J Appl Math Comput (2009) 31: 447–457 DOI 10.1007/s12190-008-0223-4
JAMC
A characterization of the group 2 Dn (2), where n = 2m + 1 ≥ 5 M.R. Darafsheh · A. Mahmiani
Received: 4 October 2008 / Revised: 3 December 2008 / Published online: 6 January 2009 © Korean Society for Computational and Applied Mathematics 2008
Abstract In this paper it is proved that the group 2 Dn (2), where n = 2m + 1 ≥ 5, can be uniquely determined by its order components. More precisely we will prove that if G is a finite group and OC(G) denotes the set of order components of G, then OC(G) = OC(2 Dn (2)) if and only if G ∼ = 2 Dn (2). A main consequence of our result is the validity of Thompson’s conjecture for the group under consideration. Keywords Prime graph · Order component · Linear group Mathematics Subject Classification (2000) 20D06 · 20D60
1 Introduction For a positive integer n, let π(n) be the set of all prime divisors of n. If G is a finite group, we set π(G) = π(|G|). The Gruenberg-Kegel graph of G, or the prime graph of G, is denoted by GK(G) and is defined as follows. The vertex set of GK(G) is the set π(G) and two distinct primes p and q are joined by an edge if and only if G contains an element of order pq. We denote the connected components of GK(G) by π1 , π2 , . . . , πs(G) , where s(G) denotes the number of connected components of GK(G). If the order of G is even, the notation is chosen so that 2 ∈ π1 . It is clear that the order of G can be expressed as the product of the numbers m1 , m2 , . . . , ms(G) , where π(mi ) = πi , 1 ≤ i ≤ s(G). If the order of G is even and s(G) ≥ 2, according to our notation m2 , . . . , ms(G) are odd numbers. The positive integers m1 , m2 , . . . , ms(G) M.R. Darafsheh () School of Mathematics, College of Science, University of Tehran, Tehran, Iran e-mail:
[email protected] A. Mahmiani Islamic Azad University, Aliabad-e-Katool, Iran
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M.R. Darafsheh, A. Mahmiani
are called the order components of G and OC(G) = {m1 , m2 , . . . , ms(G) } is called the set of order components of G. It is a natural question to ask: If the finite groups G and H have the same order components does it follow G is isomorphic to H ? For many simple groups H with the number of order components s(H ) at least 2, the answer to the above question is affirmative. However if s(H ) = 1 the answer is negative. The simple groups Bn (q) and Cn (q) where n = 2m ≥ 4 and q is odd, have the same order components but they are not isomorphic. Hence it is natural to adopt the following definition. Definition 1 Let G be a finite group. The number of non-isomorphic finite groups with the same order components as G is denoted by h(G) and is called the h-function of G. For any natural number k we say the finite group G is k-recognizable by its set of order components if h(G) = k. If h(G) = 1 we say that G is characterizable by its set of order components or briefly G is a characterizable group. In this case G is uniquely determined by the set of its order components. Obviously for any finite groups G we have h(G) ≥ 1. The components of the Gruenberg-Kegel graph GK(P ) of any non-abelian finite simple group P with GK(P ) disconnected are found in [19] from which we can deduce the component orders of P . These information which will be used in proving our main result are listed in Tables 1, 2 and 3. In [17] and [18] it is proved that if n = 2m ≥ 4, then h(Bn (q)) = h(Cn (q)) = 2 for q odd and h(Bn (q)) = h(Cn (q)) = 1 for q even. Apart from the families Bn (q) and Cn (q), n = 2m ≥ 4, q odd. The following groups have been proved to be characterizable by their order components by various authors. All the sporadic simple groups in [2], P SL2 (q); 2 Dn (3) where 9 ≤ n = 2m + 1 is not a prime; 2 Dp+1 (2); in [3, 4, 20], respectively. Lp+1 (2); 2 Dp (3), where p ≥ 5 is a prime number not of the form 2m + 1; in [9] and [10], respectively. Some projective special linear (unitary) groups have been characterized in a series of articles in [12–14] and [15]. A few of the alternating or symmetric groups are proved to be characterizable by their order components in [1] and [16]. Based on these results we put forward the following conjecture. Conjecture 1 Let P be a non-abelian finite simple group with s(P ) ≥ 2. If G is a finite group and OC(G) = OC(P ), then either G ∼ = P or G ∼ = Bn (q) or Cn (q) where n = 2m ≥ 4 and q is an odd number or G ∼ B (3), C (3), where p is an odd prime = p p number. In fact in [11] it has been proved that if a finite group G has the same set of order components as Bp (3), p an odd prime, then G ∼ = Bp (3) or Cp (3). In this way the last part of Conjecture 1 is proved positively. A motivation for characterizing finite groups by the set of their order components is the following conjecture due to J.G. Thompson. Conjecture 2 (Thompson) For a finite group G let N (G) = {n ∈ N | G has a conjugacy class of size n}. Let Z(G) = 1 and M be a non-abelian finite simple group satisfying N (G) = N (M). Is it true that G ∼ = M?
A characterization of the group 2 Dn (2), where n = 2m + 1 ≥ 5
449
Table 1 The order components of finite simple groups P with s(P ) = 2 (p an odd prime) P
Restrictions on P
m1
m2
An
6 < n = p, p + 1, p + 2;
n! 2p
p
one of n, n − 2 is not a prime p
Ap−1 (q)
(p, q) = (3, 2), (3, 4)
q(2)
Ap (q)
(q − 1) | (p + 1)
q(
2 A (q) p
p
(q + 1) | (p + 1),
q(
(p, q) = (3, 3), (5, 2) 2 A (2) 3
Bn (q)
i i=1 (q − 1)
p+1 2 ) (q p+1
q(2)
2A p−1 (q)
p−1
p−1
p−1
i i=2 (q − 1)
i i i=1 (q − (−1) )
p+1 2 ) (q p+1
×
− 1)
p−1
− 1)
n = 2m ≥ 4, q odd
Bp (3)
q p +1 q+1
5 n−1
2i i=1 (q − 1) 2 p−1 3p (3p + 1) i=1 (32i − 1) 2 2i q n (q n − 1) n−1 i=1 (q − 1) 2 p−1 q p (q p + 1) i=1 (q 2i − 1) p−1 q p(p−1) i=1 (q 2i − 1) q p(p+1) (q p + 1)
q
q p −1 q−1 q p +1 (q+1)(p,q+1)
i i i=1 (q − (−1) )
26 .34 n2
q p −1 (q−1)(p,q−1)
(q n − 1)
Cn (q)
n = 2m ≥ 2
Cp (q)
q = 2, 3
Dp (q)
p ≥ 5, q = 2, 3, 5
Dp+1 (q)
q = 2, 3
2 D (q) n
n = 2m ≥ 4
2 D (2) n
n = 2m + 1 ≥ 5
2n(n−1) (2n + 1)(2n−1 − 1) 2i × n−2 i=1 (2 − 1)
2D p+1 (2)
5 ≤ p = 2m − 1
2 D (3) p
5 ≤ p = 2m + 1
2p(p+1) (2p + 1)(2p+1 + 1) p−1 × i=1 (22i − 1) p−1 3p(p−1) i=1 (32i − 1)
2 D (3) n
9 ≤ n = 2m + 1 = p
G2 (q)
2 < q ≡ (mod 3), = ±1
p−1 ×(q p+1 − 1) i=1 (q 2i − 1) 2i q n(n−1) n−1 i=1 (q − 1)
q n +1 2 3p −1 2 q n +1 (2,q−1) q p −1 (2,q−1) q p −1 q−1 q p −1 (2,q−1) q n +1 (2,q+1) 2n−1 + 1
2p − 1 3p +1 4 3n−1 +1 2
1 3n(n−1) (3n + 1)(3n−1 − 1) 2 2i × n−2 i=1 (3 − 1) q 6 (q 3 − )(q 2 − 1)(q + )
q 2 − q + 1
q 12 (q 6 − 1)(q 2 − 1)q 4 − q 2 + 1
(q 4 + q 2 + 1)
q 24 (q 8 − 1)(q 6 − 1)2 (q 4 − 1)
q4 − q2 + 1
2 F (2) 4
211 .33 .52
13
E6 (q)
q 36 (q 12 − 1)(q 8 − 1)(q 6 − 1)
q 6 +q 3 +1 (3,q−1)
3 D (q) 4
F4 (q)
q odd
×(q 5 − 1)(q 3 − 1)(q 2 − 1) 2 E (q) 6
q >2
q 36 (q 12 − 1)(q 8 − 1)(q 6 − 1)
q 6 −q 3 +1 (3,q+1)
×(q 5 + 1)(q 3 + 1)(q 2 − 1) M12
26 .33 .5
J2
27 .33 .52
7
Ru
214 .33 .53 .7.13
29
11
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M.R. Darafsheh, A. Mahmiani
Table 1 (Continued) P
Restrictions on P
m1
m2
He
210 .33 .52 .73
17
McL
27 .36 .53 .7
11
Co1
221 .39 .54 .72 .11.13
23
Co3
210 .37 .53 .7.11
23
Fi22
217 .39 .52 .7.11
13
HN
214 .36 .56 .7.11
19
In [5] it is proved that if s(M) ≥ 3, then the above conjecture holds. Also in [5] it is proved that if G and M are finite groups with s(M) ≥ 2, Z(G) = 1, N (G) = N(M), then |G| = |M|, in particular s(M) = s(G) and OC(G) = OC(M). Therefore if the simple group M is characterizable by the set of its order components, then the Thompson’s conjecture holds for M. There is another conjecture due to W. Shi and J. Bi which states: Conjecture 3 Let G be a group and M a finite simple group. Then G ∼ = M if and only if (a) |G| = |M| and (b) πe (G) = πe (M) where πe (G) denotes the set of order elements of G. Clearly conditions (a) and (b) above imply OC(G) = OC(M). Therefore if the group G is characterizable by its order components, then we will deduce G ∼ =M and Conjecture 3 is true for M. According to the main theorem of this paper which is stated below, Conjectures 2 and 3 are true for the simple groups 2 Dp (3) where p = 2m + 1, p ≥ 5 is a prime number. In this paper we consider the Steinberg group 2 Dn (2) where n = 2m + 1 ≥ 5 and prove that this group is characterizable by its order components. Another name for this group is P − 2n (2). More precisely we will prove: Main Theorem If a finite group G has the same set of order components as 2 Dn (2), n = 2m + 1 ≥ 5, then G ∼ =2 Dn (2).
2 Preliminary results The structure of finite groups with disconnected Gruenberg-Kegel graph follows from Theorem A of [21] which will be stated below: Lemma 1 Let G be a simple group with s(G) ≥ 2. Then one of the following holds: (1) G is either a Frobenius or 2-Frobenius group.
A characterization of the group 2 Dn (2), where n = 2m + 1 ≥ 5
451
Table 2 The order components of finite simple groups P with s(P ) = 3 (p an odd prime) P
Restrictions on P
m1
m2
m3
An
n > 6, n = p, p − 2
n! 2n(n−2)
p
p−2
q −
q
q+ 2
q
q −1
q +1
A2 (2)
8
3
7
2 A (2) 5
215 .36 .5
7
11
3p−1 +1 2
3p +1 4
2p + 1
2p+1 + 1
q2 − q + 1 √ q − 3q + 1
q2 + q + 1 √ q + 3q + 1 q4 − q2 + 1 q 2 + 2q 3 + q √ + 2q + 1
are primes A1 (q)
3 < q ≡ (mod 4), = ±1
A1 (q)
q > 2, q even
2 D (3) p
p = 2m + 1 ≥ 5
2D p+1 (2)
p = 2n − 1, n ≥ 2
2.3p(p−1) (3p−1 − 1) p−2 × i=1 (32i − 1) 2p(p+1) (2p − 1)
G2 (q)
q ≡ 0 (mod 3)
q 6 (q 2 − 1)3
2 G (q) 2
q = 32m+1 > 3
q 3 (q 2 − 1)
F4 (q)
q even
q 24 (q 6 − 1)2 (q 4 − 1)2
2 F (q) 4
q = 22m+1 > 2
q 12 (q 4 − 1)(q 3 + 1)
q4 + 1 q 2 − 2q 3 + q √ − 2q + 1
263 .311 .52 .73 .11.13.
73
127
757
1093
E7 (2)
p−1 × i=1 (22i − 1)
17.19.31.43 E7 (3)
223 .363 .52 .73 .112 .132 . 19.37.41.61.73.547
M11
24 .32
5
11
M23
27 .32 .5.7
11
23
M24
210 .33 .5.7
11
23
J3
27 .35 .5
17
19
HiS
29 .32 .53
7
11
Suz
213 .37 .52 .7
11
13
Co2
218 .36 .53 .7
11
23
Fi23
218 .313 .52 .7.11.13
17
23
F3
215 .310 .53 .72 .13
19
31
F2
241 .313 .56 .72 .11.13.
31
47
17.19.23
K (2) G has a normal series 1 H K G such that H is a nilpotent π1 -group, H G G K is a non-abelian simple group, K is a π1 -group, | K | divides |Out( H )| and any K odd order component of G is equal to one of the odd order components of H .
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M.R. Darafsheh, A. Mahmiani
Table 3 The order components of finite simple groups P with s(P ) > 3 P
Restrictions
m1
m2
m3
26
3
5
m4
m5
m6
on P A2 (4) 2 B (q) 2
q = 22m+1
q2
q −1
7
q−
√
2q
+1
>2
q+
√ 2q
+1
2 E (2) 6
236 .39 .52 .72 .11
13
17
19
E8 (q)
q 120 (q 20 − 1)
q 10 −q 5 +1 q 2 −q+1
q 10 +q 5 +1 q 2 +q+1
q8 − q4
q ≡ 2, 3
×(q 18 − 1) (mod 5)
+1
×(q 14 − 1) ×(q 12 − 1) ×(q 10 − 1) ×(q 8 − 1) ×(q 4 + 1) ×(q 4 + q 2 + 1)
M22
27 .32
5
7
11
J1
23 .3.5
7
11
19
ON
29 .34 .5.73
11
19
31
LyS
28 .37 .56 .7.11
31
37
67
Fi24
221 .316 .52 .73 .
17
23
29
41
59
71
q 10 −q 5 +1 q 2 −q+1
q 10 +q 5 +1 q 2 +q+1
q8 − q4
11.13 246 .320 .59 .76 .
F1
112 .133 17.19.23.29.31.47 E8 (q)
q ≡ 0, 1, 4 (mod 5)
q 120 (q 18 − 1) ×(q 14 − 1)
q 10 +1 q 2 +1
+1
×(q 12 − 1)2 ×(q 10 − 1)2 ×(q 8 − 1)2 ×(q 4 + q 2 + 1) J4
221 .33 .5.7.113
23
29
31
37
43
To deal with the first case in the above lemma we need the following results which are taken from [6] and [2], respectively. Lemma 2 (a) Let G be a Frobenius group of even order with kernel and complements K and H , respectively. Then s(G) = 2 and the prime graph components of G are π(H ) and π(K). (b) Let G be a 2-Frobenius group of even order. Then s(G) = 2 and G has a K G G | = m2 , |H || K | = m1 and | K | divides normal series 1 H K G such that | H K | H | − 1 and H is a nilpotent π1 -group.
A characterization of the group 2 Dn (2), where n = 2m + 1 ≥ 5
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Lemma 3 Let G be a finite group with s(G) ≥ 2. If H G is a πi -group, then s(G) ( j =1,j =i mj ) | (|H | − 1). The following result of Zsigmondy [22] is important in some number theoretical considerations. Lemma 4 Let n and a be integers greater than 1. There exists a prime divisor p of a n − 1 such that p does not divide a i − 1 for all 1 ≤ i < n, except in the following cases. (1) n = 2, a = 2k − 1, where k ≥ 2, (2) n = 6, a = 2. The prime p in Lemma 4 is called a Zsigmondy prime for a n − 1. Remark 1 If p is a Szigmondy prime for a n − 1, then p > n. Because if p ≤ n, then n = kp + r, 0 ≤ r < p, and we can write a n − 1 = a r (a kp − a k ) + a k+r − 1. Since (p, a) = 1 we have a p ≡ a (mod p), hence a kp ≡ a k (mod p), therefore p | a k+r − 1. By assumption about p we must have k + r ≥ n which implies k ≥ kp, hence k = 0. Therefore n = r < p contradicting p ≤ n. Next we consider the Steinberg groups 2 Dn (2) where n = 2m + 1 ≥ 5. Us1 2 n(n−1) n 2i ing [7] we have | Dn (q)| = (4,q n +1) q (q + 1) n−1 i=1 (q − 1) and for n > 3 all these groups are simple. The outer authomorphism group of 2 Dn (q) has order (4, q n + 1).f , where q = r f , r is a prime number. By [19] if n = 2m + 1 ≥ 5, then s(2 Dn (2)) = 2. Therefore in this case the prime graph of 2 Dn (2) has two components. by Table 1 are: m1 = 2n(n−1) (2n + 1)(2n−1 − 1) The 2itwo order components n−1 + 1. The components of the prime graph of 2 D (2) × n−2 (2 − 1) and m = 2 2 n i=1 2i − 1)) and π = π(2n−1 + 1). (2 are π1 = π(2(2n + 1)(2n−1 − 1) n−2 2 i=1 3 Proof of Main Theorem Let G be a finite group with OC(G) = {m1 , m2 }, where m1 and m2 are the order components of the group 2 Dn (2), where n = 2m + 1 ≥ 5. In order to use Lemma 1, first we will prove the following lemma. Lemma 5 If G is a finite group with OC(G) = {m1 , m2 }, then G is neither a Frobenius nor a 2-Frobenius group. Proof First we will prove that G is not a Frobenius group. If G is a Frobenius group with complement H and kernel K, then by Lemma 2 we have OC(G) = {|H |, |K|}. Since |H | | |K| − 1 we have |H | < |K| from which it follows that |K| = m1 and |H | = m2 . If r is a Zsigmondy prime for 22(n−2) − 1, then r | 2n−2 + 1 and the order of a Sylow r-subgroup S of G, as well as K, is a divisor of 2n−2 + 1. Since K is a nilpotent normal subgroup of G, we obtain S G and by Lemma 3, m2 | |S| − 1. But m2 = 2n−1 + 1 and since |S| | 2n−2 + 1 a contradiction is obtained.
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M.R. Darafsheh, A. Mahmiani
In the next step we will show that G is not a 2-Frobenius group. Suppose in the contrary G is a 2-Frobenius group. Then by Lemma 2(b) there is a normal series 1 K G K | = m2 and | K | | (| H |− H K G for G such that H is a nilpotent π1 -group, | H n−1 2(n−2) − 1, we 1) = 2 . Now arguing as above by taking a Zsigmondy prime r for 2 G |, hence by Lemma 2, r | |H | and by Lemma 3, m2 | |S| − 1, where observe that r | K S is a Sylow r-subgroup of H , a contradiction. Therefore G cannot be a 2-Frobenius group and the lemma is proved. Proof of Main Theorem By the Lemmas 1 and 4, if G is a finite group with OC(G) = K is OC(2 Dn (2)), then there is a normal series 1 H K G for G such that H G a non-abelian simple group, H and K are π1 -group and H is nilpotent. Moreover G K |K | divides |Out( H )| and the odd order component of G is one of the odd order K K components of H and s( H ) ≥ 2. K Since P = H is a non-abelian simple group with disconnected Gruenberg-Kegel graph, by the classification of finite simple groups we have one of the possibilities in Tables 1, 2 or 3 for P . In the following we deal with these groups. Case 1 P ∼ =2 A3 (2), 2 F4 (2) , A2 (2), A2 (4), 2 A5 (2), E7 (2), E7 (3), 2 E6 (2), or one of the 26 sporadic simple groups listed in Tables 1, 2 or 3. The odd order component of 2 Dn (2) is m2 = 2n−1 +1, m ≥ 2. Therefore m2 ≥ 17, can be a and by inspection we find out that only the groups H e, J3 , 2 E6 (2) and F i24 candidate for P with an odd order components m2 = 17. In this case we have n = 5 and the order of the simple groups P mentioned above should satisfy |P | | |2 D5 (2)| which is not the case. Case 2 P ∼ = An and either n = p, p + 1, p + 2, one of n or n − 2 is not prime; or n = p, p − 2 are both prime, where p > 6 is a prime number. By Tables 1 and 2, the odd order components of An are p or(and) p − 2. If m2 = 2n−1 + 1 = p, then the largest power of 2 dividing |Ap | is [ p2 ] + [ p4 ] + · · · = 2n−1 − 1 > n(n − 1) if n ≥ 5, hence |P | |G| in this case. If m2 = 2n−1 + 1 = p − 2 we get a contradiction in a similar manner. Case 3 P ∼ = E6 (q) or 2 E6 (q), q > 2. 6 ±q 3 +1 In this case we have q(3,q∓1) = 2n−1 + 1. If (3, q ∓ 1) = 1, then q 3 (q 3 ± 1) = n−1 which is impossible. If (3, q ∓ 1) = 3, then q ≡ ±1 (mod 3) and q 3 (q 3 ± 1) = 2 n−1 + 2. Since n = 2m + 1, m ≥ 2, it can be verified that 2n−1 ≡ 4 or 7 (mod 9), 3.2 hence in any case 3.2n−1 + 2 ≡ 5 (mod 9). Since q ≡ ±1 (mod 3) we obtain q 3 ≡ ±1 (mod 9) and q 3 (q 3 ± 1) ≡ 2 (mod 9), contradicting q 3 (q 3 ± 1) = 3.2n−1 + 2. Case 4 P ∼ = F4 (q), q odd 3 D4 (q), G2 (q), 2 < q ≡ (mod 3), = ±1. In this case we have q 4 − q 2 + 1 = 2n−1 + 1 or q 2 − q + 1 = 2n−1 + 1 and both cases are clearly impossible. Case 5 P ∼ = 2 Dn (3), 9 ≤ n = 2m + 1 = p; 2 Dp (3), 5 ≤ p = 2m + 1; 2 D (2), n = 2m + 1 ≥ 5. n In these cases we obtain the following equalities respectively: 3n −1 = 2n + 1, −1 −1 p n−1 n n−1 n n and 2 = 2 . If 3 = 2 + 1, then by [8] we have n = 3, 3 −3=2 n = 2 which is not the case. The equality 3p − 3 = 2n−1 is clearly impossible. K | = |G| which implies H = 1 and If 2n −1 = 2n−1 , then n = n, hence |P | = | H 2 ∼ ∼ G = P = Dn (2). This is what we have claimed in our main theorem. Therefore
A characterization of the group 2 Dn (2), where n = 2m + 1 ≥ 5
455
we continue case by case investigation of simple groups with disconnected prime graph to rule out any other possibilities. Case 6 P ∼ = Dp+1 (q), Dp (q), Cp (q), q = 2, 3, Dp (5), Bp (3), Bn (q), n = 2m ≥ 4, q odd. In all of the cases a contradiction in reached either by [8] or immediate calculation. Case 7 P ∼ = Ap (q), (q − 1) | (p + 1). = 2 Ap (q), (q + 1) | (p + 1), or P ∼ p +1 p −1 In these cases we have qq+1 = 2n−1 + 1 or qq−1 = 2n−1 + 1, respectively. Therefore we obtain q p−1 − q p−2 + · · · − q + 1 = 2n−1 + 1 or q p−1 + q p−2 + · · · + q + 1 = 2n−1 + 1, hence q(q p−2 − q p−3 + · · · − 1) = 2n−1 or q(q p−2 + q p−3 + · · · + 1) = 2n−1 . Both of these equations are impossible. Case 8 P ∼ = Cn (q), n = 2m ≥ 2. = 2 Dn (q), n = 2m ≥ 4, or P ∼ n ∼2 Dn (q), n = 2m ≥ 4. We have q +1 = 2n−1 + 1. If q is First assume P = (2,q+1)
odd, we obtain q n − 2n = 1 which by [8] is impossible. If q is even we will obtain q n = 2n−1 . Assuming q = 2f we will obtain f n = n − 1 hence f 2m = 2m , and (n −1) n n(n−1) |2 which implies n − 1 ≤ n. Thereconsidering |P | | |G| we deduce q m m fore 2 − 2 ≤ 2, and since m ≥ 2 we deduce m ≤ m. Hence f = 2m−m . Now G | = t. Then t | |Out(P )| = f , and since G = we will use Lemmas 1 and 4. Let | K n −1 2i G K 2 | K || H ||H | we obtain t|G||H | = | Dn (2)| which implies tH q n (n −1) i=1 (q − n−2 2i n(n−1) n n−1 1) = 2 (2 + 1)(2 − 1) i=1 (2 − 1). Let r be a Zsigmondy prime for 22n − 1, which exists because n > 3. Therefore r | 2n + 1. Suppose r | q 2i − 1 for some 1 ≤ i ≤ n − 1. Then 2f i ≥ 2n which implies f i ≥ n or 2m−m i ≥ 2m + 1. But i ≤ n − 1 = 2m − 1 < 2m , from which it follows m−m m m−m m i < 2 contradicting 2 i ≥ 2 + 1. Since r is odd we must have r | |H |. 2 Clearly the order of a Sylow r-subgroup S of G, and hence of H , is a divisor of 2n + 1. Since H is a nilpotent normal subgroup of G we must have S G, hence by the Lemma 3 we obtain m2 = 2n−1 + 1 | |S| − 1 implying |S| ≥ 2n−1 + 2. Since we n have |S| | 2n + 1, hence |S| < 2 2+1 < 2n−1 + 2, contradicting |S| ≥ 2n−1 + 2. This final contradiction rules out the possibility P ∼ =2 Dn (q) where n = 2m ≥ 4. m Next we consider the case P ∼ = Cn (q), n = 2 ≥ 2. In this case we have
q n +1 (2,q−1)
= 2n−1 + 1. If q is odd, then we obtain q n − 2n = 1 which is impossible
by [8]. If q is even, then we obtain q n = 2n−1 and with a similar consideration as above we reach a contradiction. We omit the details of the conclusion. Case 9 P ∼ =2 Ap−1 (q). = Ap−1 (q), (p, q) = (3, 2), (3, 4) or P ∼ p q −1 The odd order component of Ap−1 (q) is (q−1)(p,q−1) and that of 2 Ap−1 (q) is p q +1 q p −1 . We will give the details in the case of P ∼ = = Ap−1 (q) with (q+1)(p,q+1)
(q−1)(p,q−1)
2n−1 + 1. The case of P ∼ =2 Ap−1 (q) is similar. p −1 If (p, q − 1) = 1, then qq−1 = 2n−1 + 1 which implies q(q p−2 + · · · + q + 1) = n−1 p−2 + · · · + q + 1 are coprime a contradiction is obtained. 2 . Since q and q Therefore in what follows we assume (p, q − 1) = p, and hence p | q − 1. We have q p − 1 = p(q − 1)(2n−1 + 1)
(1)
456
M.R. Darafsheh, A. Mahmiani
If p = 2, then from (1) we obtain q = 2n + 1. Since q is a prime power and n ≥ 5, by [8] there is no solution for q = 2n + 1. Therefore we assume p is an odd prime and from (1) we obtain q p−1 + q p−2 + · · · + q + 1 = p.2n−1
(2)
For each k ∈ N, we can write: k−2 k (q − 1)k−i + k(q − 1) + 1 q = (q − 1 + 1) = i k
k
(3)
i=0
Since p | q − 1, using (3) we obtain q p−1 + q p−2 + · · · + q + 1 − p = a multiple p−1 p−1 p−1 , hence (q − 1) k=1 k is a multiple of p 2 + (q − 1) k=1 k. But k=1 k = p(p−1) 2 of p 2 . Therefore q p−1 + q p−2 + · · · + q + 1 − p = p.2n−1 must be a multiple of p 2 which is a contradiction. Hence the case P ∼ = Ap−1 (q), (p, q) = (3, 2), (3, 4) is impossible. In the next step we will consider simple groups P with s(P ) = 3. Now we will go through those groups in Table 2 which are not considered so far. Case 10 P ∼ = A1 (q), 3 < q ≡ (mod 4), = ±1. n−1 +1 or q+1 = 2n−1 +1, The odd order components of P are q and q+ 2 . If q = 2 2 n−1 + 1 from which then by [8] a contradiction is obtained. Hence we assume q−1 = 2 2 it follows that q = 2n + 3. Since n is of the form 2m + 1 ≥ 5, we obtain 5 | 2n + 3, and so q is a power of 5, say q = 5f . But then q ≡ 1 (mod 4) contradicting = −1 and q ≡ −1 (mod 4). Case 11 P ∼ = A1 (q), q > 2, q even. The odd order components are q − 1 and q + 1. If q − 1 = 2n−1 + 1, then q = 2(2n−1 + 1) cannot be a power of 2. Therefore we will assume q + 1 = 2n−1 + 1 m G from which it follows q = 2n−1 = 22 , where n = 2m + 1. Now we set | K |=t from which it follows that |G| = t|H ||P |, where by the Lemma 1 and [8] we have t | |Out(P )| = 2m . Considering orders of G and P we obtain 2
t|H | = 2(n−1) (2n + 1)
n−2
(22i − 1)
(4)
i=1
Now if r is a Zsigmondy prime for 22n − 1, then r | 2n + 1, hence by (4) r | |H |. But the order of a Sylow 2-subgroup of G, hence of H , is a divisor of 2n + 1 and from the Lemma 3 it follows that m2 = 2n−1 + 1 | |S| − 1. But r | 2n + 1 implies n r ≤ 2 2+1 < 2n−1 + 1, a contradiction. Therefore the case P ∼ = A1 (q), q > 2, q even is ruled out. Case 12 P ∼ =2 Dp (3), p = 2m + 1 ≥ 5, 2 Dp+1 (2), p = 2n − 1, n ≥ 2. p−1 p If P ∼ =2 Dp (3), then 3 2 +1 = 2n−1 + 1 or 3 4+1 = 2n−1 + 1 and both cases are ruled out easily. If P ∼ =2 Dp+1 (2), then 2p + 1 = 2n−1 + 1 or 2p+1 + 1 = 2n−1 + 1, hence p = n − 1 or p = n − 2. Since n = 2m + 1 we obtain p = 2m or p = 2m − 1. Obviously p = 2m is impossible. If p = 2m − 1, then m = n . Therefore we have P ∼ = 2 Dp+1 , p = 2m − 1 and G ∼ = 2 Dn (2), where n = 2m + 1. Now the order consideration reveals that |P | |G|, a contradiction.
A characterization of the group 2 Dn (2), where n = 2m + 1 ≥ 5
457
Case 13 In the remaining cases of Tables 2 and 3, the odd order components of P is a number of the form qf (q) + 1. If qf (q) + 1 = 2n−1 + 1, then we obtain qf (q) = 2n−1 . This will imply that q is a power of 2. But examination of each case shows that q and f (q) are coprime, a contradiction. Finally the main theorem is proved. Acknowledgement The first author would like to thank the research council of the university of Tehran for financial supports through grant #6101031-1-02.
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