# Structures_HW6A Essay

## March 20, 2019

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

1 0

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0 1

;

1 0

0 1

;

1 0

0 1

;

1 0

0 1

.

### Show that Gis isomorphic to Z

2 x

Z

2.

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

A

1

=

a b

1

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

## 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.

a b

1 0

0 1

.

b a

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(

n=

g

n M

n 2

GL

2R

.

2(

a 0

0 a

## such that a6

= 0. 3.

Theorem (3.4.1) Show that the multiplicative group Zx

10 is isomorphic to the

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

4.

10 .

## Let (n ) = an

(mod 10), where n2 Z

Z x

10 = 1

;3 ;7 ;9( mod 10)

= 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] >.

(mod 16). 3

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