 and m ⩾ n − 3. By [32, Table 4], we have t(p, S) ⩽ 3, and similarly to the last case, we conclude that p = 2, 5, 13.

• If p = 2, then S ≅ U_{m}(2^{α}). If m = 7, then t(S) = 4, which implies that t(G) = 5 and so n = 9, 10. Therefore 757 ∈ ρ(2, G) ⊆ π(S). On the other hand e(757, 2) = 756 = 2^{2} × 3^{3} × 7 and since m = 7, the order of U_{7}(2^{α}) shows that 9 × 7 | α. Now π(2^{63} − 1) ⊈ π(G), which is a contradiction. Therefore m ⩾ 8, and so π(2^{8} − 1) ⊆ π(S), which implies that 17 ∈ π(S) and since e(17, 3) = 16, we get that n ⩾ 16. This implies that m ⩾ 13 and so 31 ∈ π(2^{10α} − 1) ⊆ π(S). We note that e(31, 3) = 30 and e(31, 2) = 5. Hence e(31, 2^{α}) | 5. We know that if ν(e(x, 2^{α})) ⩽ m − 10, then 31 ∼ x in Γ(S), by Lemma 2.7. Therefore t(31, S) ⩽ 10. Now we determine t(31, G). As e(31, 3) = 30 similarly to the above, we conclude that if e(x, 3) ⩽ m − 30, then 31 ∼ x in Γ(G). Let C = {n − 29, …, n}. So 30 divides exactly one element of C. Also since 30 = 2 × 3 × 5, there are at most 8 elements in C such that 30/i is an integer. Thus 22 ⩽ t(31, G). Therefore 21 ⩽ t(31, G) − 1 ⩽ t(31, S) ⩽ 10, which is a contradiction.

• If p = 5, then S ≅ U_{m}(5^{α}). Since m ⩾ 7, we have 449 ∈ π(S). Also e(449, 3) = 448 and e(449, 5) = 14. Now similarly to the above we get a contradiction.

• If p = 13, then similarly to the above by using e(157, 3) = 78 and e(157, 13) = 6, we get a contradiction. Case 3. Let S ≅ D_{m}(q), where q = p^{α}, p ≠ 3. Then t(S) ⩾ t(G) − 1 implies that m ⩾ 5. Similarly to the last cases, we conclude that p = 2, 5 or 13.

If p = 2, then we note that e(31, 2) = 5. Therefore e(31, 2^{α}) | 5. Thus for every x ∈ π(S), such that 2η(e(x, 2^{α})) ⩽ 2n − 10 − 2, we have x ∼ 31 in Γ(S). Therefore t(31, S) ⩽ 12. On the other hand, as we mentioned above t(31, G) ⩾ 22, which is a contradiction. Similarly if p = 5, then we use t(31, S) and if p = 13, then we use t(157, S) and similarly to the above we get a contradiction. Case 4. Let S ≅ B_{m}(q) or S ≅ C_{m}(q), where q = p^{α} and p ≠ 3.

So (3m + 5)/4 ⩾ [(3m + 5)/4] = t(S) ⩾ t(G) − 1 = [(n + 1)/2] − 1 ⩾ 4. Thus similarly m ⩾ 4 and p = 2, 5 or 13. For p = 5, we note that e(31, 3) = 30 and e(31, 5) = 3. By Lemma 2.5 we know that if η(e(x, 5^{α})) ⩽ m − 3, then x ∼ 31 in Γ(S). Therefore t(31, S) ⩽ 6, which is a contradiction since t(31, G) ⩾ 22.

If p = 13, then using e(157, 3) = 78 and e(157, 13) = 6, we get a contradiction. If p = 2, then S ≅ B_{m}(2^{α}). Now we note that e(17, 3) = 16, e(17, 2) = 8. Then

e(17, 2^{α}) | 8 and so η(e(17, 2^{α})) ⩽ 4. If η(e(x, 2^{α})) ⩽ m − 4, then x ∼ 17 in Γ(S) by Lemma 2.5. So only 8 elements, where η(e(x, 2^{α})) > m − 4 may not be joined 17 in Γ(S). Hence the independent set which contains 17 has at most 9 elements in Γ(S). On the other hand, if r_{i} ≁ 17 in Γ(G), then m − 16 < i ⩽ m; and i/16, 16/i are not integers. Then 16 divides one of the numbers in [m − 15, m] ∩ ℤ. Also at most 5 numbers in this interval can divide 16. So at least 10 elements are not adjacent to 17 in Γ(G). Therefore ρ(17, G) has at least 11 elements and we get a contradiction since 10 ⩽ t(17, G) − 1 ⩽ t(17, S) ⩽ 9. Case 5. Let S ≅ ^{2}D_{m}(q), where q = p^{α}, p ≠ 3. Therefore (3m + 4)/4 ⩾ t(S) ⩾ 4 implies that 3m ⩾ 2n − 6 and m ⩾ 4. Now we consider the following cases.

• Let n ⩾ 11. Then B′ = {r′_{i} | n − 5 ⩽ i ⩽ n} is an independent set in Γ(G). Since t(p, S) ⩽ 4, we conclude that p is joined to at least two elements of B in Γ(G). In each case similarly to the previous cases we conclude that p = 2, 5, 11 or 13.

If p = 2, then since e(31, 3) = 30 and e(31, 2) = 5, similarly to the last cases we get a contradiction. Also we know that e(31, 3) = 30, e(31, 5) = 3; e(7321, 3) = 1830, e(7321, 11) = 8 and e(157, 3) = 78, e(157, 13) = 6. Hence for p = 5, 11 and 13 we get a contradiction.

• Let n = 9 and S ≅ ^{2}D_{m}(q), where p ∈ π(PSL_{9}(3)) ∖ {3}.

If p ∈ {2, 13, 41, 757, 1093}, then π(p^{8} − 1) ⊈ π(G). Also for p ∈ {5, 7, 11}, we see that π(p^{6} − 1) ⊈ π(G).

• If n = 10, then similarly we get a contradiction. Case 6. In this case we prove that S is not isomorphic to an exceptional simple group. Let S ≅ F_{4}(q), E_{6}(q) or ^{2}E_{6}(q), where q = p^{α} and p ∈ π(G). Then t(S) ⩽ 5 and so 9 ⩽ n ⩽ 12. Easily we can see that for each 3 ≠ p ∈ π(G), π((p^{8} − 1)(p^{12} − 1)) ⊈ π(G), which is a contradiction since π((p^{8} − 1)(p^{12} − 1)) ⊆ π(S).

If S ≅ E_{7}(q), where q = p^{α}, then t(S) = 8 and so t(G) ⩽ 9. Therefore 9 ⩽ n ⩽ 18. Similarly to the last case for each 3 ≠ p ∈ π(G), we can get a contradiction.

If S ≅ E_{8}(q), where q = p^{α}, then 9 ⩽ n ⩽ 24 and for each p ∈ π(PSL_{n}(3)) we have π(p^{10} − p^{5} + 1) ⊈ π(G), which is a contradiction.

If S ≅ ^{2}F_{4}(2^{2n+1} ), then 9 ⩽ n ⩽ 12. If n = 9 or n = 10, then 757 ∈ ρ(2, G) and so 757 ∈ π(^{2}F_{4}(2^{2n+1} )). We know that e(757, 2) = 756 and so 756 | 12(2n + 1). Therefore 7 | (2n + 1), and so π(2^{7} − 1) ⊆ π(S) ⊆ π(G), which is a contradiction.

If n = 11, 12, then 3851 ∈ ρ(2, G) and similarly we get a contradiction, since e(3851, 2) = 3850.

If S ≅ ^{2}B_{2}(2^{2n+1} ), then similarly we get a contradiction. Step 3. Now we consider the simple groups of Lie type over GF(3^{α}). In the sequel, we use r′_{k} for a primitive prime divisor of (3^{α})^{k} − 1. Case 1. Let S ≅ PSL_{m}(q), where q = 3^{α}. By [32, Table 6], r_{n−1} ∈ π(S) or r_{n} ∈ π(S). Also m ⩾ n − 3. (I) Let n be odd and so r_{n} ∈ ρ(2, S) = {2, r′_{m}, r′_{m−1} }.

• If r_{n} = r′_{m}, then n | αm and so n ⩽ αm. On the other hand, using Zsigmondy’s Theorem, we conclude that αm ⩽ n, since π(S) ⊆ π(G). Therefore αm = n.

Also we know that m ⩾ n − 3. If α ⩾ 2, then n = αm ⩾ 2m ⩾ 2n − 6, which implies that 6 ⩾ n and this is a contradiction. Thus α = 1, and so m = n. Therefore S = PSL_{n}(3).

• If r_{n} = r′_{m−1} , then n | α(m − 1) and so n ⩽ α(m − 1). Also by Zsigmondy’s Theorem, α(m − 1) ⩽ n. Hence α(m − 1) = n. On the other hand, (3^{αm} − 1) | |S| and since π(S) ⊆ π(G), we conclude that αm ⩽ n, which is a contradiction. (II) Let n be even and so r_{n−1} ∈ π(S). Therefore r_{n−1} ∈ ρ(2, S) = {2, r′_{m}, r′_{m−1} }.

• If r_{n−1} = r′_{m−1} , then (n − 1) | α(m − 1) and α(m − 1) ⩽ n, since π(S) ⊆ π(G).

Hence α(m − 1) = n − 1. We know that m ⩾ n − 3. If α ⩾ 2, then

n − 1 = α(m − 1) ⩾ 2(m − 1) ⩾ 2(n − 3) − 2 ⩾ 2n − 8.

Hence n ⩽ 7, which is a contradiction. Thus α = 1 and so m = n. Therefore S ≅ PSL_{n}(3).

• If r_{n−1} = r′_{m}, then (n − 1) | αm. Also αm ⩽ n, since π(S) ⊆ π(G). Therefore αm = n − 1. If α ⩾ 2, we get that n − 1 = αm ⩾ 2m ⩾ 2n − 6. Thus n ⩽ 5, which is a contradiction. Thus α = 1 and so m = n − 1. Therefore S ≅ PSL_{n−1} (3). So r_{n} ∈ π(K) ∪ π(Ḡ/S). Also we note that π(Ḡ/S) ⊆ π(Out(S)) = {2}. So r_{n} ∈ π(K). We note that there exists a Frobenius subgroup of PSL_{n−1} (3) of the form 3^{n−2} : (3^{n−2} − 1)/d, where d = (n − 1, 2). On the other hand, r_{n} ≁ r_{n−1} in Γ(G). So by Lemma 2.9, we conclude that r_{n} is joined to r_{n−2} in Γ(G), which is a contradiction. Case 2. Let S ≅ U_{m}(q), where q = 3^{α}. Then m ⩾ n − 3 and r_{n} ∈ π(S) or r_{n−1} ∈ π(S). (I) Let n be odd and so r_{n} ∈ π(S). Using [32, Table 4] we must consider four cases, since r_{n} ∈ {r′_{2m} , r′_{2m−2} , r′_{m}, r′_{m/2} }.

• If r_{n} = r′_{2m}, then m is odd by [32, Table 4]. Also similar to the previous cases, n | 2αm. On the other hand, since π(S) ⊆ π(G), we conclude that 2αm ⩽ n. Therefore 2αm = n, which is a contradiction since n is odd.

• If r_{n} = r′_{2m−2} , then m is even and n | 2α(m − 1). Also since π(S) ⊆ π(G), we get that 2α(m − 1) ⩽ n. Thus n = 2α(m − 1), which is a contradiction since n is odd.

• If r_{n} = r′_{m}, then 4 | m, by [32, Table 4]. Also n | αm. Thus n = αm, a contradiction since n is odd.

• Let r_{n} = r′_{m/2} . Thus n | αm/2, which implies that n ⩽ αm/2 ⩽ n, since π(S) ⊆ π(G). Therefore n = αm/2. Hence αm = 2n and r_{αm} ∈ π(S) ⊆ π(G), which is a contradiction. (II) Let n be even and so r_{n−1} ∈ π(S).

• Let r_{n−1} = r′_{2m}. Thus (n−1) | 2αm. So n−1 ⩽ 2αm ⩽ n. Therefore 2αm = n−1, which is a contradiction, since n is even.

• Let r_{n−1} = r′_{2m−2} . So (n − 1) | 2α(m − 1). So similarly to the above, we conclude that 2α(m − 1) = n − 1, which is a contradiction since n is even.

• Let r_{n−1} = r′_{m}. So n − 1 = αm. If α ⩾ 2, then n − 1 = αm ⩾ 2m ⩾ 2n − 6. Therefore 5 ⩾ n, which is a contradiction. If α = 1, then m = n − 1. Therefore S = U_{n−1} (3). Since n − 1 is odd, we conclude that (3^{n−1} + 1) | |S|. Hence r_{2(n−1)} ∈ π(G), which is a contradiction.

• Let r_{n−1} = r′_{m/2} . Thus m is even and n − 1 = αm/2. We know that n − 1 is odd and so α is odd. If α ⩾ 3, then n − 1 ⩾ 3m/2 ⩾ 3(n − 3)/2. Therefore 7 ⩾ n, which is a contradiction. If α = 1, then m = 2n − 2. So r_{2n−2} ∈ π(S) ⊆ π(G), which is a contradiction. Case 3. Let S ≅ B_{m}(q), where q = 3^{α}. Since t(S) ⩾ t(G) − 1. We have 3m > 2n − 11. Also ρ(2, S) = {2, r′_{m}, r′_{2m} }. (I) Let n be odd and so r_{n} ∈ ρ(2, G). Therefore r_{n} = r′_{m} or r_{n} = r′_{2m}.

• Let r_{n} = r′_{m}. Then m is odd by [32, Table 6]. Also n | αm. Hence n ⩽ αm ⩽ n, since π(S) ⊆ π(G). Thus n = αm. Obviously α is odd. If α ⩾ 5, then n = αm ⩾ 5m ⩾ \frac{10}{3} n − \frac{55}{3} . So 55 ⩾ 7n, which is a contradiction. If α = 1, then n = m. So S ≅ B_{n}(3^{α}). Hence r_{2n} ∈ π(S) ⊆ π(G), which is a contradiction. If α = 3, then n = 3m. Hence S ≅ B_{n/3} (27). Thus π(27^{2n/3} − 1) = π(3^{2n} − 1) ⊆ π(S) ⊆ π(G), which is a contradiction.

• Let r_{n} = r′_{2m}. So m is even. Therefore similarly to the above, we conclude that n = 2αm, which is a contradiction since n is odd. (II) Let n be even and so r_{n−1} ∈ ρ(2, G). Then r_{n−1} = r′_{m} or r_{n−1} = r′_{2m}. Similarly to the above we get a contradiction. Case 4. Let S = D_{m}(q), where q = 3^{α}. Similarly, we conclude that if m ≢ 3 (mod 4), then 3m ⩾ 2n − 2 and if m ≡ 3 (mod 4), then 3m > 2n − 4, since t(S) ⩾ t(G) − 1. Therefore in each case we have 3m > 2n − 4. We know that ρ(2, S) = {2, r′_{m−1} , r′_{m}, r′_{2m−2} }. Also if r′_{m−1} ∈ ρ(2, S), then m is even and if r′_{m} ∈ ρ(2, S), then m is odd. (I) If n is odd, then r_{n} ∈ π(S).

• Let r_{n} = r′_{m}. So n and m are odd. Similarly to the above, we conclude that n = αm. If α = 1, then n = m. So S ≅ D_{n}(3). Hence r_{2n−2} ∈ π(S) ⊆ π(G), which is a contradiction. If α ⩾ 2, then n = αm ⩾ 2m ⩾ 4(n − 2)/3, which is a contradiction.

• Let r_{n} = r′_{m−1} . So n is odd and m is even. Similarly, we conclude that n = α(m − 1). Now since π((3^{α})^{2(m−1)} − 1) ⊆ π(S) ⊆ π(G), we get a contradiction.

• Let r_{n} = r′_{2m−2} . Then n | 2α(m − 1), and so n = 2α(m − 1), which is a contradiction since n is odd. (II) If n is even, then r_{n−1} ∈ π(S).

• Let r_{n−1} = r′_{m−1} . Hence (n − 1) | α(m − 1) and α(m − 1) ⩽ n. Hence n − 1 = α(m − 1). Then r_{2α(m−1)} ∈ π(S) ⊆ π(G), which is a contradiction.

• Let r_{n−1} = r′_{m}. So n is even and m is odd. Similarly to the above, we conclude that (n − 1) | αm. Thus αm = n − 1 and so α is odd. Also we know that 3m > 2n − 4. If α ⩾ 3, then n − 1 = αm ⩾ 3m ⩾ 2n − 4. Therefore n ⩽ 3, which is a contradiction. If α = 1, then n − 1 = m and so S ≅ D_{n−1} (3). Hence r_{2n−2} ∈ π(S) ⊆ π(G), which is a contradiction.

• If r_{n−1} = r′_{2m−2} , then n−1 = 2(m−1)α, which is a contradiction since n is even. Case 5. Let S ≅ ^{2}D_{m}(q), where q = 3^{α}.

Similarly to the above, we conclude that 3m > 2n − 10. By [32, Table 6] it follows that if r′_{2m−2} ∈ ρ(2, S), then m is odd. (I) If n is odd, then r_{n} ∈ π(S). Hence r_{n} = r′_{2m} or r_{n} = r′_{2m−2} . If r_{n} = r′_{2m} or r′_{2m−2} , then similarly to the above, we conclude that n = 2αm or 2α(m − 1), which is a contradiction since n is odd. (II) If n is even, then r_{n−1} ∈ π(S). If r_{n−1} = r′_{2m}, or r′_{2m−2} , then n − 1 = 2αm or 2α(m − 1), which is a contradiction since n − 1 is odd. Case 6. Let S ≅ F_{4}(q), where q = 3^{α}. Since t(S) = 5, t(G) ⩽ 6 and so 9 ⩽ n ⩽ 12. We know that π(3^{12α} − 1) ⊆ π(F_{4}(q)). Therefore α = 1 and so q = 3, n = 12. Now r_{11} ∈ ρ(2, G) ⊆ π(S), which is a contradiction.

Similarly, we conclude that S can not be isomorphic to E_{6}(q) and ^{2}E_{6}(q). Case 7. Let S ≅ ^{2}G_{2}(3^{2m+1} ), where m ⩾ 1. Since t(S) = 5, we get that 9 ⩽ n ⩽ 12. Similarly to the previous case if n ⩾ 11, then we get a contradiction since 3 ∈ ρ(S), for each independent set ρ(S). Therefore t(G) = 5 and so n = 9 or n = 10.

Now Zsigmondy’s Theorem implies that 6(2m + 1) ⩽ 10, which is a contradiction.

Step 4. In this step we prove that S is not isomorphic to a sporadic simple group. If S ≅ J_{4}, then 43 | |S| and since e(43, 3) = 42 we have n ⩾ 42. So t(G) ⩾ [\frac{42+1}{2} ] = 21, which is a contradiction since t(J_{4}) = 7. For the rest of sporadic simple groups t(S) ⩽ 5 and so 9 ⩽ n ⩽ 12. Hence {757, 1093} ∩ π(S) ≠ ∅, which is a contradiction.

Therefore the quasirecognition of PSL_{n}(3), where n ⩾ 9, is proved. □

Theorem 3.2. If Γ(G) = Γ(PSL_{n}(3)), where n ⩾ 9, then PSL_{n}(3) ⩽ G/N ⩽ Aut(PSL_{n}(3)), where N is a 3-group for even n and N is a {2, 3}-group for odd n.

Proof. By Theorem 3.1, we know that PSL_{n}(3) ⩽ G/N ⩽ Aut(PSL_{n}(3)). Similarly to [20], we can assume that N is an elementary abelian p-group for some p ∈ π(G). Now we prove that PSL_{n}(3) acts faithfully on N. For this reason, we prove that C = C_{G}(N) ⩽ N. Since C is a normal subgroup of G, if C ≰ N, then CN/N is a nontrivial normal subgroup of G/N. As the proof of the main theorem in [20] shows that socle(G/N) ≅ PSL_{n}(3) and so CN/N has a subgroup isomorphic to PSL_{n}(3). Therefore r_{n−1} , r_{n} ∈ π(PSL_{n}(3)) implies that r_{n−1} , r_{n} divide the order of CN/N ≅ C/(C ∩ N). Hence p ∼ r_{n} and p ∼ r_{n−1} in Γ(G), which implies that p = 1, by Lemma 2.6. Therefore C ⩽ N and PSL_{n}(3) acts faithfully on N. Also PSL_{n}(3) contains Frobenuis subgroups of the form 3^{n−1} : (3^{n−1} − 1)/(n, 2) and 3^{n−2} : (3^{n−2} − 1)/(n − 1, 2). Hence if p ≠ 3, then using Lemma 2.2 it follows that p ∼ r_{n−1} and p ∼ r_{n−2} in Γ(G). Therefore p = 2 or p = 3, using Lemma 2.6. Now if n is even, then 2 ≁ r_{n−1} , which is a contradiction. Therefore if n is odd, then N is a {2, 3}-group and if n is even, then N is a 3-group. □

Corollary 3.1. Let Γ(G) = Γ(PSL_{n}(3)), where n ⩾ 9. Then G/N ≅ PSL_{n}(3)

or PSL_{n}(3) · 2, the extension of PSL_{n}(3) by the graph automorphism, where N is a 3-group, if n is even and N is a {2, 3}-group, if n is odd.

Proof. We know that using the notations of [7], f = 1, g = 2 and d = (n, 2). By the assumption, we know that PSL_{n}(3) ⩽ Ḡ := G/N ⩽ Aut(PSL_{n}(3)). Let S = PSL_{n}(3). Then Ḡ/S ⩽ Aut(S). Now if ϕ is a diagonal automorphism of S, and ψ is a graph automorphism of S, then S · ϕ and S · (ϕψ) have elements of orders 2r_{n−1} and 2r_{n}, which is a contradiction, since in the prime graph of G we have 2 ≁ r_{n−1} if n is even and 2 ≁ r_{n} if n is odd. Therefore Ḡ ≅ S or S · ψ, the extension of PSL_{n}(3) by the graph automorphism. □

Theorem 3.3. Let ω(G) = ω(PSL_{n}(3)), where n ⩾ 9. Then G ≅ PSL_{n}(3) or PSL_{n}(3) · 2, the extension of PSL_{n}(3) by the graph automorphism.

Proof. Using Corollary 3.1 we know that if n is even, then G/N ≅ PSL_{n}(3) or PSL_{n}(3)·2, where N is a 3-group. Now using Lemma 2.10 it follows that N = 1. Therefore G ≅ PSL_{n}(3) or PSL_{n}(3) · 2.

Similarly if n is odd, then N can not be a 3-group. Also as we stated in the proof of Theorem 3.2, PSL_{n}(3) has a Frobenius subgroup of the form 3^{n−1} : (3^{n−1} − 1). Now if 2 | |N|, then using Lemma 2.9 we get that 2(3^{n−1} − 1) ∈ ω(PSL_{n}(3)), which is a contradiction by [6]. Therefore in each case we have G ≅ PSL_{n}(3) or PSL_{n}(3) · 2. □

Remark 3.1. In [29], it is proved that h(PSL_{3}(3)) = ∞. Also PSL_{4}(3) and PSL_{5}(3) are recognizable by spectrum (see [10, 25]). In [11], it is shown that h(PSL_{6}(3)) = 2. In [9], it is proved that h(PSL_{7}(3)) = 2 and h(PSL_{8}(3)) = 1. Also in [8] for each prime number p > 3, the following conjectures arise.

Conjecture 1. If p ≡ 1 (mod 3), then PSL_{p}(3) is 2-recognizable by spectrum. Conjecture 2. If p ≡ 2 (mod 3), then PSL_{p}(3) is recognizable by spectrum.

Acknowledgement. The authors would like to thank the referees.

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Department of Mathematics, Faculty of Science (Received 13 02 2013) Qom University of Technology, Qom, Iran

    khosravibahman@yahoo.com

Department of Mathematics

Institute For Advanced Studies in Basic Sciences

Zanjan 45137-66731, Iran

    behnam_kho@yahoo.com

Shahid Sattari Aeronautical University of Science and Technology

P.O. Box 13846-63113, Tehran, Iran

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