Acta Mathematica Sinica, English Series Jul., 2008, Vol. 24, No. 7, pp. 1117–1126 Published online: June 15, 2008 DOI: 10.1007/s10114-007-6143-7 Http://www.ActaMath.com

Acta Mathematica Sinica, English Series The Editorial Office of AMS & Springer-Verlag 2008

Characterizability of the Group 2 Dp (3) by its Order Components, Where p ≥ 5 is a Prime Number Not of the Form 2m + 1 M. R. DARAFSHEH School of Mathematics, College of Science, University of Tehran, Tehran, Iran E-mail: [email protected] Abstract The author will prove that the group 2 Dp (3) can be uniquely determined by its order components, where p = 2m + 1 is a prime number, p ≥ 5. More precisely, if OC(G) denotes the set of order components of G, we will prove OC(G) = OC(2 Dp (3)) if and only if G is isomorphic to 2 Dp (3). A main consequence of our result is the validity of Thompson’s conjecture for the groups under consideration. Keywords

prime graph, order component, linear group

MR(2000) Subject Classification 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) 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 that 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, in Received March 16, 2006, Accepted March 30, 2007

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Darafsheh M. R.

short, 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 [1], from which we can deduce the component orders of P. This information, which will be used in proving our main result, is listed in Tables 1, 2 and 3. Table 1 The order components of finite simple groups P with s(P ) = 2 (p an odd prime) P An Ap−1 (q)

Restrictions on P 6 < n = p, p + 1, p + 2; one of n, n − 2 is not a prime (p, q) = (3, 2), (3, 4)

Ap (q)

(q − 1) | (p + 1)

2

Ap−1 (q)

2

Ap (q)

2

A3 (2) Bn (q) Bp (3) Cn (q)

(q + 1) | (p + 1), (p, q) = (3, 3), (5, 2) n = 2m ≥ 4, q odd n = 2m ≥ 2

Cp (q)

q = 2, 3

Dp (q)

p ≥ 5, q = 2, 3, 5

Dp+1 (q)

q = 2, 3

2

Dn (q) 2 Dn (2)

n = 2m ≥ 4 n = 2m + 1 ≥ 5

2

Dp (3)

5 ≤ p = 2m + 1

2

Dn (3)

9 ≤ n = 2m + 1 = p

G2 (q) 3 D4 (q)

2 < q ≡ (mod 3),  = ±1

F4 (q)

q odd

2

F4 (2) E6 (q) 2

E6 (q)

M12 J2 Ru He M cL Co1 Co3 F i22 HN

q>2

m1

m2 p

n! 2p

p  i q (2) p−1 i=1 (q − 1) p+1  p+1 i q ( 2 ) (q − 1) p−1 i=2 (q − 1) p  i i q (2) p−1 i=1 (q − (−1) ) p+1 q ( 2 ) (q p+1 − 1) p−1 i i i=1 (q − (−1) ) 6 4 2 .3  2 2i q n (q n − 1) n−1 − 1) i=1 (q  2 p−1 2i p p 3 (3 + 1) i=1 (3 − 1)  2 2i q n (q n − 1) n−1 − 1) i=1 (q  2 p−1 2i p p q (q + 1) i=1 (q − 1)  2i q p(p−1) p−1 − 1) i=1 (q

1 q p(p+1) (q p + 1) (2,q−1)  p+1 2i (q − 1) p−1 − i=1 (q  n−1 2i n(n−1) q (q − 1) i=1 n(n−1) n n−1

1)

2 + 1)(2 − 1) n−2 (2 2i (2 − 1) i=1  2i 3p(p−1) p−1 i=1 (3 − 1)

1 n(n−1) n 3 (3 + 1)(3n−1 − 1) 2  n−2 2i i=1 (3 − 1) q 6 (q 3 − )(q 2 − 1)(q + ) q 12 (q 6 − 1)(q 2 − 1) (q 4 + q 2 + 1) q 24 (q 8 − 1)(q 6 − 1)2 (q 4 − 1) 211 .33 .52 q 36 (q 12 − 1)(q 8 − 1)(q 6 − 1) (q 5 − 1)(q 3 − 1)(q 2 − 1) q 36 (q 12 − 1)(q 8 − 1)(q 6 − 1) (q 5 + 1)(q 3 + 1)(q 2 − 1) 26 .33 .5 27 .33 .52 214 .33 .53 .7.13 210 .33 .52 .73 27 .36 .53 .7 221 .39 .54 .72 .11.13 210 .37 .53 .7.11 217 .39 .52 .7.11 214 .36 .56 .7.11

(q p −1) (q−1)(p,q−1) (q p −1) (q−1) (q p +1) (q+1)(p,q+1) (q p +1) (q+1)

5

(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) n−1

2

+1

(3p +1) 4 (3n−1 +1) 2

q 2 − q + 1 q4 − q2 + 1 q4 − q2 + 1 13

(q 6 +q 3 +1) (3,q−1) (q 6 −q 3 +1) (3,q+1)

11 7 29 17 11 23 23 13 19

Characterizability of the Group 2 Dp (3) by Its Order Components

1119

Table 2 The order components of finite simple groups P with s(P ) = 3 (p an odd prime) P An

Restrictions on P

m1

m2

m3

n > 6, n = p, p − 2

n! 2n(n−2)

p

p−2

q−

q

(q+) 2

q

q−1

q+1

3

7

are primes A1 (q)

3 < q ≡ (mod 4),  = ±1

A1 (q)

q > 2, q even

A2 (2) 2

8 15

A5 (2)

6

2 .3 .5

7

2

Dp (3)

p = 2m + 1 ≥ 5

2.3p(p−1) (3p−1 − 1)  p−2 2i i=1 (3 − 1)

2

Dp+1 (2)

p = 2n − 1, n ≥ 2

2p(p+1) (2p − 1)  p−1 2i i=1 (2 − 1)

(3

11 p−1

2

+1)

(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

G2 (q)

q ≡ 0(mod 3)

q 6 (q 2 − 1)3

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

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

5

11

11

23

G2 (q)

F4 (q)

E7 (2)

17.19.31.43 E7 (3)

223 .363 .52 .73 .112 .132 . 19.37.41.61.73.547 4

2

M11

2 .3

M23

27 .32 .5.7 10

3

M24

2 .3 .5.7

11

23

J3

27 .35 .5

17

19

7

11

11

13

9

2

3

HiS

2 .3 .5

Suz

213 .37 .52 .7 18

6

3

Co2

2 .3 .5 .7

11

23

F i23

218 .313 .52 .7.11.13

17

23

F3

215 .310 .53 .72 .13

19

31

31

47

F2

41

13

6

2

2 .3 .5 .7 .11.13. 17.19.23

1120

P A2 (4) 2 B2 (q) 2

E6 (2) E8 (q)

M22 J1 O N LyS F i24 F1 E8 (q) J4

Darafsheh M. R.

Table 3 The order components of finite simple groups P with s(P ) > 3 Restrictions m1 m2 m3 m4 m5 on P 26 3 5 √ 7 √ 2m+1 q=2 > 2 q2 q−1 q − 2q q + 2q +1 +1 236 .39 .52 .72 .11 13 17 19 q 10 −q 5 +1 q 10 +q 5 +1 q ≡ 2, 3 q 120 (q 20 − 1)(q 18 − 1) q8 − q4 q 2 −q+1 q 2 +q+1 14 12 (mod 5) (q − 1)(q − 1) +1 (q 10 − 1)(q 8 − 1) (q 4 + 1)(q 4 + q 2 + 1) 27 .32 5 7 11 23 .3.5 7 11 19 29 .34 .5.73 11 19 31 28 .37 .56 .7.11 31 37 67 221 .316 .52 .73 .11.13 17 23 29 246 .320 .59 .76 .112 .133 . 41 59 71 17.19.23.29.31.47 10 5 10 5 q −q +1 q +q +1 q 10 +1 q ≡ 0, 1, 4 q 120 (q 18 − 1)(q 14 − 1) q8 − q4 q 2 −q+1 q 2 +q+1 q 2 +1 (mod 5) (q 12 − 1)2 (q 10 − 1)2 +1 (q 8 − 1)2 (q 4 + q 2 + 1) 221 .33 .5.7.113 23 29 31 37

m6

43

In [2–3] 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 [4]; P SL2 (q); 2 Dn (3) where 9 ≤ n = 2m + 1 is not a prime; 2 Dp+1 (2); in [5–6] and [7], respectively. Some projective special linear (unitary) groups have been characterized in a series of articles in [8–11]. A few of the alternating or symmetric groups are proved to be characterizable by their order components in [12] and [13]. On the basis of these results we put forward the following conjecture: 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 ∼ Bp (3), Cp (3) where p is an odd prime number. and q is an odd number, or G =

Conjecture 1

A motivation for characterizing finite groups by the set of their order components is the following conjecture due to Thompson: Conjecture 2 (Thompson) For a finite group G let N (G) = {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? In [14] it is proved that, if s(M ) ≥ 3, then the above conjecture holds. Also in [14] 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 Shi and Bi, which states: Conjecture 3

Let G be a group and M a finite simple group. Then G ∼ = M if and only if :

Characterizability of the Group 2 Dp (3) by Its Order Components

1121

(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 Dp (3) where p = 2m + 1, p ≥ 5 is a prime number, and prove that this group is characterizable by its order components. Another name for this group is P Ω− 2p (3). More precisely, we will prove: Main Theorem If a finite group G has the same set of order components as 2 Dp (3), p = 2m + 1, p ≥ 5 a prime, then G is isomorphic to 2 Dp (3). 2

Preliminary Results

The structure of finite groups with disconnected Gruenberg–Kegel graph follows from Theorem A of [15], 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; (2) G has a normal series 1  H  K  G such that H is a nilpotent π1 -group, K/H is a non-abelian simple group, G/K is a π1 -group, |G/K| divides |Out(K/H)| and any odd order component of G is equal to one of the odd order components of K/H. To deal with the first case in the above lemma we need the following results, which are taken from [16] and [4], 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 normal series 1  H  K  G such that |K/H| = m2 , |H| |G/K| = m1 and |G/K| divides |K/H| − 1 and H is a nilpotent π1 -group. s(G) Lemma 3 Let G be a finite group with s(G) ≥ 2. If H  G is a πi -group, then ( j=1,j=i mj )| (|H| − 1). The following result of Zsigmondy [17] 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 an − 1 such that p does not divide ai − 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 an − 1. Remark 1 If p is a Zsigmondy prime for an − 1, then p > n. Because if p ≤ n, then n = kp + r, 0 ≤ r < p, and we can write an − 1 = ar (akp − ak ) + ak+r − 1. Since (p, a) = 1 we have ap ≡ a(mod p), hence akp ≡ ak (mod p), therefore p | ak+r − 1. By assumption about p we

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Darafsheh M. R.

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 (q). Using [18] we have 2 Dn (q) = (4,qn1 +1) q n(n−1) n−1 (q n + 1) i=1 (q 2i − 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 [1] if p > 5 is a prime number not of the form 2m + 1, then s(2 Dp (3)) = 2. Therefore in this case the prime graph of 2 Dp (3) has two components. The two order components by Table 1 are: p−1 p m1 = 3p(p−1) i=1 (32i − 1) and m2 = 3 4+1 . The components of the graph GK(2 Dp (3)) are p−1 p π1 = π(3 i=1 (32i − 1)) and π2 = π( 3 4+1 ). 3

Proof of Main Theorem

We assume G is a finite group with OC(G) = {m1 , m2 }, where m1 and m2 are the order components of the group 2 Dp (3), where p ≥ 5 is a prime number not of the form 2m + 1. We will use Lemma 1. Therefore we start with 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 assume G is a Frobenius group with complement H and kernel K and derive a contradiction. By Lemma 2 we have OC(G) = {|H| , |K|}. Since |H| | |K| − 1 we must p−1 p have |H| < |K| , hence |K| = m1 = 3p(p−1) i=1 (32i − 1) and |H| = m2 = 3 4+1 . Let r be a p−1 Zsigmondy prime for 32(p−1) − 1, which exists by Lemma 4 because p ≥ 5. Then r | 3 2 +1 and from the order of 2 Dp (3) we observe that the order of a Sylow r-subgroup S of G, and hence p−1 K, is a divisor of 3 2 +1 . Since K is a nilpotent normal subgroup of G we deduce that S  G, p and using Lemma 3 the divisibility relation m2 | |S| − 1 is obtained. Considering m2 = 3 4+1 p−1 and |S| | 3 2 +1 a contradiction is obtained. Next assume that G is a 2-Frobenius group. By Lemma 2 (b) there is a normal series 1  H  K  G for G such that H is a nilpotent π1 -group, |K/H| = m2 and |G/K| | p−1 p (|K/H| − 1) = 3 4−3 = 3(3 4 −1) . Therefore |G/K| | 3p−1 − 1. Again if r is a Zsigmondy prime for 32(p−1) − 1, then r  3p−1 − 1, hence r  |G/K| , and from Lemma 2 we deduce that r | |H| . Since a Sylow r-subgroup of H has order rk and H is a nilpotent normal subgroup of G, using p Lemma 3 we deduce that m2 = 3 4+1 | r k − 1. But r k | 3p−1 + 1 and a contradiction is obtained. Therefore G is not a 2-Frobenius group and the lemma is proved. The following lemma is useful in our further investigations. We remind ourselves that, for a prime number r and a positive integer n, nr denotes the r-part of n, i.e. n = mnr where (m, r) = 1. p−1 Lemma 6 Let r be a prime divisor of i=1 (32i − 1). Then the largest power k of r is such   p−1 p−1 that r k | 2 Dp (3) satisfies r k | r 4(p−1) if r = 2 and r k | ((32s − 1)r )[ s ] r [ s(r−1) ] , where r = 2 and s is the least positive integer for which r | 32s − 1. Proof It is easy to verify that (32l −1)2 = 8(l)2 . Hence when l varies in the interval 1 ≤ l ≤ p−1, p−1 p−1 p−1 we obtain 2k | 8p−1 .2[ 2 ]+[ 4 ]+··· . Therefore, k ≤ 3(p−1)+ p−1 2 + 4 +· · · < 3(p−1)+p−1 = 4(p − 1), implying k < 4(p − 1).

Characterizability of the Group 2 Dp (3) by Its Order Components

1123

p−1 Now let r be an odd prime divisor of i=1 (32i − 1). Let s be the least positive integer for which r | 32s − 1. Let r | 32l − 1 and assume l = sv + u, where 0 ≤ u < s. Then 32l − 1 = 32sv+2u − 1 = (32sv − 1)32u + 32u − 1, and we deduce that r | 32u − 1. Therefore u = 0 and we obtain l = sv. But it is easy to verify that (32l − 1)r = (32sv − 1)r | (32s − 1)r (v)r , and p−1 p−1 therefore r k divides ((32s − 1)r )[ s ] r [ s(r−1) ] and the lemma is proved. By Lemmas 1 and 4, if G is a finite group with OC(G) = OC(2 Dp (3)), then there is a normal series 1  H  K  G for G such that K/H is a non-abelian simple group, H and G/K are π1 -groups and H is nilpotent. Moreover |G/K| divides |Out(K/H)| and the odd order component of G is one of the odd order components of K/H and s(K/H) ≥ 2. Since P = K/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. ∼ A2 (2), A2 (4), 2 A3 (2), 2 A5 (2), E7 (2), E7 (3), 2 E6 (2), 2 F4 (2) or one of the 26 Case 1 P = sporadic simple groups listed in Tables 1, 2 or 3. p

7

The odd order component of 2 Dp (3) is m2 = 3 4+1 ≥ 3 4+1 = 547. But by Tables 1–3, only p E7 (2) has two odd order components 757 and 1093 meeting the above range. Since 3 4+1 = 757 or 1093 has no solution for p, we obtain a contradiction. ∼ An and either n = p , p + 1, p + 2, one of n or n − 2 is not prime, or Case 2 P = 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 and(or) p − 2. If p − 2 = then 3p + 9 = 4p , which implies 9 | p , a contradiction. 

p

3p +1 4 ,



Therefore 3 4+1 = p . The largest power of 3 dividing p is [ p3 ] + [ p9 ] + · · · , and since   p p p . Since 3 −1−2p > p(p−1) p = 3 4+1 = 3p−1 +3p−2 +· · ·+1 we obtain: [ p3 ]+[ p9 ]+· · · = 3 −1−2p 4 4 holds for p ≥ 5 we deduce that |Ap |  |G| , and a contradiction is obtained. Case 3 P ∼ = E6 (q) or 2 E6 (q) with q > 2. In this case we have q 6 +q 3 +1 (3,q−1) ,

3p +1 4

=

q 6 +q 3 +1 (3,q−1)

or

3p +1 4

=

q 6 −q 3 +1 (3,q+1) .

We deal with the case

3p +1 4

=

the other case is similar.

If (3, q −1) = 1, then 3(3p−1 −1) = q 3 (q 3 +1). Obviously q cannot be a power of 3, hence q ≡ ±1(mod 3) from which it follows that q 3 ≡ ±1(mod 9). But then from 3(3p−1 − 1) = q 3 (q 3 + 1) we will obtain: −3 ≡ 1(mod 9) if q ≡ −1(mod 3) and −3 ≡ 0(mod 9) if q ≡ −1(mod 3), a contradiction. p+1 If (3, q − 1) = 3, then q 3 (q 3 + 1) = 3 4 −1 . Since (q 3 (q 3 + 1))2 divides the order of P ∼ = E6 (q)  p−1 p+1 p 2i p+1 − 1 | (3 + 1) i=1,i= p+1 (3 − 1). Let r be a Zsigmondy prime for 3 − 1. Then we get 3 2

by the order of G we have either r | 3p + 1 or r | 32i − 1 for some i, 1 ≤ i ≤ p − 1, i = p+1 2 . If p 2i i r | 3 + 1, then r | 2, a contradiction. If r | 3 − 1, then 2i > p + 1, and r | 3 + 1. Therefore r | 3p+1 + 3i = 3i (3p+1−i + 1), from which it follows that r | 3p+1−i + 1. Hence r | 32(p+1−i) − 1,

so 2(p + 1 − i) ≥ p + 1, implying 2i ≤ p + 1, a contradiction. Finally we deduce that the case P ∼ = E6 (q) is impossible. The impossibility of the case P isomorphic to 2 E6 (q), q > 2, is treated similarly. Case 4 P ∼ = F4 (q), q odd, 3 D4 (q), G2 (q), 2 < q ≡ (mod 3),  = ±1.

1124

Darafsheh M. R. p

p

In these cases we have 3 4+1 = q 4 − q 2 + 1 (for F4 (q) and 3 D4 (q)) and 3 4+1 = q 2 − q + 1 p for G2 (q). We rule out the possibility 3 4+1 = q 4 − q 2 + 1, because the other case is similar. From the last equality we obtain 3(3p−1 − 1) = 4q 2 (q 2 − 1). Obviously 3  q. Hence q ≡ ±1, ±2, ±4(mod 9) from which it follows that q 2 ≡ 1, 4, 7(mod 9). Now calculating the equality 3(3p−1 − 1) = 4q 2 (q 2 − 1) modulo 9, a contradiction is obtained. Case 5 P is isomorphic to one of the groups 2 Dn (3), 2 Dp (3), 2 Dn (2). p

n−1

p

p

n−1

In these cases we obtain 3 4+1 = 3 2 +1 , 3 4+1 , 2n−1 + 1, respectively. If 3 4+1 = 3 2 +1 , p then 2.3n−1 + 1 = 3p , a contradiction. If 3 4+1 = 2n−1 + 1, then 3p − 3 = 2n+1 , a contradiction. 3p +1 4

p

= 3 4+1 , then p = p , hence |K/H| = |G| implying H = 1 and K = G. Therefore G is isomorphic to 2 Dp (3) and this is the case we are trying to establish in our main theorem. ∼ Dp +1 (2), Dp +1 (3); Dp (2), Dp (3), Dp (5), p ≥ 5; 2 Dn (q), n = 2m ≥ 4. Case 6 P =

If

If P ∼ = Dp +1 (2), then 3 4+1 = 2p − 1 from which the equality 3p + 5 = 2p +2 is obtained. The largest power of 2 dividing the order of Dp +1 (2) is p (p + 1), and by Lemma 6 we must  have p (p + 1) ≤ 4(p − 1). From 3p + 5 = 2p +2 we obtain p + 2 > p, hence p + 1 > p − 1, and p p the inequality p (p + 1) ≤ 4(p − 1) does not hold. If P ∼ = Dp +1 (3), then 3 4+1 = 3 2−1 which  results in 3p + 3 = 2.3p , a contradiction. ∼ Dp (q), q = 2, 3, 5. The cases q = 2, 3 are treated as above, hence we Now we assume P = p







p p  continue with P ∼ = Dp (5). In this case we will obtain 3 4+1 = 5 4−1 , which results in 5p −3p = 2.   p−1   Since |P | | 2 Dp (3) we should have 5p (p +1) | i=1 (32i − 1) where p (p + 1) is the highest . From power of 5 dividing |P | . Now using Lemma 6 with s = 2 we obtain p (p + 1) ≤ 5(p−1) 8 p  p p this inequality we obtain p < 2 and this contradicts 5 − 3 = 2. n

p

q +1 Finally if P is isomorphic to 2 Dn (q), n = 2m ≥ 4, then 3 4+1 = (2,q+1) . If q is even, then p−1 n n p − 1) = 4.q , implying 3 | q, a contradiction. If q is odd, then 3 − 1 = 2q n , and from 3(3 2     Dn (q) | 2 Dp (3) we obtain 3p − 1 | p−1 (32i − 1). Now a Zsigmondy prime argument leads i=1 to a contradiction. ∼ Cp (2), Cp (3); Cn (q), n = 2m ≥ 2; Bp (3), Bn (q), n = 2m ≥ 4, q odd. Case 7 P =

In all of these cases the equations obtained are similar to those of Case 6 and can be dealt with in the same way. We omit the details here. Case 8 P is isomorphic to one of the groups 2 Ap (q), (q + 1) | (p + 1), (p , q) = (3, 3), (5, 2); 2 Ap −1 (q); Ap (q), (q − 1) | (p + 1); Ap −1 (q), (p , q) = (3, 2), (3, 4). The odd order components p

p

p

p

+1 q +1 q −1 q −1 of the above groups are: qq+1 , (q+1)(p  ,q+1) , q−1 and (q−1)(p ,q−1) , respectively. Letting p m2 = 3 4+1 equal the above expression, we have reached a contradiction. In the following we p p give only the details for P ∼ = Ap (q), (q − 1) | (p + 1). In this case we have q −1 = 3 +1 from q−1

4

which it follows that: 

4q p − (q − 1)3p = q + 3.

(1)

By (1) q cannot be a power of 3 because p, p ≥ 3. Hence q = rf , r = 3 a prime. From p−1 |Ap (q)| | |G| using Lemma 6 we obtain 12 p (p + 1) ≤ t[ p−1 s ] + [ s(r−1) ] where s is the least positive integer for which r | 32s − 1 and (32s − 1)r = r t . Hence we obtain p (p + 1) ≤ 2(t + 1)(p − 1).

(2)

Characterizability of the Group 2 Dp (3) by Its Order Components

1125

Using the above inequality we obtain  1 p < (−1 + 1 + 8(t + 1)(p − 1)). (3) 2   Now from (q −1) | (p +1) we obtain p +1 ≥ q −1, and considering (2) yields (q −2)(q −1) ≤ 2(t + 1)(p − 1), from which it follows that  1 q < (3 + 1 + 8(t + 1)(p − 1) ). (4) 2 If q = 2, then s = 1, r = 2, t = 3, and using the equality (1) we get p > 5. Now using the   inequality (3) we obtain p < 12 (−1 + 1 + 32(p − 1) ) < p. Since, by (1), we have 2p +2 − 3p = 5, p < p is a contradiction to this equality. If q = 5, similarly we obtain a contradiction. 

Therefore we assume q > 5, and using equation (1) we obtain 4q p − (q − 1)3p > 0 from which   , because q > 5. Therefore, using the inequality (3) from q p > 3p , it follows that q p > 3p√ we obtain q > 32p/(−1+ 1+8(t+1)(p−1)) , contradicting (4). Therefore the case P ∼ = Ap (q) is impossible. In the following case we deal with simple groups P with 3 components, which are listed in Table 2. Case 9 P ∼ = A1 (q), 3 < q ≡ (mod 4),  = ±1. According to Table 2, the order components of A1 (q) are q and q+ 2 . First we assume p = 3 4+1 . If  = −1, then 2q = 3p + 3 = 3(3p−1 + 1), implying q = 3, which is a contradiction p because p ≥ 5. If  = 1, then we will obtain q = (3 2−1) . Since |A1 (q)| | |G| we must have p−1 3p − 1 | i=1 (32i − 1). If p is a Zsigmondy prime for 3p − 1, then there is 1 ≤ i ≤ p − 1 such that p | 32i − 1, hence 2i ≥ p. Therefore p | 32i − 3p = 3p (32i−p − 1), implying p | m2i−p − 1, so 2i − p ≥ p. q+ 2

p

Hence we can have only m2 = 3 4+1 = q in both cases  = ±1. Now let |G/K| = t which, by Lemma 1, t | |Out(P )| = 2f where q = r f , r a prime number. Therefore |G| = t |K/H| |H| = p−2 3t (3p + 5) |H| . If p is a Zsigmondy prime t |P | |H| , implying 3p(p−1) (3p−1 + 1) i=1 (32i − 1) = 32 for 32(p−1) −1, then clearly p | 3p−1 +1, and by Remark 1, p > 2(p−1). Since 3p +5 and 3p−1 +1 are coprime we obtain p | t |H| which implies p | f or p | |H| . But from 3p + 1 = 4r f we obtain 3r f + r f − 3p = 1. Since r > 3, if f ≥ p, then r f − 3p ≥ 0 and so 3r f ≤ 1, which is impossible.  Therefore f < p < p2 + 1 and p | f cannot happen. Hence p | |H| , implying that the order of a Sylow p -subgroup S of H is a divisor of 3p−1 + 1. Since H is a nilpotent normal subgroup p of G, we obtain S  G, and by Lemma 3, m2 = 3 4+1 | |S| − 1 which is impossible because p−1 |S| | (3 2 +1) . Finally we have proved that the case P ∼ = A1 (q), 3 < q ≡ (mod 4),  = ±1 is impossible. Case 10 P ∼ = A1 (q), q > 2, q even. The odd order components of P are q − 1 and q + 1. Considering m2 = q ± 1 and similarly to Case 9, we have proved this case is impossible and we omit the details. Case 11 P is isomorphic to 2 Dp (3), p = 2m + 1 ≥ 5. In this case we have either m2 =

 −1

3p

2  (3p +1) , 4

+1



or

(3p +1) . 4

If m2 =

 −1

3p

2

+1



then 2.3p −1 + 1 = 3p ,

which is impossible, and if m2 = we will obtain p = p which is not the case because we are assuming that p is not of the form 2m + 1.

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Darafsheh M. R.

Case 12 P is isomorphic to 2 Dp +1 (2), p = 2n − 1 ≥ 3. The odd order components of P    are 2p + 1 and 2p +1 + 1. If we let m2 equal any of the above numbers we obtain that 2p or p−1  2p +1 equals 3(3 4 −1) , which is impossible. Case 13 P ∼ = A2 (4). This is a simple group with s(P ) = 4, and the odd order components are 3, 5 and 7. An easy computation shows the impossibility of m2 = 3, 5 or 7. Case 14 For the rest of groups P in Table 2 with s(P ) = 3 and in Table 3 with s(P ) > 3, the odd order components are of the form qf (q) + 1 where f (q) is a polynomial with coefficients in Z. If m2 = qf (q) + 1, then we obtain 4qf (q) = 3(3p−1 − 1). In the next step, considering the appropriate groups, the impossibility of the equations is deduced easily. The main theorem is proved now. Added in proof. Part of Conjecture 1 has been proved in [19]. Acknowledgements This research was done at the Mathematics Department of the University of North Carolina at Charlotte, USA, while the author had a visiting position in 2006–2007. The author deeply appreciates the kindness and hospitality of the chair and the cochair of the UNCC Mathematics Department. The support of the University of Tehran is also acknowledged. References [1] Kondratev, A. S.: On prime graph components of finite simple groups. Math. USSR Sbornik, 190(6), 787–797 (1989) [2] Khosravi, B., Khosravi, B., Khosravi, B.: The number of isomorphism classes of finite groups with the set of order components of C4 (q). Appl. Algebra Engrg. Comm. Comput., 15, 349–359 (2005) [3] Khosravi, A., Khosravi, B.: r−recognizability of Bn (q) and Cn (q) where n = 2m ≥ 4. J. Pure Appl. Algebra, 199, 149–165 (2005) [4] Chen, G. Y.: A new characterization of sporadic simple groups. Algebra Colloq., 3(1), 49–58 (1996) [5] Chen, G. Y.: A new characterization of P SL2 (q). Southeast Asian Bull. Math., 22, 257–263 (1998) [6] Chen, G. Y.: 2 Dn (3) (9 ≤ n = 2m + 1 not a prime) can be characterized by its order components. J. Appl. Math. Comput., 19(1-2), 353–362 (2005) [7] Shi, H., Chen, G. Y.: 2 Dp+1 (2) (5 < p = 2m − 1) can be characterized by its order components. Kumamoto J. Math., 18, 1–8 (2005) [8] Iranmanesh, A., Alavi, S. H., Khosravi, B.: A characterization of P SL(3, q) where q is an odd prime power. J. Pure. Appl. Algebra, 170(2–3), 243–254 (2002) [9] Iranmanesh, A., Alavi, S. H., Khosravi, B.: A characterization of P SL(3, q) for q = 2n . Acta Mathematica Sinica, English Series, 18(3), 463–472 (2002) [10] Iranmanesh, A., Khosravi, B., Alavi, S. H.: A characterization of P SU (3, q) for q > 5. Southeast Asian Bull. Math., 26(2), 33–44 (2002) [11] Khosravi, A., Khosravi, B.: A new characterization of P SL(p, q). Comm. Algebra, 32, 2325–2339 (2004) [12] Alavi, S. H., Daneshkhah, A.: A new characterization of alternating and symmetric group. J. Appl. Math. Comput., 17(1), 245–258 (2005) [13] Khosravi, A., Khosravi, B.: A new characterization of some alternating and symmetric groups. Int. J. Math. Math. Sci., 45, 2863–2872 (2003) [14] Chen, G. Y.: On Thompson’s conjecture. J. Algebra, 185, 184–193 (1996) [15] Williams, J. S.: Prime graph components of finite groups. J. Algebra, 69, 487–513 (1981) [16] Chen, G. Y.: On Frobenius and 2-Frobenius groups. J. Southwest China Normal Univ., 20(5), 485–487 (1995) [17] Zsigmondy, K.: zur theorie der potenzreste. Monatsh. Math. Phys., 3, 256–284 (1892) [18] Conway, J. H., Curtis, R. T., Norton, S. P., Parker, R. A., Wilson, R. A.: Atlas of Finite Groups, Clarendon Press, Oxford, 1985 [19] Darafsheh, M. R.: On non-isomorphic groups with the same set of order components. J. Korean. Math. Soc., 45, 137–150 (2008)