Finite Groups of Fourth-Power Free Order

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 B C D or .M .A B C //:E, where A; B; C and D are all nilpotent, M is perfect,W andW EW is abelian. Further,W W 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 square- free 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 A B with A; B both cyclic. W 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 non- nilpotent group of cube-free order has the form A B C or PSL.2; p/ .A B/; W W W 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 A B C D or .M .A B C //:E; W W W W W 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. 276 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 . G ; 6/ 1, then G L, or L M . j j D D W (2) If . G ; 2/ 1, then G L, or L M , or L M G3. j j D D W W W (3) If . G ; 3/ 1, then G L, or L M , or L M G2. j j D D W W W Each case does occur. Here are examples with three nilpotent layers:

3 3 2 2 Z Z31 Z3 < AL.1; 5 / and Z D10 < AL.1; 11 /. 5 W W 11 W 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 L, or L M , D W (ii) G has three nilpotent layers, and has the form

(i) L M G2 or L M G3 W W W W or 2 (ii) .L Z / .M Q8/ Z3, 3 W W (iii) G has four nilpotent layers, and has the form

2 L M G3 G2 or L .M Z / G3 Z2: W W W W 2 W W We remark that each case listed in the theorem indeed occurs. Here are a few of basic examples for three nilpotent layer groups:

2 2 3 1 2 3 2 Zp D2q; Zp S3; Zp 3 C Q8; Zp A4; Z3 Q8 Z3; W W W C W W W W 3 3 3 Z Q8 Z3; Z Z7 Z3; Z Z31 Z3: p W W 2 W W 5 W W The groups

3 3 3 3 2 .Z11 Z5/ S3 Z Z Z3 Z2 and Z S4 Z Z Z3 Z2 W o D 11 W 5 W W p W D p W 2 W W 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 P divides p3 and Q divides q3, with p; q prime: j j j j P , P Q, W P Q R, with R dividing 23 or 33, W W j j 2 3 P Z R Z2, with R dividing 3 , W 2 W W j j P Q R S, with R dividing 33 and S dividing 23. W W W j j j j 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; Ä 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 .M .A B C //:E, where A; B; C are nilpotent subgroups of odd order, D W W E is abelian, and M is a perfect subgroup satisfying one of the following W (i) M is almost quasi-simple: f A7, J1, PSL.2; 8/, or PSL.2; p /, where p is odd and f 3, 3 Ä 3:A6, 3:A7, SL.2; p/ with p 5, or SL.2; p / with p odd, (ii) M ASL.2; p/ Z2 SL.2; p/, where p 5, D D p W (iii) M F SL.2; 5/, where F is nilpotent with each Sylow r-subgroup isomor- D W2 2 3 phic to Z , Z Zr or Z , with either r 5, or r 1 .mod 10/, r r W r D Á ˙ (iv) M Z3 S, where m is square-free, and one of the following holds: D m W (a) S PSL.2; 7/, and r 11 and r3 1 .mod 7/ for each prime divi- D Á sor r of m,

(b) S 3:A6, and r 1 or 19 .mod 30/ for each prime divisor r of m, D Á (c) S A5, and r 1 .mod 10/ for each prime divisor r of m. D Á ˙ The paper is organized as follows: In Section 2, we study automorphism groups of the groups of order pd with p prime and d 3. Then in Sections 2 and 3, we Ä 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 C. H. Li and S. Qiao

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 B, and then we W write A B C A .B C/ and A B C D A .B .C D//: W W D W W W W W D W W W 3 In what follows, Z denotes Z Z Z with l a positive integer, 3:A6 denotes l l l l a perfect group which has central factor group A6.

2 Automorphism groups of p-groups

Let G be a group. The Fitting subgroup F F.G/ is the largest nilpotent normal WD subgroup of G. If G is soluble, then the Fitting subgroup satisﬁes CG.F / F . Ä Hence

G=F G=.CG.F /F / Out.F / Out.F1/ Out.Ft /; D Ä D where F F1 Ft is such that each Fi is a Sylow pi -subgroup of F . D 3 Now Fi has order dividing pi , 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 p3. Then either P is abelian, or P is one of the two non-abelian groups of order p3. For p 2, the only non-abelian groups of 3 D order p 8 are Q8, the quaternion group, and D8, the dihedral group of order 8. D 3 2 If p is odd, then a non-abelian group of order p is Z Zp or Z 2 Zp. Note that p W p W 2 1 2 Z Zp has exponent p and is usually denoted by p , p W C 2 1 2 C 2 Zp Zp has exponent p and is denoted by p C . W For a p-group P of order at most p3, 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 Zpn with p > 2, then D Aut.P / Zpn 1.p 1/ D I if P Z2n with n 2, then D Aut.P / Z2 Z2n 2 : D For P elementary abelian, that is, P Z2 or Z3, we have Aut.P / GL.2; p/ D p p D or GL.3; p/, respectively. For the other abelian groups of order p3, we have the following lemma.

Lemma 2.1. Let P Zp2 Zp. Then Aut.P / is soluble, and a Hall p0-subgroup D 2 of Aut.P / is isomorphic to Zp 1. Finite groups of fourth-power free order 279

Proof. Write P Z 2 Zp a b . It is easily shown that each automor- D p DW h i h i phism ˛ of P has the form ´ a ai bj ; ˛ ! W b akbl ; ! where 1 i p2 1 with .i; p/ 1, 0 j p 1, k 0; p; : : : ; .p 1/p , Ä Ä D Ä Ä 2 ¹ º and 1 l p 1. The automorphism ˛ is uniquely determined by the integers Ä Ä i; j; k; l. Counting the tuples .i; j; k; l/, we conclude that Aut.P / p.p 1/:p:p:.p 1/ p3.p 1/2: j j D D On the other hand, P has two automorphisms and , deﬁned by a ai ; b b; W ! ! a a; b bj ; W ! ! where 1 i p2 1 and 1 j p 1. For suitable choices of i and j , we Ä Ä 2 Ä Ä have ; Zp 1, which is a Hall p0-subgroup of Aut.P /. h i D Clearly, Aut.P / Aut.P /pAut.P /p , a product of two nilpotent subgroups, so D 0 Aut.P / is soluble. Next, assume that P is non-abelian. Then P has order p3. It is well known that Aut.Q8/ S4, and Aut.D8/ D8. For p odd, we will identify the subgroups of D D 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].

1 2 Lemma 2.2. Let P p Z 2 Zp, with p > 2. Then Aut.P / is soluble, and D C D p W a Hall p -subgroup of Aut .P / is cyclic of order p 1 which acts non-trivially on 0 Z.P /. 2 The automorphism group of the non-abelian group P Z Zp is determined D p W by the symplectic group GSp.2; p/ GL.2; p/ in the following way. It is known D that P a; b; c ap bp cp 1; Œa; b c ; D h j D D D D i and the commutator subgroup

P Œx; y x; y P c ˆ.P / Zp: 0 D h j 2 i D h i D D Now Œx; y can be viewed as a symplectic form over P=P Z2, and since ˆ.P / 0 D 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. Not- ing that Sp.2; p/ SL.2; p/ and GSp.2; p/ GL.2; p/, we have the following D D statement. 280 C. H. Li and S. Qiao

2 Lemma 2.3. Let P Z Zp be non-abelian of exponent p. Then D p W Out.P / GL.2; p/: D As mentioned before, we also need structural information about subgroups of 2 3 2 Out.P /. For P Z , Z , or Z Zp, there are many subgroups of Out.P / which D p p p W 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 Z2 or Z2, then D 2 3 Aut.P / GL.2; 2/ S3 or Aut.P / GL.2; 3/ Q8 S3: D D D D W 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 Z Zm or H Zp .Z Zm/, where l and m divide p 1, D l D W l (ii) H L.1; p2/, and H satisfies one of the following statements: Ä 2 (a) H Zl Z2i , where l .p 1/, 0 i 3, D W j 2 Ä Ä (b) H Z Q8, where l .p 1/ and l is odd, D l W j (iii) H GL.1; p/ S2, and H satisfies one of the following statements: Ä o 2 (a) H A Z i , where A Z , 0 i 3, D W 2 Ä p 1 Ä Ä 2 (b) H A Q8, where A Zp 1 and A is odd, D W Ä j j (iv) H Z Q8 Z3, where l .p 1/ and l is odd. D l W j Proof. We inspect the subgroups H of GL.2; p/ listed in [1, Lemma 3.3 and The- orem 3.4]. If H is reducible, then H Zp .Zp 1 Zp 1/, and so H is as in 2 Ä W part (i). Suppose that H L.1; p / Zp2 1 Z2 is irreducible. Let H2 be a Ä D W Sylow 2-subgroup of H. If H2 has only one subgroup of order 2, then H2 is cyclic or Q8, and so H Z Q8 or Z Z i , 0 i 3. If H2 has at least two sub- D l W l W 2 Ä Ä groups of order 2, then it is easy to see that H Z Z2. Then we have part (ii). If D l W H L.1; p/ S2 is imprimitive, then it satisfies part (iii). Finally, assume that H Ä o is a subgroup of symplectic type, that is, H Zp 1 .Q8 Z3/, a central product. Ä ı W 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 Z3, and thus H Z Q8 Z3 W D l W since H has order fourth-power free order, as in part (iv). For P Z3, we only need subgroups of GL.3; p/ of order co-prime to p. D p Subgroups of GL.3; p/ can be found in [4, Theorem 2.2]. Finite groups of fourth-power free order 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 Z L, where l .p 1/ and L has order prime to p and is isomor- Ä l j phic to a soluble subgroup of GL.2; p/, 3 (ii) A:Z3 H A H 2;3 GL.1; p/ S3, where A Zp 1 is a non-trivial Ä D W ¹ º Ä o Ä Hall 2; 3 0-subgroup of H , and H 2;3 is a 2; 3 -subgroup of order divisi- ¹ º ¹ º ¹ º ble by 3, 3 (iii) H Zl Z3i Zp3 1 Z3, where l .p 1/, D W Ä1 2 W j (iv) H < Zp 1 .3 C :Sp.2; 3//, and H satisﬁes one of the following: ı C (a) H Zp 1 .Q8 Z3/, Ä 2 W (b) H Zp 1 .Z3 Z6/, Ä ı 1 2 W (c) H Zl .3 C Q8/, where l .p 1/ is co-prime to 3. Ä C W j Proof. We inspect the subgroups of GL.3; p/ described in [4]. 2 If H is reducible, then H is contained in Zp .Zp 1 GL.2; p//. Since p − H , W j j it follows that H satisﬁes part (i). Suppose that H is imprimitive. Then

H GL.1; p/ S3: Ä o 3 If A H GL.1; p/ 1, then H S3 which is included in part (i). So we WD \ D 3Ä may suppose that A H GL.1; p/ 1. If H=A 3 1, then H is reducible, WD \ 6D j j D contrary to the hypothesis. Thus, H A:Z3 or A:S3, as in part (ii). D Suppose that H is irreducible and contained in the normalizer of a Singer cycle, 3 p that is, H L.1; p / a Zp3 1 Z3, where a a . Let H3 be a Ä D h i W h i D W D Sylow 3-subgroup of H . If H3 has only one subgroup of order 3, then H3 is cyclic and H Z Z i , where l .p 1/ is odd. If H3 has at least two subgroups of D l W 3 j order 3, then H Zl Z3. Then part (iii) holds. D W 1 2 Finally, suppose that H is a subgroup of symplectic type. If H 3 C 3, 1 2 j \2 C j Ä then H Zp 1 .Q8 Z3/. If H 3 C 9, then H Zp 1 .Z3 Z6/. Sup- 1 Ä2 W 1j 2\ C j D Ä ı W pose 3 C H , H Zl .3 C Q8/, where l .p 1/ is co-prime to 3. C Ä Ä C W j It is known that if H is an irreducible abelian subgroup of GL.n; p/ with exponent dividing p 1, then n 1 and H is cyclic. D Noticing that a 2; 3; q -subgroup of GL.1; p/ S3 of nilpotent length 3 is ho- 2 ¹ º o momorphic to Z S3, we have the following immediate consequence. q W Lemma 2.6. Let P be a p-group of order at most p3. 2 (i) Out.P / has no subgroup isomorphic to ASL.2; 3/ Z Q8 Z3. D 3 W W (ii) If a 2; 3; q -subgroup H < Out.P / has nilpotent length 3, then H is homo- ¹ º 2 morphic to Z S3, where q .p 1/. q W j 282 C. H. Li and S. Qiao

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 A B and A is ﬁxed by H. Then there exists a sub- D group K of G such that G A K and H ﬁxes both A and K. D

3 Structural information for soluble case

All groups considered in this section are ﬁnite soluble groups of orders indivisible by p4 for any prime p.

Lemma 3.1. Let p be the largest prime divisor of the order G of G. Assume j j p 5. Then one and only one of the following statements hold: (i) Gp C G, 3 (ii) there exists a prime divisor q of G such that Gq Zq which is minimal 2 j j D normal in G, p .q q 1/, and Gp=CG .Gq/ Zp. j C C p D 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 Fp Fp be the Fitting subgroup D 0 of G. Since F 1 is self-centralized in G, we conclude that Gp does not central- 6D ize F . Suppose ﬁrst that G centralizes F . Then G does not centralize F , and as p p0 p p 3 2 Gp p , we have Fp Z or Z 2 . Let G G=Fp, and then Gp is a Sylow j j Ä Š p p D p-subgroup of G. Then Gp Zp. If Gp is normal in G, then Gp is a normal Š Sylow p-subgroup of G, which is a contradiction. Thus, Gp is not normal in G, 3 and so by the minimality of G, G has a Sylow q-subgroup Gq Zq that is a 2 Š minimal normal subgroup of G such that p .q q 1/, where Gq Syl .G/. j C C 2 q Then Fp Gq is normal in G. Clearly, C CG.Fp/ has order divisible by q. W WD Since C=Fp is normal in G, we have Gq C=Fp since Gq is minimal normal in Ä G. It follows that

Gq char Fp Gq Fp Gq C G; D W 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 p. Since Q is normal in G, the subgroup X QGp Q Gp is 6D WD D W not nilpotent. Since p > q, it follows that p divides neither q 1 nor q 1. By C Lemmas 2.1–2.5, we conclude that p .q2 q 1/ and Q Z3. Moreover, as j C C D q Finite groups of fourth-power free order 283 p > 5, we have that p2 > .q 1/2 > q2 q 1; C C C 2 2 and thus p does not divide q q 1. It follows that Gp=CGp .Q/ Zp. Since 4 C C Š q does not divide G , we conclude that Gq Q C G satisﬁes part (ii) of the j j D lemma, which is again a contradiction. This proves the lemma.

Lemma 3.2. Assume that G has exactly m prime divisors and that each prime j j divisor is at least 5. Then for each number k m there exists of size k such that Ä G G G . D W 0

Proof. By Lemma 3.1, there exists a normal Sylow p1-subgroup Gp1 , and so we get G Gp1 Gp . If k > 1, then Gp has a normal Sylow p2-subgroup Gp2 , and D W 10 10 so G Gp1 Gp2 G p1;p2 . Repeating this process, we have D W W ¹ º0

G Gp1 Gp2 Gpk G p1;p2;:::;pk G G ; D W W W W ¹ º0 D W 0 where p1; p2; : : : ; p . D ¹ kº The next lemma determines groups of orders dividing p3q3.

Lemma 3.3. Assume that G is a non-nilpotent p; q -group with p; q being distinct primes. Then one of the following holds: ¹ º (1) G has a normal Sylow subgroup, and one of the following holds:

(i) G Gp Gq is supersoluble, and q divides p 1, D W 2 3 (ii) G Gp G2, Gp=ˆ.Gp/ Z or Z , and G2 acts irreducibly on the D W Š p p factor group Gp=ˆ.Gp/, 3 3 G Z Z9 with 9 − .p 1/, or Z Z27 with 27 − .p 1/, D p W p W 2 2 1 2 (iii) G Q8 G3, Z Gq, Zp .Z Gq/, or p Gq, where q is an odd D W p W p W C W divisor of p 1, C 3 C 2 (iv) G Z Gq, where q .p p 1/ and q 3, D p W j C C 6D 2 (2) G Z Q8 Z3 ASL.2; 3/, D 3 W W D 2 (3) G Z G3 Z2, and one of the following holds: D 2 W W 2 1 2 2 (i) G Z2 3 C Z2 .Z2 Z9/ Z6, G2 is abelian, D 2 W 1 2 W D 2 W (ii) G Z2 3 C Z2 Z6 Z6, G2 is abelian, D 2 W C W D W (iii) G Z2 G3 Z2, and G=O3.G/ S4, D 2W W D (iv) G .Z2 Z9/ D6, D W 2 (v) G A4 H, where H D18, D6, D6 Z3, or Z Z2 and H satis- D D 3 W ﬁes Z.H / 1. D 284 C. H. Li and S. Qiao

Proof. By Lemma 3.1, we know that either G has a normal Sylow subgroup or G is a 2; 3 -group. ¹ º Case 1. Assume first that G Gp Gq. Assume further that G is not supersol- D W uble. 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 1 or p2 p 1, and we 2 3 1 2 C C C have Gp Q8, Zp, Zp or p C . D 2 C 3 Let q divide p p 1. Then Lemma 2.5 is satisfied, and Gp Z which C C D p is a minimal normal subgroup of G. If q 3, then part (1) (iv) of the lemma is 6D satisfied, while if q 3, then Z3 < G3=CG3 .Gp/ < Zp3 1, and so G3 Z9 or D D Z27, as in part (1) (ii). If q .p 1/ and q is odd, then part (1) (iii) is satisfied. j C Finally, assume that q .p 1/. Since Gq acts irreducibly on Gp=ˆ.Gp/, we j 2 2 conclude that q also divides p 1 or p p 1, depending on Gp=ˆ.Gp/ Z C C C Š p or Z3, respectively. If q divides p2 p 1, then q divides .p 1; p2 p 1/, p C C C C and hence q 3, as in part (1) (ii). If q divides p 1, then q divides .p 1; p 1/. D C C Hence q 2, as in part (1) (ii). D Case 2. Assume that G has no normal Sylow subgroup. By Lemma 3.1, G is a 2; 3 -group. Let F be the Fitting subgroup of G. Then F F2 F3, where ¹ º D F2 4 and F3 9. Hence F is abelian and CG.F / F . Since G is soluble, j j Ä j j Ä D G G=F Out.F / Aut.F / Aut.F2/ Aut.F3/: D Ä D D

Let F2 1. Then F F3 < G3, and CG2 .F3/ 1. Since G=F Aut.F /, D D2 D Ä we conclude that F3 Z and G3 is non-abelian. Clearly, D 3 Aut.F / GL.2; 3/ Q8 S3: D D W 2 By Lemma 2.2, we have G3 Z3 Z3. Since G3 is not normal in G, it follows D W 2 that G=F is not nilpotent, and so G=F Q8 Z3, and G Z Q8 Z3. This is D W D 3 W W part (2). 2 Next assume that F2 1. Then we have F2 Z which is minimal normal 6D Š 2 in G. If F2 ˆ.G/ 1, then F2 ˆ.G/ since F2 is minimal normal in G. Then \ 6D D G=F2 is 2-nilpotent, and thus G is 2-nilpotent, a contradiction. It follows that 2 G Z G3 Z2: D 2 W W If G2 is not abelian, then G2 D8 and G=O3.G/ S4, and this is part (3) (iii). Š Š 2 3 Suppose that G2 is abelian. Then G3 is not cyclic. It follows that G3 Z3, Z3, 1 2 1 2 2 D 1 2 Z3 Z9, 3 C or 3 C . If G3 Z3, then we have G A4 D6. If G3 3 C 1 2 C 2 D 2 D D or G3 3 C , then G .Z2 Z9/ Z6 or Z6 Z6, respectively. Observe that the D C D W W 2 group G3 Z3 Z9 yields the groups G A4 D18 and G .Z2 Z9/ D6. D 3 2 D D W Finally, if G3 Z , then G A4 Z Z2. Thus parts (3) (i), (ii), (iv) and (v) D 3 D 3 W are obtained. Finite groups of fourth-power free order 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 . G ; 6/ 1, then G A B, where A; B are both nilpotent, and j j D D W A is characteristic in G. Proof. Let F be the Fitting subgroup of G. Then G F:G. Since D G G=F Out.F / D Ä and each prime divisor of G is at least 5, it follows from Lemmas 2.1-2.5 that j j 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 G. Then G MF . Š D Suppose that F M ˆ.M /. Then there exists a maximal subgroup H of M \ 6Ä such that M .F M /H since F M C M . Thus, G FM FH , which D \ \ D D contradicts the minimal choice of M . Hence we have F M ˆ.M /. Moreover, \ Ä M=F M G=F is abelian, and hence M is nilpotent, see [9, III, Theorem 3.5]. \ Š In order to prove the theorem, we may suppose that M is not divisible by the cube j j of any prime. Since F M ˆ.M /, .F M/r ˆ.M /r ˆ.Mr /, we have \ Ä \ Ä D Mr b Z 2 . Below, we try to ﬁnd a characteristic nilpotent subgroup A and D h i D r a nilpotent subgroup B such that G A B. D W If F M 1, we are done. Thus suppose that F M 1. Let r be a prime \ D 2 \ 6D divisor of F M . Then Fr has order r . Let Fr a Z 2 , .F M/r Zr j \ j D h i D r \ D which is normal in G. Since Mr b Z 2 , it follows that D h i D r r r a b .F M/r h i D h i D \ 3 which is centralized by Mr . Since Gr r and Z 2 Fr C Gr , we conclude 0 j j D r D that Gr Z 3 or Fr Zr . As ˆ.Fr / is centralized by Mr , by [6, Theorem 5.2.4], D r W 0 Fr , and so Gr , is centralized by Mr . Then we write G Fr .Gr Mr /, Gr Mr 0 D 0 0 0 is nilpotent. We could consider F instead of F since F is a characteristic nilpo- r0 r0 2 tent subgroup in G. If Fr Z , since F is nilpotent, we conclude that D r F A .F M/ D \ for some nilpotent subgroup A such that G FM AM , A and Ar are both D D normal in G by Lemma 2.7. We write Ar a Zr , Mr b Z 2 and .F M/r Zr . Let Gr be D h i D D h i D r \ D a Sylow r-subgroup of G. Then Gr > Fr , and Gr Ar Mr Zr Zr2 . D Š r 2 Assume that NG. a / CG. a /. Clearly, Or .G/ Fr a; b Z . Sup- h i 6D h i D D h i D r pose that a ai brj , with .j; r/ 1, such that 0 D D o.a / r and NG. a / CG. a /: 0 D h 0i 6D h 0i 286 C. H. Li and S. Qiao

Let x NG. a / CG. a /. Then x can be written as x vwu, where u A, 2 h 0i n h 0i D 2 v Mr and w Mr . Thus, there exist integers j ; i such that 1 j ; i < p 2 2 0 0 0 Ä 0 0 and aij 0 brjj 0 .a /j 0 .a /x .ai brj /x x 1ai brj x D 0 D 0 D D u 1w 1v 1ai brj vwu u 1w 1ai brj wu D D u 1w 1ai wbrj u u 1ai 0 brj u D D ai 0 brj : D It follows that brjj 0 brj , and hence j 1. So x centralizes a , which is a D 0 D 0 contradiction. Therefore, a is the unique subgroup of G of order r such that h i NG. a / CG. a /, and so Ar a is a characteristic subgroup of G, as de- h i 6D h i D h i sired. Now assume that NG. a / CG. a /. Since a is normal in G, it follows that h i D h i h i ŒGr ;Mr 1, and hence G AM Ar .Gr Mr /, where Ar Fr which 0 D D D 0 0 0 D 0 is characteristic in G. Repeating this process, we obtain that G A1B1 such that D A1;B1 are both nilpotent, A1 B1 1, and each Sylow subgroup of A1 is a char- \ D acteristic subgroup of G. So G A1 B1, and A1 is characteristic in G. D W Lemma 4.2. Let G P H, where P is a Sylow p-subgroup of G. Suppose that D W N < P is non-identity and normal in G such that G=N .P=N / H. Then Š P L M with L N , and G L .M H/. D W Ä D W Proof. By Lemma 2.7 and Lemmas 2.1–2.5, we only need to consider the cases 2 P Z 2 Zp, Z 2 Zp or Z Zp. If N ˆ.P /, by [9, III, Theorem 3.18], D p p W p W D H centralizes P , and the lemma trivially holds. Thus we suppose that N ˆ.P /. p 6D Let P a b Z 2 Zp. Suppose that N a b . Without loss, D h i h i D p 6D h i h i we may suppose N a , or N b . If N ap , then N ˆ.P /, contrary Ä h i D h i D h i D to our hypothesis. If N a , by Lemma 2.7, H normalizes some subgroup of P D h i which has trivial intersection with a , say b . In this case, H centralizes b , and h i h i h i so G a . b H/. If N b , similarly, we may suppose H normalizes a , D h i W h i D h i h i and so G b . a H/. Now suppose that N ap b . Since H acts on D h i W h i D h i h i N and ˆ.P /. ap / is normalized by H, by Lemma 2.7, there is a subgroup D h i of N which is normalized by H, say b . Then since H acts on P and b is h i h i normalized by H, again by Lemma 2.7, there is a subgroup which is normalized by H, say a . It follows that G b . a H/. h i D h i W h i p Let P a b Z 2 Zp. Since N ˆ.P /, we have N a b . As D h i W h i D p W 6D D h i h i H normalizes ap , by Lemma 2.7, H normalizes some subgroup different from h i ap , say b . Then consider the action of H on P P=ˆ.P / Z2. Since b h i h i D Š p h i is normalized by H , there is a subgroup A of P such that P A b and H D h i Finite groups of fourth-power free order 287

normalizes A. Then A Z 2 since each subgroup of P of order p is contained Š p in N . Since P H b .A H/ and ˆ.P / ˆ.A/; W D h i W D by [9, III,Theorem 3.18], H centralizes A, thus H centralizes ˆ.P /. Z.P //. By D Lemma 2.2, H centralizes P , and the lemma holds trivially. 2 The case P Z Zp is dealt with in a similar way. D p W

Lemma 4.3. Assume that G G 2;3 G 2;3 . Then there exist nilpotent groups D ¹ º0 W ¹ º L;M such that one of the following holds:

(i) G 2;3 G3 G2 and G L M G3 G2, ¹ º D W D W W W (ii) G 2;3 G2 G3 and G L M G3, ¹ º D W D W W 2 2 (iii) G 2;3 Z2 G3 Z2 and G L .M Z2/ G3 Z2, ¹ º D W W D W W W 2 2 (iv) G 2;3 Z3 Q8 Z3 and G .L Z3/ .M Q8/ Z3. ¹ º D W W D W W In particular, if . G ; 2/ 1, then G L M or L M G3, while if . G ; 3/ 1, j j D D W W W j j D then G L M or L M G2. D W W W Proof. By Lemma 4.1, G 2;3 L M with L characteristic in G. We analyse ¹ º0 D W the structure of G according to the structure of G 2;3 . ¹ º If G 2;3 G3 G2, then ¹ º D W

G G 2;3 G 2;3 L M G3 G2: D ¹ º0 W ¹ º D W W W

Let G 2;3 G2 G3, which is homomorphic to A4. Then ¹ º D W G L M G2 G3: D W W W Suppose that Mp is not centralized by G2 for some prime p dividing M . Then j j Mp G2 G3 has nilpotent length 3. It follows from Lemma 2.6 that Mp central- W W izes L. Write M M M such that is maximal subject to that, for each D 0 p , Mp is not centralized by G2. Then 2 G L M G2 G3 .L M / .M G2/ G3 A B G3: D W W W D W 0 W D W W 2 Suppose now that G 2;3 Z2 G3 Z2, which is homomorphic to S4. If ¹ º D W W 2 2 Mp Z G3 Z2 .Mp Z / G3 Z2; W 2 W W 6D 2 W W then M centralizes L. Write M M M such that is maximal subject to p 0 D 2 that, for each p , Mp is not centralized by Z . Then 2 2 2 2 G L M G 2;3 .L M / .M Z2/ G3 Z2 A .B Z2/ G3 Z2: D W W ¹ º D W 0 W W D W W W 288 C. H. Li and S. Qiao

2 Finally, let G 2;3 Z3 Q8 Z3 ASL.2; 3/. By Lemma 2.6, for each Sylow ¹ º D W W D 2 p-subgroup Gp for p 5, we have Gp G 2;3 .Gp Z3/ Q8 Z3. Thus, W ¹ º D W W 2 G L M G 2;3 .Z3 .L M // Q8 Z3: D W W ¹ º D W W W Consider the subgroup L M Q8 Z3. If a Sylow p-subgroup Mp is not central- W W W ized by Q8, it follows from Lemma 2.6 that Mp centralizes L. Write M in the form M M M such that M centralizes L, and M is centralized by Q8. D 0 0 Therefore, L M Q8 Z3 .L M / .M Q8/ Z3; W W W D W 0 W 2 and thus, G .L M Z / .M Q8/ Z3. D 3 W 0 W Lemma 4.4. Assume that G is not normal in G. Then one of the following 2;3 0 statements hold: ¹ º 3 (i) Z G2 C G and G A B or G A B G3, 2 D D W D W W 3 (ii) Z G3 C G and G A B or G A B G2, 3 D D W D W W where A; B are suitable nilpotent subgroups of G.

Proof. Choose the set of primes to be maximal such that G G G subject D W 0 to 2; 3 . Then we have G G 2;3 . We note that may be empty, in this … 0 6D ¹ º0 case, G 1. Moreover, any Sylow p-subgroup of G with p 5 is not normal D 0 in G . By Lemma 3.1, one and only one of the following statements holds: 0 3 (a) G2 C G , and Z G7 G which is homomorphic to 0 2 W Ä 0 3 3 Z Z7 AGL.1; 2 /; 2 W D 3 (b) G3 C G , and Z G13 G which is homomorphic to 0 3 W Ä 0 3 3 Z Z13 < AGL.1; 3 /: 3 W 3 First suppose that (a) holds and let Z G2 C G , as in (a). Suppose that G2 2 Š 0 is not normal in G. Then G2 does not centralize some non-trivial Sylow subgroup 3 P of G . We consider the subgroup P Z2 G7. Since Aut.P / has no section iso- 3 3 W W 3 morphic to Z Z7, we get P Z G7 P .Z G7/, which is a contradiction. 2 W W 2 W D W 2 Consequently, G2 E G. Then G G2 G2 . D W 0 Suppose that G2 is 3-nilpotent. By Lemma 4.3, G2 L M or L M G3, 0 0 D W W W where L and M are nilpotent. If G2 L M G3, then we have 0 D W W G G2 G2 G2 L M G3 .G2 L7 / .G7 M/ G3 A B G3: D W 0 D W W W D 0 W W D W W Thus, we suppose that G2 L M . Then G G2 L M . Let C CG.G2/. 0 D W D W W D Then G=C Z7, or Z7 Z3. Hence Š W G G2 L M .G2 L7 / L7 M .G2 L7 / .G7 M3 / M3: D W W D 0 W W D 0 W 0 W Finite groups of fourth-power free order 289

If M3 1, then G A B, as desired. Suppose that M3 1. Let D D W 6D X .G2 L7 / .G7 M3 /: D 0 W 0 Then a Sylow 3-subgroup X3 has order at most 9. Since X2 G2 C X, it follows D that X X X . Then G .G L / .G M / G . 3 30 2 3;7 0 7 30 3 D D ¹ º W W 3 Now we suppose that G2 is not 3-nilpotent. Then G2 has section Z Z13. 0 0 3 W By induction, G is normal in G or G L M for some suitable nilpotent 3 20 20 D W 3 subgroups L;M . Let G2 L M . If L3 G3, then G2 has no section Z Z13, 0 D W 6D 0 3 W a contradiction. Thus L3 G3, and then G .G2 L7 / .G7 M/. Next we D D 0 W let G C G . Suppose that G is not normal in G. Then G does not centralize 3 20 3 3 3 a Sylow p-subgroup P of G . Consider the subgroup P Z3 G13. Since Aut.P / 3 3 W W 3 has no section isomorphic to Z Z13, we have P Z G13 P .Z G13/, 3 W W 3 W D W 3 which is a contradiction. Consequently, G3 E G, and so

3 3 3 3 G .Z2 Z3/ L M .Z2 Z3 L 7;13 / .L7 L13/ M: D W W D ¹ º0 W W By Lemma 4.2, there are nilpotent subgroups A1;A2;B1;B2 such that

L7 A1 B1; D L13 A2 B2; D 3 3 G .Z2 Z3 L 7;13 A1 B1/ .A2 B2 M /: D ¹ º0 W 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 G is fourth-power free. j j If G is nilpotent, then G is as in part (1) of Theorem 1.1. If . G ; 6/ 1, then j j D by Lemma 4.1, G is as in part (1) of Theorem 1.1. Assume that G is normal in G. Then G is determined in Lemma 4.3, which 2;3 0 ¹ º 2 leads to the following statements. If G 2;3 G3 G2 or Z2 G3 Z2, then G has ¹ º D W W W four nilpotent layers, and is as in part (3) of Theorem 1.1. If G 2;3 G2 G3 or 2 ¹ º D W Z Q8 Z3, then G has three nilpotent layers, and is as in part (2) of Theorem 1.1. 3 W W Finally, if G is not normal in G, then Lemma 4.4 shows that G satisﬁes 2;3 0 part (1) or (2). ¹ º

5 Semi-direct products

Here we study the semi-direct products appeared in Theorem 1.1.

Lemma 5.1. Let G .X Y/ Z and let Z normalize X and Y . Then G is a sub- D W group of .X Z/ .Y Z/. W W 290 C. H. Li and S. Qiao

Proof. An element g of G has the form .x; y; z/, which corresponds to an element .x; z y; z/ of the group .X Z/ .Y Z/. I W W Lemma 5.2. Let G X .Y Z/. Then G is a subgroup of .X Y/ .X Z/. D W W W Proof. An element g of G has the form .x; y; z/, which corresponds to an element .x; y x; z/ of the group .X Y/ .X Z/. I W W Applying Lemmas 5.1 and 5.2 repeatedly leads to the next conclusion.

Lemma 5.3. Suppose that G .X1 X2/ .Y1 Y2/ .Z1 Z2/, and suppose D W W that Xi Yj Zk is a subgroup of G, 1 i; j; k 2. Then G is a subgroup of the QW W Ä Ä group 1 i;j;k 2.Xi Yj Zk/. Ä Ä W W Proof of Corollary 1.2. Let G be a group of fourth-power-free order. Then G sat- isﬁes the conditions of Theorem 1.1. In the following, P and Q denote groups of order dividing p3 and q3, respectively, where p; q are primes. If G L M with L;M both nilpotent, then D W Y G .P Q/: Ä W For three-nilpotent-layer groups, by Theorem 1.1, basic group G has the form 2 Z G3 Z2, or L M G2, or L M G3. If G L M G2, then 2 W W W W W W D W W Y G .P Q G2/; Ä W W while if G L M G3, then D W W Y G .P Q G3/: Ä W W

Finally, consider four-nilpotent-layer groups. If G L M G3 G2, then D W W W Y G .P Q G3 G2/; Ä W W W 2 while if G L Z G3 Z2, then D W 2 W W Y 2 G .P Z G3 Z2/: Ä W 2 W W We next give more precise characterizations of the ‘basic’ groups of fourth- power-free order, all of which have at most four primes dividing the order. If G is a p; q -group, then the structure of G is determined in Lemma 3.3. We study the ¹ º structure of groups of orders divisible by three or four primes in the rest of this section. Finite groups of fourth-power free order 291

Lemma 5.4. Let G be a 2; p; q -group with p; q 5. Assume that G has at least three nilpotent layers. Then¹ º

G Gp Gq G2; D W W 2 which is homomorphic to Zp H , where H is a non-nilpotent 2; q -subgroup of 2 W ¹ º L.1; p /, GL.1; p/ S2, or GL.1; p/ S3. o o Proof. By Theorem 1.1 and Lemma 3.1, we have G Gp Gq G2. Suppose that 3 2 D W W p < q. By Lemma 3.1, Gp Z , and q .p p 1/, and by Lemma 2.5, we D p j C C conclude that G=CG.Gp/ Zp2 p 1 Z3: Ä C C W Hence G2 centralizes Gp. Let C CG .Gp/. It follows that D q .Gq G2/=C .Gq=C / G2: W Š By Lemma 4.2, we have Gq R Q such that R C and Q is centralized by G2. D W So G .Gp R/ .Q G2/, which is a contradiction since G has three nilpotent D W layers. Thus, G Gp Gq G2 with p > q. Let G G=ˆ.Gp/ Gp Gq G2. Since D W W D D W W G has three nilpotent layers, it follows from Lemma 4.2 that H G=C .Gp/ WD G is not nilpotent. By Lemmas 2.4 and 2.5, we have that H is a 2; q -subgroup of 2 ¹ º L.1; p /, GL.1; p/ S2, or GL.1; p/ S3. o o Lemma 5.5. Let G be a 3; p; q -group with p; q 5. Assume that G has at least ¹ º 3 three nilpotent layers. Then G Zp Gq G3, and one and only one of the following statements hold: D W W 2 (i) q .p p 1/ and G=CG.Gp/ Z i Z3 with i 3, j C C Š q W Ä 2 3 (ii) q .p 1/ and G=CG.Gp/ Zq Z3, Z Z3 or Z Z3. j D W q W q W Proof. By Theorem 1.1, we have G Gp Gq G3. By Lemma 3.1, either p > q, 3 2 D W W 3 or p < q and Gp Zp, q .p p 1/, and Gq=CGq .Gp/ AL.1; p /. D j C C Ä 3 Suppose that p > q. By Lemmas 2.4-2.5, we conclude that Gp Z , and ei- D p ther G=CG.Gp/ GL.1; p/ Z3 with q .p 1/ Ä o j or 3 G=CG.Gp/ AL.1; p /: Ä 2 For the former, we conclude that G=CG.Gp/ Zq Z3, Z Z3 or Zq Z3. For Š o q W W the latter, we have G=CG.Gp/ Zqi Z3, where i 3. 3 D 2 W Ä 3 For p < q and Gp Z , q .p p 1/ and Gq=CG .Gp/ AL.1; p /, D p j C C q Ä it follows that Gq=CG .Gp/ Z i Z3 with i 3. q D q W Ä 292 C. H. Li and S. Qiao

Lemma 5.6. Let G be a 2; 3; p -group that has at least three nilpotent layers. Then one and only one of¹ the followingº statements hold: 3 3 3 (1) G Z G7 G3, and G is homomorphic to AL.1; 2 / Z Z7 Z3, D 2 W W D 2 W W (2) G Gp G 2;3 , and one of the following holds: D W ¹ º 3 1 2 (i) G Zp 3 C Q8, D W C W 2 (ii) G Gp G3 G2, and G is homomorphic to Z S3 or .Zp Z3/ S2, D W W p W W o 2 (iii) G Gp Q8 G3, and G is homomorphic to Z Q8 Z3, D W W p W W 3 2 3 (iv) G Z Z G3 Z2, and G is homomorphic to Z S4. D p W 2 W W p W Proof. Suppose that Gp is not normal in G. By Lemma 3.1, for some prime divi- 3 2 sor q of G , Zq Gq C G and p .q q 1/. Now q 2 or 3, and further, if j j D 3 j C C D 3 q 2, then p 7 and G2 Z2; if q 3, then p 13 and G3 Z3. D D 3 D D D D Suppose that G Z G13 G2. Then by Lemma 2.5, we have D 3 W W 3 G=CG.G3/ AL.1; 3 /: Ä

Let C CG .G3/. Then D 13

G13 G2=C .G13=C / G2: W Š

By Lemma 4.2, we have G13 L M such that L C and M is centralized D W by G2. Thus, G .G3 L/ .M G2/; D W which is a contradiction since G has at least three nilpotent layers. 3 Assume that G Z G7 G3. Then by Lemma 2.5, we have D 2 W W

G=CG.G2/ Z7 Z3: D W

Finally, assume that G is of the form G Gp G 2;3 . The structure of G 2;3 D W ¹ º ¹ º is determined in Lemma 3.3. Analysing the candidates for G 2;3 leads to the conclusion in part (2). ¹ º

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 .G/ 4. Then G has the form Gp Gq G3 G2. j j D W W W Proof. This directly follows from Theorem 1.1 because G has four prime divi- j j sors. Finite groups of fourth-power free order 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 f order 4 or 8 is A7,J1, PSL.2; 8/, or PSL.2; p /, where p is odd and f 3, see Ä [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 3 G=Z.G/. It follows that G 3:A6, 3:A7, SL.2; p/ with p 5, or SL.2; p / with D p odd. Assume that G is almost simple. Then either G is simple, or G is one of the groups: PL.2; 8/, or PGL.2; p/, or PSL.2; p3/: with o./ 6. h i j Finally, if G is almost quasi-simple but not listed above, then G SL.2; p3/:3. D 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: f A7, J1, PSL.2; 8/, or PSL.2; p /, where p is odd and f 3, Ä 3 3:A6, 3:A7, SL.2; p/ with p 5, or SL.2; p / with p odd, PL.2; 8/, PGL.2; p/, or PSL.2; p3/: with o./ 6, or SL.2; p3/:3. h i j Lemma 6.2. Suppose that G has a normal quasi-simple subgroup S. Then G has a soluble group C such that either G S C , or G is one of the following groups: D (i) .PSL.2; 8/ C /:3, with G=C PL.2; 8/, Š (ii) .PSL.2; p/ C /:2, with G=C PGL.2; p/, Š (iii) .PSL.2; p3/ C /: , with o./ 6 and h i j G=C PSL.2; p3/: PL.2; p3/; Š h i Ä (iv) .SL.2; p3/ C /:3, with G=C SL.2; p3/:3 < L.2; p3/. Š Proof. The centralizer CG.S/ is soluble, and S CG.S/ Z.S/. By Lemma 6.1, \ D the centre Z.S/ 1, Z2 or Z3, and Z.S/ is a Sylow group of S. Thus, we may D write CG.S/ C Z.S/, and SCG.S/ C S. D D To complete the proof, we assume that G C S. Then G=CG.S/ is an al- 6D most simple group that is not a simple group. From Lemma 6.1 we conclude that 294 C. H. Li and S. Qiao

3 G=CG.S/ PL.2; 8/, PGL.2; p/, or PSL.2; p /: with o./ 6. It is now D h i j 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 F.G/ F1 Ft ; D D where Fi is a Sylow subgroup. Then G=F Out.F / Out.F1/ Out.Ft /, Ä D 2 3 1 2 and thus Out.Fi / is insoluble for some i. Hence Fi Z , Z or p , for some D p p C prime p. C Recall that a group X is called perfect if X equals its commutator subgroup X 0. Lemma 6.3. Let P Z2 or p1 2, with p 5. Then a non-trivial perfect sub- D p C group S of Out.P / is quasi-simple,C and further we have either S SL.2; p/ or D S SL.2; 5/ with p 1 .mod 10/. Moreover, if H is an insoluble subgroup D Á ˙ of Out.P / of fourth-power-free order, then S H Zp 1 S. Ä Ä ı Proof. First, for the case P Z2, we have either S SL.2; p/ or S SL.2; 5/ D p D D and p 1 .mod 10/; see, for example, [1, Theorems 3.4–3.5]. Á ˙ If P p3, we only need to identify quasi-simple subgroups of Out.P / of j j D 2 order co-prime to p. Thus, for P Z Zp non-abelian, by Lemma 2.3, D p W Out.P / GSp.2; p/; D and so S SL.2; 5/, with p 1 .mod 10/. D Á ˙ Lemma 6.4. Let P Z3, and let S be a non-trivial perfect subgroup of Out.P / D p such that . S ; p/ 1. Then S is quasi-simple, and one of the following holds: j j D S PSL.2; 7/, where p 11 and p3 1 .mod 7/, D Á S 3:A6, where p 1 or 19 .mod 30/, D Á S A5, where p 5 or p 1 .mod 10/. D D Á ˙ If H is an insoluble subgroup of Out.P / of fourth-power-free order, then

S H Zp 1 S: Ä Ä ı Proof. As S is perfect and . S ; p/ 1, we have that p is odd and S SL.3; p/. j j D Ä Write Z Z.SL.3; p// Z3. Then SZ=Z PSL.3; p/. If SZ=Z has no non- WD Ä Ä trivial 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. Finite groups of fourth-power free order 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 . S ; p/ 1, j j D by Lemma 2.4, SZ=Z SL.2; 5/. If Z3 Z S, then we have that A5 has a D D Ä covering group with centre of order 6, which is a contradiction. Thus, SZ=Z S SL.2; 5/: D D 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 ASL.2; p/ Z2 SL.2; p/, where p > 5, D D p W (ii) G .Z2 Z3 B/ SL.2; 5/, where D p1:::pr q1:::qs W 1 2 1 2 1 2 B .r1/ C .r2/ C .rt / C ; D C C C with pi ; qj ; r 1 .mod 10/, with possible exception that one of pi is 5, l Á ˙ 3 (iii) G Zp1p2:::pr S, where the numbers pi are distinct primes, and one of the followingD holds:W 3 (a) S PSL.2; 7/, where pi 11 and pi 1 .mod 7/, D Á (b) S 3:A6, where pi 1 or 19 .mod 30/, D Á (c) S A5, where pi 1 .mod 10/. D Á ˙ Proof. Let R be the largest soluble normal subgroup of G. Then G R:G, where D 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/ G. By Lemma 6.1, we have D Z.S/ Zc, where c 3. D Ä Since G is fourth-power free, we get R2 2 and R3 9. Then R is not a j j j j Ä j j Ä 2; 3 -group since G is perfect and F .G/ F .G/ is self-centralized in G. Since ¹ º D 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/: Ä We may suppose that CG.Rp/ G. Because G is perfect, G=CG.Rp/Rp is a 6D non-trivial perfect subgroup of Out.Rp/. By Lemmas 6.3–6.4, we conclude that

G=CG.Rp/Rp S; Š where S is as in Lemmas 6.3–6.4. Let N Rc . Then D 0 N Rp Np Np Np : D 0 D 0 296 C. H. Li and S. Qiao

As G is perfect, so is G=Rp. By induction, we have that N is nilpotent. It then follows that R N Z.S/, and G N S. Since G=Np Np S is a perfect D W D W 0 Š W 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 SL.2; p/ with p 7, then by Lemma 6.3, we conclude that D G Z2 SL.2; p/: D p W 2 Let S SL.2; 5/. By Lemma 6.3, p 5 or p 1 .mod 10/, and Np Zp, 1 2 D 3 D Á ˙ D p C , or Zp. It follows that C pi qj rk Y 2 Y 1 2 Y 3 N Zp q C Zr ; D C p p1 q q1 r r1 D D D where p1; : : : ; pi ; q1; : : : ; qj ; r1; : : : ; rk are all of the prime divisors of N . 3 j j Suppose now that S PSL.2; 7/. By Lemma 6.4, Np Zp, with p > 11 and 3 D 3 D 3 p 1 .mod 7/. Thus, N Zn with n square-free, and G Zn PSL.2; 7/. Á D 3 D W For S 3:A6, Lemma 6.4 tells us that Np Zp, with p 1 or 19 .mod 30/. D 3 D3 Á Thus, N Zn with n square-free, and G Zn 3:A6. D D 3W Finally, for S A5, by Lemma 6.4, Np Zp, with p 1 .mod 10/. Thus, 3 D 3 D Á ˙ N Z with n square-free, and G Z A5. D n D n W Proof of Theorem 1.3. Let G be a ﬁnite insoluble group of fourth-power-free order. Let M G , the smallest normal subgroup of G such that G=M is soluble. D 1 Then M is perfect, and so M is one of the groups listed in Lemmas 6.1 and 6.5. Let C CG.M / C G, and let E G=.MC /. Then D D E Out.M / and G .M C /:E: Ä D ı Clearly, we have M C Z.M /. If Z.M / 1, then MC M C . Suppose \ D D D that Z.M / 1. Then M is quasisimple but not simple, or M is the group in (ii) 6D of Lemma 6.5. Let M be quasisimple but not simple. Then M is 3:A6, 3:A7, 3 SL.2; p/, SL.2; p /. For the former two cases, MC M C3 , while for the D 0 latter two cases, MC M C2 . Let D 0 M .Z2 Z3 r1 2 r1 2 r1 2/ SL.2; 5/: D p1:::pr q1:::qs 1C 2C t C W Then Z.M / Zr :::r . Thus we get M C Zr :::r . Clearly, M C is a Hall D 1 t \ D 1 t \ subgroup of C . Then

MC M C r1;:::;rt : D ¹ º0 Thus we have G .M C0/:E where C0 C . D Ä Below we prove that E is abelian. If M is a quasi-simple group, then by Lemma 6.2, E is abelian. Finite groups of fourth-power free order 297

Thus, we assume that M is not a quasi-simple group. Then M satisﬁes Lem- ma 6.5, and hence M N S, where N is nilpotent and S is quasi-simple. Let D W X G=C0. Then X M:E .N S/:E, and N is the Fitting subgroup of X. D Š D W Since 16 does not divide G , either E is odd and X N:.S E/, or S A5 j j j j D D and X N:.S:2 E2 /. Let p be a prime divisor of N . D 0 j j If N is a p-group, then S:E Out.N /. By Lemmas 6.3 and 6.4, we conclude Ä that S:E Zp 1 S or Zp 1 SL.2; 5/:2, and so E is abelian. Ä ı ı Assume that N is not a p-group. Let N Np Np . Then D 0 CX .N / CX .Np/ CX .Np /: D \ 0 Let B1 CS:E .Np/ and B2 CS:E .Np /. Then D D 0 X=Np .Np B1/:S:E1 and X=Np .Np B2/:S:E2; 0 D D 0 and by induction, both E1 and E2 are abelian. Moreover, we have B1 B2 1, \ D E=B1 E1 and E=B2 E2. Thus, B1 E , and B2 E , and so D D 0 0 1 B1 B2 E : D \ 0 Then E is abelian.

Bibliography

[1] D. M. Bloom, The subgroups of PSL.3; q/ for odd q, Trans. Amer. Math. Soc. 127 (1967), 150–178. [2] H. Dietrich and B. Eick, On the groups of cube-free order, J. Algebra 292 (2005), 122–137. [3] K. Doerk and T. Hawkes, Finite Soluble Groups, Walter de Gruyter, Berlin, 1992. [4] D. L. Flannery and E. A. O’Brien, The linear groups of small degree over finite fields, Internat. J. Algebra Comput. 15 (2005), 467–502. [5] O. E. Glenn, Determination of the abstract groups of order p2qr; p, q, r being distinct primes, Trans. Amer. Math. Soc. 7 (1906), 137–151. [6] D. Gorenstein, Finite Groups, Harper and Row, 1968. [7] D. Gorenstein, Finite Simple Groups: An Introduction to Their Classification, Plenum Press, New York, 1982. [8] D. Hölder, Die Gruppen der Ordnungen p3, pq2, pqr, q4, Math. Ann. 43 (1893), 301–412. [9] B. Huppert, Endliche Gruppen, Springer-Verlag, Berlin, 1967. [10] B. Huppert and N. Blackburn, Finite Groups III, Springer-Verlag, Berlin, 1982. [11] H. L. Lin, On groups of order p2q, p2q2, Tamkang J. Math. 5 (1974), 167–190. 298 C. H. Li and S. Qiao

[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, Springer- Verlag, Berlin, 1992. [14] Z. X. Wen, On ﬁnite groups of order 23pq, Chinese Ann. Math. Ser. B 5 (1984), 695–710. [15] A. E. Western, Groups of order p3q, 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]