ALGEBRAIC STRUCTURE ( TMA 3033 )

ASSIGNMENT 2

GROUP A

LECTURER : PUAN NORASHIQIN BT MOHD IDRUS

GROUP MEMBERS :

NAME MATRIC NUMBER

DESIGA A/P VESWANAHAN D20061026152

STEPHANIE KARMINI A/P EMMANUEL GERARD D20061026158

VERONICA SELVI A/P MARATHA MUTHU D20061026163

NAGESWARI A/P YELLAPA APPARAO D20061026166

QUESTION 1

Let Zm and Zn be two groups of integer mod m and n respectively. Define

by ( , ) , with the operation + defined on by

( , ) ( , ) (( ) mod , ( ) mod ) where ( , ) (m n m n m nZ Z a b a Z b Z Z Z

a b c d a c m b d n a b c

, )

we say that is the external direct sum of Zm and Zn.m nd Z Z

QUESTIONS:

(a) Prove that this algebraic structure gives us a group.

(i) closed under operation

Is ? and is [ ]

Since and then, and since and then,

Thus , .

Hence is closed under operation .

( , ) ( , ) m na b c d Z Z ( , ) ( , ) (( ) mod , ( ) mod )a b c d a c m b d n m nZ Z

ma Z mc Z ( ) moda c m nb Z nd Z

( ) modb d n

( , ) ( , ) (( ) mod , ( ) mod ) m na b c d a c m b d n Z Z

m nZ Z

(ii) Associative

( , ) ( , ) ( , ) ( , ) ( , ) ( , )a b c d e f a b c d e f

LHS = ( , ) ( , ) ( , )

= (a+c) mod , (b+d) mod + (e, f)

= ( ) mod , ( ) mod

a b c d e f

m n

a c e m b d f n

RHS = (a, b)+ (c, d)+(e, f)

= ( ) ( ) mod , ( ) mod

= ( ) mod , ( ) mod

a b c e m d f n

a c e m b d f n

LHS=RHS

a, c, e and b, d, f

(a, b)+(c, d)+(e, f)

is associative

m n

m n

m n

Z Z

Z Z

Z Z

(iii) Identity

a b

a b

a b a b

a b

Let e , e are the identities of and respectively.

then e , e is the identity of

Since, (a+b) + e , e = (a+b) = e , e (a+b)

LHS = (a+b) + e , e

m n

m n

Z Z

Z Z

a b

a b

a b

= ( e ) mod , ( e ) mod

= a mod , b mod

RHS = e , e (a+b)

= (e ) mod , (e ) mod

= a mod

a m b n

m n

a m b n

, b modm n

LHS=RHS,

therefore has an identitym nZ Z

(iii) Inverse An inverse of is (a, b) -1 -1(a , b )

-1

-1

m n

-1 -1 -1 -1a b

-1 -1 -1 -1

-1

, a

, b

where Z , Z is group of integer

(a, b) + (a , b ) = e , e (a , b ) + (a, b)

LHS = (a, b) + (a , b ) RHS = (a , b ) + (a, b)

= (a + a

m m

n n

a Z Z

b Z Z

-1 -1 -1

a b a b

)mod , (b + b )mod = (a + a) mod , (b + b) mod

= e mod m, e mod = e mod m, e mod

m n m n

n n

LHS = RHS,

therefore has an inverse

Hence, is a group, which is a external direct sum

m n

m n

Z Z

Z Z

(b)

(a) List all the elements of

(b) Show whether cyclic is cyclic or not?

2 2Z Z

2

2 2

0,1

( , ) ,

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

m n m n

Z

Z Z a b a Z b Z

Z Z

2 2Z Z

2 2

2 2 2

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

0,0 (0 ,0 ) m, n (0,0)

0,1 (0 ,1 ) m, n (0,0), (0,1)

1,0 (1 ,0 ) m, n (0,0), (1,0)

1,1 (1 ,1 ) m, n (0,0), (1,1)

is not a cyclic group as no element of

m n

m n

m n

m n

Z Z

Z

Z

Z

Z

Z Z Z Z

2 2 2 generate Z Z

Does not generate

2 2Z Z

(c)

gcd(2, 2) = 2 non-cyclic gcd(3, 4) = 1 cyclic

gcd(2, 3) = 1 cyclic gcd(3, 5) = 1 cyclic

gcd(2, 4) =2 non-cyclic gcd(2, 5) = 1 cyclic

gcd(3,6) = 3 non-cyclic gcd(3,7) =1 cyclic

gcd(3,8) = 1 cyclic gcd(3, 9) = 3 non-cyclic

gcd(4, 6) = 2 non-cyclic

We can check the other basis gcd and find that for the basis with the gcd 1 it is cyclic.The basis will only be cyclic if the gcd (m,n) = 1

(m,n) Character (m,n) Character

(2,2) Non cyclic (3,6) Non cyclic

(2,3) Cyclic (3,7) Cyclic

(2,4) Non cyclic (3,8) Cyclic

(2,5) Cyclic (3,9) Non cyclic

(3,4) Cyclic (4,6) Non cyclic

(3,5) Cyclic

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

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

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

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

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

(e,e) (e,a) (a,e) (a,a)

(e,e) (e,e) (e,a) (a,e) (a,a)

(e,a) (e,a) (e,e) (a,a) (a,e)

(a,e) (a,e) (a,a) (e,e) (e,a)

(a,a) (a,a) (a,e) (e,a) (e,e)

2 2Z Z 2 2V V

and are different but isomorphic The one to one correspondence

(0,0) can substitute by (e,e) (0,0) (e,e)

(0,1) can substitute by (e,

G G

a) (0,1) (e,a)

(1,0) can substitute by (a,e) (1,0) (a,e)

(1,1) can substitute by (a,a) (1,1) (a,a

)

transforms to and ; and are isomorphic to each other because there exist an isomorphism

from the Cayley Table to the Klein Group.

G G G G

(d) By using Cayley Table, show that is isomorphic to the Klein four group, V2 2Z Z

2 2 2 2

2 2 2 2

:

: where G = and

Z Z V V

G G Z Z G V V

(c)(a)

3 2

3 2

3 2

(3,2)

0,1,2 0,1

(0,0), (0,1), (1,0), (1,1), (2,0), (2,1)

Z Z

Z Z

Z Z

4 4

4

4 4

(4,4)

0,1,2,3

{(0,0), (0,1), (0,2), (0,3), (1,0), (1,1), (1, 2), (1,3), (2,0), (2,1), (2,2),

(2,3),(3,0),(3,1),(3,2),(3,3)}

Z Z

Z

Z Z

3 4

3 4

3 4

(3,4)

0,1,2 0,1,2,3

{(0,0), (0,1), (0,2), (0,3), (1,0), (1,1), (1, 2), (1,3), (2,0), (2,1), (2,2), (2,3)}

Z Z

Z Z

Z Z

6 2

4 2

6 2

(6,2)

0,1,2,3,4,5 0,1

{(0,0), (0,1), (1,0), (1,1), (2,0), (2,1),(3,0),(3,1),(4,0)(4,1),(5,0)(5,1)}

Z Z

Z Z

Z Z

(b)3 2

4 4

3 4

6 2

order of (3,2) = 6 = (3 2)

order of (4,4) = 16 =(4 4)

order of (3,4) = 12 =(3 4)

order of (6,2) = 12 =(6 2)

From this we can make the conjecture that the or

Z Z

Z Z

Z Z

Z Z

der of

The Proof :

= (m) (n)

= mn

m n

m n m n

Z Z mn

Z Z Z Z

QUESTION 2

The group of quaternions denoted by Q4 consists of the following elements ,

is defined by the following Cayley’s table :

2 3 2 3, , , , , , ,e a a a b ba ba baQ4

eee

ee

ee

ee

e

a

aa

aa

aa

a

a

a

2a

2a

2a

2a

2a

2a2a

2a

2a

2a

3a

3a

3a

3a

3a

3a3a

3a3a

3a

b

bb

bb

b

b

bb

b

baba

ba

baba

baba

baba

ba

2ba2ba

2ba2ba

2ba

2ba

2ba

2ba2ba

2ba3ba 3ba

3ba3ba

3ba

3ba3ba

3ba

3ba3ba

Q4

Questions :

Use the table to determine each of the following (Show your answer) :

a) The center of Q4.

b) All cyclic subgroups of Q4.

c) cl(a) & cl(b) where cl(a) is defined as below :

let G be a group. For each , defined the conjugacy class of a, cl (a) as

a G

1( )cl a xax x G

4a THE CENTER OF Q

4 4

4

The center of is a commutative subgroup of ,

denoted by

and define by :

Q Q

Z(Q )

4 4 4:Z(Q ) = a Q aq = qa, q Q

Therefore,

From the Cayley 's table :

This is because

2

2 3 2

3 3

3

ae = a = ea

aa = a = aa

aa = a = a a

aa = e = a a

ab = ba ba ba = ba

:a

2 2 2

2 3 2

2 2 2 2

2 3 3 2

2 2 2

2 3 2

2 2 2 2

2 3 3 2

a e = a = ea

a a = a = aa

a a = e = a a

a a = a = a a

a b = ba = ba

a ba = ba = ba a

a ba = b = ba a

a ba = ba = ba a

:2a

This is because

3 3 3

3 3

3 2 2 3

3 3 2 3 3

3 3 3 3

a e = a = ea

a a = e = aa

a a = a = a a

a a = a = a a

a b = ba ba ba = ba

:3a

This is because 3

be = b = eb

ba = ba ab ab = ba

:b

This is because2

ba e = ba = e ba

ba a = ba a ba a ba = b

:ba

This is because

2 2 2

2 3 2 2

ba e = ba = e ba

ba a = ba a ba a ba = ba

:2ba

This is because

3 3 3

3 3 3 2

ba e = ba = e ba

ba a = b a ba a ba = ba

:3ba

4

4

From the Cayley 's table, it is shown that is the

commutative subgroup of which means

and identity is the center of .

2

2

a

Q

a e Q

4Therefore, 2Z Q e,a

4

b

ALL CYCLIC

SUBGROUP OF Q

4 4, nFor any q Q the subgroup H x Q x q for n Z

4.is called thecyclic subgroup of Q

, :Thus fromthetable

1

2 3 2 32 4

2 2 23

3 3 3 24 2

2 25

2 36

2 2 2 27

3 3 3 28

1, , , , , , , sin 1

( ) ,

( ) , , ,

, , ,

( ) , , ,

( ) , , ,

( ) , , ,

n

n

n

n

n

n

n

n

H e e n Z e

H a a n Z a a a e a a a e ce Q

H a a n Z a e

H a a n Z a a a e H

H b b n Z b a ba e

H ba ba n Z ba a ba e

H ba ba n Z ba a b e

H ba ba n Z ba a ba 6e H

c

cl(a) & cl(b)

14i ( ) cl a qaq q Q

1

1 3

2 2 1 2 2

3 3 1 3

1 2 3

1 3 3

2 2 2 3

3 3 1 3 3

1)

2)

3) ( )

4) ( )

5) ( ) ( )

6) ( ) ( ) ( ) ( )

7)( ) ( ) ( ) ( )

8) ( ) ( ) ( ) ( )

eae a

aaa aaa a

a a a a aa a

a a a a aa a

bab b a ba a

ba a ba ba a ba a

ba a ba ba a b a

ba a ba ba a ba a

1

1 3 2

2 2 1 2 2

3 3 1 3 2

1 2

1 3 2

2 2 2

3 3 1 3 2

1)

2)

3) ( )

4) ( )

5) ( ) ( )

6) ( ) ( ) ( ) ( )

7) ( ) ( ) ( ) ( )

8) ( ) ( ) ( ) ( )

ebe b

aba aba ba

a b a a ba b

a b a a ba ba

bbb b b ba b

ba b ba ba b ba ba

ba b ba ba b b b

ba b ba ba b ba ba

14ii ( ) cl b qbq q Q

THE END