Subgroup:à
Suppose
(G,o) be a group and H⊆G where H≠ Ø .. Then H is called Subgroup of g if
1.
H is
stable for operation “o”.
- (H,O) is also a Group.
In other words ,
H is called subgroup of G if it is satisfy all condition of Group…
Complex of a Groupà
Suppose (G,o) be
a Group ,then a subset of H of G is
called complex iff a,b∈H
=> aob∈H for all a,b ∈H
Proper and improper subgroupà
If a group (G,o) have two subgroup (G,o) and ({e},o) then that
subgroup is called improper or trival .. and different from it that is
others subgroup is called proper
(non-trival subgroup)…
Some
example of subgroup:- let (Z,+) be a set of integer.let (E,+) be a set denoted
to even numbers and (O,+) be set that
denoted to odd numbers.set of even numbers is a subgroup of Z ie (E,+) is a
subgroup of Z but (O,+) is not a subgroup of Z because it does not satisfies
closure law because if 1,3 ∈O then 1+3=4 does not belogs to O.and we
can also say (E,+) is a proper subgroup of Z.
* (Z,+) is a subgroup of set of all rational numbers.and
it is also a proper subgroup of (Q,+).
Theoram-
For a
subgroup , of a non-empty set H⊆G the necessary and sufficient condition is
that
a∈H ,b∈H
=>ab-1∈H where b-1 is
inverse of b.
proof-
The
condition is necessary:à let H be a subgroup of G.
Then to showà a∈H ,b∈H =>ab-1∈H
We know that a H is a subgroup then it have a inverse of each
element i.e. if à a∈H=> a-1∈H.
And H should be closure with respect to .(dot)
Therefore a∈H , b∈H =>
a∈H,b-1∈H (inverse low)
=> ab-1∈H (closure low) proved I condition…
The
condition is sufficientà
Let H be a non-empty set of G s.t. a∈H ,b∈H =>ab-1∈H……………..(i)
Then to showà H is a subgroup of G.
For this we have to show four exioms.
1
(identity)à a∈H ,a∈H =>aa-1∈H
from(i)
=>e∈H
Therefor
H have a identity element.
II(inverse)à
Let a is element of H
Then e∈H ,a∈H =>ea-1∈H
=>a-1∈H
i.e. H
contains inverse of each element.
III(closure)à
Let
a,b∈H then
a∈H ,b∈H =>
a∈H ,b-1∈H
=>a (b-1 )-1∈H
=>ab∈H
Therefor H is closure.
IV(associate)à H satisfies associate
law because
for all a,b,c∈H=>
a,b,c∈H
=>a,b,c∈G
=>a(bc)=(ab)c
for all a,b,c ∈H
H
satisties all condition of group Therefore H is a Subgroup.
hence proved
Q 2 A
non-empty set H⊆G is a subgroup of G iff
(i)a∈H
,b∈H =>ab∈H
(ii) a∈H=>a-1∈H
where a-1 is the
inverse of a
Proof:-
Suppose H be a subgroup of G .
To show:- (i)a∈H ,b∈H =>ab∈H
(ii) a∈H=>a-1∈H
Since H is a subgroup of G so it is
satisfies all condition of group.
(i)a∈H
,b∈H =>ab∈H (closure in H)
(ii) a∈H=>a-1∈H (inverse in H). first part prove
Conversaly:-
Let H be a non-empty
set of G.and it is satisfies two conditions (i) and (ii)
To
show:- H shall be subgroup.
1.
closure:- a∈H ,b∈H =>ab∈H .
. . . . . from (i)
H is satisfies
closure low.
2. H satisfies associate law because
for all a,b,c∈H => a,b,c∈H H⊆G
=>a,b,c∈G
=>a(bc)=(ab)c
for all a,b,c ∈H
3. Identity
low => form (ii)
a∈H=>a-1∈H
a∈H , a-1∈H =>aa-1∈H from
(i)
ð e∈H
therefore
indentity satisfies identity low.
4. inverse low:- a∈H=> a-1∈H from(ii)
So H have inverse of each element.
H satisfies all condition of group
therefore H is a subgroup of G.
Hence
proved.
Th 3. For
a subgroup , of a non-empty set H⊆G the necessary and sufficient condition HH-1⊆H.
Proof:- necessary
condition:-
suppose H be a subgroup of G.
Let
ab-1∈HH-1 be a orbitary element .
a∈H , b∈H => a∈H,b-1∈H (inverse low in H ,because H is a subgroup)
=>ab-1∈H
So, ab-1∈HH-1 => ab-1∈H
, for all a,b in H
Therefore HH-1⊆H….
Suffiecient
condition:-
Let H be a non-empty set such
that HH-1⊆H.
To show
:- H shall be a subgroup of G.
a∈H , b∈H => ab-1∈HH-1 (by the definition of HH-1 )
=> ab-1∈H (HH-1⊆H)
i.e. a∈H , b∈H => ab-1∈H
for all a,b in H.
therefore H is a Subgroup of
G.
hence
proved……………
Thà if H and K are two subgroups then H∩K
also a subgroup.
Proofà H∩K is a subgroup of G
since H and K are two subgroups. Let e∈H, e∈K => e∈H∩K . i.e. H∩K≠ Ø.
Since
H is subgroup of G then a∈H
,b∈H =>ab-1∈H …………(1)
Since K is a subgroup of G then a∈K
,b∈K =>ab-1∈K ………(2)
Now, let a∈H∩K ,b∈H∩K=>a,b∈H∩K
=>a,b∈H
and a,b∈H∩K
=> ab-1∈H and ab-1∈K by (1) and (2)
=> ab-1 ∈H∩K
for all a,b ∈H∩K
i.e. a∈H∩K ,b∈H∩K=> ab-1 ∈H∩K
for all a,b ∈H∩K
therefore H∩K is a subgroup of G..
Theoramà prove that union of two subgroup H and K
is a subgroup of G iff they contains each others.
Proofà
Firstly
To showà H∪K is a subgroup of G.
let H and K are two subgroups.and let they contains each others. i.e. H⊆k
and K⊆H.
then H∪K=H or K . and H and K are
subgroup therefore H∪K is a subgroup of G.
conversalyà let H∪K is a subgroup of G
to showà H⊆k and K⊆H.
assume
contrary- HéK and KéH.
now HéK=>
there exist a∈H then a∉K. ……..(1)
KéH => there exist b∈K then b∉H. ……..(2)
From (1) and (2)
a∈H then a∉K=>a∈ H∪K.
b∈K then b∉H=>b∈ H∪K.
so, a∈ H∪K , b∈ H∪K=>ab ∈ H∪K (since H∪K is a subgroup then
closure)
let ab=c ∈ H∪K=> ab=c ∈ H or, ab=c ∈ K
now
ab=c ∈ H=>b=c-1a∈ H ,
but b∉H this is a contradiction.
ab=c ∈ K => a=cb-1∈ H
but a∉K this a contradiction.
Therefore H⊆k and K⊆H.
Hence
proved
if u like this then give us feedback and follow ..........
if u like this then give us feedback and follow ..........
No comments:
Post a Comment