Two Theorems about Nilpotent Subgroup

In the paper, we introduce some concepts and notations of Hall π-subgroup etc, and prove some properties about finite p-group, nilpotent group and Sylow p-subgroup. Finally, we have proved two interesting theorems about nilpotent subgroup.

In this paper, we introduced some concepts and notations such as Hall π-subgroup and so on.Using concepts, terms and notations in group theory, we have proved some properties about finite group, nilpotent group and Sylow p-subgroup, and proved two interesting theorems about nilpotent subgroup in these properties.
Let π be a set of some primes and the supplementary set of π in the set of all primes be notated π', When π contains only one prime p we notate π and π' as p and p', When all prime factor of integer n be in π we called n as a π-number, If the order H of G's subgroup be a π-number we called H as a π-subgroup.
Definition 1.If H be a π-subgroup of G and : G H be a π'-number, we called H as a Hall π-subgroup of G.
Lemma 1.A nontrivial finite p-group has a nontrivial center.
Proof.Let 1 be the class equation [1] of the group; n i divides p m and hence is a power of p.If the center were trivial, only n i would equal 1 and , which is impossible since .[2] if it has a central series [2], that is, a normal series A nilpotent group of class 0 has order l of course, while nilpotent groups of class at most 1 are abelian.Whereas nilpotent groups are obviously soluble, an ex-ample of a non nilpotent soluble group is 3 (its centre is trivial).The great source of finite nilpotent groups is the class [3]   The proof can be found in Reference [1].
Then H is subnormal in G and there is a series G p q  where p and q are unequal primes; iii) there is a unique Sylow p-subgroup P and a Sylow q-subgroup Q is cyclic.Hence and .G QP  P G  Proof.i) Let G be a counterexample of least order.If N is a proper nontrivial normal subgroup, both N and G N are soluble, whence G is soluble.It follow that G is a simple group.
Suppose that every pair of distinct maximal subgroups of G intersects in 1.Let M be any maximal subgroup: then certainly [10] every pair of which intersect trivially.Hence the conjugates of M account for exactly Since each nonidentity element of G belongs to ex-actly one maximal subgroup, n -1 is the sum of integers lying strictly between 1 2 n  and n -1.This is plainly impossible.
It follows that there exist distinct maximal subgroups M 1 and M 2 whose intersection I is nontrivial.Let M 1 and M 2 be chosen so that I has maximum order I N M M M     , which contradicts the maximality of I .
ii) Let ; also the since .Hence P 1 P i is nilpotent and thus . This means that all Sylow subgroup of G are normal, so G is nilpotent.By this contradiction k = 2 and  iii) Let there be a maximal normal subgroup M with index [6] q.Then the Sylow p-subgroup P of M is normal in G and is evidently also a Sylow p-subgroup of G.In an insoluble group [3] Hall π-subgroups, even if they exist, may not be conjugate: for example, the simple group PSL (2, 11) of order 660 has subgroups isomorphic with D 12 and A 4 : these are nonisomorphic [10] Hall   2,3 -subgroups and they are certainly not conjugate.However the situation is quite different when a nilpotent Hall π-subgroup is present.
Theorem 2. Let the finite group G possess a nilpotent Hall π-subgroup H. Then every π-subgroup of G is contained in a conjugate of H.In particular all Hall π-subgroups of G are conjugate.
Proof.Let K be a π-subgroup of G.We shall argue by induction on K , which can be assumed greater than l.By the induction hypothesis a maximal subgroup of K is contained in a conjugate of H and is therefore nilpotent.If K itself is not nilpotent, Theorem 1 may be applied to produce a prime q in π dividing K and a Sylow q-subgroup Q which has a normal complement L in K.Of course, if K is nilpotent, this is still true by Lemma 5. Now write

Lemma 3 .
is nilpotent by induction on G .By forming the preimages of the terms of a central series of G G  under the natural homomorphism [5] G G G   and adjoining [6] 1,we arrive at a central series of G.  The class of nilpotent groups is closed under the formation of subgroups, images, and finite direct products.
I cannot be normal in G; thus N is proper and is contained in a maximal subgroup M. Then 1 1

1
then is a proper subgroup of G and hence is contained in a maximal subgroup of G, say M. Then Therefore each Sylow subgroup of G is normal and there is exactly one Sylow p-subgroup for each prime p since all such are conjugate.The product of all the Sylow subgroups is clearly direct and it must equal G.
 ; however this contradicts Lemma 4. v)→i) by Lemma 2 and Lemma 3. Theorem 1. Assume that every maximal subgroup of a finite group G itself is not nilpotent.Then: i) G is soluble; ii)m n [8] B. Li and A. Skiba, "New Characterizations of Finite Supersoluble Groups," Science in China Series A: Mathematics, Vol.51, No. 5, 2008, pp.827-841.