Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak
Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak
Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak
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UNIT-I<br />
9<br />
Solutions:<br />
1. CGbg≠ a φ,because e∈C G<br />
(a).<br />
Let x,y<br />
C G<br />
(a). Then<br />
bg bbg<br />
bg g bg<br />
1 d<br />
∈−<br />
1<br />
1<br />
Zx xy xy , y G, ax x∈<br />
− C=<br />
∈<br />
−1 −1<br />
Ga ΙxC ⇔ Za yG<br />
C ax a=<br />
Thus a=<br />
, ∀y xa a∈∀ Gx xy, a,<br />
∈hence<br />
Gx a -1 , x∈C y G<br />
(a), hence C G<br />
(a) is subgroup of G<br />
a∈G<br />
G<br />
Remarks:<br />
−1 −1<br />
= y dx axiy<br />
−1<br />
so C<br />
= y a ycΘ<br />
x∈C Gbg<br />
a h G<br />
(a) is the set of all elements of G commuting with a.<br />
we call C<br />
= a cΘ<br />
y∈<br />
G<br />
(a), the centralizer of a<br />
CGbg<br />
a h<br />
xy ∈C a<br />
2.<br />
.<br />
Let x, y ∈Zbg G . From above<br />
G<br />
bg<br />
Also, x a x = x a x<br />
x ∈C a<br />
i b g<br />
−1<br />
−1<br />
−1 −1<br />
d i d i<br />
−1<br />
G<br />
bg<br />
= x x ax x Θ x ∈CG<br />
a<br />
−1 −1<br />
= dx x i adx x i<br />
= e a e<br />
= a<br />
d i<br />
−1 −1<br />
c<br />
bgh<br />
hence<br />
Note that<br />
Thus Z (G) is a subgroup.<br />
Definition<br />
Z (G) is called the center of the group G.<br />
3. let x<br />
a<br />
c<br />
b<br />
= F H G I K J ∈ centre of GL (2, IR)<br />
d<br />
∴ x commutes with all non-singular 2 x 2 matrices, So in particular x commutes with