Maximal Subgroups of Finite Groups

In finite groups maximal subgroups play a very important role. Results in the literature show that if the maximal subgroup has a very small index in the whole group then it influences the structure of the group itself. In this paper we study the case when the index of the maximal subgroups of the groups have a special type of relation with the Fitting subgroup of the group.

1. INTRODUCTION P.Hall proved that a finite group with the property that its maximal subgroups have index a prime or square of a prime is solvable.J.Kohler studied in detail finite groups with the property of Hall mentioned above (see [5]).B.Huppert [3] proved that if every maximal subgroup has index prime, then the group is supersolvable.O.U.Kramer [6] proved that if a finite solvable group G has the property that for every M <" G, [F(G) F(G) N M] 1 or prime, then G is supersolvable.In this paper we consider groups with the following property: For every M <o G, [F(G) F(G) N M] I, a prime or square of a prime.We consider only finite groups.
G) denotes the intersection of all nonnormal maximal subgroups of G. M < G means that M is a maximal subgroup of G. Consider the exponents in the orders of the chief factors of a chief series of a solvable group G.For each prime p c (G), the maximal such exponent is denoted by rp(G), called the p-ram of G. r(G) max rn(G) p c (S) }, is called the ram of G.
We mention below the following known results for easy reference.L 2.1 (Kohler [5], Lemma 3.3): Let G be an irreducible subgroup of GL(2, p) with IGI odd.Then G is cyclic and IGI divides (p2 i).THEOREM 2.2 (Huppert [3], Satz I): Let G be solvable.Let pn be the highest power of p that occurs in a maximal chain of G. Then r (G) n.P T[IEORI 2.3 (Gaschtz [I], Satz 13): For a finite group G, F(G)/@(G) is a direct product of minimal abelian normal subgroups of G/@(G).I 2.4 (Kohler [5], Lemma, p.440):If G is a subdlrect product of primitive solvable groups on a prime or prime square number of letters, then r (G) 2 for every p c (G). P TIRI 2.5 (Gaschtz [i] Satz 15): If r (S/(G)) < 2, then rp(G) 2. p We prove the following lemma in a general setting.
3.1: Let G be a solvable group with #(G) i.Let F F(G).
i For every M <-G let IF F N M] p p an arbitrary prime and i Z 0.
Then every G-chlef factor of F has order qJ q an arbitrary prime and j e max i [F F N M] p p an arbitrary prime }.PROOF: Since @(G) i, it follows from Theorem 2.3 that F(G) is the direct product of minimal normal subgroups of G.This means that for each G-chlef factor H/K of F there exists a minimal normal subgroup S of G which lles in F with S H/K.Then there is a maximal subgroup M of G with MS G and M N S i.
ISI. Hence the lemma is proved.

RI:
The condition @(G) I is needed in the hypothesis of Lemma 3.1 as can be easily seen from the example of Huppert [4], Beisplel 2, p.140. ).By induction on IGI we can conclude that r (G/@(G)) < 2 and hence r (G) < 2. So assume that @(G) I. P P By Lemma 3.1 it suffices to show that chief factors of G/F are of order a prime or prime square.By Theorem 2.3 F H 1 x H 2 x x Hr where H i are minimal abelian normal subgroups of G. Since (G) i, for every H i there exists M.
Hence by hypothesis on [F F M i] we 2 conclude that IHil p or p for some p 8 (G).If IHil p, then G/CG(Hi) is 2 cyclic.If IHil= p then G/CG(Hi) is an irreducible subgroup of GL( Hr shows that CG(F) =[._, CG(Hi).Since G is solvable, CG(F) < F. Since F is abelian, F < CG(F).Thus F CG(F).Therefore G/F G/CG(F) .xG/CG(Hi) implies that G/F is abelian.This means that all chief factors of G/F are of prime order.This completes the proof of the theorem.W is a normal, elementary abellan p-subgroup of order 4 in HW.

EXAMPLE (Kohler
H <,HW. H fl W I, W F(HW). HW has the property that for every 2 N <, HW [F(HW) F(HW) fi N] I, p or p for some p (HW).

4
[HW HI p4.:HW: :H: :W: 2 6 p Thus, we see that in Theorem 3.2 we require that G is of odd order.
However, for groups of even order we have the following theorem.
THEO 3.3: Let G be a solvable group.Let F F(G).For every pi, i 0, I or 2. If a Sylow 2-subgroup of G centralizes every Sylow q-subgroup of G for all q, q odd, then rp(G) 2 for all p w(G).PROOF: As in the proof of Theorem 3.2 we can assume that @(G) I and with IHil p or p for p (G). r If IHil p, then G/CG(Hi) is cyclic.If IHil p2 with p odd, then also we can conclude as in the proof of Theorem 3.2 that G/CG(Hi) is cyclic.If IHil 4, then G/CG(Hi) is an irreducible subgroup of GL(2, 2) S 3 and hence is cyclic.The rest of the proof now follows as in the proof of Theorem 3.2.This completes the proof of the theorem.
THEOIM 3.4: Let G be a solvable group of odd order.Let F F(G).For every M <, G, let [F F 8 M] pl, i 0, I or 2. Then every subgroup of G has the same property.
PROOF: It is enough to show that every maximal subgroup of G has the same property as G.By Theorem 3.2 we have r (G) _< 2 for all primes p c (G).Now P applying Theorem 2.2 we see that p2 is the highest power of p that occurs as the index in a maximal chain of G for some p c (G) corresponding to r (G) 2.
P Let H be any maximal subgroup of G. Let K <," H. Therefore [H K] p 2 or p for some p e (H).If F(H) K, then clearly [F(H) F(H . Thus H has the same property as the group G.This completes the proof of the theorem.
RMARK: Theorem 3.4 can be modified as in Theorem 3.3 for the even order case.
THEORI 3.5: Let G be a solvable group of odd order.G has the property that for every M <a G, IF(G) F(G) N M] pi for i 0, I or 2 if and only if G/ (G) is isomorphic to a subdirect product of primitive solvable groups on a prime or prime square number of letters.
PROOF: Assume that G/(G) has the above property.By Lemma 2.4, r (G/ (G)) _< 2. By Theorem 2.5 we then have r (G) _< 2. Hence G has the P P required property.Now suppose that G has the property mentioned in the statement of the theorem.By Theorem 3.2, r (G) <_ 2. Using Theorem 2.2 we see that every P maximal subgroup has index either a prime or square of a prime.Let be a permutation representation on the conjugacy classes of maximal subgroups of G. Let be the restriction of to one of these conjugacy classes.Since M < G, NG(M) G, and thus is the identity.If M NG(M), then is 2 primitive of degree [G M] p or p for some p 8 (G).Hence G/ker() is a subdirect product of primitive solvable groups on prime or prime square number of letters, x 8 ker() if and only if x 8 NG(M) for every M <o G and M nonnormal in G. Hence ker() (G).This completes the proof of the theorem.

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July 1, 2009 First Round of Reviews October 1, 2009 Let K A 1 * A 2 where * denotes the free product.IAII IA21 p A.I elementary abelian with A i < a i b i >.