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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

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