………………………………………………………………………………………….

R{0 }

≅ R

as there is no other ideal between {0 }∧R

{0 }is themaximal ideal → R is a field ,

9. gcd ( o (a ) , o (b ) )=1∧ab=ba → o (ab )=o ( a ) o (b )

10.

by Cauch y ' s theorem ,∃a , b∈G suchthat o ( a )=3∧o (b )=2

H=¿a>¿ {e , a ,a2 }

i (H )=63=2

clearly b∉H as b ≠ e ,a∨a2 ; (if b=a2 →b2=a4→ e=a )

the only ¿cosets of H∈G are He=H∧Hb

G=He∪Hb=H ∪Hb={e , a ,a2 ,b ,ab ,a2b }

ba=a2 bas ba ≠ e , a , a2….

aba=b

Hence G= {e , a ,a2 , b , ab , a2 b}where ba=a2b∨aba=b

S3= {e ,a , a2 , b , ab , a2 b}=¿a , b>¿

where a=(123 )∧b=(12 )

henceG≅ S3

…………………………………….

Let G be a finite group and p be a prime. If p divides the order of G, then G has an element of order p.

Z2 × Z2={e , a ,b ,ab } where a2=b2=(ab )2=e∧a b=ba

if H is a proper subgroup →o (H )=2( prime); (o ( H ) ≠1∨4 )

H= {e ,a }, {e ,b }, {e , ab }

every proper subgroup of Z2× Z2 iscyclic , but Z2 × Z2is not cyclic

S3 has proper subgroups of order 2∧3 only

proper subgroups are cyclic→ hence abelian

but S3 is not abelian .

θ (1 ) must bea generator of Z15

number of generators of Z15=φ (15 )

henceθ (1 )has φ (15 )=8 choices .

φ ( pq )=pq (1− 1p )(1−1

q )=( p−1 ) (q−1 )

……………………………………………………………………………………..

9 (0,1 )+ A=(0,0 )+ A=A → (0,1 )+ A is an element of order 9

φ : Z × Z → Z × Z2

st φ (a ,b )=(b−2 a , [ a ]2 )

then ker φ=¿ (2,4 )>¿

{a=2 , b=4 → b=2 a

→ b−2a=0∧[a ]2=0 }.

………………………………………………………………………………………………..

SUBRING:

a ,b∈S → a−b∧ab∈S

……………………………………………………………………………………………..

INFINITE Integral domain with FINITE characteristic: D = Z2 [x]

as 2.1 = 0; char D = 2;

If 1 has additive order n, then char R is n.

…………………………………………………………………………………………..

d∨n∧G is acyclic group of order n

then number of elements of order d=φ (d) .

……………………………………………………………….

Normal subgroups: {e}, S3 and H = {e, (123), (132)} as i (H) = o (S3/H) = 6/3 = 2.

every subgroup of index 2 is NORMAL.

S3=¿a , b∨a3=e=b2∧ba=a2 b>¿

a=(123 )∧b=(12 )

H=¿a>¿ {e , a ,a2 }→ i (H )=2→ H isnormal∈S3

subgroups of order 2are not normal

K= {e , b }, a∈S3∧b∈K → a−1 ba=ab∉K .

S3 : possible subgroups

{e } , {e ,b } , {e , ab }, {e , a2 b } , {e ,a , a2 }, S3 .

………………………………………………….

……………………

Aut (Z8) = Z8x (or U8)

Z8x = {1,3,5,7}; maximum order of any element is 2;

no element has order 4, hence Z8x is not cyclic. Aut (Z8) is not cyclic even though Z8 is cyclic.

…………………………………………………………………………….

…………………………………………………..

(1, 0) + < (2,2) > has infinite order , whereas in Z2xZ2 there is no element with infinite order.

……………………………………………………………………………………………..

…………………………………………

a =1,b=4 b = 4a or [b – 4a]8 = [0]8

φ : Z4 × Z8 → Z8

st φ ( [a ]4 , [b ]8 )=[ b−4 a ]8

then ker φ=¿([ 1 ]4 , [4 ]8)> .

or

Any group of order 4 is isomorphic to Z4 or K4 – Klein’s 4 group, which are Abelian.

……………………………………………………………………………………………..

H = <2> + <2>;

G/H ~ Z2+Z2;

G/K -> order is 4

(1,1) + K has order 4. hence G/K is cyclic and is isomorphic to Z4.

gcd ( a ,an−1 )=1∧a<an−1∀ n>1 ; {for n=1 , result is trivial }

∴ a∈U an−1

[ a ]n= [an ]= [1 ]∈U an−1

if m<n s. t [ a ]m= [1 ]→ am≡ 1mod (a¿¿ n−1)¿

(a¿¿n−1)¿¿

hence o (a )=n

………………………………………………………………………………………………………………..

An ideal I is a prime ideal if and only if I ≠ R, and whenever ab ∈ I, either a ∈ I or b ∈ I.………………………………………………………………………

if P and Q are both prime ideals, would P∩Q be a prime ideal? NO

(2) is a prime ideal in Z, because 2 is a prime number. Similarly (3) is prime ideal

(2)∩(3) = < lcm(2,3) > = < 6 >; as 6 is not a prime number, (6) is not a prime ideal.

…………………………………………………………….

In Z, prime ideals correspond to the ideals < p > generated by prime numbers. as Z is a PID, prime ideal iff maximal ideal.

Z¿ p>¿=Z p¿

¿ Z ,maximal ideals (¿ prime ideals ) are< p>where p is prime .

……………………………………………………………..

……………………………………………………………..

Theorem: Let G be a group such that every element of G other than the identity has order 2 . Then G is abelian.

……………………………………………………………………………..

Aut (K) ~ S3

a->a, b->b, ab->ab

a->a, b-> ab, ab->b

………………..

a->b, b->a, ab->ab

a->b, b->ab, ab->a

……………..

a->ab, b->b, ab->a

a->ab, b->a, ab->b

……………………………..

similarly for Z2+Z2

Aut (Z2+Z2) ~ S3

………………………………………………………………………………………..

G=¿a ,b∨a2=b2=(ab )2=e ;ab=ba>¿K 4

Aut ( K4 )≅ S3

as Z2× Z2≅ K 4 → Aut (Z¿¿2 × Z2)≅ Aut (K ¿¿4 )≅ S3 .¿¿

……………………………………………………………………………………………………..

number of Automorphism on Z 9 × Z 16 Zm × Zn ≃ Zmn when m and n are relatively primeZ9 × Z16 ≃ Z144Aut (Zm × Zn) ≃ Aut (Zmn) = Umn……………………………………………………………………………………………………………………………………..Aut (G × H) ≅ Aut (G) × Aut (H); when G and H are finite groups of relatively prime orderAut (Zm × Zn) ≃ Aut (Zm) × Aut (Zn), whenever m and n are relatively prime.……………………………………………………………………………………………………………………………………..φ (n )=n∏ (1− 1

pi )prime factorize n=p1

a p2b p3

c … ... ………………………………………………………………………..

φ :G1 ×G2→G1

H×

G2

K

st φ ( x , y )=( Hx , Ky )

Then Ker φ=H × K .

……………………………………..

If G is cyclic then Aut(G) is cyclic?

No, consider Z8

G=Z8

Aut (G )=Aut (Z8 )=U8= {1,3,5,7 }⨂mod 8

but U 8is not cyclic as there isno element of order 4.

………………………………………………………………………………………………………………………..

If G is abelian then Aut(G) is abelian?

No

let G=Z2× Z2∨Klei n' s4 group

G is abelian

Aut (G )=Aut (Z2 × Z2 )≅ S3 , which is not abelian .

………………………………………………………………….

U n is cyclic iff :

1¿n=1 , n=2 , n=4∨¿

2¿n=pk∨n=2 pk where p is any odd primei . e p=3 ,5 ,7 ….∧k∈N .

…………………………………………………………………………………………………………………….

if p is prime → U p is a cyclic group withorder φ ( p )=p−1

∴U p≅ Z p−1.

……………………………………………………………………………………………………………

I = 6Z, E = 2Z

E/I = {0+6Z, 2+6Z, 4+6Z} is group with order 3(prime)Hence E/I is cyclic group with order 3 any cyclic group of order 3 is isomorphic to Z3Hence E/I ≃ Z3 ……………………………………………………………

Even though M is maximal ideal, R/M is not a field because R is not a commutative ring.…………………………………………………………………………………………………………ONLY in a commutative ring RM is maximal ↔ R/M is a field.

……………………………………………………………………………………………………………………Let f (x )∈Z [ x ] . Let p bea prime∧let f p ( x )be thereduction of f ( x )mod p

If f p ( x ) is irreducibleZ p

∧deg f p ( x )=deg f ( x ) , then

f ( x ) is irreducibleQ

as deg f p ( x )=deg f ( x ); p should not divide the leading coefficient of f (x)

f 2 ( x ) is irreducibleZ2

→ f (x ) is irreducibleQ

any Isomorphismmaps an identity element ¿an identity element

……………………………………………………………………………………………………………………….

characteristic of a ring is the additive order of the identity element (1, 1)

f ( x )∈Z [ x ] .

If f ( x ) is irreducibleZ

, thenf ( x )is irreducibleQ……………………………………………………..

if I is a maximal ideal of R, then I[x] need not be a maximal ideal of R[x].I = 2Z, R = Z; R/I ≃ Z2, a field. Therefore I is maximal ideal of R

R[x]/I[x] ≃ R/I [x] Z[x]/2Z[x] ≃ Z/2Z[x] ≃ Z2[x], which is not a field.Hence 2Z[x] is not a maximal ideal of Z[x], even though 2Z is a maximal ideal of Z.…………………………………………………………………………………If I is a prime ideal of R <=> I[x] is a prime ideal of R[x].……………………………………………………………………………………..

I=¿ f ( x )>is a principal ideal .

if I=¿ f (x )>is maximal ideal of Z [ x ] , then

f ( x ) is irreducibleZ

take any prime p whichdoesnot divide theleading coefficient of f ( x )

deg f p ( x )=deg f ( x ) where f p ( x ) is obtained by taking modulo pof coefficients of f ( x )

since f ( x ) is irreducibleZ

→ f p ( x ) is irreducibleZ p

∴< p , f (x )>is amaximal ideal of Z [ x ]

¿ f ( x )>⊂< p , f ( x )>¿

hence I=¿ f ( x )>cannot be amaximal ideal

∴no principa l ideal<f ( x )>can be maximal .

…………………………………………………………………………….Z [ x ]

¿ p , f ( x )>¿≅Z p[ x ]

¿ f p ( x )>¿which is a field¿¿

¿ p , f ( x )> isa maximal ideal of Z [ x ]

f p ( x ) should be irreducibleZ p

………………………………………………………………………………………

R has only one maximal ideal i.e. {0}.

maximal ideals of R x R : {0} x R and R x {0}.

R R{0 } R

≅ R{0 }

⨁ RR≅ R⨁ {0 }≅ R → field

……………………………………………………………………………………..

let the given ringsbe isomorphic

∃an ontohomomorphism φ: Z [ x ] → Z [ x ]¿x2+7>¿ s . t φ ( x )=n x+m+¿ x2+7>¿¿

where n ,m∈Z

2 x2+7∈ ker φ → φ (2 x2+7 )=0+¿ x2+7>¿

2 φ ( x )2+7 φ (1 )=0+¿ x2+7>¿

2 (nx+m )2+7 ≡ 0→2 (n2 x2+m2+2 nmx )+7≡ 0. but x2 ≡−7

−14n2+2 m2+7+4 nm x≡ 0→−14 n2+2m2+7 ≡0∧4 nm≡0

these equations arenot satisfied of any n , m∈Z

Thusno onto homomorphismφ has kernel (2x2+7 )

………………………………………………………………………………………………………………………………

let x , y∈Z [ i ]∧ y ≠ 0

let xy=a+bi where a ,b∈Q

∃ c ,d∈Z such that|a−c|≤ 12

,|b−d|≤ 12

xy=(a−c )+ (b−d )i+(c+d i )

x=(c+d i ) y+[ ( (a−c )+ (b−d ) i ) y ] . {x=t y+r ; t , r∈Z [ i ] }

show N (r )<N ( y )

………………………………………………………………….

define N (a+bi )=a2+b2

then for any x , y∈Z [ i ] → N ( xy )=N (x ) ≥ N ( x ) as N ( y )≥ 1 (¿N ( y )>0 )

N ( x )≤ N (x ) N ( y )

…………………………………………..

Z [√5 ] ,2 isirreducible but not prime .

−4= (1+√5 ) (1−√5 )

2∨(1+√5 ) (1−√5 )

but 2 does not divide either (1+√5 )∨ (1−√5 )

………………………………………………………………………………………………………………………

if α ∈Z [ i ] , α is an irreducibleelement iff N (α ) isa primenumber .

if p∈Z , p is anirreducible element iff p isa primenumber∧p≡3mod 4

……………………………………………………………………………………..

3+√3 is not an irreducibleelement∈Z [√3]

3+√3=√3 (1+√3 ) canbe expressed as product of nonunits

……………………………………………………………………………………………….

show that ideal generated by3∧1+√−5 is not principalideal∈Z [√−5 ]. Hence Z [√−5 ] isnot a PID .

α is a unit iff N (α )=1 .let α ∈Z [ i ] ,

α=a+bi;

N (α )=a2+b2=1 ;

a=± 1, b=0

a=0 , b=±1

hence the group of units {±1 , ±i }

………………………………………………..

J=I +¿17> ,then I ⊂J⊂ Z [ x ] . 17∈ J∧17∉ I

Let J=Z [ x ] .then 1∈ J ,1=xf ( x )+17 b(x )

1=17 b0 whereb0∈Z

…………………………………………………………………

I={a+b√−5|a+b is even }=¿2 ,−1+√−5>is amaximal ideal

φ : Z [√−5 ] → Z2

s . t φ (a+b√−5 )=[ a+b ]2

then ker φ=I

∴ Z [√−5 ]I

≅ Z2

……………………………………………………………………………………….

………………………………

32+Z=1

2+Z ;

RZ

={r+Z|0≤ r<1∧r∈R }

define φ: RZ

→ S

s. t φ (r+Z )=e i 2πr

S is a group of complex numbers withunit modulus→ ˚group

………………………………………………………………………………………………………………..

φ1 ( x )=x ;φ2 (x )=−x.

………………………………………………………………………..

I=¿5 , x>; then Z [ x ]I

≅Z5 [ x ]

¿ x>¿≅ Z5 ¿

show that Z [ x ]I

≅ Z5

φ : Z [ x ] → Z5

st φ ( f ( x ) )=f (0 )mod 5 ;

…………………………………………………………….

H=¿ (123 )>¿{e , (123 ) ,(132)}

o ( S3

H )=63=2

S3

H={He , H (12 ) }∧H (12 )2=He=H

H (12 )=H (13 ) as (12 ) (13 )−1=(12 ) (13 )=(132 )∈H

similarly H (12 )=H (23 ) . also H (123 )=H∧H (132 )=H

HenceS3

His acyclic group of order 2.

………………………………………………………..

RZ≅ { z∈C :|z|=1 }

……………………………………………………………………………………..

φ :2Z →3 Z

φ (2 )=3 j ;

4=22∧4=2+2

φ ( 4 )=φ (2 )2∧φ ( 4 )=2 φ (2 )

¿>φ (2 )2=2 φ (2 ) .

the only homomorphism between2 Z∧3 Z isthe trivial homomorphismi . eφ ( x )=0 ∀ x ,

which is neither 1−1 nor onto.

φ :2Z →4 Z

φ (2 )=4 j ;

4=22∧4=2+2

φ ( 4 )=φ (2 )2∧φ ( 4 )=2 φ (2 )

¿>φ (2 )2=2 φ (2 )

16 j2=8 j .

j=12∉Z .

only j=0 is possible .Only homomorphismis thetrivial homomorphismφ ( x )=0 ∀ x ,

which is not isomophic .

I=¿ x2−2>¿

Q [ x ]I

={a+bx+ I : a ,b∈Q }

φ : Q [ x ]I

→Q [√2 ] suchthat

φ (a+bx+ I )=a+b√2

……………………………………………………………………………………………………..

………………………………………..

solution 1:

I=2Z

¿ I [ x ] , coefficients belong ¿ I=2 Z , i . e coefficients are even

consider J=¿2 , x>¿

I [ x ]⊂ J⊂Z [ x ]

as 2+x∈ J but 2+x∉ I [ x ]∴ J ≠ I [ x ]

1∈Z [ x ] but 1∉ J ∴ J ≠ Z [ x ]

hence I [ x ] isnot a maximal ideal .

solution 2:

Z [ x ]2 Z [ x ]

≅ Z2Z

[ x ]≅ Z2 [ x ]

clearly Z2 [ x ] is not a field as xdoes not possess an inverse .

2 Z [ x ] cannot be amaximal ideal of Z [ x ] .

solution 3:

show J=¿2 , x>is maximal ideal of Z [ x ] .

…………………………………………………………………………………………………………………………………………………………..

I=Z⨁ {0 }

Z⨁ZZ⨁ {0 }

≅ ZZ⨁ Z

{0 }≅ {0 }⨁Z ≅ Z

…………………………………………………………

φ : Z⨁Z → Z suchthat

φ (a ,b )=b

then Ker φ=Z⨁ {0 }

Z⨁ZZ⨁ {0 }

≅ Z

Z is an integral domain∧not a field

Z⨁ {0 }is primeideal but not a maximal ideal ;

……………………………………………………………………..

we know that → Qis field of quotientsof Z

all fields of quotientsof a givenintegral domain D are isomorphic

But Q∧R are not isomorphic since Qis countable∧R isnot countable

Aliter : if R is a field of quotients of Z , then every element of R must be a ratioof two integers

but √2 cannot be expressed as the ratioof two integers .

………………………………………………………………………………………………..

field of quotients of a field F is isomorphic to FField of quotients of R must be isomorphic ¿ R ;but C is not isomorphic ¿ R

Let φ be an isomorphism¿C ¿ R

Then φ (−1 )=−φ (1 )=−1

−1=φ (−1 )=φ (i2 )=φ (i )2>0 which isnonsense

φ ( i )∈R , φ (i )2>0

……………………………………………………………………..

A4 cannot have an element of order 6

A4≤ S4

we show that S4 doesnot haveany element of order 6

decomposition into productsof disjoint cycles → various possiblities

(4 cycle ) → order 4

(3cycle ) (1 cycle ) → order 3

(2cycle ) (2 cycle ) →order 2

(2cycle ) → order 2

(1cycle ) → order 1

S4 has elements of orders 1 ,2 , 3 ,4

S4 has no element of order 6 → A4 hasno element of order 6

……………………………………………………………………………………..

There is no subgroup of order 6 in A4 ………………………………………………………………………………

Let H be a subgroup of order 6 → H must be isomorphic ¿either Z6∨S3

as A4 doesnot haveany element of order 6 , H cannot be isomorphic ¿ Z6

hence H ≅ S3

S3 has3elements of order 2→ (12 ) , (13 )∧(23)

A4 hasonly 3 elements of order 2 → (12 ) (34 ) , (13 ) (24 )∧(14 ) (23 )

as H ≅ S3 , H should also contain 3elements of order 2 ;∧H ≤ A4

∴order 2elements of A4 must be∈H

But elements of order 2∈S3 donot commute where they commute∈H

hence H≇ S3

as H is neither isomorphic ¿Z6 nor ¿S3 , H doesnot exist

A4 doesnot have ansubgroup of order 6 .

………………………………………………………………………………………………………………………………………..

6∨o ( A4 ) but A4 doesnot have an subgroup of order 6

converse of Lagrange ' s theoremis not true .

……………………………………………………………………………………………………………………………………….

Let p bea primenumber∧G be a group of order 2 p

thenG is cyclic (i . eG≅Z2 p )∨G≅ D p

where Dp= {a , b|ap=e=b2∧aba=a }

D3≅ S3: a → (123 )∧b→(12) .

……………………………………………………………………………………………………………………………………………

any groupof order 6 (2× 3 ) isisomorphic ¿ Z6∨D3(¿ S3)

…………………………………………………………………………..

let f : R → Sbe ani somorphism

φ : R [ x ] → S [ x ]

s . t φ (r 0+r1 x+r2 x2+… )=f (r 0)+ f (r1)x+f (r2) x2+…

…………………………………………..

G⨁H isabelian iff G∧H are abelian

As S3is not abelian , S3⨁Z2 is not abelian .

S3⨁ Z2 haselements of order 6 , whereas A4 do not haveanelement of order 6

( (123 ) ,1 )∈S3⨁ Z2→ order=lcm (3,2 )=6.

…………………………………………………………………………………………

a=2 , b=1 , c=2

∴a=2b , c=2 b∨a−2b=0 , c−2b=0 ;

for ker

;

…………………

Z × Z¿ (3,1 )>¿≅Z .¿

…………………………………………………………………………..

let mbe theleast positiveinteger s . t am∈H →<am>⊆H

let b∈H , thenb∈G=¿a>∴b=ak

k=m q+r where r=0∨r<m

if r ≠ 0→r<m

r=k−mq→ ar=ak a−mq → ar=b (am )−q∈H

ar∈H∧r<m→contradiction

hence r=0→ k=mq→ ak=amq

b=(am )q∈<am>¿

H ⊆<am>¿

∴H=¿am>¿

……………………………………………………………………

¿ Zn , for eachk∨n ,< nk>is a unique subgroup of order k

……………………………………………………………………………..

if d∨n ,then∈a cyclic group of order n ,number of elements of order d is ϕ (d )

Let b∈G∧b∉H , if o (b )=1 thenb=e∈H∴o (b )≠ 1

o (b )≠ 5as otherwise K=¿b>would be another subgroup of order 5

hence o (b )=25 ,∴G=¿b> .

…………………………………………………………………………………….

¿ax−1>¿R [ x ] i . e 1∈<ax−1>¿

an=0 →1=1−an xn .

……………………….

φ (n )=n+ I where I=¿2−i√3>¿

onto :a+b i √3+ I =a+2 b+ I=φ (a+2 b ) as i √3≡2

¿7>⊆Ker φ∧Ker φ⊆<7>.

………………………………………………………………………………………………………………………………………………

…………………………………………………………

¿a field , every nonzero element is a unit → every field is aUFD

UFD :any nonzero element∈R is either aunit∨canbe writtenas a product of finite number

of irreducible elements of R .

…………………………………………………………………………….

Why is Z [√−n ]not aUFD∀n≥ 3

¿canbe easily show that 2is irreducible .

………………

If n is even , then2∣(√– n )2=−n

but 2does not divide √−n , so2 is anot a prime element

but 2 is irreducibleelement .

¿a UFD , all irreducibles are prime , so this showsZ [√−n] is not aUFD .

……………….

if n is odd ,then 2∣(1+√−n)(1−√−n)=1+n=even

without dividing either of the factors , so again2 is anot a primeelement .

This argument works equally well for n=3∧greater

but fails for n=1 ,2

¿∈Z! [√−1 ]∧Z [√−2 ] areUFDs

…………………………………………………………………………………………………

Show that Z [√−2 ] is a PID

simply show Z [√−2 ] is aEuclidean Domain

ED→ PID →UFD .

………………………… …………

¿ show that 2 is irreducible

let 2=( a+b√−n ) (c+d √−n )

4=( a2+n b2 ) (c2+nd2 )

as a ,b are integers→ ( a2+nb2 ) ≠ 2 , ( c2+nd2 ) ≠ 2

hence either (a+b√−n )∨(c+d √−n ) is aunit

2 isirreducible ..

…………………………………………………………………………………………..

Consider the ideal J=¿2 , 1+√−5>¿<α >for some α∈Z [√−5 ]

2∈ J →2=αβ → N (α )∨N (2 )=4

1+√−5∈ J →1+√−5=αγ → N (α )∨N (1+√−5 )=6

N (α )∨gcd ( 4,6 )=2

N (α )=1∨2.

N (α )≠ 2hence N (α )=1→ α=± 1→ J=¿1>¿

but this is nonsense . show by writing1 as a combination of 2∧1+√−5

Then take mod 2.

………………………………………………………………………………………………….

g∨ f ∈Z [ x ]→ f =g q , whereq∈Z [ x ]

g=c g0 where c=c ( g )∧g0is primitive

q=d q0 whered=c (q )∧q0is primitive

but Gauss Lemma , g0 q0 is again primitive

f =cd g0 q0

cd=1→c=± 1asc∈Z .

…………………………………………………………………………………………………………………..

Gauss Lemma

1¿ product of two primitive polynomials is primitive

2¿ if f ( x ) irreducibleZ

, then f ( x ) is irreducibleQ

.

……………………………………………………………………………………………………………………

………………………………………………………………………

an even cycleis odd permutation

an odd cycle is even permutation

…………………………………………………………………………………………………..

60=4× 3 ×5 as 4 , 3∧5are relatively prime

U 60=U 4 ×U3 ×U 5≅ Z2⨁ Z2⨁ Z4

U 4 is cyclic group of φ (4 )=2 →U 4≅ Z2

U p is cyclic iff p is prime→U p≅ Z p−1 aso (U p )=φ ( p )=p−1

U 3≅ Z2∧U5≅ Z4 .

………………

Z2⨁ Z2⨁ Z4 isnot cyclic as there is noelement of order 16

the greatest order possible islcm (2,2,4 )=4.

………………………………………………………………………

Group of order 4 is isomorphic to Z4 or K4

Let G be a group of order 4; by Lagrange theorem, any element of G has order 1, 2 or 4.

If G has an element of order 4 then G is cyclic.

Assume G has no elements of order 4, then G has 3 elements of order 2. Then G is Abelian.

So G is Abelian and has 3 elements of order 2 and 1 element of order 1 (the identity), it follows that G is isomorphic to the Klein four group.

both Z4 or K4 are abelian any group of order 4 is always abelian

D2≅ K4≅ Z2× Z2

……………………………………………………………………………………………………………………………………………

o ( a )=p ;o (b )=q ;

gcd ( p , q )=¿1∧ab=ba¿

Then o (ab )=o ( a ) o(b) .

………………………………………………………………………………………………………………………………

if G is an abelian group of order pq, where p and q are distinct prime numbers, then G is cyclic.

o (G )=pq∴∃a , b∈G s.t o (a )=p ,o (b )=q

now G is abelian→ ab=ba

gcd (¿o (a ) , o (b ))=1∨gcd ( p , q )=1¿

o ( ab )=o (a ) o (b )=pq

G=¿ab>.

……………………………………………………………………………………………………………………………………….

Abelian group of order 6 is cyclic and isomorphic to Z6.

non-Abelian group of order 6 is isomorphic to S3.

any groupof order 6 is isomorphic ¿ Z6∨D3≅ S3

…………………………………………………………………………………………………………………………………………………….

case1 : G1 x G2 is abelian as G1 and G2 are abelian (as they are cyclic)

case 2: Let G1 ×G2 be cyclic , generated by c=(a ,b )∧o (c )=mn

let gcd (m ,n )=d>1

consider cmnd =(a , b )

mnd =((am )

nd , (bn )

md )=( e1 , e2 )

¿ mnd

<mn asd>1

this is a contradiction asmn isthe least positiveinteger s .t cmn=(e1 ,e2 )

∴d=1∨gcd (m, n )=1

……………………………………………………………………………

D3= {a , b|a3=e=b2 ;aba=b∨ba=a2b }={e ,a , a2 , b , ab , a2 b }

S3= {id , (12 ) , (23 ) , (13 ) , (123 ) , (132 ) }

a ↔ (123 )∧b↔ (12 )

Then S3≅ D3

……………………………………………………………………………..

A non-empty subset S of R is a subring if a , b∈S→ a−b∈S∧ab∈S .

S is closed under subtraction and multiplication.

………………………………………………………..

Z3={0 , 1,2 }

squares= {0 ,1 }

a2+b2=0+0 ,0+1, 1+0 ,1+1

a2+b2=0 →a=0∧b=0

………………………………………………………………………………

Z7 →squares : {0 ,1 ,2 ,4 }

a2canbe 0 ,1 , 2∨4.Similarly for b2

a2+b2=0 iff a=0∧b=0;

Z5 →squares : {0 ,1 , 4 }

a2+b2=0 even whena2=1∧b2=4

A=[1 −22 1 ]doesnot have aninverse asdet A=12+22=5=0

……………………………………………………………………………………………………

s . t φ (a+bx+c x2+… .. )=[ a ]4+ [ b ]4 x+ [ c ]4 x2+……

canbe easily show that φ is an ontohomomorphism

Ker φ=¿ 4>¿ 4 Z [ x ]={4 f ( x )|f (x )∈Z [ x ] }

Z [ x ]

¿4>¿=Z4 [ x ]∨ Z [ x ]4 Z [ x ]

=Z4 [ x ] ¿

……………………………………………………………………………..

Think of Z2 x Z2 as a 2-dimensional vector space.

Then, Aut (Z2 x Z2) can be thought as the vector space of all linear transformations from Z2 x Z2 to itself whose determinants are nonzero.

i.e. think of these linear transformations as 2 x 2 matrices.

only 6 of them have nonzero determinant. They are

(1 00 1) ,(1 0

1 1) ,(1 10 1) ,(0 1

1 0) ,(1 11 0) ,(0 1

1 1)Now, there are only two groups of order 6: Z6 (abelian) or S3 (non-abelian). Since it may be checked quickly that the matrix multiplication among these elements is NOT commutative, we have one conclusion: Aut (Z2 x Z2) is isomorphic to S3.

(1 10 1) implies that (1,0 ) → (1,0 )∧(0,1 )→ (1,1 )

…………………………………………………………………….

Z2⨁ Z2={(0,0 ) , (1,0 ) , (0,1 ) , (1,1 ) }

any automorphism φ is determined by values of φ (1,0 )∧φ (0,1 )

always φ (0,0 )= (0,0 )

hence φ (1,0 ) has3 choices

after choosing φ (1,0 ) , φ (0,1 ) is ¿2choices

∴6 automorphismsare possible

φ 1 : (1,0 ) → (1,0 ) ; (0,1 ) → (0,1 )

φ 2 : (1,0 ) → (1,0 ) ; (0,1 ) → (1,1 ) .

φ 3 : (1,0 ) → ( 0,1 ); (0,1 ) → (1,0 )

φ 4 : (1,0 ) → (0,1 ) ; (0,1 ) → (1,1 ) similarly φ 5 , φ6

but φ 2φ 3 ≠ φ 3 φ 2 .→ Aut ( Z2⨁Z2 ) is not abelian

any groupof order 6 isisomorphic ¿ either Z6∨S3

Aut (Z2⨁Z2 )≅ S3 .

……………………………………………………………………………………..

S.t every group G of order 4 is isomorphic to either Z4 or K4.

let x∈G → o ( x )∨o (G ) → o ( x )=1 ,2∨4

if o (x )=4 , thenG=¿ x>is cyclic

if o (x ) ≠ 4 →o ( x )=2∀ x ≠ e→G is abelian

the only noncyclic abelian group of order 4 is K4

G≅ K4≅ Z2⨁ Z2

………………………… ………

S3

A3≅ Z2as S3/ A3has only 2elements .

………………………………………………………………………

K 4={e ,a ,b , ab } where a2=b2=e; ab=ba , (ab )2=e

φ : K 4→Z2⨁ Z2 , st

φ (e )= (0,0 ) ,φ (a )=(1,0 ) , φ (b )=(0,1 ) , φ (ab )=(1,1 )

φ (ab )=φ (a )+φ (b ).

K 4≅ Z2⨁Z2

………………………..

also K4 can be defined as

K 4={e , a ,b ,c } where a2=b2=c2=e

ab=c , bc=a , ca=b .

σ∗(1,2 )=( σ (1 ) ,σ (2))∗σ

(σ (1 ) , σ (2 ) )=(1,2)

( σ (1 ) , σ (2))=(2,1)

ℑφ≅ Z2 hasorder 2∧ℑφis subgroup of Z6

subgroup of Z6that has order 2 is3 Z6∨¿3>.

………………………………….

asWis subspace :u ,v∈W →au+bv∈W for any scalar a , b

a ,b∈R

5−x∧x∈W → 15

(5−x+x )∈W →1∈W .

…………………………………………………………………………………………………

………………………………

Ker f = {c1+c2e− x|c1 , c2∈R }=span (1 , e−x )

1∧e−x arelinearly independent → basis

dim( Ker f )=2.

………………………………………………………………………………

……………………..

B=P−1 AP where A=[T ]E∧B=[ T ]s∧¿

P is changeof basis ¿E ¿S

if S={u1 , u2 }→ P= [u 1u 2 ]

diagonalize A ¿get D=P−1 AP →¿P get the basis S={u1 , u2}

henceu1=(−2 ,1 )∧u2=(3 ,1 ) give therequired basis vectors

……………………………………….

…………………………………..

φ (1 ) should bean idempotent →φ (1 )=0∨φ (1 )=1

φ (n )=0∨φ (n )=n

……….

……………………

……………………………..

Let G1 ×G2=¿ (a ,b )> , o (a , b )=mn.

Let gcd (m, n )=d>1

(a ,b )mnd =(e1, e2 )

but mnd

<mn,which is a contradiction as mn isthe least positive integer .

…………………………………………………………………………………………………..

The only homomorphisms from a Field F to any ring R are zero homomorphism and injective homomorphism.

φ : F → R is anyhomomorphism

Kerφ isideal of F → Kerφ= {0 }∨Kerφ=F

∴φis 1−1∨φ ( x )=0 ∀ x∈F .

………………………..R[x] is an integral domain iff R is an integral domainLet f(x) = a0 + a1x + ⋯ + an xn

g(x) = b0 + b1x + ⋯ + bm xm. If these are not 0, then we can assume an, bm ≠ 0. Thus the highest-degree term of f⋅g is anbm xm+n ≠ 0, so f⋅g ≠ 0. Thus R[X] has no zero divisors.………………………………………………

………………………………………………………………….

Q= {±1 , ±i ,± j , ± k } , N=Z (Q )= {−1, 1 }

QN

={ N ,∋, Nj , Nk }

as∋¿N (−i )= {±i }

( ¿ )2=N i2=N (−1 )=N

o (¿ )=2=o ( Nj )=o ( Nk )

QN≅ Z2× Z2.

………………………………………………………………….

φ :C → R beany isomorphis m

φ (1 )=1∧φ (−1 )=−φ (1 )=−1

−1=φ (−1 )=φ (i2 )=φ (i )2

φ ( i )∈R → φ (i )=x , x is some real number

−1=x2 , where x is some realnumber

this is impossible ∀ x∈R .

…………………………….

I=¿2 ,−1+√−5>¿

φ : Z [√−5 ] → Z2

st φ ( a+b√−5 )=[a+b ]2

…………………………………….

………………………………….

…………………………..

o ( β ) cannot be1∨3

if o (β )=3→ β3=1 →o ( β3 )=1 which is not possible

again o ( β )=21 not possible

if o (β )=21

then β= (21cycle )∨(3 cycle ) (7 cycle )

thus only possibility is o (β )=7.

………………………

φ :2Z →Z4

st φ (2 n )= [ n ]4

if 2n∈ker φ↔ φ (2n )= [0 ] ↔ [ n ]4= [0 ]4

↔ nis amultiple of 4

↔ 2n is a multiple of 8↔ 2n∈8 Z

ker φ=8 Z

…………………………………………If Gis a group s . t

1¿ab=ba

2¿o (a )=m ,o (b )=n , gcd ( m,n )=1

Then o (ab )=o ( a ) o (b )

Proof :

let A=¿ a>¿B=¿b>¿

as|A|∧|B|are relatively prime → A ∩ B= {e }

let c=ab∧o (c )=i

c i=( ab )i=aib i=e asab=ba

a i=b−i∈ A ∩ B={e }

∴ai=e→ m|i . similarly n|i

hence lcm ( m,n )|i→mn|i

cmn=( ab )mn=e→i∨mn

∴i=mn

o ( c )=mn

o ( ab )=o (a ) o(b).

……………………………………………………..

o ( an )= o (a )gcd ( n , o (a ) )

→ o ( β2 )= o (β )gcd (2, o ( β ) )

=15

o ( β2 )=lcm (5,3 )=15

gcd ( 2, o ( β ) )=1∨2→ o ( β )=15∨30

o ( β )=30 is not possible∈S9

o ( β )=15→ β15=e→ β16=β

β=( β2 )8=(13579 )8 (268 )8=(17395 ) (286 )

as (13579 )∧(268 ) are disjoint , they commute .

………………………………………………………………finite non-commutative ring.

infinite non-commutative ring without unityM 2 (2 Z )

…………………………………………………………..finite noncommutative ring without unityM 2 (2 Z8 )

……………………………………………..M n (R ) , the set of all n× n matrices

R,is anoncommutative ring

if R hasunity 1, then M n ( R ) also hasunity I n

if R doesnot have unity ,then M n ( R ) also doesnot have unity

if R is finite , M n ( R )is finite

if R isinfinite , M n ( R ) is alsoinfinite .

…………………………………………….Show that every group of order 15 is cyclicby cauch y ' s theorem ,∃ elements a∧bof order 3∧5

A=¿a>¿ B=¿ b>¿

ab ≠ e ,ab≠ a ,ab≠ a2 .

clearly ab∉ A∧ab∉B

Now ab is not∈either A∨B .Hence order of abca n' t be 1, 3 , 5.

Hence order of ab is15.∧ab generates the group ..

…………………………..if G is an abelian group of order pq, and p & q are distinct primes, then G is cyclic.p|ord (G )∧q|ord (G )

by cauch y ' s theorem ,∃ elements a∧bof order p∧q

ab=ba∧(o ( a ) ,o (b ) )=1 → o (ab )=o (a ) o (b )=pq

henceG=¿ab> .

…………………………………Any group of order 15 is cyclic.A5 has no element of order 15, since maximum order of any element in A5 is 5.Hence A5 has no subgroup of order 15……………….

similarly A5 has no subgroup of order 30 as this subgroup contains an element of order 15……………………………………………….

o ( a2 )= o ( a )gcd (2 , o (a ) )

=o (a )→ gcd (2 , o (a ))=1

o ( a )=odd .

……………………………………………….Let G bea group of order pq , for some primes p∧q ,

p<q∧p∤q−1

thenG is cyclic∧G≅Z pq≅ Zp⨁ Zq

…………………………………………………………………………………..

¿ R , every element (≠ e ) has infinite order .Thereis no element having order 2

¿ R¿ , however (−1 ) has order 2.

……………

Z=¿1>, whereas Qis not cyclic

let Q be cyclic→Q=¿ a>¿ {na , n∈Z }, as we are∈additive group

a2∈Q ,but a

2∉<a>¿

henceQ cannot becyclic .

……………………………………

…………………………………………..

………………………….

if f :G → H is an isomorphism

F : Aut (G )→ Aut (H )

st F (α )=f ∘α ∘ f −1 whereα :G→ Gis an isomorphism

f ∘α ∘ f −1 is an automorphism¿ H ¿H .

……………..¿ show F is bijective→ just find aninverse function

F : Aut ( H )→ Aut (G )

st F ( β )=f −1∘β ∘ f

then F F=I

F is invertible →one−one∧onto .

………………………………………………………………………………………………………………………………………….every groupG of order 2 p is either cyclic∨dihedral

i .e G≅ Z2 p∨G≅ Dp

ex :any group of order 6 isisomorphic ¿ Z6∨D3

any groupof order 4 is isomorphic ¿ Z4∨D2≅ K4≅ Z2× Z2(hence abelian) .

note : D p= {a ,b|ap=e=b2∧aba=b }

( groups of order 4,6,10,14 ) are either cyclic∨dihedral

……………………………………………………………………………………………………if p>q are primes∧q doesnot divide p−1 , thenevery group of order pq is cyclic

group of order 15 is cyclic (5.3)……………………………………………………………………………………….

I⊆ J⊆F [ x ]

as F [ x ] isa PID → J=¿ f ( x )>¿

p ( x )∈ I ⊆J=¿ f ( x )>¿

∴ p ( x )=f ( x ) g (x )

p ( x ) is irreducible→ either f ( x )∨g ( x )has degree 0

case 1: deg f ( x )=0→ f ( x )=c where c∈F

as F is a field → every nonzero element is aunit

c is aunit of F , henceunit of F [ x ] also

c∈ J → J=F [ x ]

case 2: degg (x )=0→ g ( x )=d where d∈F

p ( x )=d f ( x )→< p ( x )>¿< f ( x )>¿

∴ I=J

…………………………………………………………………………………………………………………………

R :ring of continuous , realvalued functions on unit interval [ 0,1 ]

φ : R → R s. t φ ( f ( x ) )=f (12 ); φis anonto homomorphism

let a∈R → thena=a( 12 )+ a

2=φ(ax+ a

2 );

let I=ker φ={ f ( x )|f (12 )=0 }

then RI≅ R

…………..similarly ψ : R → R s. t ψ (f ( x ) )=f ( 1

3 );ψ is an onto homomorphism

let J=ker ψ={f ( x )|f ( 13 )=0 }

then RJ≅ R

………………∴ R

I≅ R

Jbut clearly I ≠ J

x−12∈ I but doesnot belong¿ J .

……………………………………..

…………………………………………………….

if α isaunit →∃ β∈Z [i ] s . t αβ=1

N (αβ )=1→ N (α ) N ( β )=1

as N (α ) isan i nteger∧N (α )>0→ N (α )=1

………………………

…………………………………………….

………………………………………….

if p≡1 mod 4 →∃m,n∈Z s . t p=m2+n2= (m+¿ ) (m−¿ )

neither of the factors is a unit as Normis greater than1

Thus p isred ucible∈Z [ i ].

………………………

p ≡3 mod 4

let p be reducible→ p= (m❑1+i n1 ) ( m2+i n2 )

neither factor is a unit → norm>1→ (m❑12+n1

2 )>1∧(m22+n2

2 )>1.

taking norm bothsides

p2=(m❑12+n1

2) (m22+n2

2 )

p=(m❑12+n1

2)=(m22+n2

2 )

by Fermat ' s theorem→ p≡ 1mod 4 , whichis a contradiction

.

…………………………………………………………………………………………………….

o (G )=pk mwhere p∤m, p is prime

number of Sylow p−subgroups=np , then

np ≡1mod p∧np∨m

particular case :o (G )=pq ,where p∧q are distinct primes

np ≡1mod p∧np∨q

nq≡ 1mod q∧nq∨p .

…………………………………………………………………………………………

………………………………………………………………………………………………………………………….

sp∨q∧s p≡ 1mod p

sq∨p∧sq ≡1 mod q .

………………………………………

¿class equation →|G|=pn→ Z (G )≠ {e }

also Z (G ) ≤G →o (Z (G ) )∨o (G ) .

…………………………………………………………………………………………………………….

let d=gcd (a ,b )

¿ show → aZ+b Z=d Z∨¿a>+¿b>¿<d>¿

step1 :d=ax+by∈<a>+¿b>¿

¿d>⊆<a>+¿b>¿

step2 : let l∈<a>+¿b>¿

l=ia+ jb , d=gcd (a ,b ) →d|a∧d|b

a=md∧b=nd

l=ℑd+ jn d→ d∨l →l∈<d>¿

¿a>+¿b>⊆<d> .

…………………………………………………………………………………………………Z [√−n ] not aUFD ∀n ≥3

…………………………………………………………………………………………………………….σ be a non identity permutations . t σ ( i )= j ; i ≠ j

as n≥ 3∃k ≠ i , j

let τ= ( j k )∈Sn

τ σ ( i )=τ ( j )=k∧σ τ ( i )=σ ( i )= j

∴ τ σ ≠ σ τ

hence for every nonidentity permutation∈Sn , there∃some element not commuting with¿

∴Z ( Sn ) must be trivial