Structures_HW6A Essay

HW 6A

Melody Mott

March 20, 2019

1. Theorem (3.4#12) LetGbe the following set of matrices over R:

1 0

0 1

;

1 0

0 1

;

1 0

0 1

;

1 0

0 1

.

Show that Gis isomorphic to Z

2 x

Z

2.

Proof. Adirect product ofG

1 and

G

2, denoted

G

1 x

G

2 is the set of all

ordered pairs ( x

1; x

2) such that

x

1 2

G

1and

x

2 2

G

2.

If we can show that G is isomorphic and Z

2 x

Z

2 is isomorphic, then it follows

that the mapping of Gto Z

2 x

Z

2 is also isomorphic.

First, we will show that Gis isomorphic.

One way we can show that Gis isomorphic is to take the inverse of each

matrices in G.

A

1

=

a b

c d

1

= 1 ad

bc

d b

c a

1 0

0 1

1

= 1 (1)(0)

(0)(0)

1 0

0 1

=

1 0

0 1

;

1 0

0 1

1

= 1 (

1)(1) (0)(0)

1 0

0 1

=

1 0

0 1

;

1 0

0 1

1

= 1 (1)(

1) (0)(0)

1 0

0 1

=

1 0

0 1

;

1 0

0 1

1

= 1 (

1)( 1) (0)(0)

1 0

0 1

=

1 0

0 1

So we see that Gis isomorphic.

Now, we construct an addition table for Z

2 x

Z

2

1

+ (0,0) (1,0) (0,1) (1,1)

(0,0) (0,0) (1,0) (0,1) (1,1)

(1,0) (1,0) (0,0) (1,1) (0,1)

(0,1) (0,1) (1,1) (0,0) (1,0)

(1,1) (1,1) (0,1) (1,0) (0,0)

We see in this table that the order of mapping from row to column do not e ect

the outcome. This is shown by the diagonals, which all have the same ordered

pairs.

Since Gand Z

2 x

Z

2 are individually isomorphic, the mapping of

Gto Z

2 x

Z

2

is also isomorphic. 2.

Theorem (3.2.22) Compute the center of GL

2(

R ), denoted by Z(G ) is given

by Z(G ) = fx 2 G :xg =gx for all g2 Gg.

Proof. For all elements g2 G, we want to show that xg=gx when x2 G.

Let a; b; c; d 2R.

Let M =

a b

c d

and I =

1 0

0 1

.

So M I =

b a

d c

and I M=

c d

a b

.

M I =bc ad and I M=cb da

So M I =I M because multiplication is commutative in GL

2(

R ).

Repeating this process for each element of GL

2(

R ), we

nd that M g

n=

g

n M

for all g

n 2

GL

2R

.

Therefore M is the center of GL

2(

R ), where M=

a 0

0 a

such that a6

= 0. 3.

Theorem (3.4.1) Show that the multiplicative group Zx

10 is isomorphic to the

additive group Z

4. De

ne

:Z ! Zx

10 by

([ n]

4 ) = [

a]n

10 and

nd a generator

[ a ]

10 of

Zx

10 .

Proof. We will show that the multiplicative group Zx

10 is isomorphic to the

additive group Z

4.

First, we will

nd a generator [ a]

10 .

Let (n ) = an

(mod 10), where n2 Z

Z x

10 = 1

;3 ;7 ;9( mod 10)

Let a= 7 and n2 Z

Then 7 1

= 7, 7 2

= 9, 7 3

= 3, and 7 4

= 1.

So (n ) = 7 n

(mod 10), where n2 Z.

Therefore, Z

4 !

Zx

10 . 2

4.

Theorem (3.3.1) FindH KinZx

16 , if

H=< [3] >and K=< [5] >.

Proof. We will compute H K. //H K =an

(mod 16). 3