normal subgroup

NORMAL SUBGROUP

Definition 2 -11.
A subgroup (H ,* ) of the group (G , * ) is said to be . NORMAL SUBGROUP ( or INVARIANT ) in ( G , * ) if . and only if every left coset of H in G is also a right coset . of H in G .

Definition 2_11/ .
A subgroup(H,*)is normal subgroup of the group(G ,*)if . and only if a *H = H * a for every a ? G .

Examples .
If H={ f1 (x) = x , f4(x)=(x-1) / x, f6 (x) =1/(1-x)} , show that . (H , o ) is normal .
If K = { f1 (x) = x , f2 (x) = 1/ x } , show that (K , o ) is not . normal .

Exercise.
Find all normal subgroups of S3 and Z10 .

Theorem 2 -15 .
The subgroup (H ,*) is normal subgroup of the group . (G , *)If and only if for each a ? G,
a * H * a-1 ?H

Proof .
Suppose that H is normal , we must prove that . a * H * a-1 ?H.
Let a* h1 * a-1 ? a* H * a-1.
Since a * H = H * a
Then there exists h2 ? H such that
a* h1 = h2 * a .
a* h1 *a-1 = h2
Hence a* H *a-1 ?H . 34
Conversely , suppose a* H *a-1 ?H , we must prove that
a* H = H * a.
Let a * h ? a* H.
Since a* H *a-1 ? H ,
Then a* h* a-1 = h1 for some h1 ? H .
thus
a * h = (a * h * a-1 ) * a = h1* a .
but h1 * a ? H * a .
then a * H ?H * a .
Similarlly H * a a * H .
Then a * H = H * a .


Exercise .
Show that ( cent G , * ) is a normal subgroup of each . group ( G , * ) .

Proof .
Let a? G .
a* cent G = { a * c | c? cent G }
= { c * a | c ? cent G }
= cent G * a .
Then cent G is normal .











35
QUOTINRT GROUP

Definition 2 -12 .
If ( H , * ) is a normal subgroup of the group ( G , * ) . . then we shall denoted the collection of distant coset of H . in G by G/ H :
G / H = { a * H ? a ? G }.
We will denoted the binary operation on G /H by
and define as :
(a* H ) ( b * H ) = ( a * b ) * H .
Exercise .
Show that ( G / H , ) is group .
Proof.
First we must prove that ? is well define .
Suppose that
a * H = a1 * H and b *H = b1 * H .
hence, a-1 * a1 , b-1 * b1?H
Since H is normal in G , then
x * H * x-1 ? H for every x in G .
Then,
b-1 * H * b = b-1 * H * (b-1)-1? H ,
so b-1 * (a-1 * a1 ) * b ?H
but
(a * b)-1 * (a1 * b1) = ( b-1 * (a-1 * a1) * b) * ( b-1 * b1) ? H ?
hence
( a * b ) * H = ( a1 * b1 ) * H .
Then ? is well – defined .
It is clear that G / H ? ??
G/H is closed under ??
? is associative ?
H is identity of G/H ?
(a * H )-1 = a-1 * H ?

36