J.Venkateswara Rao, M.V.Ramanjaneyulu, M.Sandhya
Prof. Of Mathematics,Mekelle
University,Mekelle,Ethiopia.
Asst.Prof. of
Mathematics,K.L.University,vaddeswaram-522502.
Lecturer
of Mathematics,T.J.P.S.College,Guntur-522006.
ABSTRACT
This
paper studies about the notion of indecomposable groups. It is proved that if a
group is a simple then it is an indecomposable group. If G is a p-group of class two, then obtained
a necessary and sufficient condition for a group to be indecomposable and also
centre of a group is cyclic. If group is non abelian group and the centre of
group is not contained in the intersection of maximal sub groups of group then
it is a factor direct product group.
Key words: cyclic groups, centre of a group, finite
Abelian group,p-group.
1.PRELIMINARIES
Definition 1.1
[3,4] A group A is said to be an ingroup of a group G if one of the
following holds:
a)
A is isomorphic to a subgroup of G.
b)
A is isomorphic to a factor group of G.
c)
A is isomorphic to a factor group of a
subgroup of G.
The
notation
will be used for the above with
denoting that A is an ingroup of G but A is
not isomorphic with G. In this case A is said to be a proper ingroup of G.
REMARKS[5]
·
. Every homomorphic image of group G is
isomorphic to some quotient group of G.
·
.If
is a homomorphism and onto with kernel then
[
is the homomorphic image of G under f]
Definition 1.2:
[3,4] Let
be a set of groups. A group G is said to be
rank 0 if
for some
. G is of rank 1 if G is not of
rank 0 and
with A and B of rank 0 and
with A and B of rank 0. In general G is of
rank n. If G has not been assigned rank n-k for k>0 and
with C and D both of ranks less than n. The
set of groups with assigned rank is called the closure of
and is denoted by
.
Definition 1.3
[3,4]Let G be a group and let
be the set of all proper ingroups of G. G is
called decomposable if
.
Definition
1.4 [3,4]A group G is called a factordirect product of A and B if for some
N, G
(A
B)/N and 1
A, B
G.
P-group
If a group G has order pm
where p is a prime number and m is a positive integer, then
we say that G is a p-group.
SOLABLE GROUP[5]
A
group is said to be Solvable group if
for some
positive integer ‘k’.
Nilpotent group[5]A
group G is said to be Nilpotent if
for some ‘m’
the smallest ‘m’ such that
is called the
class of Nilpotency of G.
SIMPLE group [5]
A
group G is called SIMPLE if it has no proper normal sub groups .
Remarks [5]
·
A
group G is abelian if and only if Z(G)=G.
·
A group of prime order has no proper
normal sub groups .
·
Every cyclic group is an abelian group.
·
Every group of prime order is ‘simple’.
·
Every group of prime order is cyclic. But
converse is not true .
·
Prime power group is solvable.
·
Every Nilpotent group is solvable.
·
G is simple
G
has no normal sub groups other than G and {e}.
·
Let G be a finite group. Then the order
of any sub group of G divides the order of G.
Theorem 1[3,4] Let
G=
with H1 a sub direct product A1,
B1≠1. Then there exist factor groups A,B,H and N of A1, B1,
, and
respectively such that G
and there exist R,S
H such that R
S=R
N=S
N=1.
Assume in addition that G is not a sub direct product and that G is not
isomorphic to a factor group of A or B. Then G contains a normal abelian
p-group.
Proof .Consider the
element
. Since all of these subgroups are
normal
may be written as RN or NR.
Assume first that
. Since the lattice of normal
subgroups of a group is modular it follows that
Now if
and
then
and hence
is a subdirect product. If
then
and hence G is isomorphic to a factor group of
A. If NS = N then
and G is isomorphic to a factor group of B.
Hence If
then G is either a subdirect product or G is
isomorphic to a factor group of A. But it is easy to see that
if and only if
which holds if and only if
. For if
then there exist r, s and n such that
and hence
and
. Note that
and
for otherwise
and
contrary to Lemma 1.6. Hence if
and
then
and
.
The lattice diagram in Figure 1
below illustrates the relationships of the groups that have been discussed
above. All of the argument above and some of the argument that follows will
verify that the unions and intersections given in the diagram are indeed
correct.
Figure 1
(ii) XY = XZ = YZ = XYZ
Ore[10] has shown that under conditions
(i) and (ii) the group XY is abelian.
Consider
Also
Therefore
Hence we have exhibited a normal subgroup of
H/N=G isomorphic with X, which is abelian. Now if X had composite order then it
would contain two characteristic subgroups with trivial intersection. Since a
characteristic subgroup of a normal subgroup is normal it would follow that G
is not sub-directly indecomposable. Hence X is a p-group. This completes the
proof of the theorem.
Theorem 2. [3,4]Let
be a set of groups. If
then
.
Proof Let G be of
rank 0 in
. This means that
We now consider all the
possibilities for
a)
Let
Clearly
. Since
the modular law applies and since
Hence
But
. Hence
is
isomorphic to a factor group of subgroup of
and
.
The
proceeding argument may be illustrated by the lattice diagram in Figure-2.
Figure
– 2
a)
The second case to consider is G = A/N.
Let Ia = aZ(A) for a ÎA and IaN =
aNZ(A/N). We will show that the mapping:
Ia
IaN of
A/Z(A) into G/Z(G) is a homomorphism.
Suppose
that Ia = 1 then a Z(A) = Z(A) and a Î Z(A). Consider IaN
= aNZ(A/N). Since a ÎZ(A), aNÎ Z(A/N). Hence IaN
= 1.
Now
let
and
b)
Finally if
and
then
is
isomorphic to a factor group of a subgroup of
and
is
isomorphic to a factor group of
Hence
.
This
completes the argument for G of rank 0. Notice that the above proves that if
then
.
Now
if G is of rank
then we assume that for all elements of
of rank less than n the theorem holds. But
with C and D of rank less than n.
Hence
. But
. Hence since
The proof is then complete by
induction.
Corollary 1: [3,4]Let
G be a nilpotent group. If G is decomposable then G contains a proper in group
whose class (length of upper central series) is the same as the class of G.
Proof If
the class of G is n, written c(G) = n1 and Zr(G) is the rth
element of the upper central series, then
and
. If
and the class of A is less than n, then
for all
. Therefore
. But it follows from the theorem
that if
then
. Since
this is a contradiction. Hence G is either not
decomposable or it contains an ingroup with class n.
Corollary 2:[3,4]
Let G be a soluble group. If G is decomposable then G contains a proper
in-group whose derived length equals that of G.
Proof Let {Aα}
be the set of proper in-groups of G. If the derived length of G is d, then G(d)=
1 but
. But it follows from the theorem
that
, and if the derived length of
every proper ingroup of G is less than d then
, a contradiction. Hence either G
is indecomposable or it contains a proper ingroup whose derived length equals
that of G.
Remarks [3,4,6,7]
·
If G is a nilpotent group then
.
·
If G is a solvable group then
.
·
If G is nilpotent and N
G
then
.
·
If G is nilpotent and the class of G is
n then
.
·
If G is solvable and N
G
then
.
·
IfG is solvable and the class of G is n
then
.
Corollary 1:
If G is a simple group then G is indecomposable.
Proof: Suppose
G is simple group. If it has no proper
sub groups and no proper normal sub groups and no proper in group then G is not
decomposable. therefor G is indecomposable.
Corollary2:
Let G be a p-group. G is a sub direct product if and only if Z(G) is not cyclic.
Proof: G
is p- group. Let us suppose G is sub direct product .therefore G has a normal
sub groups .By the LAGRANGE THEOREM.(Let G be a finite group. Then the order of
any sub group of G divides the order of G.). Z(G) is also normal sub group of
G. Therefore The order of Z(G) is also divides group of order G. The order of
Z(G) is not prime .Therefore Z(G) is not
cyclic .Conversely suppose that Z(G) is
not cyclic. Therefore the order of Z(G) is not prime. Z(G) has a proper normal
sub groups .Therefore the order of Z(G)
is also divides group of order G. Therefore the order of G is not a prime. Therefore G has a proper normal sub
groups. Therefore G is sub direct product.
Corollary 3:
If G is a non-abelian group and
then G
is a factor direct product.
Proof: Since
is the intersection of all maximal subgroups
of G, then there exists one maximal subgroup M of G such that
.Since
M is maximal and Z(G) is normal in G, it follows that G = MZ(G). But these
groups permute element wise and
, hence
G is a factor direct product.
Note
If
G is a non-abelian group and
then G
is not a factor direct product.
Corollary 4:
Let G be a p-group of class two. G is indecomposable if and only if
is cyclic and G may be generated by two
elements.
Proof Suppose G is indecomposable. Therefore G is simple. G is simple
G has no normal
sub groups other than G and {e}. A group of prime order has no proper normal
sub groups. Therefore G has a prime order group. Every group of prime order is
cyclic. Therefore G is cyclic. Z(G) is sub group of G and Z(G)=G. Since G is cyclic. Z(G) is cyclic
Conversely
suppose that Z(G) is cyclic and Z(G)=G. Since given G is a p-group and G is
cyclic .G has no proper normal sub groups . Therefore G has prime order. Therefore
G is simple . Therefore G is indecomposable.
CONCLUSION
It follows from Theorem 1. that a
decomposable groups is either a sub direct product or it contains a normal
abelian p-groups. At this stage, however, it is not clear how useful it is to
know that groups with no normal abelian p-subgroup are either decomposable or
sub direct product. Certainly this question might be profitably pursued. One
might begin with composite order groups which have abelian Sylow subgroups.
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