………………………………………….

…………………………………………………………….

………………………………………….

………………………………………………………….

……………………………………………………………………………..

…………………………………………………………

………………………………….

………………………..

……………………………………………….

……………………………..

…………………………………………

………………………….

…………………..

Z+Z/{0}+Z is isomorphic to Z

as Z is an integral domain, {0}+Z is a prime ideal but not maximal.

similarly, Z+Z/Z+{0} is isomorphic to Z

……………………………………………….

…………………………

all homomorphism from Z x Z to Z

(1,0) -> 0, 1,0,1

(0,1) -> 0, 0,1,1

(n,m)=n(1,0)+m(0,1)

……………………………………………

……………………..

…………………………

……………………………………………….

I = (3,x)

a0 + a1x + a2x2 +……. [ a0 ]3

kernel = (3,x)

……………………………………………………

maximal ideal of Z3 = <0> as Z3 is a field

maximal ideal of Z4 = <2> as 4=22

required maximal ideals are : <0> + Z4 , Z3 + <2>

………………………………………………………

…………………….……………………………………………………

Zn is a cyclic group of order n.

G/Z is cyclic G is Abelian Z = G

…………………………………………..

If 1 has additive order n, then the characteristic of R is n.

Also char(Z × Zm) = lcm (0, m) = 0.

2. characteristic of the ring Z4 ⊕ 4Z = lcm (4, 0)

char Z4 = 4, Char 4Z = 0.

……………………………………………………………..

Z3[x]/A = {a + bx + A | a, b belong to Z3}; A = < x2 +1 >

a + bx + A a + bi

…………………………………………………………………

O(ᶲ(1,0)) | O(1,0) = p;

O(ᶲ(1,0))= 1 or p;

every element of Zp has an order 1 or p.

O (0) =1 ; O(1) = p; O(2) = p as p is prime

ᶲ(1,0) can be any element of Zp, hence has p choices.

……………………………….

D12 has order 2 x 12=24

a12 = e; b2 = e; o(a)=12

D12 and S4 are not isomorphic.

……………………………………………….

G × H ≃ H × G

………………………..

S3 = { e, a, a2, b, ab, a2b }

where a = (123) and b = (12); a3 = e, b2 = e; aba = b.

………………………………………………………………

all

a generator must be mapped to a generator.

Phi (1) must be a generator = 1, 5, 7, 11.

4 isomorphisms.

……………………….

,……………..

…………………………….

Let |G| = 15. If G has only one subgroup of order 3 and only one of order 5, prove

that G is cyclic. Generalize to |G| = pq, where p and q are prime.

ans:

If |H| = 3, H = {e, a, a2}; since 3 is prime, H is cyclic and each element has an order 3 (except e).

Number of elements of order 3 = 2 (i.e. a and a2).

No other element can have order 3, otherwise there would be two subgroups of order 3.

Similarly, subgroup of order 5 has 4 elements of order 5.

There is 1 identity element.

15 – (2 + 4 + 1) = 8 elements have order 15.

|G| = 15, there are elements with order 15; therefore G is cyclic.

………..

Suppose |G| = pq, where p and q are prime, and G has only

one subgroup of order p, and one of order q.

Every element of G has order 1, p, q, or pq.

If H is the one subgroup of order p, then H has p – 1 elements of order p.

If K is the one subgroup of order q, then K has q – 1 elements of order q.

These, together with the identity, account for

1 + (p − 1) + (q − 1) = p + q − 1 elements.

p + q − 1 < pq.

Therefore, there is at least one element of G which is not in H or K.

If such an element had order p or q, it would generate a second such subgroup. Therefore this element must have order pq, and so G is cyclic.

………………………

1 element of order 1 p-1 elements of order p q-1 elements of order 1 pq > p + q + 1 = 1 + (p-1) + (q-1). 

………………………………………………………………

…………………………………………………………..

let g(x) be an element of I, such that g(x) has the minimum degree.

to show than I = <g(x)>

since g(x) ϵ I

let f(x) ϵ I. Then, by the division algorithm, we may write

f(x) = g(x)q(x) + r(x), where r(x) = 0 or deg r(x) < deg g(x).

r(x) ≠ 0 case is not possible.

hence r(x) = 0; f(x) ϵ <g(x)>

…………………………………….

…………………………………………..

aj is a generator iff gcd(n, j) = 1.

if k /n, then < an/k > is a subgroup of order k.

…………………….

if k divides n, then < n/k > is a subgroup of Zn of order k.…………………

if d/n, number of elements of order d is PHI (d)…………….

……………….

…………………………………………….

…………………….

o (a )=m .c∈<a>∩<b>¿c∈<a>→c=ak→cm=akm=e→o (c )∨o(a)……………………

in a cyclic group, only one unique subgroup of a given order exists.

……………………………………………………………………….……………..

………………..

can be easily showed that ᶲ is one-one, onto and homomorphism.…………………………………………………

|(a,b)| = lcm (|a|,|b|)

in D6, a6 = e. there is an element of order 6.

in A4 , non-identity permutation (123) or (432) i.e a 3 cycle. or (12)(34)order of any element is either 1, 3, 2.there is no element in A4 of order 6.…………………………………………………………………………

…………………………………………………

if d/n, then o(d) = n/d o(10) = 40/10 = 4 in Z1o.

………………..

…………………..

…………………………………

…………………..

……………..

…………………….

define θ : < k > Zns.t θ(km) = [m]n

or

any element of <k>/<kn> is : mk + <kn> = m (k + <kn>)

hence <k>/<kn> = < k + <kn> >.

………………………….

………………..

…………………………

let x ϵ kerx mod m = 0 mod m m|x. also n|x

m|x and n|x. hence lcm (m,n)|x x ϵ < lcm (m,n) >

………………………

……………………

|θ(1,0)| divides |(1,0)|=p;|θ(1,0)| can be 1 or p.every element in Zp has order p or 1. and θ(1,0) ϵ Zp.

θ(1,0) can be any element of Zp have p choices.θ(0,1) can have p choices.there are p2 homomorphisms.…………………………………………………

D6hasanelement of order 6 ,but A4doesnot haveany element of order6……………………….

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

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

……………………………………………………………………………

clearly ᶲ is isomorphism………………………………………………………………………………………………

……………………………………..

………………………………………………..

………………………………………………………………………………

……………………………………………………….

………………………….

I = < 2, x >Z[x]/< 2, x > is isomorphic to Z2

…………………………………………………………………….

Z3[x] is an integral domain but not a field. Z3 is a field, but Z3[x] is not a field.………………………………………………………………………………………….

p(x) ϵ ker; < p(x) > is contained in ker.………………………………………………………………..