Applied Mathematics, 2011, 2, 562-564
doi:10.4236/am.2011.25074 Published Online May 2011 (http://www.SciRP.org/journal/am)
Copyright © 2011 SciRes. AM
Two Theorems about Nilpotent Subgroup
Department of Mat hematics, Zunyi Nor m al C oll e ge, Zu nyi , Chin a
Received February 3, 2011; revised March 21, 2011; acce p ted Mar ch 23, 2011
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 theo-
rems about nilpotent subgroup.
Keywords: Hall π-Subgroup, Sylow p-Subgroup, Normalizer, Nilpotent Group
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
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 :GH
be a π'- number, we called H as a Hall π-subgroup of G.
Lemma 1. A nontrivial finite p-group has a nontrivial
Proof. Let 1 be the class equation 
of the group; ni divides pm and hence is a power of p. If
the center were trivial, only ni would equal 1 and
, which is impossible since .
Definition 2. A group G is called nilpotent  if it has
a central series , that is, a normal series
such that i
GG is contained in the center of i
all i. The length of a shortest central series of G is the
nilpotent class of G.
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  of groups whose orders  are prime pow-
Lemma 2. A finite p-group is nilpotent.
Proof. Let G be a finite p-group of order > 1. Then
Lemma 1 shows that 1G
. Hence GG
tent by induction on G. By forming the preimages of
the terms of a central series of GG
under the natural
homomorphism  GGG
and adjoining  1,
we arrive at a central series of G.
Lemma 3.The class of nilpotent groups is closed un-
der the formation of subgroups, images, and finite direct
The proof can be found in Reference .
Lemma 4. Let P be a Sylow p-subgroup  of a finite
ii) If , then
is a Sylow p-subgroup of
N and PNN is a Sylow p-subgroup of GN.
Proof. i) Let
we have x
. Obviously P and Px are Sylow p-sub-
group of H, so
for some . Hence
. It follows that
ii) In the first place ::PNPNP N , which is
prime to p. Since PN
is a p-subgroup, it must be a
Sylow p-subgroup of N. For PNN the argument is
Lemma 5. Let G be a finite group. Then the following
properties are equivalent:
Foundation Item: Project supported by Natural Science Foundation
(13116339) of China; Natural Science Foundation (2075) of Sci-
ence and Technology Department of Guizhou; Natural Science Founda-
tion (069) of Education Department of Guizhou; Science Re-
search item(2010028) of Zunyi Normal College.
i) G is nilpotent;
ii) every subgroup of G is subnormal ;
iii) G satisfies the normalizer  condition;
L. J. ZENG563
iv) every maximal subgroup  of G is normal;
v) G is the direct product of its Sylow subgroups.
Proof: i)→ii) Let G be nilpotent with class c. If
G, then 1ii
and H is subnormal in G in c steps.
G. Then H is subnormal in G and
there is a series 01n
. If i is the
least positive integer such that
iii)→iv) If M is a maximal subgroup of G, then
NM, so by maximality
iv)→v) Let P be a Sylow subgroup of G. If P is not
normal in G, then is a proper subgroup of G
and hence is contained in a maximal subgroup of G, say
G; however this contradicts Lemma 4.
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 sub-
groups is clearly direct and it must equal G.
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;
Gpq 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 . GQPPG
Proof. i) Let G be a counterexample of least order. If
N is a proper nontrivial normal subgroup, both N and
GNare 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:
NM If Gn and
then M has nm conjugates  every pair of which
intersect trivially. Hence the conjugates of M account for
nontrivial elements. Since m ≥ 2, we have
in addition it is clear that
Since each nonidentity element of G belongs to ex-
actly one maximal subgroup, n – 1 is the sum of integers
lying strictly between 1
and n – 1. This is plainly
It follows that there exist distinct maximal subgroups
M1 and M2 whose intersection I is nontrivial. Let M1 and
M2 be chosen so that I has maximum order . Write
NNI. Since M is nilpotent,
Lemma 5, so that 1
. Now I cannot be normal
in G; thus N is proper and is contained in a maximal
subgroup M. Then 11
, which con-
tradicts the maximality of
ii) Let 1
Gpk, where and the i
are distinct primes. Assume that . If M is a maxi-
mal normal subgroup, its index is prime since G is solu-
ble; let us say
. Let i be a Sylow
pi-subgroup of G. If , then i and, since M is
nilpotent, it follows that i; also the since .
Hence P1Pi is nilpotent and thus
PP1 (by Lemma
5). It follows that
G and 1. This
means that all Sylow subgroup of G are normal, so G is
nilpotent. By this contradiction k = 2 and
We shall write 2
iii) Let there be a maximal normal subgroup M with
index  q. Then the Sylow p-subgroup P of M is nor-
mal in G and is evidently also a Sylow p-subgroup of G.
Let Q be a Sylow q-subgroup of G. Then G = QP. Sup-
pose that Q is not cyclic. If
Q, then ,
since otherwise QGP, which is cyclic . Hence
P is nilpotent and ,1gP. But this means that
, a nilpotent group. Hence Q
In an insoluble group  Hall π-subgroups, even if
they exist, may not be conjugate: for example, the simple
group PSL (2, 11) of order 660 has subgroups isomor-
phic with D12 and A4: these are nonisomorphic  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 con-
tained 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
, 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
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 12
where H1 is the unique Sy-
low q-subgroup of H. Since , the induction hy-
Copyright © 2011 SciRes. AM
L. J. ZENG
Copyright © 2011 SciRes. AM
 C. W. Curtis and I. Reiner, “Methods of Representation
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 J. E. Roseblade, “A Note on Subnormal Coalition Cla-
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LH because L is a q'-group.
NNL contains 1,
serve  C. F. Miller III, “On Group Theoretic Decision Problems
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No. 68, Princeton University Press, Princeton, 1971, pp.
:H is not divisible by q; hence 1
Sylow q-subgroup of N and by Sylow’s Theorem
QH for some
LL and, using
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