Algebra Assignment 2
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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
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QUESTION 1
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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:
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(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
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(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
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(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
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(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
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(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
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(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
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(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
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(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
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(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
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QUESTION 2
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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
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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
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4a THE CENTER OF Q
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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 :
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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
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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
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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
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This is because 3
be = b = eb
ba = ba ab ab = ba
:b
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This is because2
ba e = ba = e ba
ba a = ba a ba a ba = b
:ba
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This is because
2 2 2
2 3 2 2
ba e = ba = e ba
ba a = ba a ba a ba = ba
:2ba
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This is because
3 3 3
3 3 3 2
ba e = ba = e ba
ba a = b a ba a ba = ba
:3ba
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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
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4
b
ALL CYCLIC
SUBGROUP OF Q
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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
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c
cl(a) & cl(b)
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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
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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
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THE END