Groups with complete prime graph connected components

J. Group Theory 7 (2004), 373–384 Journal of Group Theory( de Gruyter 2004

Groups with complete prime graph connected components

Maria Silvia Lucido and Ali Reza Moghaddamfar

(Communicated by Y. Segev)

Abstract. Let G be a finite group and peðGÞ the set of all element orders of G. We prove thatif peðGÞ ¼ peðMÞ where G is a finite group and M is a finite simple group except PSLð3; 3Þ,PSUð3; 3Þ, PSpð4; 3Þ or A10, then G is non-soluble. We also classify the finite simple groups Gfor which the connected components of the prime graph are cliques.

1 Introduction

In one of his last papers [23], Suzuki studied the prime graph, using the methodsintroduced by W. Feit and J. G. Thompson in their Odd Order Paper [9] and modi-fied by H. Bender and G. Glauberman in [2].

The prime graph GðGÞ of a finite group G is defined as follows. The set of ver-tices of GðGÞ is the set pðGÞ of primes dividing jGj and two distinct primes p; q arejoined by an edge if there is an element in G of order pq. The number of connectedcomponents of GðGÞ is denoted by tðGÞ, and the vertex sets of the connected com-ponents are denoted by pi ¼ piðGÞ, i ¼ 1; 2; . . . ; tðGÞ. If 2 A pðGÞ we always assumethat 2 A p1.

In his paper, Suzuki proved the following result:

Proposition 1 ([23], Theorem B). Let G be a finite simple group whose prime graph

GðGÞ is not connected and let D be a connected component of GðGÞ whose vertex set

does not contain 2. Then D is a clique.

This proposition had already been verified by J. S. Williams in [24], using the clas-sification of the finite simple groups. However, Suzuki proved Proposition 1, andProposition 2 below, without using the classification and, as far as we know, hispaper is the first on the prime graph which does not use the classification.

Proposition 2 ([23], Theorem C). Let G be a finite simple group whose prime graph

GðGÞ is not connected and let D be a connected component of GðGÞ with vertex set piwhere id 2. Then G has a nilpotent pi-subgroup that is isolated in G.

Brought to you by | Penn State - The Pennsylvania State UniversityAuthenticated | 10.248.254.158

Download Date | 8/11/14 1:21 AM

The papers of Suzuki and Williams do not consider the connected component D1

of GðGÞ with vertex set p1. In [10], the groups admitting a p1-Hall subgroup werestudied, and it was shown, for example, that there are simple groups admitting asimple non-Abelian p1-Hall subgroup.

In this paper we characterize the finite simple groups such that D1 is a clique. Thiscondition is equivalent to saying that p1 is complete, i.e., any pair of vertices of p1 isjoined by an edge. Mazurov proved that if G is a soluble group with tðGÞ > 1 thenthe connected component D1 of GðGÞ is a clique, as shown in Proposition 3.

Both in [10] and here, we use the classification of finite simple groups.

Theorem 1. Let G be a finite simple group. Then all the connected components of

GðGÞ are cliques if and only if G is one of the following: A5, A6, A7, A9, A12, A13;M11, M22, J1, J2, J3, HS; PSLð2; qÞ with q > 2, SzðqÞ with q ¼ 22mþ1, PSpð4; qÞ,G2ð3kÞ, PSLð3; qÞ where q is a Mersenne prime, PSUð3; qÞ where q is a Fermat prime,PSLð3; 4Þ, PSUð3; 9Þ, PSpð6; 2Þ, PSUð4; 3Þ, PSUð6; 2Þ, Oþ

8 ð2Þ, 3D4ð2Þ.

We use Theorem 1 to prove a general result on the set of element orders of a finitegroup.

For a finite group G, we denote by peðGÞ the set of all element orders of G andby pðGÞ the set of all prime factors of jGj. It is clear that the set peðGÞ is closedand partially ordered by divisibility, and hence it is uniquely determined by mðGÞ,the subset of its maximal elements. Denote by mi ¼ miðGÞ the set of all n A mðGÞ suchthat each prime divisor of n belongs to pi.

Let G be a finite simple group whose prime graph GðGÞ is not connected. Then byProposition 2, jmiðGÞj ¼ 1 for 2c ic tðGÞ.

We denote by hðpeðGÞÞ the number of isomorphism classes of groups H withpeðHÞ ¼ peðGÞ and we say that a group G is characterizable if hðpeðGÞÞ ¼ 1 and non-

recognizable if hðpeðGÞÞ ¼ y. Soluble groups are non-recognizable (see [14]). How-ever some simple groups are known to be characterizable (see [1], [3], [12], [13], [14],[15], [18], [25]). In establishing this, the first step is to prove that if M is a finite simplegroup and G is a finite group such that peðGÞ ¼ peðMÞ, then G is not soluble. Herewe prove that this is the case for every simple group except PSLð3; 3Þ, PSUð3; 3Þ,PSpð4; 3ÞGPSUð4; 2Þ and A10.

Theorem 2. Let M be a finite simple group, except PSLð3; 3Þ, PSUð3; 3Þ, PSpð4; 3Þ andA10, and G be a group such that peðGÞ ¼ peðMÞ. Then G is non-soluble.

If M is isomorphic to PSLð3; 3Þ, PSUð3; 3Þ, PSpð4; 3Þ or A10, then hðpeðMÞÞ ¼ y.

2 Main results

In this section we prove Theorem 1 and Theorem 2. First we state a propositionabout soluble groups due to Mazurov [15].

Proposition 3 ([15], Lemmas 5 (a), (b) and 7 (1)). Let G be a finite soluble group whose

Maria Silvia Lucido and Ali Reza Moghaddamfar374



prime graph GðGÞ is not connected and let D be a connected component of GðGÞ. ThenG has an element of order

Qr AD r, and therefore D is a clique.

Let ni be the only element of miðGÞ, for id 2. We use the classification of finitesimple groups and consider first the case when G is a sporadic or an alternatinggroup. The following statement is easily proved using the Atlas [7].

Lemma 1. Let G be a sporadic simple group. Then p1ðGÞ is complete if and only if G is

one of the following: M11, M22, J1, J2, J3, HS.

Before considering alternating groups, we recall two useful lemmas:

Lemma 2 (see [25]). Let An be the alternating group on n elements and

m ¼ pa11 pa2

2 . . . pass , where p1; p2; . . . ; ps are distinct primes and a1; a2; . . . ; as; s

are natural numbers. Then An has an element of order m if and only if

pa11 þ pa2

2 þ � � � þ pass c n for odd m and pa1

1 þ pa22 þ � � � þ pas

s c n� 2 for even m.

Lemma 3 (see [12]). If nd 6 then there are at least sðnÞ primes pi such that12 ðn� 1Þ < pi < n. Here

sðnÞ ¼ 6 for nd 48,

sðnÞ ¼ 5 for 42c nc 47,




sðnÞ ¼ 1 for 6c nc 13.

Lemma 4. Let G be an alternating group of degree nd 5. Then p1ðGÞ is complete if

and only if n ¼ 5; 6; 7; 9; 12; 13.

Proof. If nc 37, the result is straightforward. If n > 37, by Lemma 3 there are dis-tinct primes p1; p2; p3; p4 such that

12 ðn� 1Þ < p1 < p2 < p3 < p4 < n:

Since tðAnÞc 3, and piðAnÞ contains exactly one prime for id 2, we concludethat p1; p2 A p1ðGÞ. On the other hand, since p1 þ p2 > n, by Lemma 2 we getp1 p2 B peðGÞ. Therefore the connected component p1ðGÞ is not complete.

We suppose now that G is a finite simple group of Lie type defined over a fieldGFðqÞ of order q ¼ pk, for some prime p. An element x of G is called semisimple ifand only if x is a p 0-element. We now quote some useful lemmas.

Groups with complete prime graph connected components 375



Lemma 5 (see [20]). Let a; b A PSLðn; qÞ where q ¼ pm, nd 4, jaj ¼ p and ab ¼ ba.Then pðjbjÞJ pðSLðn� 2; qÞÞ.

Lemma 6 (Zsigmondy’s Theorem, [26]). Let qd 2 and nd 3 and suppose that

ðq; nÞ0 ð2; 6Þ. Then qn � 1 has a ‘primitive’ prime divisor, that is, a prime divisor

which does not divide q j � 1 for j ¼ 1; 2; . . . ; n� 1.

We denote by qn one of the primitive prime divisors of qn � 1. It is an easy conse-quence of Lemma 6 that if qn divides qm � 1 then md n. In the following case-by-case analysis, we choose primitive prime divisors qm; qn of q

n � 1 and qm � 1, and weprove that there exist elements of orders qn and qm such that G has no element oforder qnqm. Such an element of order qnqm would be semisimple and therefore wouldbe contained in a maximal torus (see [4]). Thus we look for primes qn; qm A p1ðGÞsuch that qnqm does not divide the order of a maximal torus of G.

The orders of the finite simple groups of Lie type can be found in [4], and theorders of their maximal tori can be deduced from [5].

Lemma 7. Let G be a finite simple group of Lie type. Then the connected component

p1 of GðGÞ is complete if and only if G is one of the following groups: PSLð2; qÞ withq > 2, SzðqÞ with q ¼ 22mþ1, PSpð4; qÞ, G2ð3kÞ, PSLð3; qÞ where q is a Mersenne

prime, PSUð3; qÞ where q is a Fermat prime, PSLð3; 4Þ, PSUð3; 9Þ, PSpð6; 2Þ,PSUð4; 3Þ, PSUð6; 2Þ, Oþ

8 ð2Þ, 3D4ð2Þ.

Proof. We consider the di¤erent Lie types in turn.

Case 1. GGAlðqÞGPSLlþ1ðqÞ. The orders of maximal tori areQki¼1ðqri � 1Þ

ðq� 1Þðl þ 1; q� 1Þ ; ðr1; . . . ; rkÞ A Parðl þ 1Þ:

We choose ql�1 and ql�2 in p1ðGÞ. For ld 5, we have ðl � 1Þ þ ðl � 2Þd l þ 2 and sono maximal torus has order divisible by ql�1ql�2.

If l ¼ 4 we choose q4 and q3. If l ¼ 3, we choose q4 A p1ðGÞ. Then by Lemma 5 wehave q4 S p.

If q ¼ 2, we cannot apply Zsigmondy’s Theorem for l ¼ 7; 8; in this case, wechoose 17; 31 A p1ðGÞ.

For the simple groups PSLð2; qÞ with q ¼ pn, we have

mðPSLð2; qÞÞ ¼ p;q� 1

d;qþ 1

d

� �;

where d ¼ ðq� 1; 2Þ. Thus p1ðGÞ is complete whenever q > 3.We study together the groups PSLð3; qÞ and PSUð3; qÞ, and, in order to

unify our treatment, we introduce the following notation. For e A fþ;�g we letPSL eð3; qÞ ¼ PSLð3; qÞ if e ¼ þ; and PSLeð3; qÞ ¼ PSUð3; qÞ if e ¼ �. We write q� e

for q� e1.




The set of element orders peðPSL eð3; qÞÞ for e ¼G can be found in [1] and [8]; wehave

mðPSLeð3; qÞÞ ¼ q� e; 13 pðq� eÞ; 13 ðq2 � 1Þ; 13 ðq2 þ eqþ 1Þ� �

if d ¼ 3;

fpðq� eÞ; q2 � 1; q2 þ eqþ 1g if d ¼ 1;

(

where q ¼ pn is odd and d ¼ ð3; q� eÞ, and

mðPSL eð3; 2nÞÞ ¼ 4; 2n � e; 23 ð2n � eÞ; 13 ð22n � 1Þ; 13 ð22n þ e2n þ 1Þ� �

if d ¼ 3;

f4; 2ð2n � eÞ; 22n � 1; 22n þ e2n þ 1g if d ¼ 1;

(

where d ¼ ð3; 2n � eÞ, except when ðe; nÞ A fðþ; 1Þ; ðþ; 2Þg.First assume that q is odd. From the form of mðGÞ it is easy to see that rS p for

all odd primes r A pðqþ eÞ. Hence p1ðGÞ is complete if and only if pðqþ eÞ ¼ f2g.Finally, for e ¼ þ, q is a Mersenne prime and for e ¼ �, either q ¼ 9 or q is a Fermatprime.

Next suppose that q is even and ðe; nÞ0 ðþ; 1Þ; ðþ; 2Þ. In this case, for all primest A pð2n þ eÞ we have tS 2. On the other hand, for all primes r A pð2n þ eÞ ands A pð2n � eÞ, we have r@ s. Thus p1ðGÞ is complete if and only if pð2n þ eÞ ¼ q.The latter equation has no solution for nd 2. When ðe; nÞ ¼ ðþ; 1Þ, we havePSLð3; 2ÞGPSLð2; 7Þ, which has already been considered. If ðe; nÞ ¼ ðþ; 2Þ, then wehave mðPSLð3; 4ÞÞ ¼ f3; 4; 5; 7g, and so p1ðPSLð3; 4ÞÞ ¼ f2g is complete.

Case 2. GGBlðqÞGPW2lþ1ðqÞ. The order of a maximal torus is of the form

Qki¼1ðqri � 1Þ

Qmj¼1ðqsj þ 1Þ

ð2; q� 1Þ ðr1; . . . ; rk; s1; . . . ; smÞ A ParðlÞ:

We choose q2ðl�1Þ and q2ðl�2Þ in p1ðGÞ. If ld 4 then ðl � 1Þ þ ðl � 2Þd l þ 1 and sono maximal torus has order divisible by q2ðl�1Þq2ðl�2Þ. If l ¼ 3 we choose q6 and q4.

Case 3. GGClðqÞGPSpð2l; qÞ. This case is similar to Case 2. For q ¼ 2 we cannotapply Zsigmondy’s Theorem when l ¼ 3; 4; 5. If l ¼ 4, we choose 5; 7; if l ¼ 5 wechoose 11; 7. If l ¼ 3, then p1ðPSpð6; 2ÞÞ is complete.

We must also consider the case when l ¼ 2, so that G ¼ PSpð4; qÞ. By [11] and[24] we have p1ðGÞ ¼ pðqðq2 � 1ÞÞ and p2ðGÞ ¼ pððq2 þ 1Þ=dÞ; where d ¼ ð2; q� 1Þ.Since G contains L2ðqÞ � L2ðqÞ as a subgroup, it is easy to see that p1ðGÞ is complete.

Case 4. GGDlðqÞGPWþ2lðqÞGOþ

2lðqÞ. The order of a maximal torus is of the form

Qki¼1ðqri � 1Þ

Qmj¼1ðqsj þ 1Þ

ð4; ql � 1Þ ðr1; . . . ; rk; s1; . . . ; smÞ A ParðlÞ; m even:

We choose q2ðl�1Þ and q2ðl�2Þ in p1ðGÞ. If ld 4 then ðl � 1Þ þ ðl � 2Þd l þ 1 and so




no maximal torus has order divisible by q2ðl�1Þq2ðl�2Þ. If q ¼ 2, we cannot applyZsigmondy’s Theorem when l ¼ 4; 5. If l ¼ 5, we choose q8 ¼ 17; q3 ¼ 7. For l ¼ 4we have that p1ðOþ

8 ð2ÞÞ is complete.

Case 5. In the following table, for the simple group G of type indicated in the firstcolumn, we give the corresponding primes qn; qm A p1ðGÞ with qnqm B peðGÞ.

G primes

E6ðqÞ q5, q8

E7ðqÞ q5, q8

E8ðqÞ q8, q9

F4ðqÞ q3, q4

2E6ðqÞ q8, q10

2F4ðqÞ q2, q4

Case 6. GGG2ðqÞ. In [19] it was proved that q3 S p and q6 S p. Therefore, ifq1 1 ðmod 3Þ, we choose q3 A p1ðGÞ and if q1�1 ðmod 3Þ, we choose q6. If q ¼ 3k,then p1ðGÞ ¼ pð3ðq2 � 1ÞÞ. By [6] we have q2 � 1 A peðGÞ and by [22] we havePSLGð3; qÞ ,! G2ðqÞ, which implies that 3ðpG 1Þ A peðGÞ. Now it is easy to see thatp1ðGÞ is complete.

Case 7. GG2 AlðqÞGPSUlþ1ðqÞ. The orders of the maximal tori areQki¼1ðqri � 1Þ

Qmj¼1ðqsj þ 1Þ

ðqþ 1Þðqþ 1; l þ 1Þ ðr1; . . . ; rk; s1; . . . ; smÞ A Parðl þ 1Þ; ri even; sj odd:

If l is odd, we choose ql�1 and q2ðl�2Þ in p1ðGÞ. If ld 5 then ðl � 1Þ þ ðl � 2Þd l þ 2and so no maximal torus has order divisible by ql�1q2ðl�2Þ. If l ¼ 3 and q0 2; 3 wechoose q4 and q6. If q ¼ 3 then q6 B p1ðGÞ and in fact p1ðPSUð4; 3ÞÞ is complete. Ifq ¼ 2 then PSUð4; 2ÞGPSpð4; 3Þ has already been described.

If l is even, then we choose q2ðl�1Þ and ql�2 in p1ðGÞ. If ld 6 thenðl � 1Þ þ ðl � 2Þd l þ 2 and so no maximal torus has order divisible by q2ðl�1Þql�2.If l ¼ 4 we choose q4 and q6. If l ¼ 2, then we have G ¼ PSUð3; qÞ, which has al-ready been considered.

When q ¼ 2, Zsigmondy’s Theorem does not apply for l ¼ 4; 5; 7; 8. For l ¼ 7; 8we can choose 17; 43 A p1ðGÞ. For l ¼ 4, we can choose 2; 5. For l ¼ 5 we have thatp1ðPSUð6; 2ÞÞ is complete.

Case 8. GG 2B2ðqÞG SzðqÞ. In this case p1ðSzðqÞÞ ¼ f2g and it is therefore complete.

Case 9. GG 2DlðqÞGPW�2lðqÞ. The orders of the maximal tori are




Qki¼1ðqri � 1Þ

Qmj¼1ðqsj þ 1Þ

ð4; ql þ 1Þ ðr1; . . . ; rk; s1; . . . ; smÞ A ParðlÞ; m odd:

If q0 2; 3, we choose q2ðl�1Þ and q2ðl�2Þ in p1ðGÞ. For ld 4 we haveðl � 1Þ þ ðl � 2Þd l þ 1 and so no maximal torus has order divisible by q2ðl�1Þq2ðl�2Þ.

If q ¼ 2, and l ¼ 4 we choose 5; 7; if q ¼ 2 and l ¼ 5 we choose 11; 7.If q ¼ 2; 3 and l0 2m þ 1 for some md 2, then again q2ðl�1Þ and q2ðl�2Þ are in

p1ðGÞ. If q ¼ 2 and l ¼ 2m þ 1 for some md 3, then q2ðl�2Þ and q2l are in p1ðGÞ. Ifq ¼ 3 and l ¼ 2m þ 1 is not a prime then q2ðl�1Þ and q2l are in p1ðGÞ. If q ¼ 3 andl ¼ 2m þ 1 is a prime then ld 5, and ql�1 and q2ðl�2Þ are in p1ðGÞ.

Case 10. GG 3D4ðqÞ. We choose q3 and q6 in p1ðGÞ. Then q3q6 does not divide the

order of any maximal torus of G. For q ¼ 2 we have that p1ð3D4ð2ÞÞ is complete.

Case 11. GG 2G2ðqÞGRðqÞ with q ¼ 32mþ1 > 3. By [3] we have

mðGÞ ¼�6; 9; q� 1; 12 ðqþ 1Þ; q�

ffiffiffiffiffi3q

pþ 1; qþ

ffiffiffiffiffi3q

pþ 1

�:

Hence for all primes 20 r A pðq� 1ÞU p 12 ðqþ 1Þ� �

we have 3r B peðGÞ. Thus p1ðGÞis complete if and only if pðq� 1Þ ¼ p 1

2 ðqþ 1Þ� �

¼ f2g, and hence we must have32mþ1 � 1 ¼ 2k for some k. But this equation has no solution for md 1. Thereforep1ðGÞ is not complete.

This completes the proof of Lemma 7.

Lemmas 1, 4 and 7 and the classification of the finite simple groups complete theproof of Theorem 1.

To prove Theorem 2, we observe that if G is a soluble group, then tðGÞc 2 (see[24]). Therefore we only have to consider simple groups M such that tðMÞc 2. Werecall the following easy lemma.

Lemma 8. Let M be a finite group and jpðMÞjd 3. If there exist primes r; s; t A pðMÞsuch that ftr; ts; rsgV peðMÞ ¼ q, then M is non-soluble.

Proof. Suppose that M is soluble and let H be a Hall fr; s; tg-subgroup of M. ThenH is soluble and tðHÞd 3, contradicting the above statement.

We begin with the case tðMÞ ¼ 1, that is, pðGÞ ¼ p1ðGÞ. The groups in which GðGÞis not connected are listed in [24] and [11] and we shall refer to the list throughout theproof of Lemma 9.

Lemma 9. If G is a finite simple group and GðGÞ is connected, then there exist three

primes r; s; t satisfying the hypothesis of Lemma 8, unless G ¼ A10.

Proof. We consider first the alternating group An of degree nd 5. By Lemma 3,if nd 18 there exist three primes p1; p2; p3 such that 1

2 ðn� 1Þ < p1 < p2 < p3 < n,




and p1; p2; p3 satisfy the conditions of Lemma 8. Therefore only the groups A10;A16

remain, and in A16 the primes 7; 11; 13 satisfy the conditions of Lemma 8.Let G be a simple group of Lie type. As before, qn denotes a primitive prime divi-

sor of qn � 1.If G ¼ PSLðl þ 1; qÞ, then ld 3 and by Lemma 5 the primes p, qlþ1, ql satisfy the

conditions of Lemma 8 unless q ¼ 2 and l ¼ 6; 7. But in both of these cases GðGÞ isnot connected.

If G ¼ PW2lþ1ðqÞ or G ¼ PSpð2l; qÞ, then the primes q2l , ql and q2ðl�1Þ satisfy theconditions of Lemma 8.

If G ¼ PWþ2lðqÞ and ld 5 then ql , q2ðl�1Þ and q2ðl�2Þ are distinct and satisfy the con-

ditions of Lemma 8. If l ¼ 4, then q0 2, and therefore q4, q3 and q6 are distinct andsatisfy the conditions of Lemma 8.

If G ¼ E7ðqÞ then q5, q8 and q9 satisfy the conditions of Lemma 8.If G ¼ PSUlþ1ðqÞ and l is odd with ld 5, then q2l , ql�1 and qlþ1 satisfy the con-

ditions of Lemma 8, unless q ¼ 2 and l ¼ 5; 7. If q ¼ 2 and l ¼ 5, then GðGÞ is notconnected. If l ¼ 7, then G ¼ PSUð8; 2Þ, and the primes 17; 31; 43 satisfy the con-ditions of Lemma 8. If l ¼ 3 and q0 2, the primes p, q4 and q6 satisfy the conditionsof Lemma 8 (see [16]). If l ¼ 3 and q ¼ 2, then GðGÞ is not connected.

If G ¼ PSUlþ1ðqÞ and l is even, then ld 8, and the primes ql , q2ðlþ1Þ and q2ðl�1Þsatisfy the conditions of Lemma 8.

If G ¼ PW�2lðqÞ, then q2l , q2ðl�1Þ and q2ðl�2Þ satisfy the conditions of Lemma 8.

This completes the proof of Lemma 9.

If tðMÞ ¼ 2 and p1ðMÞ is not complete, then there exist primes r; s A p1ðMÞ andt A p2ðMÞ satisfying the conditions of Lemma 8. Therefore G is non-soluble. Thuswe only have to consider the groups M such that tðMÞ ¼ 2 and p1ðMÞ is com-plete, that is, the following groups appearing in Theorem 1: J2, A9, A12, PSpð4; qÞ,PSLð3; qÞ where q is a Mersenne prime, PSUð3; qÞ where q is a Fermat prime orq ¼ 9, PSpð6; 2Þ, Oþ

8 ð2Þ, 3D4ð2Þ.The following simple groups are characterizable: A9, A12, PSUð3; 9Þ, 3D4ð2Þ. The

simple groups J2, PSpð6; 2Þ and Oþ8 ð2Þ are not characterizable; however it has been

proved that if peðGÞ ¼ peðMÞ, then G is non-soluble (see [13], [15], [18] and [21]).Therefore we only have to consider the groups PSpð4; qÞ, PSLð3; qÞ where q is aMersenne prime and PSUð3; qÞ where q is a Fermat prime.

Case 1. MGPSpð4; qÞ. We observe that

mðGÞ ¼ 12 ðq

2 þ 1Þ; 12 ðq2 � 1Þ; pðqþ 1Þ; pðq� 1Þ

� �ð1Þ

if q ¼ pn, p0 2; 3, and

mðGÞ ¼ q2 þ 1

ð2; q� 1Þ ;q2 � 1

ð2; q� 1Þ ; pðqþ 1Þ; pðq� 1Þ; p2� �

; ð2Þ

if q ¼ pn, p ¼ 2 or 3. Therefore GðGÞ is not connected. If G is soluble, this implies




that G is either a Frobenius group or a 2-Frobenius group. By Proposition 3, we getthat either q� 1 ¼ 2a or qþ 1 ¼ 2b for some a; b.

Suppose that G is a Frobenius group, with kernel K and complement C. Since K isnilpotent, if p1ðGÞ ¼ pðKÞ we obtain an element x in K of order 1

2 pðq2 � 1Þ, contra-dicting (1) and (2). Therefore pðCÞ ¼ p1ðGÞ and the center of C contains an elementof order 2. If q ¼ 3 f , then C contains an element of order 2p2, contradicting (2).

We can now suppose that pd 5. Then q� 1 ¼ 2a implies that q ¼ p is a Fermatprime, while qþ 1 ¼ 2b implies that q ¼ p is a Mersenne prime. Since C is a solvableFrobenius complement, C has a normal subgroup N such that C=N is isomorphic toa subgroup of S4 and such that all Sylow subgroups of N are cyclic (see for example[17, p. 196]). This implies that

N ¼ hx; y j xm ¼ 1; yn ¼ 1; xy ¼ xri;

with m odd and ðm; nðr� 1ÞÞ ¼ 1. The subgroup M ¼ hxi is normal in C. Sincep0 2; 3, the prime p divides jNj. Therefore a p-Sylow subgroup P of C is cyclic oforder p. If p divides m, then P is normal in C. If p does not divide m then, since m

is odd, m divides 12 ðpþ 1Þ if p is a Fermat prime or 1

2 ðp� 1Þ if p is a Mersenneprime. Since p > 1

2 ðpþ 1Þ > 12 ðp� 1Þ, we have that P centralizes M. Then, since MP

is normal in C, we have also that P is normal in C. By (1) there is a subgroup ofprime order r, with r dividing 1

2 ðpþ 1Þ or 12 ðp� 1Þ respectively, which centralizes P.

Since the Sylow r-subgroups of C are cyclic and conjugate, any subgroup of orderr centralizes P.

Let 2m be the exponent of a Sylow 2-subgroup of C. By (1) there is a subgroup oforder 2m which centralizes P. Moreover, all subgroups of order 2m are conjugate andtherefore they all centralize P. Since there exist a subgroup R of order r and a sub-group D of order 2m such that ½R;D� ¼ 1, we conclude that ½R;DP� ¼ 1, which con-tradicts (1).

If G is a 2-Frobenius group, then we may write G ¼ ABC, in the notation of [15,Lemma 7]. Here B is a cyclic group of order ðq2 þ 1Þ=ð2; q� 1Þ and C is cyclic oforder t dividing ðq2 � 1Þ=ð2; q� 1Þ if q is odd and dividing q2 if q is even. Moreover

G contains an element of order t �Y

r A pðAÞr: ð�Þ

If q is even, then A has an element of order q2 � 1, and so C is a Sylow 2-subgroup of G which is cyclic of order 4. Then by ð�Þ there exists an element of order4 �

Qr A pðAÞ r, contradicting (2).

For q ¼ 3 we have PSpð4; 3ÞGPSUð4; 2Þ. We can therefore suppose that 3 < q

and that q is odd. Then a Sylow p-subgroup of G is contained in A.By ð�Þ, either q� 1 or qþ 1 is a power of 2, and t is either 1

2 ðqþ 1Þ or 12 ðq� 1Þ.

First suppose that t ¼ 12 ðqþ 1Þ. By ð�Þ, G has an element of order

1

2ðqþ 1Þ � p �

Yr A pðq�1Þ

r:




If qþ 1 ¼ 2m, this implies that q ¼ 3, against our hypothesis. If q� 1 ¼ 2m, then t isodd. By ð�Þ, we can suppose that there exist an element c A C of order 1

2 ðqþ 1Þ andan element x A A of order p such that ½x; c� ¼ 1. By (1) and the fact that C is a cyclicp 1

2 ðqþ 1Þ� �

-Hall subgroup of G, there exists an element a A A of order q� 1 suchthat ½a; c� ¼ 1. Since A is nilpotent we have ½a; x� ¼ 1 and therefore xac has order12 pðq� 1Þðqþ 1Þ, which implies that q ¼ 3, contradicting our hypothesis.

The case when t ¼ 12 ðq� 1Þ is similar.

Case 2. MGPSLeð3; qÞ with qd 4. We observe that in the cases considered we have

mðPSL eð3; qÞÞ ¼ p� e; 13 pðp� eÞ; 13 ðp2 � 1Þ; 13 ðp

2 þ epþ 1Þ� �

: ð3Þ

This implies that G has an element of order 8.Since GðGÞ is not connected, if G is soluble then G is either a Frobenius group or a

2-Frobenius group (see [24]).Suppose that G is a Frobenius group with kernel K and complement C. Since the

kernel is nilpotent we have pðCÞ ¼ p1ðGÞ. Moreover ZðCÞ has only one element z oforder 2. Let F ¼ FitðCÞ, let S be a Sylow 2-subgroup of C and P a Sylow p-subgroupof C of order p, and let Q ¼ O2ðCÞ.

If p divides jF j, then S acts on P and S=CSðPÞ is a subgroup of AutðPÞ and so hasorder 2. Then CSðPÞ has a cyclic group of order 4, contradicting (3).

Therefore p divides jC=F j. If Q is cyclic, then G=F is Abelian and thereforejQjd 4. If Q is a quaternion group, then jQjd 8. Since P acts on Q and CPðQÞ ¼ 1the group P is isomorphic to a subgroup of AutðQÞ which has order divisible only bythe primes 2; 3. This contradiction proves that G cannot be a Frobenius group.

If G is a 2-Frobenius group, then we write G ¼ ABC, in the notation of [15,Lemma 1]. Then B is a cyclic group of order 1

3 ðp2 þ epþ 1Þ and jCj divides13 ðp� e1Þðpþ e2Þ. Therefore A has an element of order p and an element of order12 ðpþ e1Þ. But A is nilpotent and so there must also be an element of order12 pðp� e1Þ contradicting (3).

This concludes the proof of Theorem 2.

It was proved in [14] that if M is isomorphic to PSUð3; 3Þ, PSUð4; 2ÞGPSpð4; 3Þor A10, then hðpeðMÞÞ ¼ y. We conclude with an example showing that the conclu-sion of Theorem 2 does not hold for M ¼ PSLð3; 3Þ.

Example. We recall that mðPSLð3; 3ÞÞ ¼ f6; 8; 13g. We define the soluble group H tobe a non-split extension of Q8 by S3. Then mðHÞ ¼ f6; 8g. Now H has an irreduciblefixed-point free representation of degree 2 over a field of characteristic 13. Thus thereexists an elementary Abelian group K of exponent 13 on which the action of H isfixed-point free. If G ¼ KH, then G is soluble and mðGÞ ¼ f6; 8; 13g ¼ mðPSLð3; 3ÞÞ.

Acknowledgments. We thank the referee for very useful remarks and ProfessorCarlo Casolo for a helpful suggestion concerning the proof of Theorem 2.




References

[1] M. R. Aleeva. On the composition factors of finite groups having the same set of elementorders as the group U3ðqÞ. Siberian Math. J. 43 (2002), 195–211.

[2] H. Bender and G. Glauberman. Local analysis for the odd order theorem. London Math.Soc. Lecture Note Ser. 188 (Cambridge University Press, 1985).

[3] R. Brandl and W. J. Shi. A characterization of finite simple groups with Abelian Sylow2-subgroups. Ricerche Mat. 42 (1993), 193–198.

[4] R. W. Carter. Finite groups of Lie type: conjugacy classes and complex characters (J. Wiley& Sons, 1985).

[5] R. W. Carter. Conjugacy classes in the Weyl group. Compositio Math. 25 (1972), 1–59.[6] B. Chang. The conjugate classes of Chevalley groups of type ðG2Þ. J. Algebra 9 (1968),

190–211.[7] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker and R. Wilson. Atlas of finite

groups (Clarendon Press, 1985).[8] M. R. Darafsheh and A. R. Moghaddamfar. Recognition of simple groups PSLð3; qÞ by

their element orders. Acta Math. Sci. 23 (2003), to appear.[9] W. Feit and J. G. Thompson. Solvability of groups of odd order. Pacific J. Math. 13

(1963), 775–1029.[10] E. Jabara and M. S. Lucido. Finite groups with Hall coverings. J. Austral. Math. Soc., to

appear.[11] A. S. Kondrat’ev. Prime graph components of finite simple groups. Mat. Sb. 180 (1989),

787–797.[12] A. S. Kondrat’ev and V. D. Mazurov. Recognition of alternating groups of prime degree

from their element orders. Siberian Math. J. 41 (2000), 294–302.[13] V. D. Mazurov. Characterizations of finite groups by sets of orders of their elements.

Algebra and Logic 36 (1997), 23–32.[14] V. D. Mazurov. Recognition of finite groups by a set of orders of their elements. Algebra

and Logic 37 (1998), 371–378.[15] V. D. Mazurov. Recognition of finite simple groups S4ðqÞ by their element orders. Alge-

bra Logic 41 (2002), 93–110.[16] S. Nozawa. On the characters of the finite general unitary group Uð4; qÞ. J. Fac. Sci. Univ.

Tokyo Sec. IA Math. 19 (1972), 257–293.[17] D. S. Passman. Permutation groups (W. A. Benjamin Inc., 1968).[18] C. E. Praeger and W. J. Shi. A characterization of some alternating and symmetric

groups. Comm. Algebra 22 (1994), 1507–1530.[19] W. J. Shi. Pure quantitative characterization of finite simple groups. I. Progr. Natur. Sci. 4

(1994), 316–326.[20] W. J. Shi and J. Bi. A characteristic property for each finite projective special linear

group. In Groups—Canberra 1989, Lecture Notes in Math. 1456 (Springer-Verlag, 1990),pp. 171–180.

[21] W. J. Shi. A new characterization of the sporadic simple groups. In Group theory (Sin-gapore 1987) (de Gruyter, 1989), pp. 531–540.

[22] E. Stensholt. Certain embeddings among finite groups of Lie type. J. Algebra 53 (1978),136–187.

[23] M. Suzuki. On the prime graph of a finite simple group—an application of the methodof Feit–Thompson–Bender–Glauberman. In Groups and combinatorics—in memory of

Michio Suzuki, Adv. Stud. Pure Math. 32 (Mathematical Society of Japan, 2001), pp.41–207.




[24] J. S. Williams. Prime graph components of finite groups. J. Algebra 69 (1981), 487–513.[25] A. Zavarnitsin and V. D. Mazurov. Element orders in coverings of symmetric and alter-

nating groups. Algebra and Logic 38 (1999), 159–170.[26] K. Zsigmondy. Zur Theorie der Potenzreste. Monatsh. Math. Phys. 3 (1892), 265–284.

Received 6 February, 2003; revised 29 September, 2003

M. S. Lucido, Dipartimento di Matematica e Informatica, Universita di Udine, via delleScienze 200, I-33100 Udine, ItalyE-mail: [email protected]

A. R. Moghaddamfar, Department of Mathematics, Faculty of Science, Imam-Ali University,P.O. Box 13145-689, Tehran, IranE-mail: [email protected]




Documents

Groups with complete prime graph connected components