Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

10
ADVANCED ABSTRACT ALGEBRA
F
F
HG I K J F H G I K J ∈
1 1 1 0
, GL 2,
0 1 1 1
1 1
1 1
HG I 0 1K J F =
H G
a b I K JF H G I x
0 1K J =
c d
F
b IRg
Fa a + b
HG
I + K J c c d
I
HG KJ
1 1 1 1
HG I 0 1K J F =
H G I 0 1K JF H G
a b I K J F+ a c b+
d
x
=
c d c d
(1) and (2) gives
F
a a + b
HG I c c + dK J =
F+ a c b + d
c d
HG
Hence c = 0, a = d
I
KJ
.... ( 1)
.... ( 2)
Similarly,
gives
b = 0.
Therefore x
F a
= where a o a
H G 0 I aK J , ≠ , ∈R
0
Remark:
is a scalar matrix and so commutes with all 2 x 2 matrices, (nonsingular or not) Hence Z (GL(Z,R), the
center of GL (2, R) Consists of all nonzero scalar matrices.
This can be generalised that the center of GL (n, R), the general linear group of nonsingular n × n matrices
over IR, consists of all nonzero scalar matrices.
Coset of a subgroup H in G:
Let G be a group and H be a subgroup of G. For any a ∈G, Ha = {ha / h H}. This set is called right coset
of H in G. As e H, so a = e a Ha. Similarly aH = {ah / a h} is called left coset of H in G, containing a.
Some simple but basic results of Cosets:
Lemma 1: Let H be a subgroup of G and let a, b
Then
1. a
Ha
2. Ha = H
a
H
3. Either two right cosets are same or disjoint i.e. Ha = H b or
4. Ha = H b
5.
Proof:
b a -1
H
1. a = ea ∈Ha e ∈H
2. Let
i.e. there is one-one correspondence between two right Cosets
, Now h
Θb
g
G.
due to closure in H.
∈
⇔
a∀
H F x G