International Journal of Scientific & Engineering Research Volume 2, Issue 10, Oct-2011 1
ISSN 2229-5518
On subgroups of a finite p-groups
A. D. Akinola
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Let G be a finite p-group. Most authors have worked on p-groups among which we will mention few. Y. Berkovich [6] have worked on subgroups and epimorphic images of finite p- groups. Y. Berkovich [4] worked on finite p groups with few minimal nonabelian subgroups. Y.Berkovich [5] worked on subgroups of finite p-groups. Y. Berkovich [1] also woraked on abelian subgroup of a p-group G. Z. Janko [2,3] worked on element of order at most 4 in finite 2-groups and On finite nonabelian 2 groups all of whose minimal nonabelian subgroups are of exponent 4.In this paper we give an answer to some of the questions post by Y.Berkovich in [1].
2.1 Definition
metacyclic group. Dihedral groups and generalized quaternion groups are examples of metacyclic groups.
2.5. Definition
A group G is said to be minimal nonmetacyclic if G is not metacyclic but all of its proper subgroups are metacyclic.
2.6. Definition
The length of lower central series of G, that
is the greatest integer c for which c (G) > {1}
is called the class of G. The class of a p-group is a measure of the extent to which the group is non abelian. Abelian group are of class 1 and
conversely group of class 1 are abelian.
2.7. 2.7 Definition
If a group G has order
pm where p is a prime
number and m is a positive integer, then we say
The group of order p m
and class m-1 for
that G is a p-group.
2.2. Definition
Let H be a subgroup G and let a 2 G, the normalizer of H in G is denoted by N(H) defined by N(H) {a G : aHa -1 H} . It follows that the normalizer of a subgroup H is the whole
group G if and only if H is normal in G.
2.3. Definition
If x G, the centralizer of x in G, denoted by
some m 3, a p-group is said to be of maximal class where (G : (G)) = p2 ;
i 1 (G) : (G)) = p (i = 3; 4; :::;m).
3. MAIN RESULT
3.1. Theorem
Suppose a p-group G,p > 2 contains an abelian self centralizer subgroup A of
CG(x) is the set of all a G that commute with
order
p3 and |NG(A) : A| = p. Then the
x . i.e CG(x) = {a G : axa 1 = xg .It is immediate that CG(x) is a subgroup of G. Also x CG(x).
2.4. Definition
A group G which contains a cyclic normal subgroup A such that G/A is also cyclic is a
number of such subgroup in G is congruent to 1(modp).
Proof:
For H G, let q3 ( H ) denote the number of self centralizer subgroup of order p3
International Journal of Scientific & Engineering Research Volume 2, Issue 10, Oct-2011 2
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contained in H. We have that
p 2 1(mod p)
K2 is not a subgroup of D with
K K1
Let denote the set of all maximal subgroups of
not a subgroup of K 2 . It follows
G.It is known that | | 1(mod p)
enumeration principle [7]
. By hall’s
that
| K K
|1 p
| K K 2
| . Since
K K , K K
are different
q3 (G) q3 (H )(mod p)
Suppose that the
1 1 2
H
theorem has proved fall proper subgroup of G.
maximal subgroups of K 2 . We conclude
that
Take H
. By induction hypothesis
K ( K K )(K
K ) KK D
2 1 1 2 1
q3 ( H ) 0 or q3 (G) 1(mod p)
. If
contrary to the choice of K2 .Therefore
q3 (G) 1(mod p)
for all H . Then by
such K 2
does not exist.
(1) q3 (H ) | | 1(mod p) .Proving
the theorem.
Therefore suppose we may assume that some maximal subgroup of G, say H has no abelian
Therefore the number of maximal normal abelian self-centralizer subgroup of order p3 in G is congruent to 1 modulo p .
self centralizer subgroup of order
p3 . Suppose
3.2. Theorem
Let A be a subgroup of a p-group G such
that H contains a subgroup L of order p 4
and
that
CA (G) . is metacyclic . If
exponent p.Let A be a maximal abelian self
centralizer subgroup of L.Since A < L and
C ( A) A , it follows that |A| = p3 ,contrary to the what was proved in the previous
|A| =p, then G has normal subgroup of order
p p 1 .and exponent p.
Proof:
We may assume that A < Z(G). By [8]
paragraph.Therefore H has no subgroup of order
CG ( A) NG
( A) . since |A|=p. Suppose
p 4 .
that D is a normal subgroup of G of
Suppose | NG ( A) : A | p
this implies that A
exponent p. We may assume that
| D | p p 1 . and | AD | p2 .Then
is a normal subgroup of G. Also
assume that NG ( A) G
G.
then A is maximal in
CA (D) {1} .It follows that
H ACD ( A) CG ( A) that H is
Let q'3 (H )
be the number of normal abelian
metacyclic. We have CAD
(H ) H ..
self centralizer subgroup of order p3 .
in G. Since q3 (G) q'3 (G)(mod p) . it suffices to prove that q'3 (G) 1(mod p) .
Therefore we may assume that G contains a normal abelian self centralizer subgroup K1 of
Therefore by [1] AD
is of maximal class. This is a contradiction since D is not of maximal class. Therefore
| D | p p 1 . Hence the result.
3.3 Theorem
order
p3 ., K K . Set D KK . By fittings
Suppose that p-group G, p = 2 contains an
1 1
lemma, the nilpotency class of D is at most two.
abelian normal subgroup of order
p p 1 .
Therefore by [1] exp (D) = p. Considering
Then the number of nonabelian, non normal
D H
and taking into account that H has no
subgroup of order
p p 1 Is congruent to
subgroups of order p4
conclude that | D | p4
and exponent p, we
. By lemma 3 [1]
0(modp).
Proof:
Let H G .. Let q3 ( H ) denote the
qe (D) 1(mod p) . Hence the number of
number of nonabelian normal subgroup of
abelian normal self centralizer subgroup of order
p3 . in D is congruent to 1 modulo p.
order
p p 1 . contained in H. We have to
Assume that G contains a normal abelian self
3
prove that q3 ( H ) 0mod p) . Let
denote the set of all maximal subgroups of
centralizer subgroup K2
of order
p such that
G. It is known that | | 1(mod p) .
International Journal of Scientific & Engineering Research Volume 2, Issue 10, Oct-2011 3
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Take H . By induction hypothesis
maximal non normal subgroup of N such
n
q3 ( H ) 0mod p) . By [6] H contains one
that exp(A) <
p .Then | | 0(mod p) .
abelian normal subgroup of order
Therefore q3 (G) 0mod p)
theorem.
p p 1 . proving the
Proof:
Assume that N is a non normal subgroup of
G. Also let A be a maximal subgroup of N
.Let be the set of all maximal non normal
Let q'3 (H )
be the number of nonabelian, non
subgroup of N. We have to prove that
normal subgroup of order
p p 1 in G. We may
| | 0(mod p) . By sylow’s theorem, the
assume that G contains one abelian normal
number of subgroup of a group is congruent
to 1(modp).
subgroup of order subgroup of order
p p 1 . By [1] the number of
p p 1 is congruent to 1(modp).
By [6] N contains one maximal normal subgroup which implies that the number of maximal non normal subgroup of N is
Therefore q'3 (G) 0(mod p) . since by [1] G
contains one abelian normal subgroup of order
p p 1 .
3.4. Theorem
congruent to 0(modp) . i.e
| | 0(mod p) .
3.7. Theorem
Let A B G . where B is a nonabelian subgroup of a non abelian p-group
Let G be a p-group and suppose N is non normal
subgroup of a p-group G. If A is a maximal non
G, exp(B) p m . and
pm 2 , p = 2;m > 2.
normal subgroup of N then CN ( A) Z (G) .
Let be the set of all non abelian
subgroup T of G such that A < T,
Proof:
| T : A | p 2
and exp(T ) p m . Then
Assume that C CN ( A) Z (G) .Then
C N CG ( A) .Let B be non normal subgroup of N such that B/A
| | 0(mod p) .
Proof:
Let G be a 2- group of order
2m .Let G be
is a N/A non normal subgroup of exponent p in
C/A.Then B is not normal in G and B > A contrary
to the choice of A that A is maximal non normal subgroup of N. Therefore CN ( A) Z (G) . Hence the result.
3.5. Theorem
member of subgroups of G of order
2n n m .such that T is non abelian.
Let be the set of all nonabelian subgroup
T of G.
Let A be member of subgroup of G such that
| T : A | p 2 . By sylow’s theorem, the number of subgroup of a group G is
congruent to 1(modp).
Suppose that p-group G contained a subgroup M
If | T : A | p2
then | A | p n 2
of maximal class such that
3
By [6], for every value of n; n < m, G contains
CG (M ) M and
| M | p
where p = 2,
one abelian subgroup T ''
of order p n
then G is of maximal class.
Proof:
with
| T ' ': A | p2 . Therefore the number
2
| M | p3 , C
maximal class.
(M ) Z (M ) p since M is of
of T such that A < T and | T : A | p is
congruent to 0(modp). Hence the result.
Also CG (M ) Z (M ) Z (G) p
Therefore by [4] G is of maximal class since Z(G)
= p which complete the proof .
3.6. Theorem
Let A < N < G, where N is a non normal subgroup of G and A is a maximal subgroup of N,
exp( N ) pn , p n 2 . Let be the set of all
References
[1] Y. Berkovich, On Abelian subgroups of
p-groups,J. of Algebra 199,262-
International Journal of Scientific & Engineering Research Volume 2, Issue 10, Oct-2011 4
ISSN 2229-5518
280 (1998).
[2] Z.Janko,Elements of order at most 4 in finite
2-group, J. Group theory
8 (2005),683-686
[3] Z. Janko,On finite nonabelian 2-groups all of whose minimal nonabelian
subgroups are of exponent 4, J. Algebra 315 (2007) 801-808
[4] Y. Berkovich, Finite p-groups with few
minimal nonabelian subgroups, J.Algebra 297 (2006) 62-100.
[5] Y. Berkovich, On subgroups of finite p- groups, J. Algebra 224,(2000),198-
240.
[6] Y. Berkovich, On subgroups and
Epimorphic images of finite p-
Groups ,J. Algebra 248 (2002),472-553.
[7] P.Hall, A contribution to the theory of groups of prime power order, Proc. London Math. Soc. (2) 36, (1933),29-95.
[8] Y.Berkovich, Groups with a cyclic subgroups of index p,frattini subgroups, pre-print.
A. D. Akinola, Mathematics Department, College of Natural Sciences, University of Agriculture, Abeokuta,Ogun State, Nigeria.