Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 47 (2006), No. 2, 463-479.

The Number of Finite Groups Whose Element Orders is Given

Ali Reza Moghaddamfar Wujie Shi

Department of Mathematics, Faculty of Science K. N. Toosi University of Technology P. O. Box 16315-1618, Tehran, Iran e-mail: moghadam@iust.ac.ir

School of Mathematics, Soochow University Suzhou 215006, People’s Republic of China e-mail: wjshi@suda.edu.cn

Abstract. The spectrum ω(G) of a finite group G is the set of element orders of G. If Ω is a non-empty subset of the set of natural numbers, h(Ω) stands for the number of isomorphism classes of finite groups G with ω(G) = Ω and put h(G) = h(ω(G)). We say that G is recognizable (by spectrum ω(G)) if h(G) = 1. The group G is almost recognizable (resp. nonrecognizable) if 1 < h(G) < ∞ (resp. h(G) = ∞). In the present paper, we focus our attention on the projective general linear groups PGL(2, p^{n}), where p = 2^{α}3^{β} + 1 is a prime, α ≥ 0, β ≥ 0 and n ≥ 1, and we show that these groups cannot be almost recognizable, in other words h(PGL(2, p^{n})) ∈ {1, ∞}. It is also shown that the projective general linear groups PGL(2, 7) and PGL(2, 9) are nonrecognizable. In this paper a computer program has also been presented in order to find out the primitive prime divisors of a^{n} − 1. MSC 2000: 20D05 Keywords: spectrum, prime graph, projective general linear group

1. Introduction

Throughout the paper, all the groups under consideration are finite and simple groups are non-Abelian. For a group G, we denote the set of orders of all elements

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2006 Heldermann Verlag

464 W. Shi et al.: The Number of Finite Groups Whose Element Orders is Given

in G by ω(G) which has been recently called the spectrum of G. Obviously ω(G) is a subset of the set ℕ of natural numbers, and it is closed and partially ordered by divisibility, hence, it is uniquely determined by μ(G), the subset of its maximal elements.

One of the most interesting concepts in finite group theory which has recently attracted several researchers is the problem of characterizing finite groups by element orders. Let Ω be a non-empty subset of ℕ. Now, we can put forward the following questions: Is there any group G with ω(G) = Ω? If the answer is affirmative then how many non-isomorphic groups exist with the above set of element orders? Certainly, if there exists such a group, Ω must contain 1 and furthermore Ω must be closed and partially ordered under the divisibility relation. These conditions are necessary but not sufficient, for example if Ω = {1, 2, 3, 4, 5, 6, 7, 8, 9}, then there does not exist any group G with ω(G) = Ω. In fact, R. Brandl and W. J. Shi in [1] have classified all groups whose element orders are consecutive integers and in that paper they have shown that if ω(G) = {1, 2, 3, …, n}, for some group G, then n ≤ 8.

For a set Ω of natural numbers, define h(Ω) to be the number of isomorphism classes of groups G such that ω(G) = Ω, and put h(G) = h(ω(G)). Evidently, h(G) ≥ 1. Now we give a “new classification” for groups using the h function. A group G is called recognizable (resp. almost recognizable or nonrecognizable) if h(G) = 1 (resp. 1 < h(G) < ∞ or h(G) = ∞). Some list of simple groups that are presently known to be recognizable, almost recognizable or nonrecognizable is given in [13]. In particular, it was previously known that the projective general linear groups PGL(2, 2^{n}) with n ≥ 2 are recognizable and PGL(2, 2) ≅ S_{3} is nonrecognizable (see [16], Theorem 2). In [12], V. D. Mazurov proved the following result: Let P be a field which is the union of an ascending series of finite fields of orders 2^{m_{i}}, m_{i} > 1, i ∈ ℕ. If there exists a natural number s such that 2^{s} does not divide m_{i} for any i ∈ ℕ then h(PGL(2, P)) = 1. In all other cases h(PGL(2, P)) = ∞. Also he proved the following result in [11]: If p, r are odd primes, p−1 is divisible by r but not by r^{2}, and s is a natural number non-divisible by r, then h(PGL(r, p^{s})) = ∞.

Let q = p^{n} where p is a prime. In this paper, we focus our attention on the projective general linear groups PGL(2, q). The structure of Aut(L_{2}(q)) is well known, it is isomorphic to the semidirect product of PGL(2, q) by a cyclic group of order n. On the other hand we know μ(L_{2}(q)) = \{\frac{q−1}{ϵ}, p, \frac{q+1}{ϵ}\}, ϵ = (2, q − 1), and μ(PGL(2, q)) = {q − 1, p, q + 1}.

A group G is called C_{pp}-group if p is a prime divisor of |G| and the centralizer of any non-trivial p-element in G is a p-group. Evidently, the projective general linear groups PGL(2, q) where q = p^{n}, are C_{pp}-groups. In [3], the second author has classified the simple C_{pp}-groups, where p is prime and p = 2^{α}3^{β} + 1, α ≥ 0, β ≥ 0 (see Lemma 8 and Table 1). Using these results, we prove the following theorem.

Theorem 1. Let p = 2^{α}3^{β} + 1 (α ≥ 0, β ≥ 0) be a prime. Then the projective general linear groups PGL(2, p^{n}) cannot be almost recognizable. In other words, h(PGL(2, p^{n})) ∈ {1, ∞}.

In 1994, R. Brandl and W. J. Shi in [2] showed that all projective special linear groups L_{2}(q) with q ≠ 9 are recognizable and L_{2}(9) is nonrecognizable.

Here, we similarly prove that:

Theorem 2. The projective general linear group PGL(2, 9) is nonrecognizable.

For us it was interesting to face with some groups G such that μ(G) contain three consecutive natural numbers in the form {p − 1, p, p + 1} where p ≥ 5 is a prime. Such sets appear for almost simple groups PGL(2, p), where p ≥ 5 is a prime, in fact we proved in [14] that h(PGL(2, p)) ∈ {1, ∞}. For ∞ we have found an example. It has been proved in [1] that PGL(2, 5) ≅ S_{5} has ∞ for its h function. Here we also give another example of groups of type PGL(2, p) with value ∞ for its h function.

Theorem 3. There exists an extension G of a 7-group by 2.S_{4} such that μ(G) = μ(PGL(2, 7)) = {6, 7, 8}. In particular, the projective general linear group PGL (2, 7) is nonrecognizable.

Notation. Our notation and terminology are standard (see [4]). Given a group G, denote by π(G) the set of all prime divisors of the order of G. If m and n are natural numbers and p is a prime, then we let π(n) be the set of all primes dividing n, and r_{[n]} the largest prime not exceeding n. Note that π(G) = π(|G|). The notation p^{m} ∥ n means that p^{m}|n and p^{m+1} ∤ n. The expression G = K : C denotes the split extension of a normal subgroup K of G by a complement C.

2. Some preliminary results

First, we collect some results from Elementary Number Theory which will be useful tools for our further investigations in this paper. We start with a famous theorem due to Zsigmondy.

Zsigmondy’s Theorem (see [19]). Let a and n be integers greater than 1. Then there exists a primitive prime divisor of a^{n} − 1, that is a prime s dividing a^{n} − 1 and not dividing a^{i} − 1 for 1 ≤ i ≤ n − 1, except if

(1) a = 2 and n = 6, or

(2) a is a Merssene prime and n = 2.

We denote by a_{n} one of these primitive prime divisors of a^{n} − 1. Evidently, if a_{n} is a primitive prime divisor of a^{n} − 1, then a has order n modulo a_{n} and so a_{n} ≡ 1 (mod n). Thus a_{n} ≥ n + 1.

The next elementary result will be needed later.

Lemma 1. Let p and q be two primes and m be a natural number, where p, q and m satisfying one of the following conditions. Then, for every n ≥ m, there exists a primitive prime divisor p_{n} > q.

(1) p = 7, m = 5 and q = 13,
(2) p = 13, m = 5 and q = 19,
(3) p = 17, m = 4 and q = 19,
(4) p = 19, m = 7 and q = 37,
(5) p = 37, m = 7 and q = 109,
(6) p = 73, m = 5 and q = 127.

Proof. In all cases, if n ≤ q, the result is straightforward. Therefore, we may assume that n > q. Since

π(q!) ⊆ π(p ∏_{i=1}^{q−1}(p^{i} − 1)) ⊂ π(p ∏_{i=1}^{n}(p^{i} − 1)),

by Zsigmondy’s theorem we deduce that there exists a primitive prime divisor p_{n} > q, completing the proof. □

Function for finding the primitive prime divisors. In the following we submit a GAP program [5], which determines all the primitive prime divisors in the sequence a^{i} − 1 (i = 1, 2, …, n) for some a and n.

gap> PPD:=function(a,n)
    local b,i,j,s1,s2,s;
     for i in [1..n] do
     s1:=Set(Factors(a^{∧}i-1));
    s2:=[];
        for j in [1..(i-1)] do
           b:=Set(Factors(a^{∧}j-1));
           Append(s2,b);
        od;
        s:=Difference(s1,s2);
        Print(i," ",s,"\n");
    od;
end;

Using this programme we list all primitive prime divisors p_{n} for p = 7, 13, 17 and 2 ≤ n ≤ 19, in Table 1. Using Table 1, the reader can easily check the proof of Lemma 1 (1)–(3) for n ≤ q.

Lemma 2. Let p and q be two primes and m, n be natural numbers such that p^{m} = q^{n} + 1. Then one of the following holds:

(1) n = 1, m is a prime number, p = 2 and q = 2^{m} − 1 is a Mersenne prime;

(2) m = 1, n is a power of 2, q = 2 and p = 2^{n} + 1 is a Fermat prime;

(3) p = n = 3 and q = m = 2.

Proof. Well known exercise using the Zsigmondy’s theorem. □

The set ω(G) defines the prime graph GK(G) of G whose vertex-set is π(G) and two primes p and q in π(G) are adjacent (we write p ∼ q) if and only if pq ∈ ω(G).

n 7_{n}  13_{n}  17_{n}
2 –  7  3
3 19  61  307
4 5  5, 17  5, 29
5 2801  30941  88741
6 43  157  7, 13
7 29, 4733  5229043  25646167
8 1201  14281  41761
9 37, 1063  1609669  19, 1270657
10 11, 191  11, 2411  11, 71, 101
11 1123,  23, 419,  2141993519227
293459  859, 18041
12 13, 181 28393 83233
13 16148168401 53, 264031  212057,
1803647  2919196853
14 113, 911  29, 22079  22796593
15 31, 159871  4651, 161971  6566760001
16 17, 169553  407865361  18913,

184417

17 14009 103, 443, 10949,
2767631689  15798461357509 1749233,

2699538733

18 117307 19, 271, 1423,
937  5653
19 419 12865927, 229, 1103,
4534166740403 9468940004449 202607147,

291973723

Table 1. The primitive prime divisors p_{n} where p ∈ {7, 13, 17} and 2 ≤ n ≤ 19.

The number of connected components of GK(G) is denoted by t(G), and the connected components are denoted by π_{i} = π_{i}(G), i = 1, 2, …, t(G). If 2 ∈ π(G) we always assume 2 ∈ π_{1}. Denote by μ_{i}(G) the set of all n ∈ μ(G) such that π(n) ⊆ π_{i}.

The Gruenberg-Kegel Theorem (see [18]). If G is a group with disconnected graph GK(G) then one of the following holds:

(1) t(G) = 2, G is Frobenius or 2-Frobenius.

(2) G is an extension of a π_{1}(G)-group N by a group G_{1}, where S ≤ G_{1} ≤ Aut(S), S is a simple group and G_{1}/S is a π_{1}(G)-group. Moreover t(S) ≥ t(G) and for every i, 2 ≤ i ≤ t(G), there exists j, 2 ≤ j ≤ t(S) such that μ_{j}(S) = μ_{i}(G).

Lemma 3. Let S be a simple group with disconnected prime graph GK(S). Then |μ_{i}(S)| = 1 for 2 ≤ i ≤ t(S). Let n_{i}(S) be a unique element of μ_{i}(S) for i ≥ 2. Then value for S, π_{1}(S) and n_{i}(S) for 2 ≤ i ≤ t(S) are the same as in Tables 2a–2c of [13].

Proof. The simple groups S and the sets of π_{i}(S) are described in [18] and [7]; the rest is proved in Lemma 4 of [8]. The values of the numbers n_{i}(S), i ≥ 2 are listed in Table 2a–2c of [13]. □

We also use the following lemma (see [11], Lemma 1).

Lemma 4. If a group G contains a soluble minimal normal subgroup then G is nonrecognizable. In particular, if G is a soluble group then G is nonrecognizable.

The following result of V. D. Mazurov will be used several times.

Lemma 5. (see [10]) Let G be a group, N a normal subgroup of G, and G/N a Frobenius group with Frobenius kernel F and cyclic complement C. If (|F |, |N|) = 1 and F is not contained in NC_{G}(N)/N, then p|C| ∈ ω(G) for some prime divisor p of |N|.

The following lemma is taken from [16], Theorem 2.

Lemma 6. Let G be a group such that

μ(G) = μ(PGL(2, 2^{n})) = {2^{n} − 1, 2, 2^{n} + 1}.

Then, the following statements hold:

(1) If n ≥ 2, then G ≅ PGL(2, 2^{n}).

(2) If n = 1, then G ≅ S_{3} has ∞ for its h function.

We are now ready to prove the following lemma.

Lemma 7. Let G be a group such that

μ(G) = μ(PGL(2, p^{n})) = {p^{n} − 1, p, p^{n} + 1},

where p is an odd prime, n ≥ 2. Then, the following statements hold:

(1) If (p, n) ≠ (3, 2), then item (2) of the Gruenberg-Kegel theorem holds. Moreover, S is isomorphic to none of the following simple groups:

(a) alternating groups on n ≥ 5 letters,

(b) sporadic simple groups,

(c) L_{2}(p^{k}) where k ≠ n, or

(d) L_{2}(2p^{m} ± 1), m ≥ 1, where 2p^{m} ± 1 is a prime.

(2) If (p, n) = (3, 2), then there exists a soluble group G such that μ(G) = μ(PGL(2, 3^{2})).

Proof. (1) First of all, we show that G is insoluble. Assume the contrary. If π(p^{n}− 1) = {2}, then by Lemma 2 we obtain (p, n) = (3, 2) which is a contradiction. Hence, there exists a prime 2 ≠ r ∈ π(p^{n} −1). On the other hand, we consider the primitive prime divisor s = p_{2n}. Now assume that H is a {p, r, s}-Hall subgroup of G. Since G has no elements of order pr, ps and rs, it follows that H is a soluble group all of whose elements are of prime power orders. By [6], Theorem 1, we must have |π(H)| ≤ 2, which is a contradiction.

Since t(G) = 2, G satisfies the conditions of the Gruenberg-Kegel theorem. Now we show that G is neither Frobenius nor 2-Frobenius. Evidently, G can not be a 2-Frobenius group, because G is insoluble. Suppose G = KC is a Frobenius group with kernel K and complement C. Clearly C is insoluble, π(C) = π_{1}(G) = π(p^{2n} − 1), π(K) = π_{2}(G) = {p} and by [15], Theorem 18.6 C has a normal subgroup C_{0} of index ≤ 2 such that C_{0} ≅ SL(2, 5) × Z, where every Sylow subgroup of Z is cyclic and π(Z) ∩ π(30) = ∅. Therefore GK(C) can be obtained from the complete graph on π(C) by deleting the edge {3, 5}. On the other hand, if there exist primes 2 ≠ r ∈ π(p^{n} − 1) and 2 ≠ s ∈ π(p^{n} + 1), then since rs ∉ ω(G) it follows that rs ∉ ω(C). Hence, we must have Z = 1 and π(p^{2n} −1) = π(SL_{2}(5)) = {2, 3, 5} and since {2, 3, 5} ⊂ π(p^{4} −1), by Zsigmondy’s theorem we obtain that n = 2. Now, it is easy to see that π(p^{2} − 1) = {2, 3} and π(p^{2} + 1) = {2, 5}. From π(p^{2} − 1) = {2, 3}, we infer that p is a Mersenne prime or a Fermat prime. In the first case we obtain p = 7, and in the latter case p = 17. If p = 17, then 29 ∈ π(p^{2} + 1), a contradiction. If p = 7, then C contains an element of order 16, which is a contradiction.

Therefore, by the Gruenberg-Kegel theorem, G is an extension of a π_{1}(G)- group N by a group G_{1}, where S ≤ G_{1} ≤ Aut(S), S is a simple group and G_{1}/S is a π_{1}(G)-group. Now, we show that S is not isomorphic to an alternating group, a sporadic simple group, a linear group L_{2}(p^{k}) where k ≠ n or L_{2}(2p^{m}±1), m ≥ 1, where 2p^{m} ± 1 is a prime.

Before beginning we recall that in the prime graph of G the connected component π_{1}(G) consists of the primes in π(p^{n} − 1) which form a complete subsection and also the primes in π(p^{n} + 1) which forms another complete subsection. Moreover, every odd vertex in π(p^{n} − 1) is not joined to any odd vertex in π(p^{n} + 1).

(a) Assume that S ≅ A_{m}, m ≥ 5. By Lemma 3, m = p, p + 1, p + 2. Suppose S ≅ A_{p}, p ≥ 5. We have that in the prime graph GK(A_{p}) the vertex 3 is joined to 2, 5, 7, …, r_{[p−3]}. If 3 divides p − 1, then by the remark mentioned in the previous section, we conclude that 2, 3, 5, …, r_{[p−3]} belong to π(p^{n} − 1). Now, if there exists a prime s ∈ π(p^{n} + 1)\π(A_{p}) then s ∈ π(N), because A_{p} ≅ S ≤ G/N ≤ Aut(S) ≅ S_{p}. On the other hand, A_{4} = 2^{2} : 3 ≤ A_{p} and by Lemma 5 it follows that s ∼ 3 which is a contradiction. Hence, π(p^{n} + 1) ⊆ π(A_{p}). As (p^{n} − 1, p^{n} + 1) = 2 and 2, 3, 5, 7, …, r_{[p−3]} ∈ π(p^{n} − 1), the only possible cases are: π(p^{n} + 1) = {2} or π(p^{n} + 1) = {2, p − 2} in which in the latter case p − 2 is a prime. Evidently, the first case will never occur. So, we consider the case π(p^{n} + 1) = {2, p − 2}, i.e., p^{n} + 1 = 2^{l}(p − 2)^{k}. Now, if k > 1 then since (p − 2)^{k} ∈ ω(G) and (p − 2)^{k} ∉ ω(Aut(S)) = ω(S_{p}) we obtain (p − 2) ∈ π(N) and again since A_{p} contains a Frobenius subgroup of shape 2^{2} : 3 by Lemma 5, we get p − 2 ∼ 3 which is a contradiction. Finally, we have k = 1 and l > 1. Moreover 2 ∥ p^{n} − 1 which implies that n must be odd. But in this case we have p^{n} + 1 = (p + 1)(p^{n−1} − p^{n−2} + ⋯ − p + 1) = 2^{l}(p − 2) for which it follows that p^{n−1} − p^{n−2} + ⋯ − p + 1 = p − 2, giving no solution for p ≥ 5. This final contradiction shows that S ≅ A̸_{p}. The case when 3 divides p + 1, is similar. The other cases are settled similarly.

(b) Suppose S is isomorphic to one of the sporadic simple groups, for instance S ≅ J_{2}. Since p ∈ π_{2}(G), by Lemma 3 it follows that p = 7. If n ≥ 5, then we choose the primitive prime divisors 7_{n}, 7_{2n} in π(G). Evidently, 7_{2n} ∈ π(p^{n} + 1), and so G does not contain an element of order 7_{n}.7_{2n}. On the other hand since π(Aut(S)) = {2, 3, 5, 7} ⊂ π(7 ∏_{i=1}^{4}(7^{i} − 1)), it follows that 7_{n}, 7_{2n} ∉ π(Aut(S)). Therefore 7_{n}, 7_{2n} ∈ π(N), and since N is nilpotent we obtain that 7_{n}.7_{2n} ∈ ω(N), which is a contradiction. Thus n ≤ 4. If n = 4, then μ(G) = {2^{5}.3.5^{2}, 7, 2.1201}. Because, there does not exist any element of order 1201 in Aut(S), 1201 divides the order of N. Without loss of generality we may assume that N ≠ 1 is an elementary Abelian 1201-group. Now since S contains the Frobenius group A_{4} = 2^{2} : 3, from Lemma 5 we infer that G contains an element of order 1201.3, which is a contradiction. If n = 2 or 3, then 5 ∈ π(S)\π(G), which is impossible.

The other sporadic simple groups are examined similarly.

(c) Assume that S ≅ L_{2}(p^{k}), where k ≠ n. In this case we must have k < n, since otherwise by Zsigmondy’s Theorem we get p_{2k} ∈ π(S)\π(G), which is a contradiction. Now, we choose the primitive prime divisors p_{n} and p_{2n} in π(G). Since p_{2n} ∈ π(p^{n} +1), G does not contain an element of order p_{n}.p_{2n}. On the other hand, since p_{n} > n > k we have p_{n}, p_{2n} ∉ π(Aut(S)) = π(PGL(2, p^{k})⋊Z_{k}), and so p_{n}, p_{2n} ∈ π(N). Now, since N is nilpotent we obtain that p_{n}.p_{2n} ∈ ω(N) ⊂ ω(G), which is a contradiction.

(d) Suppose that S ≅ L_{2}(2p^{m} ± 1), m ≥ 1, where 2p^{m} ± 1 is a prime. By the structure of μ(G), we see that p ∈ ω(G) and p^{2} ∉ ω(G). So, if S ≅ L_{2}(2p^{m} ± 1), where 2p^{m} ± 1 is a prime and m ≥ 1, then we deduce m = 1, because in this case p^{m} ∈ ω(L_{2}(2p^{m} ± 1)) = ω(S) ⊆ ω(G). On the other hand, we know |Aut(S)| = 2^{2}p(p ± 1)(2p ± 1), where 2p ± 1 is a prime, and so π(Aut(S)) = {p, 2p ± 1} ∪ π(p ± 1). If n = 2, then (2p ± 1, |G|) = 1, which is a contradiction. Therefore n ≥ 3. Now, we consider the primitive prime divisors p_{n} and p_{2n}. Since (p_{n}, p_{2n}) = (p_{n}, p ± 1) = (p_{2n}, p ± 1) = 1, it follows that p_{n} ∉ π(Aut(S)) or p_{2n} ∉ π(Aut(S)), thus we may assume N is a p_{n}-subgroup or a p_{2n}-subgroup. First, we assume that 2p + 1 is a prime. Let P be a Sylow (2p + 1)-subgroup of S, then N_{S}(P), the normalizer of P in S, is a Frobenius group of order (2p + 1)p, with cyclic complement of order p. Now, by Lemma 5, we deduce that p_{n} ∼ p or p_{2n} ∼ p, which is a contradiction. Next, we assume that 2p − 1 is a prime. In this case, if there exists a prime s ∈ π(p^{n} + 1)\π(Aut(S)) then s ∈ π(N), because G/N ≤ Aut(S). Moreover, if Q is a Sylow (2p − 1)-subgroup of S, then N_{S}(Q) is a Frobenius group of order (2p − 1)(p − 1), with cyclic complement of order p − 1. Now, as previous case we get s.(p − 1) ∈ ω(G), which is a contradiction. Hence, π(p^{n} + 1) ⊆ π(Aut(S)). As (p^{n} − 1, p^{n} + 1) = 2, the only possible case is π(p^{n} + 1) = {2, 2p − 1}, i.e., p^{n} + 1 = 2^{l}(2p − 1)^{k} for some l and k in ℕ.

Now, if k > 1 then since (2p − 1)^{k} ∈ ω(G) and (2p − 1)^{k} ∉ ω(Aut(S)) we obtain (2p − 1) ∈ π(N). On the other hand, it is easy to see that p_{n} ∈ π(G)\π(Aut(S)), and so p_{n} ∈ π(N). Since N is nilpotent, we deduce that p_{n} ∼ (2p − 1), which is a contradiction. Finally, we have k = 1 and since (p, n) ≠ (3, 2), we obtain that l > 1. Moreover 2 ∥ p^{n} − 1 which implies that n must be odd. But in this case we have p^{n} + 1 = (p + 1)(p^{n−1} − p^{n−2} + ⋯ − p + 1) = 2^{l}(2p − 1) for which it follows that p^{n−1} − p^{n−2} + ⋯ − p + 1 = 2p − 1, giving no solution for p ≥ 3. This final contradiction shows that S ≅ L̸_{2}(2p^{m} ± 1).

(2) Consider the group H = ⟨a, b|a^{8} = b^{5} = 1, ba = ab^{2}⟩ ≅ Z_{5} : Z_{8}. For this group we have μ(H) = {8, 10}. Now, we assume that G is an extension of elementary Abelian 3-group K of order 3^{40l} by H, and the generators a, b of H act on K cyclically. Then G is a soluble group and ω(G) = ω(PGL(2, 3^{2})) = {1, 2, 3, 4, 5, 8, 10}. □

The following lemma gives a classification of simple C_{pp}-groups, where p is a prime of form p = 2^{α}3^{β} + 1, α ≥ 0, β ≥ 0.

Lemma 8. (see [3]) Let p be a prime and p = 2^{α}3^{β} + 1, α ≥ 0, β ≥ 0. Then any simple C_{pp}-group is given by Table 2.

The next lemma gives the maximal odd factors set ψ(F_{4}(q)) of μ(F_{4}(q)), q = 2^{e}.

Lemma 9. Let S ≅ F_{4}(q), where q = 2^{e}, e ≥ 1. Then ψ(S) = {q^{4} −1, q^{4} +1, q^{4} − q^{2} + 1, (q − 1)(q^{3} + 1), (q + 1)(q^{3} − 1)}.

Proof. The 2′-elements of S is contained in the maximal tori of S. From [17] we see that μ(F_{4}(q)) contains 25 maximal tori H(1), H(2), …, H(25). Since (q − 1, q^{3} + 1) = 1, (q + 1, q^{3} − 1) = 1, H(13) and H(15) are all cyclic. The conclusion holds. □

3. Main results

In this section we prove the statement of Theorems 1, 2 and 3.

Proof of Theorem 1. Let G be a group and

μ(G) = μ(PGL(2, p^{n})) = {p^{n} − 1, p, p^{n} + 1},

where p = 2^{α}3^{β} + 1 is a prime, and n is a natural numbers. If α = β = 0, then p = 2 and the result is correct by Lemma 6. Also for n = 1, the result holds by [14], and so from now on we assume that p is an odd prime and n ≥ 2. Then t(G) = 2, in fact we have

π_{1}(G) = π(p^{2n} − 1) and π_{2}(G) = {p}.

Lemma 6(1) shows that G is an extension of a π_{1}(G)-group N by a group G_{1}, where S ≤ G_{1} ≤ Aut(S), S is a simple group of Lie type (except L_{2}(p^{k}), k ≠ n and L_{2}(2p^{m} ± 1) where m ≥ 1 and 2p^{m} ± 1 is a prime) and G_{1}/S is a π_{1}(G)- group. Moreover, there exists 2 ≤ j ≤ t(S) such that μ_{j}(S) = {p}, in fact S is a simple C_{pp}-group. Using the results summarized in Table 2, we will show that S is isomorphic to L_{2}(p^{n}).

Step 1. S ≅ L_{2}(q), q = p^{n}, n ≥ 2.

In the following case by case analysis we assume that S ≇ L_{2}(p^{n}) and try to obtain a contradiction. Moreover, as S is always a C_{pp}-group for some appropriate prime p, we make use of the results summarized in Table 2 and Lemma 7 and omit the details of the argument.

Case 1. q = 3^{n}, n ≥ 2.

In this case S can only be isomorphic to one of the following simple groups: L_{2}(2^{3}), L_{3}(2^{2}). Since G does not contain an element of order 9, S can not be isomorphic to L_{2}(2^{3}). If S ≅ L_{3}(2^{2}), then since 7 ∈ π(S) we obtain that n ≥ 6. Assume first that n = 6. In this case we have π(G) = {2, 3, 5, 7, 13, 73}. Evidently 13, 73 ∉ π(Aut(S)) and 13 ≁ 73. Hence {13, 73} ⊆ π(N), and since N is nilpotent we get 13.73 ∈ ω(N), which is a contradiction. Next we suppose that n ≥ 7.

Now we choose the primitive prime divisors 3_{n} and 3_{2n} in π(G). Evidently 3_{2n} ∈ π(3^{n} + 1), and so 3_{n} ≁ 3_{2n}. Moreover, since {2, 3, 5, 7, 11, 13} = π(3 ∏_{i=1}^{6}(3^{i} − 1)), 3_{n}, 3_{2n} ∉ π(Aut(S)), and hence 3_{n}, 3_{2n} ∈ π(N). Again since N is nilpotent, N contains an element of order 3_{n}.3_{2n}, which is of course impossible.

Case 2. q = 5^{n}, n ≥ 2.

In this case we see that S can only be isomorphic to one of the following simple groups: L_{2}(7^{2}), L_{3}(2^{2}), S_{4}(3), S_{4}(7), U_{4}(3), Sz(2^{3}) or Sz(2^{5}). Since G has no element of order 25, S can not be isomorphic to L_{2}(7^{2}) or Sz(2^{5}). If S is isomorphic to one of the simple groups: L_{3}(2^{2}), S_{4}(7), U_{4}(3), or Sz(2^{3}), then 7 ∈ π(S) and so we must have n ≥ 6. Also note that

π(S) ⊂ {2, 3, 5, 7, 11, 13} ⊂ π(5 ∏_{i=1}^{6}(5^{i} − 1)).

If n = 6, then 31, 601 ∈ π(G)\π(Aut(S)) and thus 31, 601 ∈ π(N). Therefore N contains an element of order 31.601, which is a contradiction as 31.601 ∉ ω(G). For case n ≥ 7, since by Zsigmondy’s Theorem 5_{n}, 5_{2n} > 13, a similar argument with the primitive prime divisors 5_{n}, 5_{2n} ∈ π(G) also leads to a contradiction. Similarly, S can not be isomorphic to S_{4}(3).

Case 3. q = 7^{n}, n ≥ 2.

p  simple C_{pp}-groups
2  A_{5}, A_{6}, L_{2}(q) where q is a Fermat prime, a Mersenne prime or

q = 2^{m}, m ≥ 3, L_{3}(2^{2}), Sz(2^{2m+1}), m ≥ 1.

3  A_{5}, A_{6}, L_{2}(q), q = 2^{3}, 3^{m} or 2.3^{m} ± 1, which is a prime, m ≥ 1,

L_{3}(2^{2})

5  A_{5}, A_{6}, A_{7}, M_{11}, M_{22}, L_{2}(q), q = 7^{2}, 5^{m} or 2.5^{m} ± 1, which is

a prime, m ≥ 1, L_{3}(2^{2}), S_{4}(q), q = 3, 7, U_{4}(3), Sz(q), q = 2^{3}, 2^{5}.

7  A_{7}, A_{8}, A_{9}, M_{22}, J_{1}, J_{2}, HS, L_{2}(q), q = 2^{3}, 7^{m} or 2.7^{m} − 1,

which is a prime, m ≥ 1, L_{3}(2^{2}), S_{6}(2), O_{8}^{+}(2), G_{2}(q), q = 3, 19, U_{3}(q), q = 3, 5, 19, U_{4}(3), U_{6}(2), Sz(2^{3}).

13  A_{13}, A_{14}, A_{15}, Suz, F i_{22}, L_{2}(q), q = 3^{3}, 5^{2}, 13^{m} or 2.13^{m} − 1,

which is a prime, m ≥ 1, L_{3}(3), L_{4}(3), O_{7}(3), S_{4}(5), S_{6}(3), O_{8}^{+}(3), G_{2}(q), q = 2^{2}, 3, F_{4}(2), U_{3}(q), q = 2^{2}, 23, Sz(2^{3}), 3D_{4}(2), ^{2}E_{6}(2), ^{2}F_{4}(2)′.

17  A_{17}, A_{18}, A_{19}, J_{3}, He, F i_{23}, F i′_{24}, L_{2}(q), q = 2^{4}, 17^{m} or

2.17^{m} ± 1, which is a prime , m ≥ 1, S_{4}(4), S_{8}(2), F_{4}(2), O_{8}^{−}(2), O_{10}^{−}(2), ^{2}E_{6}(2).

19  A_{19}, A_{20}, A_{21}, J_{1}, J_{3}, O′N, T h, HN, L_{2}(q), q = 19^{m}

or 2.19^{m} − 1, which is a prime, m ≥ 1, L_{3}(7), U_{3}(2^{3}), R(3^{3}), ^{2}E_{6}(2).

37  A_{37}, A_{38}, A_{39}, J_{4}, Ly, L_{2}(q), q = 37^{m} or 2.37^{m} − 1,

which is a prime, m ≥ 1, U_{3}(11), R(3^{3}), ^{2}F_{4}(2^{3}).

73  A_{73}, A_{74}, A_{75}, L_{2}(q), q = 73^{m} or 2.73^{m} − 1, which is a prime,

m ≥ 1, L_{3}(2^{3}), S_{6}(2^{3}), G_{2}(q), q = 2^{3}, 3^{2}, F_{4}(3), E_{6}(2), E_{7}(2), U_{3}(3^{2}), ^{3}D_{4}(3).

109  A_{109}, A_{110}, A_{111}, L_{2}(q), q = 109^{m} or 2.109^{m} − 1, which is

a prime, m ≥ 1, ^{2}F_{4}(2^{3}).

p =  A_{p}, A_{p+1}, A_{p+2}, L_{2}(q), q = 2^{m}, p^{k}, 2p^{k} ± 1, which is a prime,
2^{m} + 1, k ≥ 1, S_{a}(2^{b}), a = 2^{c+1} and b = 2^{d}, c ≥ 1, c + d = s, F_{4}(2^{e}),
m = 2^{s} e ≥ 1, 4e = 2^{s}, O_{2(m+1)}^{−}(2), s ≥ 2, O_{a}^{−}(2^{b}), a = 2^{c+1} and b = 2^{d},

c ≥ 2, c + d = s.

Other A_{p}, A_{p+1}, A_{p+2}, L_{2}(q), q = p^{m} or 2p^{m} − 1, which is a prime,

m ≥ 1.

Table 2. Simple C_{pp}-groups, p = 2^{α}3^{β} + 1, α ≥ 0, β ≥ 0.

In this case the possibilities for S are: L_{2}(2^{3}), L_{3}(2^{2}), S_{6}(2), O_{8}^{+}(2), G_{2}(3), G_{2}(19), U_{3}(3), U_{3}(5), U_{3}(19), U_{4}(3), U_{6}(2) or Sz(2^{3}). First of all, since G has no element of order 49, S ≇ G_{2}(19) or U_{3}(19). Next, we note that π(S) ⊂ π(13!) and by Lemma 1 we see that for every n ≥ 5 there exists a primitive prime divisor 7_{n} ≥ 13. Therefore for n ≥ 5, as previous cases a similar argument with the primitive prime divisors 7_{n} and 7_{2n}, leads to a contradiction.

If n = 4, then μ(G) = {2^{5}.3.5^{2}, 7, 2.1201}. In this case S can only be L_{2}(2^{3}), L_{3}(2^{2}), S_{6}(2), O_{8}^{+}(2), U_{3}(3), U_{3}(5), or U_{4}(3) by checking their prime divisors sets. On the other hand, since the simple groups L_{2}(2^{3}), S_{6}(2), O_{8}^{+}(2) and U_{4}(3) contain an element of order 9 and 9 ∉ ω(G), S can only be L_{3}(2^{2}), U_{3}(3) or U_{3}(5). Moreover, since there does not exist any element of order 1201 in Aut(S), 1201 divides the order of N. Without loss of generality we may assume that N ≠ 1 is an elementary Abelian 1201-group. Because A_{4} = 2^{2} : 3 < A_{6} < L_{3}(2^{2}), 7 : 3 < L_{2}(7) < U_{3}(3) and A_{4} = 2^{2} : 3 < A_{7} < U_{3}(5), in all cases S contains a Frobenius group of shape 2^{2} : 3 or 7 : 3 , and so G contains an element of order 1201.3 by Lemma 5, which is a contradiction.

If n = 3, then μ(G) = {2.3^{2}.19, 7, 2^{3}.43}. In this case, S can only be L_{2}(2^{3}) by checking their element orders sets. As 43 ∉ π(Aut(S)) we have 43 ∈ π(N). Now, we may assume that N ≠ 1 is an elementary Abelian 43-group. Since 2^{3} : 7 < L_{2}(2^{3}) we get 43.7 ∈ ω(G) by Lemma 5, which is a contradiction.

If n = 2, then μ(G) = {2^{4}.3, 7, 2.5^{2}}. In this case, by checking element orders S can only be L_{3}(2^{2}), U_{3}(3) or U_{3}(5). If S ≅ L_{3}(2^{2}) or U_{3}(5), then 5 divides the order of N since 25 ∉ ω(Aut(S)). Without loss of generality we may assume that N ≠ 1 is an elementary Abelian 5-group. Since S contains a Frobenius subgroup of shape 2^{2} : 3 (in fact we have A_{4} = 2^{2} : 3 ≤ A_{6} ≤ L_{3}(4) and A_{4} = 2^{2} : 3 ≤ A_{7} ≤ U_{3}(5)), we get 5.3 ∈ ω(G) by Lemma 5, a contradiction. If S ≅ U_{3}(3), then 5 ∈ π(N), because 5 ∉ π(Aut(S)). Again, since 7 : 3 ≤ L_{2}(7) ≤ U_{3}(3) we get 5.3 ∈ ω(G) by Lemma 5, a contradiction.

Case 4. q = 13^{n}, n ≥ 2.

In this case S can only be isomorphic to one of the following simple groups: L_{2}(3^{3}), L_{2}(5^{2}), L_{3}(3), L_{4}(3), O_{7}(3), S_{4}(5), S_{6}(3), O_{8}^{+}(3), G_{2}(2^{2}), G_{2}(3), F_{4}(2), U_{3}(2^{2}), U_{3}(23), Sz(2^{3}), ^{3}D_{4}(2), ^{2}E_{6}(2) or ^{2}F_{4}(2)′. Since 13^{2} ∉ ω(G), and U_{3}(23) contains an element of order 13^{2}, S ≇ U_{3}(23). Moreover, we have π(S) ⊆ π(19!). Now since, by Lemma 1, for every n ≥ 5, there exists a primitive prime divisor 13_{n} > 19. we can consider the primitive prime divisors 13_{n} and 13_{2n}, and we get a contradiction as before cases. Henceforth, we may assume that n ≤ 4.

If n = 4, then μ(G) = {2^{4}.3.5.7.17, 13, 2.14281}. In this case, by comparing element orders, we conclude that S can only be L_{2}(3^{3}), L_{2}(5^{2}), L_{3}(3), L_{4}(3), O_{7}(3), S_{4}(5), O_{8}^{+}(3), G_{2}(2^{2}), G_{2}(3), F_{4}(2), U_{3}(2^{2}), Sz(2^{3}), ^{3}D_{4}(2), or^{2}F_{4}(2)′. In all above cases, except S ≅ F_{4}(2) , since 17, 14281 ∉ π(Aut(S)), we have 17, 14281 ∈ π(N), and so 17.14281 ∈ ω(N), which is a contradiction. If S ≅ F_{4}(2), then 14281 divides the order of N, and since S contains a Frobenius group 2^{2} : 3 (note that 2^{2} : 3 = A_{4} < S_{10} < S_{8}(2) < F_{4}(2)), G must contain an element of order 14281.3, by Lemma 5, which is not possible. If n = 3, then μ(G) = {2^{2}.3^{2}.61, 13, 2.7.157}.

In this case we have 61, 157 ∉ π(Aut(S)) and so 61, 157 ∈ π(N), hence we get 61.157 ∈ ω(N) ⊂ ω(G), which is impossible.

Case 5. q = 17^{n}, n ≥ 2.

In this case S can only be isomorphic to one of the following simple groups: L_{2}(2^{4}), S_{4}(4), S_{8}(2), F_{4}(2), O_{8}^{−}(2), O_{10}^{−}(2) or ^{2}E_{6}(2). First of all, since 5 ∈ π(S), we deduce n ≥ 4. Moreover, we have π(S) ⊆ π(19!). From Lemma 1, for every n ≥ 4, there exists a primitive prime divisor 17_{n} > 19. Now, for the primitive prime divisors 17_{n} and 17_{2n}, a similar argument as before leads to a contradiction.

Case 6. q = 19^{n}, n ≥ 2.

In this case S can only be isomorphic to one of the following simple groups: L_{3}(7), U_{3}(2^{3}), R(3^{3}) or ^{2}E_{6}(2). Evidently π(S) ⊆ π(37!). Since 7 ∈ π(S), 3|n. If n > 7, then by Lemma 1 there exists a primitive prime divisor 19_{n} > 37. Now we consider the primes 19_{n} and 19_{2n}, and we get a contradiction as previous cases. If n = 6, then we have

μ(G) = {2^{3}.3^{3}.5.7.127, 19, 2.13^{2}.181.769}.

In this case we consider the primes 127, 769 ∈ π(G), and we obtain a contradiction as before. If n = 3, then μ(G) = {2.3^{3}.127, 19, 2^{2}.5.7^{3}}. In this case S can be only L_{3}(7) or U_{3}(2^{3}), and since 5, 127 ∉ π(Aut(S)), we get a contradiction.

Case 7. q = 37^{n}, n ≥ 2.

In this case S can only be isomorphic to one of the following simple groups: U_{3}(11), R(3^{3}) or ^{2}F_{4}(2^{3}). Evidently, π(S) ⊆ {2, 3, 5, 7, 11, 13, 19, 37, 73, 109}. If n ≥ 7, then by Lemma 1 there exists a primitive prime divisors 37_{n} > 109, and hence we consider the primes 37_{n}, 37_{2n} ∈ π(G), and we get a contradiction as before. Therefore we may assume that n ≤ 6. Since

\begin{matrix}π(G) = \{2, 3, 5, 7, 13, 19, 31, 37, 43, 67, 137, 144061\}, & n = 6, \\ π(G) = \{2, 3, 11, 19, 37, 41, 4271, 1824841\}, & n = 5, \\ π(G) = \{2, 3, 5, 19, 37, 89, 137, 10529\}, & n = 4, \\ π(G) = \{2, 3, 7, 19, 31, 37, 43, 67\}, & n = 3, \\ π(G) = \{2, 3, 5, 19, 37, 137\}, & n = 2, \end{matrix}

it is easy to see that 109 ∉ π(G), and so S ≇ ^{2}F_{4}(2^{3}). Moreover, since 55 ∈ ω(U_{3}(11))\ω(G)), S ≇ U_{3}(11). Finally, if S ≅ R(3^{3}), since 13 ∈ π(R(3^{3})), we must have n = 6. Yet, in this case, we can choose the primes 67, 144061 ∈ π(G)\π(Aut(S)), and we get a contradiction as before (note that 67.144061 ∉ ω(G)).

Case 8. q = 73^{n}, n ≥ 2.

In this case S can only be L_{2}(73^{n}), L_{3}(2^{3}), S_{6}(2^{3}), G_{2}(2^{3}), G_{2}(3^{2}), F_{4}(3), E_{6}(2), E_{7}(2), U_{3}(3^{2}) or ^{3}D_{4}(3). We assume that S ≇ L_{2}(73^{n}). It is not difficult to see that π(S) ⊆ π(19!) ∪ {31, 41, 43, 73, 127}. Let n ≥ 5. Then by Lemma 1(6), 73_{n}, 73_{2n} > 127. Evidently 73_{n}.73_{2n} ∉ ω(G), as 73_{2n} ∈ π(73^{n} + 1). On the other hand, since 73_{n}, 73_{2n} ∉ π(Aut(S)), 73_{n}, 73_{2n} ∈ π(N) which implies that 73_{n}, 73_{2n} ∈ ω(N) ⊆ ω(G), a contradiction. Hence n ≤ 4. Because

\begin{matrix}ω(G) = \{2^{5}.3^{2}.5.13.37.41, 73, 2.14199121\}, & n = 4, \\ ω(G) = \{2^{3}.3^{3}.1801, 73, 2.7.37.751\}, & n = 3, \\ ω(G) = \{2^{4}.3^{2}.37, 73, 2.5.13.41\}, & n = 2, \end{matrix}

by checking the sets of element orders for each simple group, the only possibility for S is U_{3}(3^{2}), when n = 4. In this case, we consider the primes 41 and 14199121 in π(G). Since 41 ∈ π(73^{4} − 1) and 14199121 ∈ π(73^{4} + 1), 41 ≁ 14199121 and also 41, 14199121 ∉ π(Aut(S)), which implies that 41, 14199121 ∈ π(N). Now by the nilpotency of N, we obtain that 41.14199121 ∈ ω(N) ⊂ ω(G), which is a contradiction.

Case 9. q = 109^{n}, n ≥ 2.

The proof of this case follows immediately from Lemmas 7(1) and 8.

Case 10. q = (2^{m} + 1)^{n}, where 2^{m} + 1 is a prime and n ≥ 2.

In this case S can only be isomorphic to: L_{2}(2^{m}), S_{a}(2^{b}), a = 2^{c+1}, c ≥ 1, and b = 2^{d}, c + d = s, F_{4}(2^{e}), e ≥ 1, 4e = 2^{s}, O_{2(m+1)}^{−}(2), s > 1, or O_{a}^{−}(2^{b}), a = 2^{c+1}, c ≥ 2, and b = 2^{d}, c + d = s.

If S ≅ L_{2}(2^{m}), then μ(Aut(S)) = {m, 2^{m} − 1, 2^{m} + 1} = {m, p − 2, p}. First, assume that n is odd. In this case we have (p − 2, p^{n} − 1) = 1, in fact if (p − 2, p^{n} − 1) = d then d divides 2^{n} − 1, and so d | (p − 2, 2^{n} − 1) = (2^{m} − 1, 2^{n} − 1) = 2^{(m,n)} − 1 = 1. Now since π(S) ⊆ π(G), it follows that π(p − 2) ⊂ π(p^{n} + 1). Moreover, it is easy to see that 2^{m−1} + 1 divides p^{n} + 1 and (p − 2, 2^{m−1} + 1) = 3. Now we consider the primitive prime divisors

r := p_{n} ∈ π(p^{n} − 1) and s := 2_{2(m−1)} ∈ π(2^{m−1} + 1).

Evidently r, s ∉ π(Aut(S)), and so r, s ∈ π(N). From the nilpotency of N it follows that r ∼ s, which is a contradiction. Next, we suppose that n is even. In this case we have 2^{m−1} + 1 divides p^{n} − 1 and (2^{m−1} + 1, p − 2) = 1. Now, if π(p − 2) ⊂ π(p^{n} − 1) then (p − 2, p^{n} + 1) = 1 and again we consider the following primitive prime divisors

r := p_{2n} ∈ π(p^{n} + 1) and s := 2_{m−1} ∈ π(2^{m−1} − 1),

and we get r ∼ s, as before. But this is a contradiction. Therefore we must have π(p − 2) ⊆ π(p^{n} + 1). Let r ∈ π(2^{m−1} + 1) ⊆ π(p^{n} − 1). Clearly r ∉ π(Aut(S)), hence r ∈ π(N). Now since 2^{m} : 2^{m} − 1 ≤ L_{2}(2^{m}), by Lemma 5 we deduce that r(2^{m} − 1) ∈ ω(G), which is a contradiction.

If S ≅ F_{4}(2^{e}), then the maximal odd factors set ψ(Aut(S)) of μ(Aut(S)) is equal to the same set of μ(S) since |Aut(S)| = 2^{e+1}. From Lemma 9 we have

ψ(Aut(S)) = {q^{′4} − 1, q^{′4} + 1, q^{′4} − q^{′2} + 1, (q′ − 1)(q^{′3} + 1), (q′ + 1)(q^{′3} − 1)},

where q′ = 2^{e}, e ≥ 1.

In this case q^{′4}+1 = p, q^{′4}−1 = p−2. Since G is an extension of a π_{1}(G)-group N by a group G_{1}, where S ≤ G_{1} ≤ Aut(S), and μ(G) = {p^{n}−1, p, p^{n}+1}, we may get a contradiction dividing the two cases. If n ≥ 4, then the odd number \frac{1}{2}(p^{n}+1) and the odd factor of p^{n} − 1 are all greater than any number in ψ(Aut(S)). Hence we have r, s such that

r ∈ π(p^{n} + 1) and s ∈ π(p^{n} − 1),

and r, s ∉ π(Aut(S)), so r, s ∈ π(N). From the nilpotency of N it follows that r ∼ s, which is a contradiction. If n = 2, then we may infer that (p−2, p^{2} +1) = 5 and (p − 2, p^{2} − 1) = 3. It is impossible. Also we may get a similar contradiction if n = 3.

If S ≅ S_{a}(2^{b}), then the maximal odd factors set ψ(Aut(S)) of μ(Aut(S)) is equal to the same set of μ(S) since |Aut(S)| = b = 2^{d}. From [7], §3(3) we have

\{q^{′ \frac{1}{2} (a)} − 1, q^{′ \frac{1}{2} (a)} + 1\} ⊆ ψ(Aut(S)),

where q′ = 2^{b}, b ≥ 1. In this case q^{′ \frac{1}{2} (a)} + 1 = p, and q^{′ \frac{1}{2} (a)} − 1 = p − 2, since the other numbers are not primes in ψ(Aut(S)). The rest of proof is similar to the case of S ≅ F_{4}(2^{e}) by comparing the two sets of ψ(Aut(S)) and μ(G).

If S ≅ O_{2(m+1)}^{−}(2), m = 2^{s}, s > 1, then the maximal odd factors set ψ(Aut(S)) of μ(Aut(S)) is equal to the same set of μ(S) since |Aut(S)| = 2. From [7], §3(5) we have

{q^{′m+1} + 1, q^{′m} + 1, q^{′m} − 1} ⊆ ψ(Aut(S)),

where q′ = 2. In this case q^{′m} + 1 = p, and q^{′m} − 1 = p − 2. The rest of proof is similar to the above cases.

If S ≅ O_{a}^{−}(2^{b}), a = 2^{c+1}, c ≥ 2, and b = 2^{d}, c + d = s, the proof is similar.

Case 11. q = 97^{n} or q = p^{n}, where p = 2^{α}3^{β} + 1 > 109 is a prime, β ≠ 0 and n ≥ 2.

In this case S is a simple C_{pp}-group, and from Table 1 and Lemma 7, we obtain that S ≅ L_{2}(q).

Step 2. N is a 2-group.

Let P/N be a Sylow p-subgroup of S and X/N be the normalizer in S of P/N. Then X/N is a Frobenius group of order q(q − 1)/2, with cyclic complement of order (q − 1)/2. Now, by Lemma 5, we deduce that N is a 2-group.

Step 3. h(G) ∈ {1, ∞}.

First suppose that N = 1. In this case, we have S = L_{2}(q), q = p^{n}, S ≤ G ≤ Aut(S). Denote the factor group G/S by M. Obviously, M ≤ Aut(S). Therefore, every element of M is a product of a field automorphism f, whose order is a divisor of n, and diagonal automorphism d of order dividing 2. Let f ≠ 1 and r be a prime dividing the order of f. Without loss of generality, we may assume that o(f) = r. Evidently, r divides n, and we put q̄ = p^{n/r}. Denote by φ an automorphism of the field 𝔽_{q} inducing f. Since φ fixes a subfield 𝔽q̄ of 𝔽_{q}, f centralizes a subgroup \overline{S} of S isomorphic to L_{2}(q̄). But then G can not be a C_{pp}-group, which is a contradiction. Thus f = 1. Hence, we have M ≤ ⟨d⟩ and so |G/S| ≤ 2. Therefore G ≅ S or G ≅ PGL(2, q). From q +1 ∈ ω(PGL(2, q))\ω(S), we have G ≅ PGL(2, q). Thus, in this case h(G) = 1. Next, suppose that N ≠ 1. Now, by Lemma 4, we get h(G) = ∞. The proof of Theorem 1 is complete. □

Proof of Theorem 2. The proof follows immediately from Lemma 7(2) and Lemma 4. □

Proof of Theorem 3. Let H be an extension of a group of order 2 by S_{4} such that a Sylow 2-subgroup of H is a quaternion group. Then μ(H) = {6, 8}. By Lemma 8 in [9], there exists an extension G of an elementary Abelian 7-group by H, which is a Frobenius group. It follows that μ(G) = {6, 7, 8}, and then Theorem 1 follows from Lemma 4. □

Acknowledgement. The first author would like to thank the authority of K. N. Toosi University of Technology very deeply for providing the necessary facilities and their partial support to make the complete performance of this project possible. The second author supported by the National Natural Science Foundation of China (Grant No. 10171074).

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Received April 18, 2005