4-free groups

J. Group Theory 16 (2013), 275 –298 DOI 10.1515/ jgt-2012-0042

© de Gruyter 2013

Finite groups of fourth-power free order Cai Heng Li and Shouhong Qiao Communicated by Robert M. Guralnick Abstract. A characterization is given of finite groups of order indivisible by the fourth power of any prime, which shows that such a group has the following form: A W B W C W D or .M ⇥ .A W B W C //:E, where A; B; C and D are all nilpotent, M is perfect, and E is abelian. Further, the semi-direct products involved in these groups are characterized.

1 Introduction A classical result, proved by Hölder [8] in 1893, says that a finite group of squarefree order is a semi-direct product of a cyclic group by another cyclic group, namely, such a group is a semi-direct product of the form AWB

with A; B both cyclic.

Later on, some special small order groups were characterized; see for example [5, 11, 14, 15]. In particular, Dietrich and Eick [2] investigated the groups of cube-free order (that is, the order is not divisible by the cube of any prime). More recently, in [12], complementing the work of [2], it was shown that a finite nonnilpotent group of cube-free order has the form AWB WC

or

PSL.2; p/ ⇥ .A W B/;

where A; B and C are all abelian. A natural problem is to extend the cube-free case to the fourth-power-free case. A group is said to be of fourth-power-free order if its order is not divisible by the fourth power of any prime. The main purpose of this paper is to characterize such finite groups. Roughly speaking, it is shown that such a group has the form AWB WC WD

or

.M ⇥ .A W B W C //:E;

where A; B; C and D are all nilpotent, M is perfect, and E is abelian. Cai Heng Li is supported by NNSF of China (grants no. 11231008, 11171292) and an ARC Discovery Project grant. Shouhong Qiao acknowledges the support from the following grants: NNSF of China (11126098, 11201082), China Postdoctoral Science Foundation funded project (2012M521724) and FDYT (LYM11058) of Guangdong Province. The corresponding author is Shouhong Qiao.

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C. H. Li and S. Qiao

Before presenting the main result, we first state some special cases. As usual, by Gp with p prime we mean a Sylow p-subgroup of G. Corollary. Let G be a finite soluble group of fourth-power-free order. Then there exist non-trivial nilpotent groups L; M such that the following statements hold: (1) If .jGj; 6/ D 1, then G D L, or L W M .

(2) If .jGj; 2/ D 1, then G D L, or L W M , or L W M W G3 . (3) If .jGj; 3/ D 1, then G D L, or L W M , or L W M W G2 . Each case does occur. Here are examples with three nilpotent layers: Z35 W Z31 W Z3 < AÄL.1; 53 /

and

Z211 W D10 < AÄL.1; 112 /.

For the general case, we state the main results in two theorems. The first is for soluble groups. Theorem 1.1. Let G be a finite soluble group of fourth-power free order. Then one of the following holds, where L; M are non-identity nilpotent subgroups: (i) G D L, or L W M ,

(ii) G has three nilpotent layers, and has the form (i) L W M W G2 or L W M W G3

or

(ii) .L ⇥ Z23 / W .M ⇥ Q8 / W Z3 ,

(iii) G has four nilpotent layers, and has the form L W M W G3 W G2

or L W .M ⇥ Z22 / W G3 W Z2 :

We remark that each case listed in the theorem indeed occurs. Here are a few of basic examples for three nilpotent layer groups: Zp2 W D2q ;

Zp2 W S3 ;

3 2 Zp3 W 31C2 C W Q8 ; Zp W A4 ; Z3 W Q8 W Z3 ;

Zp3 W Q8 W Z3 ; Z32 W Z7 W Z3 ; Z35 W Z31 W Z3 : The groups .Z11 W Z5 / o S3 D Z311 W Z35 W Z3 W Z2

and

Zp3 W S4 D Zp3 W Z22 W Z3 W Z2

have four nilpotent layers. The semi-direct products appearing in the groups in Theorem 1.1 are essentially known, and we can easily see this in an indirect way: embedding such a group into

Finite groups of fourth-power free order

277

a larger group, which is a direct product of some ‘basic’ groups, consisting of the following groups, where jP j divides p 3 and jQj divides q 3 , with p; q prime: ✏ ✏ ✏ ✏

P , P W Q,

P W Q W R, with jRj dividing 23 or 33 , P W Z22 W R W Z2 , with jRj dividing 33 ,

P W Q W R W S, with jRj dividing 33 and jS j dividing 23 .

The structural information of these basic groups is described in Section 5. In particular, the semi-direct products involved in these groups are determined. Corollary 1.2. Let G be a group of fourth-power free order. Then G  B1 ⇥

⇥ Bn ;

where Bi is one of the basic groups described above. The next theorem characterizes the insoluble case. Theorem 1.3. Let G be an insoluble finite group of fourth-power-free order. Then G D .M ⇥ .A W B W C //:E, where A; B; C are nilpotent subgroups of odd order, E is abelian, and M is a perfect subgroup satisfying one of the followingW (i) M is almost quasi-simple: ✏ ✏

A7 , J1 , PSL.2; 8/, or PSL.2; p f /, where p is odd and f  3, 3:A6 , 3:A7 , SL.2; p/ with p 5, or SL.2; p 3 / with p odd,

(ii) M D ASL.2; p/ D Zp2 W SL.2; p/, where p

5,

(iii) M D F W SL.2; 5/, where F is nilpotent with each Sylow r-subgroup isomorphic to Z2r , Z2r W Zr or Z3r , with either r D 5, or r ⌘ ˙1 .mod 10/, (iv) M D Z3m W S , where m is square-free, and one of the following holds:

(a) S D PSL.2; 7/, and r 11 and r 3 ⌘ 1 .mod 7/ for each prime divisor r of m, (b) S D 3:A6 , and r ⌘ 1 or 19 .mod 30/ for each prime divisor r of m, (c) S D A5 , and r ⌘ ˙1 .mod 10/ for each prime divisor r of m.

The paper is organized as follows: In Section 2, we study automorphism groups of the groups of order p d with p prime and d  3. Then in Sections 2 and 3, we treat the soluble case, and prove Theorem 1.1. Corollary 1.2 is proved in Section 4. In the final section, we prove Theorem 1.3. We conclude this section by giving some notation which will be frequently used. For a set ⇡ of primes, ⇡ 0 denotes the set of the primes not in ⇡. Let G be a group. We write G⇡ for a Hall ⇡-subgroup of G; Z.G/ denotes the centre of G,

278


F.G/ denotes the Fitting subgroup of G, F ⇤ .G/ the generalized Fitting subgroup of G (see, for example, [10, X, Section 13]), ˆ.G/ the Frattini subgroup of G. A semidirect product of a group A by a group B is denoted by A W B, and then we write A W B W C D A W .B W C / and

A W B W C W D D A W .B W .C W D//:

In what follows, Z3l denotes Zl ⇥ Zl ⇥ Zl with l a positive integer, 3:A6 denotes a perfect group which has central factor group A6 .

2 Automorphism groups of p-groups Let G be a group. The Fitting subgroup F WD F.G/ is the largest nilpotent normal subgroup of G. If G is soluble, then the Fitting subgroup satisfies CG .F /  F . Hence G=F D G=.CG .F /F /  Out.F / D Out.F1 / ⇥

⇥ Out.F t /;

where F D F1 ⇥ ⇥ F t is such that each Fi is a Sylow pi -subgroup of F . Now Fi has order dividing pi3 , and thus structural information about subgroups of Out.Fi / is important in the study of G. Let P be a p-group of order at most p 3 . Then either P is abelian, or P is one of the two non-abelian groups of order p 3 . For p D 2, the only non-abelian groups of order p 3 D 8 are Q8 , the quaternion group, and D8 , the dihedral group of order 8. If p is odd, then a non-abelian group of order p 3 is Zp2 W Zp or Zp2 W Zp . Note that ✏ ✏

1C2 Zp2 W Zp has exponent p and is usually denoted by pC ,

Zp2 W Zp has exponent p 2 and is denoted by p 1C2 .

For a p-group P of order at most p 3 , the automorphism group Aut.P / and its subgroups are explicitly known. We will collect the relevant facts below. First consider the case where P is abelian. If P D Zpn with p > 2, then Aut.P / D Zpn if P D Z2n with n

1 .p

1/ I

2, then Aut.P / D Z2 ⇥ Z2n 2 :

For P elementary abelian, that is, P D Zp2 or Zp3 , we have Aut.P / D GL.2; p/ or GL.3; p/, respectively. For the other abelian groups of order p 3 , we have the following lemma. Lemma 2.1. Let P D Zp2 ⇥ Zp . Then Aut.P / is soluble, and a Hall p 0 -subgroup of Aut.P / is isomorphic to Zp2 1 .

279


Proof. Write P D Zp2 ⇥ Zp DW hai ⇥ hbi. It is easily shown that each automorphism ˛ of P has the form ´ a ! ai b j ; ˛W b ! ak b l ; where 1  i  p 2 1 with .i; p/ D 1, 0  j  p 1, k 2 π0; p; : : : ; .p 1/pº, and 1  l  p 1. The automorphism ˛ is uniquely determined by the integers i; j; k; l. Counting the tuples .i; j; k; l/, we conclude that jAut.P /j D p.p

1/:p:p:.p

On the other hand, P has two automorphisms

1/ D p 3 .p

1/2 :

and ⌧, defined by

W a ! ai ; b ! b;

⌧ W a ! a; b ! b j ; where 1  i  p 2 1 and 1  j  p 1. For suitable choices of i and j , we have h ; ⌧i D Zp2 1 , which is a Hall p 0 -subgroup of Aut.P /. Clearly, Aut.P / D Aut.P /p Aut.P /p0 , a product of two nilpotent subgroups, so Aut.P / is soluble. Next, assume that P is non-abelian. Then P has order p 3 . It is well known that Aut.Q8 / D S4 , and Aut.D8 / D D8 . For p odd, we will identify the subgroups of Aut.P / of order co-prime to p, given in the following two lemmas. The following statement can be found in [3, p. 82, Theorems 20.8–20.9]. Lemma 2.2. Let P D p 1C2 D Zp2 W Zp , with p > 2. Then Aut.P / is soluble, and a Hall p 0 -subgroup of Aut.P / is cyclic of order p 1 which acts non-trivially on Z.P /. The automorphism group of the non-abelian group P D Zp2 W Zp is determined by the symplectic group GSp.2; p/ D GL.2; p/ in the following way. It is known that P D ha; b; c j ap D b p D c p D 1; Œa; bç D ci; and the commutator subgroup

P 0 D hŒx; yç j x; y 2 P i D hci D ˆ.P / D Zp :

Now Œx; yç can be viewed as a symplectic form over P =P 0 D Zp2 , and since ˆ.P / is a characteristic subgroup, each element of Aut.P / preserves the form. It follows that Out.P / is isomorphic to GSp.2; p/, a general symplectic group. Noting that Sp.2; p/ D SL.2; p/ and GSp.2; p/ D GL.2; p/, we have the following statement.

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Lemma 2.3. Let P D Zp2 W Zp be non-abelian of exponent p. Then Out.P / D GL.2; p/:

As mentioned before, we also need structural information about subgroups of Out.P /. For P D Zp2 , Zp3 , or Zp2 W Zp , there are many subgroups of Out.P / which need considering. Since methods for analysing the cases where G is soluble or insoluble are quite different, we first give a list for soluble subgroups. If P D Z22 or Z23 , then Aut.P / D GL.2; 2/ D S3

or

Aut.P / D GL.2; 3/ D Q8 W S3 :

It is easy to identify subgroups of Aut.P /. Next, we consider the case where p 5. Lemma 2.4. Let H be a soluble subgroup of GL.2; p/ of fourth-power-free order, where p 5. Then one of the following holds: (i) H D Zl ⇥ Zm or H D Zp W .Zl ⇥ Zm /, where l and m divide p

1,

(ii) H  ÄL.1; p 2 /, and H satisfies one of the following statements: (a) H D Zl W Z2i , where l j .p 2 1/, 0  i  3, (b) H D Zl W Q8 , where l j .p 2 1/ and l is odd,

(iii) H  GL.1; p/ o S2 , and H satisfies one of the following statements: (a) H D A W Z2i , where A  Zp2

(b) H D A W Q8 , where A  Zp2

(iv) H D Zl ⇥ Q8 W Z3 , where l j .p

1,

1

0  i  3,

and jAj is odd,

1/ and l is odd.

Proof. We inspect the subgroups H of GL.2; p/ listed in [1, Lemma 3.3 and Theorem 3.4]. If H is reducible, then H  Zp W .Zp 1 ⇥ Zp 1 /, and so H is as in part (i). Suppose that H  ÄL.1; p 2 / D Zp2 1 W Z2 is irreducible. Let H2 be a Sylow 2-subgroup of H . If H2 has only one subgroup of order 2, then H2 is cyclic or Q8 , and so H D Zl W Q8 or Zl W Z2i , 0  i  3. If H2 has at least two subgroups of order 2, then it is easy to see that H D Zl W Z2 . Then we have part (ii). If H  ÄL.1; p/ o S2 is imprimitive, then it satisfies part (iii). Finally, assume that H is a subgroup of symplectic type, that is, H  Zp 1 ı .Q8 W Z3 /, a central product. If H is cyclic and irreducible, by [13, Theorem 2.3.3], H is contained in a Singer cycle, which is included in part (ii) of the lemma. If H is not cyclic, then H satisfies case 6 of [1, Theorem 3.4], then H Q8 W Z3 , and thus H D Zl ⇥ Q8 W Z3 since H has order fourth-power free order, as in part (iv). For P D Zp3 , we only need subgroups of GL.3; p/ of order co-prime to p. Subgroups of GL.3; p/ can be found in [4, Theorem 2.2].


281

Lemma 2.5. Let H be a soluble subgroup of GL.3; p/ of order co-prime to p and fourth-power free. Then one of the following holds: (i) H  Zl ⇥ L, where l j .p 1/ and L has order prime to p and is isomorphic to a soluble subgroup of GL.2; p/, (ii) A:Z3  H D A W Hπ2;3º  GL.1; p/ o S3 , where A  Zp3 1 is a non-trivial Hall π2; 3º0 -subgroup of H , and Hπ2;3º is a π2; 3º-subgroup of order divisible by 3, (iii) H D Zl W Z3i  Zp3 1 W Z3 , where l j .p 3 1/, (iv) H < Zp 1 ı .31C2 C :Sp.2; 3//, and H satisfies one of the following: (a) H  Zp 1 ⇥ .Q8 W Z3 /, (b) H  Zp 1 ı .Z23 W Z6 /, (c) H  Zl ⇥ .31C2 1/ is co-prime to 3. C W Q8 /, where l j .p

Proof. We inspect the subgroups of GL.3; p/ described in [4]. If H is reducible, then H is contained in Zp2 W .Zp 1 ⇥GL.2; p//. Since p ≠ jH j, it follows that H satisfies part (i). Suppose that H is imprimitive. Then H  GL.1; p/ o S3 :

If A WD H \ GL.1; p/3 D 1, then H  S3 which is included in part (i). So we may suppose that A WD H \ GL.1; p/3 6D 1. If jH=Aj3 D 1, then H is reducible, contrary to the hypothesis. Thus, H D A:Z3 or A:S3 , as in part (ii). Suppose that H is irreducible and contained in the normalizer of a Singer cycle, that is, H  ÄL.1; p 3 / D hai W h i D Zp3 1 W Z3 , where a D ap . Let H3 be a Sylow 3-subgroup of H . If H3 has only one subgroup of order 3, then H3 is cyclic and H D Zl W Z3i , where l j .p 1/ is odd. If H3 has at least two subgroups of order 3, then H D Zl W Z3 . Then part (iii) holds. Finally, suppose that H is a subgroup of symplectic type. If jH \ 31C2 C j  3, 1C2 2 then H  Zp 1 ⇥.Q8 W Z3 /. If jH \ 3C j D 9, then H  Zp 1 ı.Z3 W Z6 /. Suppose 31C2  H , H  Zl ⇥ .31C2 W Q8 /, where l j .p 1/ is co-prime to 3. C C It is known that if H is an irreducible abelian subgroup of GL.n; p/ with exponent dividing p 1, then n D 1 and H is cyclic. Noticing that a π2; 3; qº-subgroup of GL.1; p/ o S3 of nilpotent length 3 is homomorphic to Z2q W S3 , we have the following immediate consequence. Lemma 2.6. Let P be a p-group of order at most p 3 . (i) Out.P / has no subgroup isomorphic to ASL.2; 3/ D Z23 W Q8 W Z3 . (ii) If a π2; 3; qº-subgroup H < Out.P / has nilpotent length 3, then H is homomorphic to Z2q W S3 , where q j .p 1/.

282


We need Maschke’s theorem (see for example [3, p. 39]). Lemma 2.7. Let ⇡ be a set of primes and let a ⇡ 0 -group H act on an abelian ⇡-group G. Suppose that G D A ⇥ B and A is fixed by H . Then there exists a subgroup K of G such that G D A ⇥ K and H fixes both A and K.

3 Structural information for soluble case All groups considered in this section are finite soluble groups of orders indivisible by p 4 for any prime p. Lemma 3.1. Let p be the largest prime divisor of the order jGj of G. Assume p 5. Then one and only one of the following statements hold: (i) Gp C G, (ii) there exists a prime divisor q of jGj such that Gq D Z3q which is minimal normal in G, p j .q 2 C q C 1/, and Gp =CGp .Gq / D Zp .

In particular, G has a normal Sylow subgroup.

Proof. Suppose to the contrary that the statement of the lemma is not true. Let G be a minimal counter-example to the lemma. Assume that Gp is not normal in G. Let F D Fp ⇥ Fp0 be the Fitting subgroup of G. Since F 6D 1 is self-centralized in G, we conclude that Gp does not centralize F . Suppose first that Gp centralizes Fp0 . Then Gp does not centralize Fp , and as jGp j  p 3 , we have Fp ä Zp2 or Zp2 . Let G D G=Fp , and then Gp is a Sylow p-subgroup of G. Then Gp ä Zp . If Gp is normal in G, then Gp is a normal Sylow p-subgroup of G, which is a contradiction. Thus, Gp is not normal in G, and so by the minimality of G, G has a Sylow q-subgroup G q ä Z3q that is a minimal normal subgroup of G such that p j .q 2 C q C 1/, where Gq 2 Sylq .G/. Then Fp W Gq is normal in G. Clearly, C WD CG .Fp / has order divisible by q. Since C =Fp is normal in G, we have G q  C =Fp since G q is minimal normal in G. It follows that Gq char Fp ⇥ Gq D Fp W Gq C G; satisfying part (ii) of the lemma, which is a contradiction. Below we suppose that Gp does not centralize some Sylow q-subgroup Q of F with q 6D p. Since Q is normal in G, the subgroup X WD QGp D Q W Gp is not nilpotent. Since p > q, it follows that p divides neither q C 1 nor q 1. By Lemmas 2.1–2.5, we conclude that p j .q 2 C q C 1/ and Q D Z3q . Moreover, as


p > 5, we have that

283

p 2 > .q C 1/2 > q 2 C q C 1;

and thus p 2 does not divide q 2 C q C 1. It follows that Gp =CGp .Q/ ä Zp . Since q 4 does not divide jGj, we conclude that Gq D Q C G satisfies part (ii) of the lemma, which is again a contradiction. This proves the lemma. Lemma 3.2. Assume that jGj has exactly m prime divisors and that each prime divisor is at least 5. Then for each number k  m there exists ⇡ of size k such that G D G⇡ W G⇡ 0 . Proof. By Lemma 3.1, there exists a normal Sylow p1 -subgroup Gp1 , and so we get G D Gp1 W Gp10 . If k > 1, then Gp10 has a normal Sylow p2 -subgroup Gp2 , and so G D Gp1 W Gp2 W Gπp1 ;p2 º0 . Repeating this process, we have G D Gp1 W Gp2 W

W Gpk W Gπp1 ;p2 ;:::;pk º0 D G⇡ W G⇡ 0 ;

where ⇡ D πp1 ; p2 ; : : : ; pk º. The next lemma determines groups of orders dividing p 3 q 3 . Lemma 3.3. Assume that G is a non-nilpotent πp; qº-group with p; q being distinct primes. Then one of the following holds: (1) G has a normal Sylow subgroup, and one of the following holds: (i) G D Gp W Gq is supersoluble, and q divides p

1,

(ii) G D Gp W G2 , Gp =ˆ.Gp / ä Zp2 or Zp3 , and G2 acts irreducibly on the factor group Gp =ˆ.Gp /, G D Zp3 W Z9 with 9 ≠ .p 1/, or Zp3 W Z27 with 27 ≠ .p 1/,

1C2 (iii) G D Q8 W G3 , Zp2 W Gq , Zp ⇥ .Zp2 W Gq /, or pC W Gq , where q is an odd divisor of p C 1,

(iv) G D Zp3 W Gq , where q j .p 2 C p C 1/ and q 6D 3,

(2) G D Z23 W Q8 W Z3 D ASL.2; 3/,

(3) G D Z22 W G3 W Z2 , and one of the following holds:

(i) G D Z22 W 31C2 W Z2 D .Z22 ⇥ Z9 / W Z6 , G2 is abelian,

2 (ii) G D Z22 W 31C2 C W Z2 D Z6 W Z6 , G2 is abelian,

(iii) G D Z22 W G3 W Z2 , and G=O3 .G/ D S4 , (iv) G D .Z22 W Z9 / ⇥ D6 ,

(v) G D A4 ⇥ H , where H D D18 , D6 , D6 ⇥ Z3 , or Z23 W Z2 and H satisfies Z.H / D 1.

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Proof. By Lemma 3.1, we know that either G has a normal Sylow subgroup or G is a π2; 3º-group. Case 1. Assume first that G D Gp W Gq . Assume further that G is not supersoluble. Then the quotient Gq =CGq .Gp / is a non-trivial subgroup of Out.Gp /. By Lemmas 2.1–2.5, we conclude that q divides p 1, p C 1 or p 2 C p C 1, and we 1C2 have Gp D Q8 , Zp2 , Zp3 or pC . 2 Let q divide p C p C 1. Then Lemma 2.5 is satisfied, and Gp D Zp3 which is a minimal normal subgroup of G. If q 6D 3, then part (1) (iv) of the lemma is satisfied, while if q D 3, then Z3 < G3 =CG3 .Gp / < Zp3 1 , and so G3 D Z9 or Z27 , as in part (1) (ii). If q j .p C 1/ and q is odd, then part (1) (iii) is satisfied. Finally, assume that q j .p 1/. Since Gq acts irreducibly on Gp =ˆ.Gp /, we conclude that q also divides p C 1 or p 2 C p C 1, depending on Gp =ˆ.Gp / ä Zp2 or Zp3 , respectively. If q divides p 2 C p C 1, then q divides .p 1; p 2 C p C 1/, and hence q D 3, as in part (1) (ii). If q divides p C1, then q divides .p 1; p C1/. Hence q D 2, as in part (1) (ii). Case 2. Assume that G has no normal Sylow subgroup. By Lemma 3.1, G is a π2; 3º-group. Let F be the Fitting subgroup of G. Then F D F2 ⇥ F3 , where jF2 j  4 and jF3 j  9. Hence F is abelian and CG .F / D F . Since G is soluble, G D G=F  Out.F / D Aut.F / D Aut.F2 / ⇥ Aut.F3 /:

Let F2 D 1. Then F D F3 < G3 , and CG2 .F3 / D 1. Since G=F  Aut.F /, we conclude that F3 D Z23 and G3 is non-abelian. Clearly, Aut.F / D GL.2; 3/ D Q8 W S3 :

By Lemma 2.2, we have G3 D Z23 W Z3 . Since G3 is not normal in G, it follows that G=F is not nilpotent, and so G=F D Q8 W Z3 , and G D Z23 W Q8 W Z3 . This is part (2). Next assume that F2 6D 1. Then we have F2 ä Z22 which is minimal normal in G. If F2 \ ˆ.G/ 6D 1, then F2 D ˆ.G/ since F2 is minimal normal in G. Then G=F2 is 2-nilpotent, and thus G is 2-nilpotent, a contradiction. It follows that G D Z22 W G3 W Z2 :

If G2 is not abelian, then G2 ä D8 and G=O3 .G/ ä S4 , and this is part (3) (iii). Suppose that G2 is abelian. Then G3 is not cyclic. It follows that G3 D Z23 , Z33 , 2 1C2 Z3 ⇥ Z9 , 31C2 or 31C2 C . If G3 D Z3 , then we have G D A4 ⇥ D6 . If G3 D 3 2 2 or G3 D 31C2 C , then G D .Z2 ⇥ Z9 / W Z6 or Z6 W Z6 , respectively. Observe that the group G3 D Z3 ⇥ Z9 yields the groups G D A4 ⇥ D18 and G D .Z22 W Z9 / ⇥ D6 . Finally, if G3 D Z33 , then G D A4 ⇥ Z23 W Z2 . Thus parts (3) (i), (ii), (iv) and (v) are obtained.


285

4 Proof of Theorem 1.1 The proof of Theorem 1.1 consists of a series of lemmas. In this section, G denotes a soluble group of fourth-power-free order. Lemma 4.1. If .jGj; 6/ D 1, then G D A W B, where A; B are both nilpotent, and A is characteristic in G. Proof. Let F be the Fitting subgroup of G. Then G D F:G. Since G D G=F  Out.F /

and each prime divisor of jGj is at least 5, it follows from Lemmas 2.1-2.5 that G is abelian. Let M be a minimal preimage of G under G ! G, namely, M is a minimal subgroup of G such that MF=F ä G. Then G D MF . Suppose that F \ M 6 ˆ.M /. Then there exists a maximal subgroup H of M such that M D .F \ M /H since F \ M C M . Thus, G D FM D FH , which contradicts the minimal choice of M . Hence we have F \ M  ˆ.M /. Moreover, M=F \ M ä G=F is abelian, and hence M is nilpotent, see [9, III, Theorem 3.5]. In order to prove the theorem, we may suppose that jM j is not divisible by the cube of any prime. Since F \ M  ˆ.M /, .F \ M /r  ˆ.M /r D ˆ.Mr /, we have Mr D hbi D Zr 2 . Below, we try to find a characteristic nilpotent subgroup A and a nilpotent subgroup B such that G D A W B. If F \ M D 1, we are done. Thus suppose that F \ M 6D 1. Let r be a prime divisor of jF \ M j. Then Fr has order r 2 . Let Fr D hai D Zr 2 , .F \ M /r D Zr which is normal in G. Since Mr D hbi D Zr 2 , it follows that har i D hb r i D .F \ M /r

which is centralized by Mr 0 . Since jGr j D r 3 and Zr 2 D Fr C Gr , we conclude that Gr D Zr 3 or Fr W Zr . As ˆ.Fr / is centralized by Mr 0 , by [6, Theorem 5.2.4], Fr , and so Gr , is centralized by Mr 0 . Then we write G D Fr 0 .Gr ⇥Mr 0 /, Gr ⇥Mr 0 is nilpotent. We could consider Fr 0 instead of F since Fr 0 is a characteristic nilpotent subgroup in G. If Fr D Z2r , since F is nilpotent, we conclude that F D A ⇥ .F \ M /

for some nilpotent subgroup A such that G D FM D AM , A and Ar are both normal in G by Lemma 2.7. We write Ar D hai D Zr , Mr D hbi D Zr 2 and .F \ M /r D Zr . Let Gr be a Sylow r-subgroup of G. Then Gr > Fr , and Gr D Ar Mr ä Zr ⇥ Zr 2 . Assume that NG .hai/ 6D CG .hai/. Clearly, Or .G/ D Fr D ha; b r i D Z2r . Suppose that a0 D ai b rj , with .j; r/ D 1, such that o.a0 / D r

and

NG .ha0 i/ 6D CG .ha0 i/:

286


Let x 2 NG .ha0 i/ n CG .ha0 i/. Then x can be written as x D vwu, where u 2 A, v 2 Mr and w 2 Mr 0 . Thus, there exist integers j 0 ; i 0 such that 1  j 0 ; i 0 < p and 0 0 0 aij b rjj D .a0 /j D .a0 /x D .ai b rj /x D x 1 ai b rj x Du Du 0

0

1

w

1

1

w

1 i

D ai b rj :

v

1 i rj

a b vwu D u

a wb rj u D u

1

w

1 i rj

a b wu

1 i 0 rj

a b u

It follows that b rjj D b rj , and hence j 0 D 1. So x centralizes a0 , which is a contradiction. Therefore, hai is the unique subgroup of G of order r such that NG .hai/ 6D CG .hai/, and so Ar D hai is a characteristic subgroup of G, as desired. Now assume that NG .hai/ D CG .hai/. Since hai is normal in G, it follows that ŒGr ; Mr 0 ç D 1, and hence G D AM D Ar 0 .Gr ⇥ Mr 0 /, where Ar 0 D Fr 0 which is characteristic in G. Repeating this process, we obtain that G D A1 B1 such that A1 ; B1 are both nilpotent, A1 \ B1 D 1, and each Sylow subgroup of A1 is a characteristic subgroup of G. So G D A1 W B1 , and A1 is characteristic in G. Lemma 4.2. Let G D P W H , where P is a Sylow p-subgroup of G. Suppose that N < P is non-identity and normal in G such that G=N ä .P =N / ⇥ H . Then P D L W M with L  N , and G D L W .M ⇥ H /. Proof. By Lemma 2.7 and Lemmas 2.1–2.5, we only need to consider the cases P D Zp2 ⇥ Zp , Zp2 W Zp or Zp2 W Zp . If N D ˆ.P /, by [9, III, Theorem 3.18], H centralizes P , and the lemma trivially holds. Thus we suppose that N 6D ˆ.P /. Let P D hai ⇥ hbi D Zp2 ⇥ Zp . Suppose that N 6D hap i ⇥ hbi. Without loss, we may suppose N  hai, or N D hbi. If N D hap i, then N D ˆ.P /, contrary to our hypothesis. If N D hai, by Lemma 2.7, H normalizes some subgroup of P which has trivial intersection with hai, say hbi. In this case, H centralizes hbi, and so G D hai W .hbi ⇥ H /. If N D hbi, similarly, we may suppose H normalizes hai, and so G D hbi W .hai ⇥ H /. Now suppose that N D hap i ⇥ hbi. Since H acts on N and ˆ.P /.D hap i/ is normalized by H , by Lemma 2.7, there is a subgroup of N which is normalized by H , say hbi. Then since H acts on P and hbi is normalized by H , again by Lemma 2.7, there is a subgroup which is normalized by H , say hai. It follows that G D hbi W .hai ⇥ H /. Let P D hai W hbi D Zp2 W Zp . Since N 6D ˆ.P /, we have N D hap i ⇥ hbi. As H normalizes hap i, by Lemma 2.7, H normalizes some subgroup different from hap i, say hbi. Then consider the action of H on P D P =ˆ.P / ä Zp2 . Since hbi is normalized by H , there is a subgroup A of P such that P D A ⇥ hbi and H


287

normalizes A. Then A ä Zp2 since each subgroup of P of order p is contained in N . Since P W H D hbi W .A ⇥ H / and ˆ.P / D ˆ.A/; by [9, III,Theorem 3.18], H centralizes A, thus H centralizes ˆ.P /.D Z.P //. By Lemma 2.2, H centralizes P , and the lemma holds trivially. The case P D Zp2 W Zp is dealt with in a similar way. Lemma 4.3. Assume that G D Gπ2;3º0 W Gπ2;3º . Then there exist nilpotent groups L; M such that one of the following holds: (i) Gπ2;3º D G3 W G2 and G D L W M W G3 W G2 ,

(ii) Gπ2;3º D G2 W G3 and G D L W M W G3 ,

(iii) Gπ2;3º D Z22 W G3 W Z2 and G D L W .M ⇥ Z22 / W G3 W Z2 ,

(iv) Gπ2;3º D Z23 W Q8 W Z3 and G D .L ⇥ Z23 / W .M ⇥ Q8 / W Z3 .

In particular, if .jGj; 2/ D 1, then G D L W M or L W M W G3 , while if .jGj; 3/ D 1, then G D L W M or L W M W G2 . Proof. By Lemma 4.1, Gπ2;3º0 D L W M with L characteristic in G. We analyse the structure of G according to the structure of Gπ2;3º . If Gπ2;3º D G3 W G2 , then G D Gπ2;3º0 W Gπ2;3º D L W M W G3 W G2 : Let Gπ2;3º D G2 W G3 , which is homomorphic to A4 . Then G D L W M W G2 W G3 : Suppose that Mp is not centralized by G2 for some prime p dividing jM j. Then Mp W G2 W G3 has nilpotent length 3. It follows from Lemma 2.6 that Mp centralizes L. Write M D M⇡ ⇥ M⇡ 0 such that ⇡ is maximal subject to that, for each p 2 ⇡, Mp is not centralized by G2 . Then G D L W M W G2 W G3 D .L ⇥ M⇡ / W .M⇡ 0 ⇥ G2 / W G3 D A W B W G3 : Suppose now that Gπ2;3º D Z22 W G3 W Z2 , which is homomorphic to S4 . If Mp W Z22 W G3 W Z2 6D .Mp ⇥ Z22 / W G3 W Z2 ; then Mp centralizes L. Write M D M⇡ ⇥ M⇡ 0 such that ⇡ is maximal subject to that, for each p 2 ⇡, Mp is not centralized by Z22 . Then G D L W M W Gπ2;3º D .L ⇥ M⇡ / W .M⇡ 0 ⇥ Z22 / W G3 W Z2 D A W .B ⇥ Z22 / W G3 W Z2 :

288


Finally, let Gπ2;3º D Z23 W Q8 W Z3 D ASL.2; 3/. By Lemma 2.6, for each Sylow p-subgroup Gp for p 5, we have Gp W Gπ2;3º D .Gp ⇥ Z23 / W Q8 W Z3 . Thus, G D L W M W Gπ2;3º D .Z23 ⇥ .L W M // W Q8 W Z3 :

Consider the subgroup L W M W Q8 W Z3 . If a Sylow p-subgroup Mp is not centralized by Q8 , it follows from Lemma 2.6 that Mp centralizes L. Write M in the form M D M⇡ ⇥ M⇡ 0 such that M⇡ centralizes L, and M⇡ 0 is centralized by Q8 . Therefore, L W M W Q8 W Z3 D .L ⇥ M⇡ / W .M⇡ 0 ⇥ Q8 / W Z3 ; and thus, G D .L ⇥ M⇡ ⇥ Z23 / W .M⇡ 0 ⇥ Q8 / W Z3 .

Lemma 4.4. Assume that Gπ2;3º0 is not normal in G. Then one of the following statements hold: (i) Z32 D G2 C G and G D A W B or G D A W B W G3 ,

(ii) Z33 D G3 C G and G D A W B or G D A W B W G2 ,

where A; B are suitable nilpotent subgroups of G.

Proof. Choose the set ⇡ of primes to be maximal such that G D G⇡ W G⇡ 0 subject to 2; 3 … ⇡. Then we have G⇡ 0 6D Gπ2;3º0 . We note that ⇡ may be empty, in this case, G⇡ D 1. Moreover, any Sylow p-subgroup of G⇡ 0 with p 5 is not normal in G⇡ 0 . By Lemma 3.1, one and only one of the following statements holds: (a) G2 C G⇡ 0 , and Z32 W G7  G⇡ 0 which is homomorphic to Z32 W Z7 D AGL.1; 23 /;

(b) G3 C G⇡ 0 , and Z33 W G13  G⇡ 0 which is homomorphic to Z33 W Z13 < AGL.1; 33 /:

First suppose that (a) holds and let Z32 ä G2 C G⇡ 0 , as in (a). Suppose that G2 is not normal in G. Then G2 does not centralize some non-trivial Sylow subgroup P of G⇡ . We consider the subgroup P W Z32 W G7 . Since Aut.P / has no section isomorphic to Z32 W Z7 , we get P W Z32 W G7 D P W .Z32 ⇥ G7 /, which is a contradiction. Consequently, G2 E G. Then G D G2 W G20 . Suppose that G20 is 3-nilpotent. By Lemma 4.3, G20 D L W M or L W M W G3 , where L and M are nilpotent. If G20 D L W M W G3 , then we have G D G2 W G20 D G2 W L W M W G3 D .G2 ⇥ L70 / W .G7 ⇥ M / W G3 D A W B W G3 :

Thus, we suppose that G20 D L W M . Then G D G2 W L W M . Let C D CG .G2 /. Then G=C ä Z7 , or Z7 W Z3 . Hence G D G2 W L W M D .G2 ⇥ L70 / W L7 W M D .G2 ⇥ L70 / W .G7 ⇥ M30 / W M3 :


289

If M3 D 1, then G D A W B, as desired. Suppose that M3 6D 1. Let X D .G2 ⇥ L70 / W .G7 ⇥ M30 /: Then a Sylow 3-subgroup X3 has order at most 9. Since X2 D G2 C X , it follows that X D X3 ⇥ X30 . Then G D .G2 ⇥ Lπ3;7º0 / W .G7 ⇥ M30 / W G3 . Now we suppose that G20 is not 3-nilpotent. Then G20 has section Z33 W Z13 . By induction, G3 is normal in G20 or G20 D L W M for some suitable nilpotent subgroups L; M . Let G20 D L W M . If L3 6D G3 , then G20 has no section Z33 W Z13 , a contradiction. Thus L3 D G3 , and then G D .G2 ⇥ L70 / W .G7 ⇥ M /. Next we let G3 C G20 . Suppose that G3 is not normal in G. Then G3 does not centralize a Sylow p-subgroup P of G⇡ . Consider the subgroup P W Z33 W G13 . Since Aut.P / has no section isomorphic to Z33 W Z13 , we have P W Z33 W G13 D P W .Z33 ⇥ G13 /, which is a contradiction. Consequently, G3 E G, and so G D .Z32 ⇥ Z33 / W L W M D .Z32 ⇥ Z33 ⇥ Lπ7;13º0 / W .L7 ⇥ L13 / W M : By Lemma 4.2, there are nilpotent subgroups A1 ; A2 ; B1 ; B2 such that L7 D A1 ⇥ B1 ;

L13 D A2 ⇥ B2 ;

G D .Z32 ⇥ Z33 ⇥ Lπ7;13º0 ⇥ A1 ⇥ B1 / W .A2 ⇥ B2 ⇥ M /:

This is part (i) of the lemma. Similarly, (b) yields part (ii) of the lemma. Proof of Theorem 1.1. Let G be a group such that jGj is fourth-power free. If G is nilpotent, then G is as in part (1) of Theorem 1.1. If .jGj; 6/ D 1, then by Lemma 4.1, G is as in part (1) of Theorem 1.1. Assume that Gπ2;3º0 is normal in G. Then G is determined in Lemma 4.3, which leads to the following statements. If Gπ2;3º D G3 W G2 or Z22 W G3 W Z2 , then G has four nilpotent layers, and is as in part (3) of Theorem 1.1. If Gπ2;3º D G2 W G3 or Z23 W Q8 W Z3 , then G has three nilpotent layers, and is as in part (2) of Theorem 1.1. Finally, if Gπ2;3º0 is not normal in G, then Lemma 4.4 shows that G satisfies part (1) or (2).

5 Semi-direct products Here we study the semi-direct products appeared in Theorem 1.1. Lemma 5.1. Let G D .X ⇥ Y / W Z and let Z normalize X and Y . Then G is a subgroup of .X W Z/ ⇥ .Y W Z/.

290


Proof. An element g of G has the form .x; y; z/, which corresponds to an element .x; zI y; z/ of the group .X W Z/ ⇥ .Y W Z/. Lemma 5.2. Let G D X W .Y ⇥ Z/. Then G is a subgroup of .X W Y / ⇥ .X W Z/. Proof. An element g of G has the form .x; y; z/, which corresponds to an element .x; yI x; z/ of the group .X W Y / ⇥ .X W Z/. Applying Lemmas 5.1 and 5.2 repeatedly leads to the next conclusion. Lemma 5.3. Suppose that G D .X1 ⇥ X2 / W .Y1 ⇥ Y2 / W .Z1 ⇥ Z2 /, and suppose that XiQW Yj W Zk is a subgroup of G, 1  i; j; k  2. Then G is a subgroup of the group 1i;j;k2 .Xi W Yj W Zk /. Proof of Corollary 1.2. Let G be a group of fourth-power-free order. Then G satisfies the conditions of Theorem 1.1. In the following, P and Q denote groups of order dividing p 3 and q 3 , respectively, where p; q are primes. If G D L W M with L; M both nilpotent, then Y G .P W Q/:

For three-nilpotent-layer groups, by Theorem 1.1, basic group G has the form or L W M W G2 , or L W M W G3 . If G D L W M W G2 , then Y G .P W Q W G2 /;

Z22 W G3 W Z2 ,

while if G D L W M W G3 , then

G

Y

.P W Q W G3 /:

Finally, consider four-nilpotent-layer groups. If G D L W M W G3 W G2 , then Y G .P W Q W G3 W G2 /;

while if G D L W Z22 W G3 W Z2 , then Y G .P W Z22 W G3 W Z2 /:

We next give more precise characterizations of the ‘basic’ groups of fourthpower-free order, all of which have at most four primes dividing the order. If G is a πp; qº-group, then the structure of G is determined in Lemma 3.3. We study the structure of groups of orders divisible by three or four primes in the rest of this section.


Lemma 5.4. Let G be a π2; p; qº-group with p; q three nilpotent layers. Then

291

5. Assume that G has at least

G D Gp W Gq W G2 ; which is homomorphic to Zp2 W H , where H is a non-nilpotent π2; qº-subgroup of ÄL.1; p 2 /, GL.1; p/ o S2 , or GL.1; p/ o S3 . Proof. By Theorem 1.1 and Lemma 3.1, we have G D Gp W Gq W G2 . Suppose that p < q. By Lemma 3.1, Gp D Zp3 , and q j .p 2 C p C 1/, and by Lemma 2.5, we conclude that G=CG .Gp /  Zp2 CpC1 W Z3 : Hence G2 centralizes Gp . Let C D CGq .Gp /. It follows that .Gq W G2 /=C ä .Gq =C / ⇥ G2 : By Lemma 4.2, we have Gq D R W Q such that R C and Q is centralized by G2 . So G D .Gp ⇥ R/ W .Q ⇥ G2 /, which is a contradiction since G has three nilpotent layers. Thus, G D Gp W Gq W G2 with p > q. Let G D G=ˆ.Gp / D Gp W Gq W G2 . Since G has three nilpotent layers, it follows from Lemma 4.2 that H WD G=CG .Gp / is not nilpotent. By Lemmas 2.4 and 2.5, we have that H is a π2; qº-subgroup of ÄL.1; p 2 /, GL.1; p/ o S2 , or GL.1; p/ o S3 . Lemma 5.5. Let G be a π3; p; qº-group with p; q 5. Assume that G has at least three nilpotent layers. Then G D Zp3 W Gq W G3 , and one and only one of the following statements hold: (i) q j .p 2 C p C 1/ and G=CG .Gp / ä Zq i W Z3 with i  3,

(ii) q j .p

1/ and G=CG .Gp / D Zq W Z3 , Z2q W Z3 or Z3q W Z3 .

Proof. By Theorem 1.1, we have G D Gp W Gq W G3 . By Lemma 3.1, either p > q, or p < q and Gp D Zp3 , q j .p 2 C p C 1/, and Gq =CGq .Gp /  AÄL.1; p 3 /. Suppose that p > q. By Lemmas 2.4-2.5, we conclude that Gp D Zp3 , and either G=CG .Gp /  GL.1; p/ o Z3 with q j .p 1/ or

G=CG .Gp /  AÄL.1; p 3 /:

For the former, we conclude that G=CG .Gp / ä Zq o Z3 , Z2q W Z3 or Zq W Z3 . For the latter, we have G=CG .Gp / D Zq i W Z3 , where i  3. For p < q and Gp D Zp3 , q j .p 2 C p C 1/ and Gq =CGq .Gp /  AÄL.1; p 3 /, it follows that Gq =CGq .Gp / D Zq i W Z3 with i  3.

292


Lemma 5.6. Let G be a π2; 3; pº-group that has at least three nilpotent layers. Then one and only one of the following statements hold: (1) G D Z32 W G7 W G3 , and G is homomorphic to AÄL.1; 23 / D Z32 W Z7 W Z3 , (2) G D Gp W Gπ2;3º , and one of the following holds: (i) G D Zp3 W 31C2 C W Q8 ,

(ii) G D Gp W G3 W G2 , and G is homomorphic to Zp2 W S3 or .Zp W Z3 / o S2 ,

(iii) G D Gp W Q8 W G3 , and G is homomorphic to Zp2 W Q8 W Z3 , (iv) G D Zp3 W Z22 W G3 W Z2 , and G is homomorphic to Zp3 W S4 .

Proof. Suppose that Gp is not normal in G. By Lemma 3.1, for some prime divisor q of jGj, Z3q D Gq C G and p j .q 2 C q C 1/. Now q D 2 or 3, and further, if q D 2, then p D 7 and G2 D Z32 ; if q D 3, then p D 13 and G3 D Z33 . Suppose that G D Z33 W G13 W G2 . Then by Lemma 2.5, we have G=CG .G3 /  AÄL.1; 33 /: Let C D CG13 .G3 /. Then G13 W G2 =C ä .G13 =C / ⇥ G2 : By Lemma 4.2, we have G13 D L W M such that L C and M is centralized by G2 . Thus, G D .G3 ⇥ L/ W .M ⇥ G2 /; which is a contradiction since G has at least three nilpotent layers. Assume that G D Z32 W G7 W G3 . Then by Lemma 2.5, we have G=CG .G2 / D Z7 W Z3 : Finally, assume that G is of the form G D Gp W Gπ2;3º . The structure of Gπ2;3º is determined in Lemma 3.3. Analysing the candidates for Gπ2;3º leads to the conclusion in part (2). It remains to consider the groups which have four nilpotent layers. Lemma 5.7. Let G be a basic group of fourth-power-free order such that G has four nilpotent layers and j⇡.G/j D 4. Then G has the form Gp W Gq W G3 W G2 . Proof. This directly follows from Theorem 1.1 because jGj has four prime divisors.


293

6 Insoluble groups Let G be an insoluble group of fourth-power-free order. Let G2 be a Sylow 2-subgroup of G. Then G2 has order 4 or 8 by Burnside’s transfer theorem and the Odd Order Theorem. It is known that a non-abelian simple group of which Sylow 2-subgroups have order 4 or 8 is A7 , J1 , PSL.2; 8/, or PSL.2; p f /, where p is odd and f  3, see [7, Theorem 4.126] and [9, Chapter II, Remark 8.11]. Assume that G is quasi-simple. Then G=Z.G/ is one of the simple group listed above, where Z.G/ is a factor group of the Schur multiplier of the simple group G=Z.G/. It follows that G D 3:A6 , 3:A7 , SL.2; p/ with p 5, or SL.2; p 3 / with p odd. Assume that G is almost simple. Then either G is simple, or G is one of the groups: PÄL.2; 8/, or PGL.2; p/, or PSL.2; p 3 /:h i with o. / j 6. Finally, if G is almost quasi-simple but not listed above, then G D SL.2; p 3 /:3. For convenience, we list these almost quasi-simple groups in the following lemma. Lemma 6.1. Assume that G is an almost quasi-simple group. Then G is isomorphic to one of the following: ✏

A7 , J1 , PSL.2; 8/, or PSL.2; p f /, where p is odd and f  3,

✏

3:A6 , 3:A7 , SL.2; p/ with p

5, or SL.2; p 3 / with p odd,

✏

PÄL.2; 8/, PGL.2; p/, or PSL.2; p 3 /:h i with o. / j 6, or SL.2; p 3 /:3.

Lemma 6.2. Suppose that G has a normal quasi-simple subgroup S. Then G has a soluble group C such that either G D S ⇥ C , or G is one of the following groups: (i) .PSL.2; 8/ ⇥ C /:3, with G=C ä PÄL.2; 8/,

(ii) .PSL.2; p/ ⇥ C /:2, with G=C ä PGL.2; p/,

(iii) .PSL.2; p 3 / ⇥ C /:h i, with o. / j 6 and

G=C ä PSL.2; p 3 /:h i  PÄL.2; p 3 /; (iv) .SL.2; p 3 / ⇥ C /:3, with G=C ä SL.2; p 3 /:3 < ÄL.2; p 3 /. Proof. The centralizer CG .S/ is soluble, and S \CG .S / D Z.S /. By Lemma 6.1, the centre Z.S/ D 1, Z2 or Z3 , and Z.S / is a Sylow group of S . Thus, we may write CG .S/ D C ⇥ Z.S/, and S CG .S / D C ⇥ S. To complete the proof, we assume that G 6D C ⇥ S . Then G=CG .S / is an almost simple group that is not a simple group. From Lemma 6.1 we conclude that

294


G=CG .S/ D PÄL.2; 8/, PGL.2; p/, or PSL.2; p 3 /:h i with o. / j 6. It is now easily obtained that G is a group listed in parts (i)–(iv) of the lemma. Thus, we next assume that G has no quasi-simple normal subgroup. It follows that the Fitting subgroup F.G/ coincides with the generalized Fitting subgroup F⇤ .G/ which is non-trivial and self-centralized in G. Let F D F.G/ D F1 ⇥

⇥ Ft ;

where Fi is a Sylow subgroup. Then G=F  Out.F / D Out.F1 / ⇥ ⇥ Out.F t /, 1C2 and thus Out.Fi / is insoluble for some i . Hence Fi D Zp2 , Zp3 or pC , for some prime p. Recall that a group X is called perfect if X equals its commutator subgroup X 0 . 1C2 Lemma 6.3. Let P D Zp2 or pC , with p 5. Then a non-trivial perfect subgroup S of Out.P / is quasi-simple, and further we have either S D SL.2; p/ or S D SL.2; 5/ with p ⌘ ˙1 .mod 10/. Moreover, if H is an insoluble subgroup of Out.P / of fourth-power-free order, then S  H  Zp 1 ı S .

Proof. First, for the case P D Zp2 , we have either S D SL.2; p/ or S D SL.2; 5/ and p ⌘ ˙1 .mod 10/; see, for example, [1, Theorems 3.4–3.5]. If jP j D p 3 , we only need to identify quasi-simple subgroups of Out.P / of order co-prime to p. Thus, for P D Zp2 W Zp non-abelian, by Lemma 2.3, Out.P / D GSp.2; p/;

and so S D SL.2; 5/, with p ⌘ ˙1 .mod 10/.

Lemma 6.4. Let P D Zp3 , and let S be a non-trivial perfect subgroup of Out.P / such that .jSj; p/ D 1. Then S is quasi-simple, and one of the following holds: ✏ ✏ ✏

S D PSL.2; 7/, where p

11 and p 3 ⌘ 1 .mod 7/,

S D 3:A6 , where p ⌘ 1 or 19 .mod 30/,

S D A5 , where p D 5 or p ⌘ ˙1 .mod 10/.

If H is an insoluble subgroup of Out.P / of fourth-power-free order, then S  H  Zp

1

ı S:

Proof. As S is perfect and .jS j; p/ D 1, we have that p is odd and S  SL.3; p/. Write Z WD Z.SL.3; p//  Z3 . Then SZ=Z  PSL.3; p/. If SZ=Z has no nontrivial normal elementary abelian subgroup, by [1, Theorem 1.1], SZ=Z is isomorphic to one of A5 , or A6 , or PSL.2; 7/. Then, by [1, Lemmas 6.4–6.6] we have that S is isomorphic to A5 , or 3:A6 , or PSL.2; 7/, with p satisfies the condition stated in [1, Lemmas 6.4–6.6]. These perfect subgroups are all irreducible.


295

Suppose next that SZ=Z has a non-trivial normal elementary abelian subgroup. By [1, Theorem 7.1], SZ=Z has a normal elementary abelian p-subgroup R=Z such that SZ=R is isomorphic to a subgroup of GL.2; p/. Since .jS j; p/ D 1, by Lemma 2.4, SZ=Z D SL.2; 5/. If Z3 D Z  S , then we have that A5 has a covering group with centre of order 6, which is a contradiction. Thus, SZ=Z D S D SL.2; 5/:

However, SL.2; 5/ is a reducible subgroup of GL.3; p/. Next we determine the general perfect groups. Lemma 6.5. Assume that G is a perfect group. Then either G is quasi-simple or one of the following holds: (i) G D ASL.2; p/ D Zp2 W SL.2; p/, where p > 5,

(ii) G D .Zp21 :::pr ⇥ Z3q1 :::qs ⇥ B/ W SL.2; 5/, where B D .r1 /1C2 ⇥ .r2 /1C2 ⇥ C C

⇥ .r t /1C2 C ;

with pi ; qj ; rl ⌘ ˙1 .mod 10/, with possible exception that one of pi is 5,

(iii) G D Zp31 p2 :::pr W S, where the numbers pi are distinct primes, and one of the following holds: (a) S D PSL.2; 7/, where pi 11 and pi 3 ⌘ 1 .mod 7/, (b) S D 3:A6 , where pi ⌘ 1 or 19 .mod 30/, (c) S D A5 , where pi ⌘ ˙1 .mod 10/. Proof. Let R be the largest soluble normal subgroup of G. Then G D R:G, where G is non-abelian simple. Let S be a minimal preimage of G under G ! G. Then S is a quasi-simple group such that S=Z.S / D G. By Lemma 6.1, we have Z.S/ D Zc , where c  3. Since jGj is fourth-power free, we get jR2 j  2 and jR3 j  9. Then R is not a π2; 3º-group since G is perfect and F .G/ D F ⇤ .G/ is self-centralized in G. Since G is perfect, by Lemma 3.1, R has a normal Sylow p-subgroup Rp , p 5. Then Rp C G

and G=CG .Rp /Rp  Out.Rp /:

We may suppose that CG .Rp / 6D G. Because G is perfect, G=CG .Rp /Rp is a non-trivial perfect subgroup of Out.Rp /. By Lemmas 6.3–6.4, we conclude that G=CG .Rp /Rp ä S ;

where S is as in Lemmas 6.3–6.4. Let N D Rc 0 . Then N D Rp ⇥ Np0 D Np ⇥ Np0 :

296


As G is perfect, so is G=Rp . By induction, we have that N is nilpotent. It then follows that R D N W Z.S/, and G D N W S . Since G=Np0 ä Np W S is a perfect group, by Lemmas 6.3–6.4, S is one of the groups: SL.2; p/, SL.2; 5/, PSL.2; 7/, 3:A6 or A5 . If S D SL.2; p/ with p 7, then by Lemma 6.3, we conclude that G D Zp2 W SL.2; p/:

Let S D SL.2; 5/. By Lemma 6.3, p D 5 or p ⌘ ˙1 .mod 10/, and Np D Zp2 , 1C2 pC , or Zp3 . It follows that N D

pi Y

pDp1

Zp2

⇥

qj Y

qDq1

1C2 qC

⇥

rk Y

Z3r ;

rDr1

where p1 ; : : : ; pi ; q1 ; : : : ; qj ; r1 ; : : : ; rk are all of the prime divisors of jN j. Suppose now that S D PSL.2; 7/. By Lemma 6.4, Np D Zp3 , with p > 11 and p 3 ⌘ 1 .mod 7/. Thus, N D Z3n with n square-free, and G D Z3n W PSL.2; 7/. For S D 3:A6 , Lemma 6.4 tells us that Np D Zp3 , with p ⌘ 1 or 19 .mod 30/. Thus, N D Z3n with n square-free, and G D Z3n W 3:A6 . Finally, for S D A5 , by Lemma 6.4, Np D Zp3 , with p ⌘ ˙1 .mod 10/. Thus, N D Z3n with n square-free, and G D Z3n W A5 . Proof of Theorem 1.3. Let G be a finite insoluble group of fourth-power-free order. Let M D G 1 , the smallest normal subgroup of G such that G=M is soluble. Then M is perfect, and so M is one of the groups listed in Lemmas 6.1 and 6.5. Let C D CG .M / C G, and let E D G=.M C /. Then E  Out.M / and

G D .M ı C /:E:

Clearly, we have M \ C D Z.M /. If Z.M / D 1, then M C D M ⇥ C . Suppose that Z.M / 6D 1. Then M is quasisimple but not simple, or M is the group in (ii) of Lemma 6.5. Let M be quasisimple but not simple. Then M is 3:A6 , 3:A7 , SL.2; p/, SL.2; p 3 /. For the former two cases, M C D M ⇥ C30 , while for the latter two cases, M C D M ⇥ C20 . Let M D .Zp21 :::pr ⇥ Z3q1 :::qs ⇥ r11C2 ⇥ r21C2 ⇥

⇥ r t1C2 / W SL.2; 5/:

Then Z.M / D Zr1 :::r t . Thus we get M \ C D Zr1 :::r t . Clearly, M \ C is a Hall subgroup of C . Then M C D M ⇥ Cπr1 ;:::;r t º0 : Thus we have G D .M ⇥ C0 /:E where C0  C . Below we prove that E is abelian. If M is a quasi-simple group, then by Lemma 6.2, E is abelian.


297

Thus, we assume that M is not a quasi-simple group. Then M satisfies Lemma 6.5, and hence M D N W S , where N is nilpotent and S is quasi-simple. Let X D G=C0 . Then X ä M:E D .N W S /:E, and N is the Fitting subgroup of X. Since 16 does not divide jGj, either jEj is odd and X D N:.S ⇥ E/, or S D A5 and X D N:.S:2 ⇥ E20 /. Let p be a prime divisor of jN j. If N is a p-group, then S:E  Out.N /. By Lemmas 6.3 and 6.4, we conclude that S:E  Zp 1 ı S or Zp 1 ı SL.2; 5/:2, and so E is abelian. Assume that N is not a p-group. Let N D Np ⇥ Np0 . Then CX .N / D CX .Np / \ CX .Np0 /: Let B1 D CS:E .Np / and B2 D CS:E .Np0 /. Then X=Np0 D .Np ⇥ B1 /:S:E1

and

X=Np D .Np0 ⇥ B2 /:S:E2 ;

and by induction, both E1 and E2 are abelian. Moreover, we have B1 \ B2 D 1, E=B1 D E1 and E=B2 D E2 . Thus, B1 E 0 , and B2 E 0 , and so 1 D B1 \ B2

E 0:

Then E is abelian.

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[12] S. H. Qiao and C. H. Li, The finite groups of cube-free order, J. Algebra 334 (2011), 101–108. [13] M. Short, Primitive Soluble Permutation Groups of Degree Less than 256, SpringerVerlag, Berlin, 1992. [14] Z. X. Wen, On finite groups of order 23 pq, Chinese Ann. Math. Ser. B 5 (1984), 695–710. [15] A. E. Western, Groups of order p 3 q, Proc. Lond. Math. Soc. 30 (1899), 209–263. Received February 10, 2012; revised September 25, 2012. Author information Cai Heng Li, School of Mathematics and Statistics, Yunnan University, Kunming 650031, P. R. China; and School of Mathematics and Statistics, The University of Western Australia, Crawley, WA 6009, Australia. E-mail: [email protected] Shouhong Qiao, School of Applied Mathematics, Guangdong University of Technology, Guangzhou 510006, P. R. China; and School of Mathematics and Statistics, Yunnan University, Kunming 650031, P. R. China. E-mail: [email protected]

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