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Finite groups in which primary subgroups have cyclic cofactors∗ 1. School of Math. and Informational Science, Shandong Institute of Business and Technology, 2. Department of Mathematics, Zhejiang University, Hangzhou 310007, China In this paper, we prove the following theorem: Let G be a group, q be the largest prime divisor of |G| and π = π(G) \ {q}. Suppose that the factor group X/coreGX is cyclic for every p-subgroupX of G and every p ∈ π. Then: 1) G is soluble and its Hall {2, 3} -subgroup is normal in G and is a dispersive group by Ore;2) All Hall {2, 3}-subgroups of G are metanilpotent;3) Every Hall p -subgroup of G is a dispersive group by Ore, for every p ∈ {2, 3};4) lr(G) ≤ 1, for all r ∈ π(G).
All groups considered in this paper are finite.
If H is a subgroup of a group G, then the factor group H/coreGH is said to be the cofactor of H in G, where coreGH = ∩x∈GHx is the maximal normal subgroup of G contained in H.
The structure of the groups with given some properties of cofactors of subgroups was studied in [1-5]. In [1], the non-soluble groups in which the cofactors of maximal subgroups are either nilpotent groups or Schmidt groups were described. From the result in [1], we know that the groups with nilpotent cofactors of maximal subgroups are soluble. In [2], for an odd prime p, it was proved that the groups in which maximal subgroups have p-nilpotent cofactors are p-soluble. In [3], the groups in which cofactors of subgroups are simple groups have considered: Such groups are soluble and all Sylow subgroups of the factor group G/Z∞(G) are abelian. In [4], the author have studiedthe groups in which all biprimary subgroups have primary cofactors and also described the finite non-soluble groups in which all proper subgroups have primary cofactors or biprimary cofactors. In ∗Research is supported by a NNSF grant of China (Grant #10771180).
0Keywords: Finite groups, cofactors of subgroups, p-length, structure of groups.
0AMS Mathematics Subject Classification (2000): 20D10, 20D20.
[5], the author investigated the p-groups G in which |H/coreGH| ∈ {1, p} for every subgroup H ofG. In particular, it was proved that such groups are metaabelian.
As a continuation, in this paper, we study the groups in which some primary subgroups have cyclic cofactors. We obtain the following theorem.
Theorem. Let G be a group, q be the largest prime divisor of |G| and π = π(G) \ {q}. Suppose that the factor group X/coreGX is cyclic for every p-subgroup X of G and every p ∈ π. Then: 1) G is soluble and its Hall {2, 3} -subgroup is normal in G and is a dispersive group by Ore; 2) All Hall {2, 3}-subgroups of G are metanilpotent; 3) Every Hall p -subgroup of G is a dispersive group by Ore, for every p ∈ {2, 3}; 4) lr(G) ≤ 1, for all r ∈ π(G).
All unexplained notations and terminologies are standard. The reader is referred to [6] and [7].
Recall that a group G is called primitive if it has a maximal subgroup M such that coreGM = 1. Inthis situation, M is called a stabilizer of G (see [6]).
Let G be a group of order pα1 pα2 · · · pαr r , where p1 > p2 > · · · pr .
dispersive group by Ore if there exists Gp , G , · · · , G If π be a non-empty set of prime numbers, then we denote by π the complement of π in the set of all prime numbers. We denote by π(G) the set of all prime divisors of the order |G| of a group G.
For a p-soluble group G, we denote by lp(G) the p-length of G; denote by rp(G) the p-rank of G, i.e., rp(G) = max{n | pn is the order of some p-chief factor of G}. For a soluble group G, we denoteby d(G) the derived length of G and denotes by r(G) the rank of G, i.e., r(G) = maxp∈π(G)rp(G). IfG = 1, then let rp(G) = r(G) = 0 (see [6, Definition VI.5.2]) In order to prove our theorem, we need the following a series of Lemmas.
Lemma 2.1(see [8, Lemma 2.6]). Let G be a soluble primitive group with stabilizer M . Then: 2) F (G) = CG(F (G)) = Op(G) and F (G) is an elementary abelian p-group of order pn for some 3) G has a unique minimal normal subgroup which coincides with F (G); 5) M is isomorphic with a irreducible subgroup of GL(n, p).
Lemma 2.2. Let F be a saturated formation and G a non-simple group. If G is not in F but the factor group G/N ∈ F for all nonidentity normal subgroup N of G, then G is a primitive group.
Proof. It is obvious and hence we omit the proof.
Lemma 2.3. Let G be a p-soluble group and rp(G) ≤ 2. Then the following statements hold.
2) If p = 2 and G is S4-free, then l2(G) ≤ 1; 3) If p = 2 and G is A4-free, then G is 2-nilpotent; 5) If p > 2 and p is the least prime devisor of the order of G, then G is p-nilpotent.
Proof. We prove the lemma by induction on the order of G. It is well known that the class of all p-soluble groups with p-length ≤ k, where k is a natural number, is a saturated formation (see [6, Lemma VI. 6.9]) and that the class of all p-nilpotent groups is also a saturated formation (see [7, p.98, Example 2]). Assume that the assertion of the lemma is false. By Lemma 2.2, G is a primitive group. Then, since G is p-soluble, it is easy to see that G = [Epn]M such that coreG(M ) = 1 andCG(Epn) = Epn, where Epn is an elementary abelian p-group of order pn for some prime p. Sincerp(G) ≤ 2 by the hypothesis, n ≤ 2.
If n = 1, then M is isomorphic to some subgroup of the automorphism group of the cyclic group Ep of order p. Since the automorphism group is a cyclic group of order p − 1, the order of M is notdivided by p and so lp(G) ≤ 1. If p is the least prime divisor of |G|, then M = 1 and consequently|G| = p.
Assume that n = 2. Then, obviously, M is isomorphic to a nonidentity irreducible subgroup of S3, where S3 is the Symmetric group of degree 3. If M S4 and l2(G) = 2. If M is not isomorphic to S3, then |M | = 3 since O2(M ) = 1 and so G Suppose that p > 2. Since |GL(2, p)| = p(p2 − 1)(p − 1), the order of a Sylow p-subgroup P of G is p3. If P is abelian, then lp(G) = 1 by [6, Theorem VI.6.6]. Assume that P is non-abelian. Thenby [6, Theorem I.14.10], P is isomorphic to either a group of exponent p or a metacyclic group M3(p) := a, b | ap2 = bp = 1, ab = a1+p = [ a ] b .
If P has exponent p, then lp(G) ≤ 1 by the Theorem in [9]. If P is a Sylow p-subgroup of M of order p. But by [10, Theorem 5.4.3], the subgroup Ω1(P ) generatedby all elements of P of order p is an elementary abelian subgroup of order p2. Hence Ω1(P ) = Ep2and hence A is not existent. The contradiction shows that P Suppose that p > 2 and p is the least prime divisor of |G|. If q is a prime divisor of G and q = p, then q divides |GL(2, p)| = p(p2 − 1)(p − 1). Since p < q, q divides p + 1. This implies that p = 2 and q = 3, a contradiction. Thus the proof is completed.
Lemma 2.4. Let H and K be subgroups of a group G. If K ⊆ H, then coreGK ⊆ coreH K.
Proof. Since coreGK is the maximal normal subgroup of G contained in K, coreGK ⊆ K ⊆ H and coreGK is normal in H. However, since coreH K is the maximal normal subgroup of H containedin H, coreGK ⊆ coreH K.
Lemma 2.5. If N and H be subgroups of a group G, N Proof. Obviously, N ⊆ coreGH and (coreGH)/N is a normal subgroup of G/N contained in On the other hand, let coreG/N (H/N ) = K/N. Since K/N is normal in G/N , K is normal in G andso K ⊆ coreGH. This implies that Therefore, coreG/N (H/N ) = (coreGH)/N.
Lemma 2.6. Let G be a group and p a prime. If the cofactor of every p-subgroup of G is cyclic.
1) If H is a subgroup of G, then the cofactor of every p-subgroup of H is cyclic; 2) If N is a normal subgroup of G, then the cofactor of every p-subgroup of G/N is cyclic.
Proof. 1) Let P be an arbitrary p-subgroup of H. By Lemma 2.4, coreGP ⊆ coreH P. Since 2) Let H/N be an arbitrary p-subgroup of G/N and P is a Sylow p-subgroup of H. Then H = P N and clearly N coreGP ⊆ coreGH. Since and P/coreGP is cyclic, we obtain that H/coreGH is cyclic. Thus by Lemma 2.5, (H/N )/(coreG/N (H/N )) =(H/N )/(coreGH)/N ∼ Lemma 2.7 ([11, Theorem 1.1]). Let G be a Schmidt group. Then 1) G = [P ]Q, where P is a normal Sylow p-subgroup of G, Q is a non-normal cyclic Sylow q-subgroup of G, p and q are different primes.
2) |P/P | = pm, where m is the least natural number such that q | pm.
Example 2.8. In the symmetric group S4 of degree 4, is a minimal normal subgroup which is also an elementary abelian group of order 4.
P = {1, (12)(34), (13)(24), (14)(23), (1234), (1423), (13), (24)} is a Sylow 2-subgroup of S4. The subgroups are an elementary abelian subgroups of order 4 such that coreGHi = 1, for i = 1, 2.
Lemma 2.9. Let G be a group and p the least prime divisor of |G|. Suppose that the factor group X/coreGX is cyclic for every p-subgroup X of G. Then: 3) If p = 2 and G is A4-free, then G is 2-nilpotent; Proof. 1) Let P be a Sylow p-subgroup of G. Since P/Op(G) is cyclic, by [6, Theorem IV.2.8], G/Op(G) is p-nilpotent and so G is soluble. Let A/B be a p-chief factor of G. Then A/B is a minimalnormal subgroup of G/B. If C/B is a proper subgroup of A/B, then obviously coreG/BC/B = 1.
By Lemma 2.6, the condition of the lemma holds for all sections of G. Hence C/B is cyclic. Since C/B is elementary abelian, |C/B| = 1 or p. This implies that |A/B| = p or p2. Thus, rp(G) ≤ 2.
2-3) Suppose that the assertions are false and G is a counterexample of minimal order. By Lemma 2.6, we see that all proper subgroups of G are p-nilpotent. Then by [6, Theorem IV.5.4], G is a minimal non-nilpotent with a normal Sylow p-subgroup. Hence by [7, Theorem 3.4.11] and Lemma 2.7, G = [P ]Q, where Q is a cyclic Sylow q-subgroup of G, P is a normal Sylow p-subgroup of G and P/P is a minimal normal subgroup of G/P of order pm and m is the least natural number such that q|pm − 1. Since p < q and rp(G) ≤ 2, m = 2. Thus p = 2 and q = 3. This implies that P/Pis an elementary abelian group of order 4 and hence QP /P ≤ GL(2, 2) A4, which contradicts the fact that G is A4-free. This contradiction show that 4) By 1) and Lemma 2.3, we only need to consider the case that p = 2 and G has a section which is isomorphic to S4. But by Example 2.8, S4 has a non-cyclic subgroup of order 4 with core 1. Thiscontradicts Lemma 2.6. Thus the lemma is proved.
Lemma 2.10. Let G be a group. Suppose that X/coreGX is cyclic, for every p-subgroup of G 1) G is soluble and G is {2, 3} -closed, i.e., G has a normal Hall {2, 3} -subgroup; 2) If H is a Hall {2, 3}-subgroup of G, then H/F (H) is cyclic.
Proof. 1) By Lemma 2.9, we know that G is soluble. Let π = P \ {2, 3} and Gπ be a Hall π-subgroup of G. If Oπ(G) = 1, then by Lemma 2.6 and by induction, Gπ/Oπ(G) is normal inG/Oπ(G). It follows that Gπ is normal in G. Now assume that Oπ(G) = 1. Since the class ofall π-closed groups is a saturated formation (see [12, p.34]), By Lemma 2.2, G is primitive. Then by Lemma 2.1, we have that G = [F (G)]M , where F (G) is the unique minimal normal subgroup of G and M is a maximal subgroup of G with coreGM = 1. Because Oπ(G) = 1, F (G) is eithera 2-subgroup or a 3-subgroup. Consequently, |F (G)| ∈ {2, 4, 3, 9} since rp(G) ≤ 2 for p ∈ {2, 3}by Lemma 2.7(1). If F (G) is a 2-group, then the order of F (G) is 2 or 4 and M is isomorphic to some subgroup of the general linear group GL(2, 2). But since the order of GL(2, 2) is 6, we have that Gπ = 1. If F (G) is a 3-group, then the order of F (G) is 3 or 9 and M is isomorphic to somesubgroup of GL(2, 3). Since |GL(2, 3)| = 48, we also have that Gπ = 1. Thus, G has a normal Hall{2, 3} -subgroup.
2) Let Hp (p = 2, 3) be a Sylow p-subgroup of H. Clearly coreG(Hp) ≤ Op(H), where p ∈ {2, 3}.
Then, by the hypothesis, we see that H2/O2(H) is cyclic and H3/O3(H) is cyclic. Since, obviously,F (H) = O2(H) × O3(H), H2F (H)/F (H) H2/H2 ∩ F (H) = H2/O2(H) is cyclic. Similarly, H3F (H)/F (H) is cyclic. Hence by [6, Theorem IV. 2.8], H/F (H) has a normal 2-complement,which is a Sylow 3-subgroup of H/F (H). On the other hand, by Lemma 2.9(4), l2(G) ≤ 1 andconsequently l2(H) ≤ 1. This means that H/O2 (H) = H/O3(H) has a normal Sylow 2-subgroup.
Thus H/F (H) (H/O3(H))/(F (H)/O3(H)) has a normal Sylow 2-subgroup. This implies that H/F (H) is nilpotent. Now since every Sylow subgroup of H/F (H) is cyclic, we obtain that H/F (H) is cyclic. This completes the proof.
Lemma 2.11. Suppose that G is a {p, q}-group, where p < q. If the cofactors of all p-subgroups are cyclic, then G is metanilpotent.
Proof. It is well known that the class of all netanilpotent groups is a saturated formation (see [7, Theorem 3.1.20]). Assume that the assertion of the lemma is false and let G be a counterexample of minimal order. Then by Lemma 2.6, we see that every factor group of G is metanilpotent. Hence by Lemma 2.2, G is primitive. Then by Lemma 2.1, F (G) is a primary group. If F (G) is a p-group, then by Lemma 2.9(4), F (G) is a Sylow p-subgroup of G. If F (G) is a q-group, i.e., Op(G) = 1,then by the hypothesis, every Sylow p-subgroup is cyclic and so G is q-closed by [6, Theorem IV.2.8].
This show that in any case, G is metanilpotent. This contradiction completes the proof.
Lemma 2.12. If G is metanilpotent group, then lp(G) ≤ 1, for any prime p.
Proof. Let N be a nilpotent normal subgroup of G such that G/N is nilpotent. Then for any prime p, N = Np × Np , where Np is the Sylow p-subgroup of N and Np is the Hall p -subgroupof N . Since Np charN p-subgroup of G. Therefore lp(G) ≤ 1.
By using Lemmas in above section, we now can prove our theorem.
Proof of Theorem. 1) If q ≤ 3, then G is a primary group or a {2, 3}-group. In this case, obviously, the assertion holds. Now assume that q > 3. Then by the hypothesis of the theorem and Lemma 2.10, G is soluble and its Hall {2, 3} -subgroup K is normal in G. If K is a q-group, then, of course, K is a dispersive group by Ore. If K is not a q-group, then by Lemma 2.9, K is r-nilpotent for the least prime divisor r of |K|. Let Kr be the normal Hall r -subgroup of K. By Lemma 2.6,we see that Kr also satisfies the hypothesis of the theorem. Hence by induction, Kr is a dispersivegroup by Ore. This implies that K is a dispersive group by Ore. Therefore 1) holds.
2) This follows directly from Lemma 2.6 and Lemma 2.11.
3) If |π(G)| = 2, then it is trivial. Now let |π(G)| ≥ 3 and H be a Hall 2 -subgroup of G. Then by Lemma 2.6(1), we see then H satisfies the hypothesis of the theorem. Hence by Lemma 2.9(2), H is p-nilpotent for the least prime divisor p of H. Obviously, Hp also satisfies the hypothesis. Byinduction, we may assume that Hp is a dispersive group by Ore. It follow that H is a dispersivegroup by Ore. For a Hall 3 -subgroup of G, by using Lemma 2.9, we can similarly prove that K is a dispersive group by Ore. Thus 3) holds.
4) If q > 3, then by 1), we see that G is q-closed. Hence lq(G) = 1. Now let p = q. If p is the least prime divisor of |G|, then by Lemma 2.9(4), we have that lp(G) = 1. If p is not the least primedivisor of |G|, then p = 2 and so by Lemma 2.9(2), G is p-nilpotent. Hence lp(G) = 1 again. Nowassume that G is a {2, 3}-group. Then by Lemma 2.11, G is metanilpotent and so by Lemma 1.12, lp(G) = 1 for all p ∈ π(G). Thus 4) holds. This completes the proof.
The following result follows directly from the statement 1) of our theorem.
Corollary 4.1. If q > 3 in the conditions of Theorem, then G has a normal Sylow q-subgroup.
Recall that a natural number n is said to be square-free if p2 does not divide n for all prime p.
Corollary 4.2. Let G be a group. If the order of the cofactor of every subgroup is square-free, 1) G is soluble and the second commutator subgroup G of G is nilpotent; 4) G has a normal Hall {2, 3} -subgroup which is a dispersive group by Ore; 5) If G is A4-free, then G is a dispersive group by Ore; 6) lp(G) ≤ 1, for all primes p ∈ π(G).
Proof. If the order of the cofactor of every subgroup of G is square-free, then the order of the cofactor of every non-normal primary subgroup of G is a prime number. This shows that G satisfies the condition of Theorem. Hence, by our theorem, G is soluble and the statements (4) and (6) hold.
The statements (3) and the statement 5) can be obtained by using Lemma 2.9 again and again.
(1) We first claim that the order of the factor group G = G/F (G) is square-free. In fact, let P is a Sylow p-subgroup of G. If |P/coreGP | = 1, then P is normal in G and so P ⊆ F (G). If|P/coreGP | = p, then coreGP ⊆ F (G) and the Sylow p-subgroup P of G/F (G) is of order p. Henceour claim holds. Now, by [6, Theorem IV.2.11], we see that the commutator subgroup G of G is a cyclic Hall subgroup of G and G/G is also a cyclic group. Therefore, G is a soluble metaabelian 1 and hence G ⊆ F (G) is nilpotent.
(2) Since |Q/coreGQ| ∈ {1, q} for every q-subgroup Q of G and for every prime q ∈ π(G), by [5, Lemma 1.1 and Lemma 2.1], every Sylow subgroup of G is metaabelian. Then because G is nilpotent by 1), we see that G is also metaabelian. This induces that the derived length d(G) ≤ 4.
Remark 4.3. In [3], the author proved that the groups in which every subgroup has simple cofactor are soluble. However, in a soluble group, the condition ”the cofactor of every subgroup is simple” is equivalent to the condition ”the cofactor of every subgroup is of prime order”. Hence the structure of the groups considered in [3] is contained in our Corollary 4.2.
Remark 4.4. The groups in our Theorem and Corollary 4.2 are not necessarily supersoluble in general. For example, let p and q be two different prime numbers, q divides p + 1 and q > 2. By Golfand’s theorem in [13], we know that there exists a Schmidt group [Ep2]Zq. Obviously, the groupis not supersoluble and the orders of all cofactors of all its subgroups are square-free. This example shows that the groups of odd order in our Theorem and Corollary 2.2 maybe is not supersoluble. The alternating group A4 of degree 4 shows that the groups of even order in our Theorem and Corollary2.2 maybe is not also supersoluble.
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