PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 99(113) (2016), 193–201 DOI: 10.2298/PIM141220015M

NSE CHARACTERIZATION OF THE SIMPLE GROUP L_{2}(3^{n})

Hosein Parvizi Mosaed, Ali Iranmanesh, and Abolfazl Tehranian

Abstract. Let G be a group and π(G) be the set of primes p such that G contains an element of order p. Let nse(G) be the set of the numbers of elements of G of the same order. We prove that the simple group L_{2}(3^{n}) is uniquely determined by nse(L_{2}(3^{n})), where |π(L_{2}(3^{n}))| = 4.

1. Introduction

Let G be a group and π(G) be the set of primes p such that G contains an element of order p and π_{e}(G) be the set of element orders of G. If k ∈ π_{e}(G), then we denote by m_{k}, the number of elements of order k in G. Let nse(G) = {m_{k} | k ∈ π_{e}(G)}. In 1987, Thompson posed a problem [6, Problem 12.37] related to algebraic number fields as follows:

Thompson Problem. Let T (G) = {(k, m_{k}) | k ∈ π_{e}(G)}. Suppose that

T (G) = T (H). If G is a finite solvable group, then is H also necessarily solvable?

Up to now, no one has been able to solve this problem completely even to give a counterexample. It is easy to see that if G and H are two finite groups with T (G) = T (H), then |G| = |H| and nse(G) = nse(H). Studies on characterizations related to nse started by Shao et al. in [9]. They proved that if S is a finite simple group with |π(S)| = 4, then S is characterizable by nse(S) and |S|, i.e., S is uniquely determined by nse(S) and |S|. Also, in [4], it is proved that sporadic simple group S is characterizable by nse(S) and |S|. Moreover, there are some research on the characterization of finite simple groups by nse. For instance, in [3, 5, 8, 10], it is proved that the groups A_{5}, A_{6}, A_{7}, A_{8}, J_{1} and L_{2}(q), where q ∈ {7, 8, 11, 13} can be uniquely determined by nse. It is worth mentioning that considering the characterization of a simple group S by nse(S) is much more complicated than its characterization by nse(S) and |S|. Because when G is a group with nse(G) = nse(S), the most challenging part to show that G ≅ S, is to prove π(G) = π(S).

2010 Mathematics Subject Classification: Primary 20D60; Secondary 20D06.

Key words and phrases: Element order, the set of the numbers of elements of the same order, projective special linear group.

Communicated by Zoran Petrović.

We here consider this characterization motivated by the following problem, which appeared in [3].

Problem. Let G be a group such that nse(G) = nse(L_{2}(q)), where q is a prime power. Is G isomorphic to L_{2}(q)?

Our main purpose is to show that the problem has an affirmative answer for q = 3^{n} and |π(L_{2}(q))| = 4. In fact, we have the following main theorem.

Main Theorem. Let G be a group such that nse(G) = nse(L_{2}(3^{n})), where n, (3^{n} − 1)/2 and (3^{n} + 1)/4 are odd primes. Then G ≅ L_{2}(3^{n}).

2. Notation and preliminaries

For a natural number m, by π(m), we mean the set of all prime divisors of m, so it is obvious that if G is a finite group, then π(G) = π(|G|). A Sylow p-subgroup of G is denoted by G_{p} and by n_{p}(G), we mean the number of Sylow p-subgroups of G. Also, the largest element order of G_{p} is denoted by exp(G_{p}). Moreover, we denote by φ, the Euler totient function and by (a, b), the greatest common divisor of integers a and b.

In the following, we bring some useful lemmas which will be used in the proof of the main theorem.

Lemma 2.1. [2] Let G be a finite group and m be a positive integer dividing |G|. If L_{m}(G) = {g ∈ G : g^{m} = 1}, then m | |L_{m}(G)|.

Lemma 2.2. [7] Let G be a finite group and p ∈ π(G) − {2}. Suppose that P is a Sylow p-subgroup of G and n = p^{s}m, where (p, m) = 1. If P is not cyclic and s > 1, then the number of elements of order n is either 0, or a multiple of p^{s}.

Lemma 2.3. [11, Theorem 3] In a group of order g, the number of elements whose orders are multiples of n is either zero, or a multiple of the greatest divisor of g that is prime to n.

Lemma 2.4. [10] Let G be a group containing more than two elements. Let k ∈ π_{e}(G) and m_{k} be the number of elements of order k in G. If s = sup{m_{k} : k ∈ π_{e}(G)} is finite, then G is finite and |G| ⩽ s(s^{2} − 1).

Lemma 2.5. [1] Let p be a prime number.

(1) If p ≠ 3, then y^{2} ≡ −3 (mod p) is solvable if and only if p ≡ 1 (mod 3).

(2) The equation y^{2} ≡ −1 (mod p) is solvable if and only if p ≡ 1 (mod 4).

From now on, we assume that n, u = (3^{n} − 1)/2 and t = (3^{n} + 1)/4 are odd primes and x = 3^{n}.

Lemma 2.6. We have

(1) (u, t) = 1, (u, 3) = 1, (u, 2) = 1, (t, 3) = 1, (t, 2) = 1;

(2) (u − 1) = 2(t − 1), (u − 1, t) = 1, (t − 1, u) = 1;

(3) 4 | 1 + xu.

NSE CHARACTERIZATION OF THE SIMPLE GROUP L_{2} (3^{n} ) 195

Proof. Parts (1) and (2) are straightforward. To prove part (3), we know that t − 1 is even and hence, 8 | x − 3. Now if 4 ∤ 1 + xu, then 8 ∤ x^{2} − x + 2 and hence 8 ∤ x^{2} − x + 2 + x − 3 = (x − 1)(x + 1) = 8tu, which is a contradiction. □

Lemma 2.7. Let p be a prime number which is prime to 2, 3, t, u, t − 1.

(1) If p | x^{2} − 5x + 8, then p ∤ 3x + 1, p ∤ x^{2} + x − 4 and p ∤ x^{2} − 5.

(2) If p | x^{2} − x − 4, then p ∤ 3x + 1 and p ∤ x^{2} + 5x + 8.

(3) If 16 | x − 3, then 16 ∤ x^{2} − x + 2.

Proof. Let p | x^{2}−5x+8. If p | 3x+1, then p | 9(x^{2}−5x+8)+(−3x+16)(3x+1) = 8.11 and since (p, 2) = 1, we conclude that p = 11. Now since p | (3x + 1), we conclude that 3^{n+1} ≡ −1 (mod p). Lemma 2.5 implies 4 | (p − 1) = 10, which is a contradiction. If p | x^{2} + x − 4, then p | (x^{2} − 5x + 8) + 2(x^{2} + x − 4) = 6xu and since (p, 2) = (p, 3) = (p, u) = 1, we get a contradiction. If p | x^{2} − 5, then p | (−5x − 13)(x^{2} − 5x + 8) + (5x − 12)(x^{2} − 5) = −4.11 and since (p, 2) = 1, we conclude that p = 11. Thus 11 | x^{2}−5−11 = (x−4)(x+4) and hence 11 | (x+4) or 11 | (x − 4). If 11 | (x + 4), then 11 | 3x + 1. Thus 3^{n+1} ≡ −1 (mod 11) and hence Lemma 2.5 implies that 4 | (p − 1) = 10, which is a contradiction. If 11 | (x − 4), then 11 | (x^{2} − 5x + 8) − (x^{2} − 5) + 5(x − 4) = −7, which is a contradiction.

Let p | x^{2}−x−4. If p | 3x+1, then p | −3(x^{2}−x−4)+(x−12)(3x+1) = −32x and since (p, 3) = (p, 2) = 1, we get a contradiction. If p | x^{2} + 5x + 8, then p | (x^{2} + 5x + 8) + 2(x^{2} − x − 4) = 12xt and since (p, 2) = (p, 3) = (p, t) = 1, we get a contradiction.

If 16 | x − 3 and x^{2} − x + 2, then 16 | (x^{2} − x + 2) + (x − 3) = (x^{2} − 1) = 8tu and since (u, 2) = (t, 2) = 1, we get a contradiction. □

3. Proof of the main theorem

According to [9], we know that |L_{2}(3^{n})| = 2^{2}3^{n}tu and

nse(L_{2}(3^{n})) = {1, 3^{n}u, 8tu, (t − 1)3^{n}u, (u − 1)3^{n}2t} .

We prove the main theorem in a sequence of lemmas.

Lemma 3.1. The group G is finite and for every i ∈ π_{e}(G), we have φ(i) | m_{i} and i | ∑ _{d|i} m_{d}. Moreover, if i > 2, then m_{i} is even.

Proof. According to Lemma 2.4, it is obvious that G is a finite group. Now, if i ∈ π_{e}(G), then Lemma 2.1 implies that i | ∑ _{d|i} m_{d}. We know that the number of elements of order i in a cyclic group of order i is equal to φ(i). Thus m_{i} = φ(i)k, where k is the number of cyclic subgroups of order i in G and hence, φ(i) | m_{i}. Also, it is known that if i > 2, then φ(i) is even and since φ(i) | m_{i}, we conclude that m_{i} is even as well. □

Lemma 3.2. 2 ∈ π(G) and exp(G_{2}) ⩽ 2^{5}.

Proof. Since 3^{n}u is the only odd element of nse(G) − {1}, Lemma 3.1 yields 2 ∈ π(G) and m_{2} = 3^{n}u. If exp(G_{2}) > 2^{5}, then 2^{6} ∈ π_{e}(G) and hence by Lemma 3.1, we have 2^{5} = φ(2^{6}) | m_{2^{6}} . Lemma 2.6 now implies that 8 | t − 1. Therefore 32 | x − 3 and also according to nse(G), we have 8 divides m_{4} and m_{8}.

Thus Lemma 3.1 implies that 8 | 1 + m_{2} and hence, 16 | x^{2} − x + 2, which is impossible according to Lemma 2.7. So exp(G_{2}) ⩽ 2^{5}. □

Lemma 3.3. We have π(G) ≠ {2}.

Proof. If π(G) = {2}, then |G| = 2^{k} = 1 + 3^{n}u + k_{1}8tu + k_{2}(t − 1)3^{n}u + k_{3}(u− 1)3^{n}2t, where k, k_{1}, k_{2}, k_{3} are natural numbers. By Lemma 3.2, exp(G_{2}) ⩽ 2^{5} and hence |π_{e}(G)| ⩽ 6. Thus k_{1} + k_{2} + k_{3} ⩽ 4. If k_{1} = 1, then 3^{n} | |G|, which is a contradiction. Thus, we have (k_{1}, k_{2}, k_{3}) = (2, 1, 1), which implies that u | |G|, a contradiction. □

Lemma 3.4. We have π(G) ≠ {2, 3}.

Proof. Let π(G) = {2, 3}. We prove the lemma in the following four steps.

Step 1. In this step, we prove that 3 is the only element of π_{e}(G) such that m_{3} = 8tu and hence |G| = 1 + 3^{n}u + 8tu + k_{1}(t − 1)3^{n}u + k_{2}(u − 1)3^{n}2t, where k_{1}, k_{2} are natural numbers and k_{1} + k_{2} ⩽ |π_{e}(G)| − 3.

According to nse(G), there exists i ∈ π_{e}(G) such that m_{i} = 8tu. Since i ∈ π_{e}(G) and π(G) = {2, 3}, we have i = 2^{α} 3^{β}, where α, β ⩾ 0 and the case α = β = 0 does not happen. By Lemma 3.1, it is obvious that if i = 3, then m_{3} = 8tu. Now we are going to reach a contradiction for the other cases of α and β.

Case 1. Let α ⩾ 1 and β = 0. Thus m_{i} = m_{2^{α}} = 8tu. If 2^{α} 3 ∉ π_{e}(G), then G_{3} acts fixed point freely on the set of elements of order 2^{α} by conjugation and hence, |G_{3}| | m_{2^{α}} , which is a contradiction. Thus 2^{α} 3 ∈ π_{e}(G) and according to Lemma 3.1, we conclude that 2^{α} 3 | ∑ _{d|2^{α} 3} m_{d} and 2^{α−1} 3 | ∑ _{d|2^{α−1} 3} m_{d}. Therefore, 3 | m_{2^{α}} + m_{2^{α} 3}. Since 3 ∤ m_{2^{α}} , we have 3 ∤ m_{2^{α} 3} and hence m_{2^{α} 3} = m_{2^{α}} = 8tu which implies that 3 | m_{2^{α}} + m_{2^{α} 3} = 16tu, a contradiction.

Case 2. Let α ⩾ 1 and β = 1. Thus m_{i} = m_{2^{α} 3} = 8tu and Lemma 3.1 implies that 2^{α} 3 | ∑ _{d|2^{α} 3} m_{d} and 2^{α−1} 3 | ∑ _{d|2^{α−1} 3} m_{d}. Thus 3 | m_{2^{α}} + m_{2^{α} 3}. According to Case 1, we know that m_{2^{α}} ≠ 8tu and hence, 3 | m_{2^{α}} . Thus 3 | m_{2^{α} 3}, a contradiction.

Case 3. Let α ⩾ 0 and β ⩾ 2. If G_{3} is cyclic of order 3^{k}, then n_{3}(G) = \frac{m_{3^{k}}}{φ(3^{k})} = \frac{m_{3^{k}}}{2.3^{k−1}} . Thus according to nse(G), we have u | n_{3}(G) or t | n_{3}(G) and since n_{3}(G) | |G|, we can get a contradiction. So G_{3} is not cyclic and Lemma 2.2 now yields 9 | m_{2^{α} 3^{β}} , which is a contradiction.

Step 2. In this step, we prove that |G_{3}| = 3^{n} and exp(G_{3}) = 3.

According to step 1, we have

|G| = |G_{2}||G_{3}| = 1 + 3^{n}u + 8tu + k_{1}(t − 1)3^{n}u + k_{2}(u − 1)3^{n}2t.

Since

3^{n} | 1 + 3^{n}u + 8tu + k_{1}(t − 1)3^{n}u + k_{2}(u − 1)3^{n}2t,
3^{n+1} ∤ 1 + 3^{n}u + 8tu + k_{1}(t − 1)3^{n}u + k_{2}(u − 1)3^{n}2t,

we conclude that |G_{3}| = 3^{n}. Let exp(G_{3}) = 3^{k}, where k ⩾ 2. Then by Step 1, for every i ⩾ 0, m_{2^{i} 3^{k}} ≠ 8tu. Lemma 2.3 now implies

|G_{2}| | ∑_{i⩾0} m_{2^{i} 3^{k}} = k′_{1}(t − 1)3^{n}u + k′_{2}(u − 1)3^{n}2t,

where k′_{1}, k′_{2} ⩾ 0 are integers and since 3^{k} ∈ π_{e}(G), we have k′_{1} + k′_{2} ⩾ 1. Thus |G_{2}| | k′_{1}(t − 1)u + k′_{2}(u − 1)2t. On the other hand, |G| = |G_{2}||G_{3}| = |G_{2}|3^{n} = 1 + 3^{n}u + 8tu + k_{1}(t − 1)3^{n}u + k_{2}(u − 1)3^{n}2t, where k_{1}, k_{2} are natural numbers and k′_{1} ⩽ k_{1}, k′_{2} ⩽ k_{2}. Thus |G_{2}| = u + 3^{n} + k_{1}(t − 1)u + k_{2}(u − 1)2t. Therefore u + 3^{n} + k_{1}(t − 1)u + k_{2}(u − 1)2t ⩽ k′_{1}(t − 1)u + k′_{2}(u − 1)2t, which is a contradiction because k′_{1} ⩽ k_{1}, k′_{2} ⩽ k_{2}. So exp(G_{3}) = 3.

Step 3. In this step, we prove that 6 ∈ π_{e}(G).

Let 6 ∉ π_{e}(G). Then G_{2} acts fixed point freely on the set of elements of order 3 by conjugation and hence, |G_{2}| | 8 and since | nse(G)| = 5 and exp(G_{3}) = 3, we conclude that 8 ∈ π_{e}(G). Thus G_{2} is cyclic and n_{2}(G) = m_{8}/4. Now according to nse(G), we conclude that t or u divides n_{2}(G) and since n_{2}(G) | |G|, we can get a contradiction.

Step 4. In this step, we prove that π(G) ≠ {2, 3}.

By steps 1 and 2, we have |G_{3}| = 3^{n} and |G| = |G_{2}||G_{3}| = 1 + 3^{n}u + 8tu + k_{1}(t − 1)3^{n}u + k_{2}(u − 1)3^{n}2t, where k_{1}, k_{2} are natural numbers. Thus |G_{2}| = u+3^{n} +k_{1}(t−1)u+k_{2}(u−1)2t = 1/8p(x), where p(x) = Ax^{2}+Bx+C, A = k_{1}+2k_{2}, B = 12−4k_{1}−4k_{2} and C = −4+3k_{1}−6k_{2}. By Lemma 3.2 and Step 2, exp(G_{2}) ⩽ 2^{5} and exp(G_{3}) = 3. Thus |π_{e}(G)| ⩽ 12, which implies that k_{1} + k_{2} ⩽ 9. According to Lemma 2.3 we know that |G_{2}| | ∑ _{i⩾0} m_{2^{i} 3} = 8tu + k′_{1}(t − 1)3^{n}u + k′_{2}(u − 1)3^{n}2t, where k′_{1}, k′_{2} ⩾ 0 are integers. Thus |G_{2}| | 1/8q(x), where q(x) = A′x^{3} + B′x^{2} + C′x + D′, A′ = k′_{1} + 2k′_{2}, B′ = 8 − 4k′_{1} − 4k′_{2}, C′ = 3k′_{1} − 6k′_{2} and D′ = −8. Now by step 3, we can conclude that k′_{1} + k′_{2} ⩾ 1 and since exp(G_{2}) ⩽ 2^{5}, we have 5 ⩾ k′_{1} + k′_{2}. It is obvious that

p(x) | (AA′x + AB′− A′B)p(x) − A^{2}q(x)
= (AB′B − A′B^{2} − A^{2}C′+ AA′C)x + (AB′C − A′BC − A^{2}D′).

Thus

Ax^{2} + Bx + C ⩽ (AB′B − A′B^{2} − A^{2}C′+ AA′C)x + (AB′C − A′BC − A^{2}D′)

which implies that

Ax^{2} + (B − AB′B + A′B^{2} + A^{2}C′− AA′C)x + (C − AB′C + A′BC + A^{2}D′) ⩽ 0.

But according to k_{1} + k_{2} ⩽ 9 and 1 ⩽ k′_{1} + k′_{2} ⩽ 5, we conclude that x < 201. Thus m = 3 and k_{1} + k_{2} = 8 and |G_{2}| = 2^{10}. Since exp(G_{2}) ⩽ 2^{5} and exp(G_{3}) = 3, we conclude that exp(G_{2}) = 2^{5}. Hence Lemma 3.1 implies that

2^{10} = |G_{2}| | 1 + m_{2} + m_{4} + m_{8} + m_{16} + m_{32}
= 1 + 3^{n}u + k′′_{1}(t − 1)3^{n}u + k′′_{2}(u − 1)3^{n}2t = 1 + 351 + k′′_{1}2106 + k′′_{2}4536,

where k′′_{1}, k′′_{2}⩾ 0 are integers and k′′_{1}+ k′′_{2}= 4, which is a contradiction. □

Lemma 3.5. We have π(G) ⊆ {2, 3, t, u}.

Proof. Let p ∈ π(G) − {2, 3, t, u}. We prove the lemma in the following six steps.

Step 1. In this step, we prove that m_{p} ≠ 8tu and hence (p, t − 1) = 1.

Let m_{p} = 8tu. Then p | (1 + m_{p}) = 3^{2n}, which is a contradiction. So m_{p} ≠ 8tu and hence m_{p} ∈ {(t − 1)3^{n}u, (u − 1)3^{n}2t}. Since (p, m_{p}) = 1, we conclude that (p, t − 1) = 1.

Step 2. In this step, we prove that exp(G_{p}) = p.

Let exp(G_{p}) > p. Then p^{2} ∈ π_{e}(G). Since p(p − 1) = φ(p^{2}) | m_{p^{2}} , we conclude that p divides one of the numbers 2, 3, t, u or t − 1, which is a contradiction. So exp(G_{p}) = p.

Step 3. In this step, we prove that if q ∈ π_{e}(G) and (p, q) = 1, then qp ∈ π_{e}(G) and p | m_{q} + m_{qp}.

Let q ∈ π_{e}(G) which is prime to p. If qp ∉ π_{e}(G), then G_{p} acts fixed point freely on the set of elements of order q by conjugation and hence |G_{p}| | m_{q}, which implies that p divides one of the numbers 2, 3, t, u or (t − 1), which is a contradiction. So qp ∈ π_{e}(G). Let q = q_{1}^{s_{1}} ⋯ q_{k}^{s_{k}} , where q_{1}, …, q_{k} are distinct prime numbers and k, s_{1}, …, s_{k} are natural numbers. We prove p | m_{q} + m_{qp} by induction on s = s_{1}+⋯ +s_{k}. Let s = 1. Then we have p | 1+m_{p}+m_{q_{i}} +m_{q_{i} p} and since p | 1+m_{p}, p | m_{q_{i}} + m_{q_{i} p}. Let s = 2. Then there exist 1 ⩽ i < j ⩽ k such that q = q_{i}q_{j} or q = q_{i}^{2}. If q = q_{i}q_{j}, then we have p | 1+m_{p}+m_{q_{i}} +m_{q_{j}} +m_{q_{i} p}+m_{q_{j} p}+m_{q_{i} q_{j}} +m_{q_{i} q_{j} p} and since p | 1 + m_{p}, m_{q_{i}} + m_{q_{i} p}, m_{q_{j}} + m_{q_{j} p}, we conclude that p | m_{q_{i} q_{j}} + m_{q_{i} q_{j} p}, as desired. The case q = q_{i}^{2} is similar and we omit the details for the sake of convenience. Now, assume the statement is true for the values less than s. We have

p | ∑_{d|qp} m_{d} = ∑_{\begin{matrix}d|qp \\ d≠q,qp \end{matrix}} m_{d} + m_{q} + m_{qp}.

Moreover, according to the induction hypothesis, p | ∑ _{d|qp, d≠q,qp} m_{d}. Therefore, p | m_{q} + m_{qp}.

Step 4. m_{p} ≠ (t − 1)3^{n}u.

Let m_{p} = (t − 1)3^{n}u. Then p | 1 + m_{p} and hence p | x^{2} − 5x + 8. On the other hand, by nse(G), there is q ∈ π_{e}(G) such that (q, p) = 1 and m_{q} or m_{qp} = (u − 1)3^{n}2t. Thus by step 3, p | m_{q} + m_{qp}. Now, there are four cases:

Case 1. If {m_{q}, m_{qp}} = {(u − 1)3^{n}2t, 3^{n}u}, then p | x^{2} − 5, which is a contradiction by Lemma 2.7(1).

Case 2. If {m_{q}, m_{qp}} = {(u − 1)3^{n}2t, 8tu}, then p | x^{2} + x − 4, which is a contradiction by Lemma 2.7(1).

Case 3. If {m_{q}, m_{qp}} = {(u − 1)3^{n}2t, (t − 1)3^{n}u}, then p | 3x + 1, which is a contradiction by Lemma 2.7(1).

Case 4. If {m_{q}, m_{qp}} = {(u − 1)3^{n}2t}, then p | (u − 1)3^{n}4t, which is a contradiction according to Step 1 and Lemma 2.6(2).

Step 5. m_{p} ≠ (u − 1)3^{n}2t.

Let m_{p} = (u − 1)3^{n}2t. Then p | 1 + m_{p} and hence p | x^{2} − x − 4. On the other hand, by nse(G), there is q ∈ π_{e}(G) such that (q, p) = 1 and m_{q} or m_{qp} = (t−1)3^{n}u. Thus by Step 3, p | m_{q} + m_{qp}. Now there are four cases:

Case 1. If {m_{q}, m_{qp}} = {(t − 1)3^{n}u, 3^{n}u}, then p | 3^{n}tu, which is a contradiction.

Case 2. If {m_{q}, m_{qp}} = {(t − 1)3^{n}u, 8tu}, then p | x^{2} + 5x + 8, which is a contradiction by Lemma 2.7(2).

Case 3. If {m_{q}, m_{qp}} = {(t − 1)3^{n}u}, then p | 2(t − 1)3^{n}u, which is a contradiction according to Step 1.

Case 4. If {m_{q}, m_{qp}} = {(t − 1)3^{n}u, (u − 1)3^{n}2t}, then p | 3x + 1, which is a contradiction by Lemma 2.7(2).

Step 6. p ∉ π(G). According to steps 1, 4 and 5, m_{p} ∉ nse(G) and hence, p ∉ π(G). □

Lemma 3.6. If t ∈ π(G), then u ∈ π(G).

Proof. If t ∈ π(G), then according to Lemma 3.1, we have m_{t} is even and (m_{t}, t) = 1 and hence, according to nse(G), it is obvious that m_{t} = (t − 1)3^{n}u. Now we claim that t^{2} ∉ π_{e}(G). Suppose, contraty to our claim, that t^{2} ∈ π_{e}(G). Lemma 3.1 implies that t(t−1) = φ(t^{2}) | m_{t^{2}} and hence according to Lemma 2.6, we have m_{t^{2}} = (u−1)3^{n}2t. On the other hand, Lemma 3.1 implies that t^{2} | 1+m_{t}+m_{t^{2}} and hence (x+1)^{2} | (x+1)^{2}(6x−28)+44(x+1). Thus (x+1) | 44. So we conclude that t = 11, which is a contradiction because 4t = 3^{n} + 1. Therefore, t^{2} ∉ π_{e}(G). Now we are going to show that |G_{t}| = t. Since 2 ∈ π_{e}(G), we can assume that q is the largest element of π_{e}(G) satisfies (q, t) = 1. Thus {q} ⊆ {s ∈ π_{e}(G) : s is multiple of q} ⊆ {q, qt}. Now Lemma 2.3 implies that |G_{t}| | m_{q} or m_{q} + m_{qt}. Hence by nse(G) and Lemma 2.6, we conclude that |G_{t}| = t. So n_{t}(G) = \frac{m_{t}}{φ(t)} = 3^{n}u and since n_{t}(G) | |G|, we have u ∈ π(G). □

Lemma 3.7. We have π(G) = {2, 3, t, u}.

Proof. We first show that u^{2} ∉ π_{e}(G). If u^{2} ∈ π_{e}(G), then by Lemma 3.1, u(u − 1) = φ(u^{2}) | m_{u^{2}} . But according to Lemma 2.6 and nse(G), we can easily see that there is no choice for m_{u^{2}} . Therefore u^{2} ∉ π_{e}(G).

If u ∈ π(G), then according to nse(G) and Lemma 3.1, we can easily conclude that m_{u} = (u − 1)3^{n}2t. Since u^{2} ∉ π_{e}(G), Lemma 2.1 implies that |G_{u}| | 1 + m_{u}. If u^{2} | 1 + m_{u}, then (x − 1)^{2} | (x − 1)^{2}x − 4(x − 1) and hence (x − 1) | 4, which is a contradiction. Thus |G_{u}| = u and we have n_{u}(G) = \frac{m_{u}}{φ(u)} = 3^{n}2t. Now according to Lemmas 3.2-3.6, we have u ∈ π(G) and since n_{u}(G) = 3^{n}2t and n_{u}(G) | |G|, we conclude that {2, 3, t, u} ⊆ π(G). On the other hand, by Lemma 3.5, there is no p ∈ π(G) such that p ≠ 2, 3, t, u. Therefore, π(G) = {2, 3, t, u}. □

Lemma 3.8. We have 2u, 3u, tu, 3t, 4t ∉ π_{e}(G) but 2t ∈ π_{e}(G) and m_{2t} = m_{t}. Moreover if 6 ∈ π_{e}(G), then m_{6} ≠ 8tu

Proof. If 2u ∈ π_{e}(G), then m_{2u} = φ(2u)n_{u}(G)k, where k is the number of cyclic subgroups of order 2 in C_{G}(G_{u}). Actually, this follows from the fact that all centralizers of Sylow u-subgroups of G in G are conjugate in G. Therefore we have (u−1)3^{n}2t | m_{2u}. Thus m_{2u} = m_{u}. On the other hand, 2u | 1+m_{2}+m_{u} +m_{2u} and u | 1+m_{2}+m_{u} which implies that u | m_{2u}, a contradiction. Similarly, we can prove that 3u, tu, 3t, 4t ∉ π_{e}(G). If 2t ∉ π_{e}(G), then G_{t} acts fixed point freely on the set of elements of order 2 by conjugation and hence, |G_{t}| | m_{2}, which is a contradiction. So 2t ∈ π_{e}(G). Similarly, we can prove that m_{2t} = m_{t}. Let 6 ∈ π_{e}(G) such that m_{6} = 8tu. Since 3 | 1 + m_{2} + m_{3} and 6 | 1 + m_{2} + m_{3} + m_{6}, we conclude that 3 | m_{6}, which is a contradiction. □

Lemma 3.9. We have exp(G_{3}) = 3.

Proof. If exp(G_{3}) > 3, then 9 ∈ π_{e}(G). Since 3t, 3u ∉ π_{e}(G), we conclude that 9t, 9u ∉ π_{e}(G). Thus G_{t} and G_{u} act fixed point freely on the set of elements of order 9 by conjugation and hence, |G_{t}| | m_{9} and |G_{u}| | m_{9}. So we have tu | m_{9}, which implies that m_{9} = 8tu, a contradiction, because 6 = φ(9) | m_{9}. □

Lemma 3.10. We have exp(G_{2}) = 2.

Proof. If exp(G_{2}) > 2, then 4 ∈ π_{e}(G). Thus Lemma 3.8 implies that 4t, 4u ∉ π_{e}(G). Thus G_{t} and G_{u} act fixed point freely on the set of elements of order 4 by conjugation and hence, |G_{t}| | m_{4} and |G_{u}| | m_{4}. So tu | m_{4}, which implies that m_{4} = 8tu. On the other hand, 12 ∈ π_{e}(G) because otherwise G_{3} acts fixed point freely on the set of elements of order 4 by conjugation and hence, 3 | m_{4}, a contradiction. So 12 ∈ π_{e}(G). Now since 6 | 1 + m_{2} + m_{3} + m_{6} and 12 | 1 + m_{2} + m_{3} + m_{4} + m_{6} + m_{12}, we conclude that 3 | m_{4} + m_{12}. Thus 3 ∤ m_{12}, which implies that m_{12} = 8tu. But we have 3 | m_{4} + m_{12} = 16tu, a contradiction. □

Lemma 3.11. We have |G| = 1 + 3^{n}u + 8tu + k_{1}(t − 1)3^{n}u + k_{2}(u − 1)3^{n}2t,

where k_{1}, k_{2} are natural and 3 ⩽ k_{1} + k_{2} ⩽ 4.

Proof. According to Lemmas 3.8–3.10, we have {1, 2, 3, t, 2t, u} ⊆ π_{e}(G) ⊆ {1, 2, 3, 6, t, 2t, u} and m_{1} = 1, m_{2} = 3^{n}u, m_{3} = 8tu, m_{t} = m_{2t} = (t − 1)3^{n}u, m_{u} = (u − 1)3^{n}2t and m_{6} ∈ {(t − 1)3^{n}u, (u − 1)3^{n}2t}. So we conclude that |G| = 1 + 3^{n}u + 8tu + k_{1}(t − 1)3^{n}u + k_{2}(u − 1)3^{n}2t, where k_{1}, k_{2} are natural numbers and 3 ⩽ k_{1} + k_{2} ⩽ 4. □

Lemma 3.12. We have |G_{3}| = 3^{n}.
Proof. By Lemma 3.11, |G| = 1+3^{n}u+8tu+k_{1}(t−1)3^{n}u+k_{2}(u−1)3^{n}2t, where

k_{1}, k_{2} are natural. Now since 3^{n} | 1 + 3^{n}u + 8tu + k_{1}(t − 1)3^{n}u + k_{2}(u − 1)3^{n}2t and 3^{n+1} ∤ 1+3^{n}u+8tu+k_{1}(t−1)3^{n}u+k_{2}(u−1)3^{n}2t, we conclude that |G_{3}| = 3^{n}. □

Lemma 3.13. We have |G_{2}| = 4.

Proof. If |G_{2}| = 2, then G_{2} is cyclic and n_{2}(G) = 3^{n}u. Since |G_{3}| = 3^{n} and n_{2}(G) = |G : N_{G}(G_{2})|, we conclude that 3 ∤ |N_{G}(G_{2})|. Thus 6 ∉ π_{e}(G) and by Lemma 3.11, we have |G| = 1 + 3^{n}u + 8tu + 2(t − 1)3^{n}u + (u − 1)3^{n}2t. Lemma 2.6 now yields 4 | |G|, which is a contradiction. Therefore, |G_{2}| ⩾ 4. If |G_{2}| ⩾ 8, then by Lemmas 2.1 and 3.10, 8 | 1 + m_{2}. Thus 16 | x^{2} − x + 2. On the other hand, by Lemma 3.8, t, 2t are only elements of π_{e}(G) which are multiple, of t and m_{t} = m_{2t}.

Thus Lemma 2.3 implies that |G_{2}||G_{3}||G_{u}| | m_{t} + m_{2t} = 2(t − 1)3^{n}u. Therefore 4 | (t − 1). So 16 | (x − 3), which is a contradiction according to Lemma 2.7. □

Lemma 3.14. We have G ≅ L_{2}(3^{n}).

Proof. Since |G_{u}| = u, |G_{t}| = t, |G_{3}| = 3^{n} and |G_{2}| = 4, we conclude that |G| = |L_{2}(3^{n})| = 2^{2}3^{n}tu. Now according to [9] we have G ≅ L_{2}(3^{n}). □

Acknowledgments. The authors express their thanks to the referees for their careful reading, helpful comments and valuable suggestions for the improvement of the paper. Partial support by the Center of Excellence of Algebraic Hyperstructures and its Applications of Tarbiat Modares University (CEAHA) is gratefully acknowledged by the second author (AI).

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Department of Mathematics (Received 24 05 2014) Science and Research Branch, Islamic Azad University Tehran, Iran

h.parvizi.mosaed@gmail.com
tehranian@srbiau.ac.ir

Department of Mathematics Tarbiat Modares University Tehran, Iran (Corresponding author)

iranmana@modares.ac.ir