If a and b are integers not both zero (a≠0, b≠0) then, az + bz = dz where d = gcd(a, b)

If a and b are integers not both zero (a≠0, b≠0) then, az + bz = dz where d = gcd(a, b)

Proof:        

We define az + bz = {al + bm| l,m ϵ Z}

Let x, y ϵ az + bz

Then,

x = al1 + bm1

y = al2 + bm2, where l1, m1, l2, m2 ϵ Z

x – y = al1 + bm1 – (al2 + bm2) ϵ az + bz, where l1 – l2 ϵZ and m1 – m2 ϵZ

But, every subgroup of Z is of the form az, where a is a non-negative integer (a≥0)

Show that, d = gcd(a, b)

We have, a = a.1 + b.0 ϵ az + bz = dz

⇒a ϵ dz

a = dk1, for some k1 ϵ Z

⇒ d|a

Also, b = a.0 + b.1 ϵ az + bz = dz

⇒ bϵ dz

b = dz2, for some k2 ϵZ

⇒d|b

Also, d = d.1 ϵdz = az + bz

d ϵ az + bz

d = al + bm, where l, m ϵ Z …………………(*)

Let C be any other common divisor of a and b

i.e., c|a and c|b

But then c|al and c|bm

c|(al+bm)

c|d [from equation(*)]

Therefore, d = gcd(a, b)

Let G be a group. A non-empty subset H of G is a subgroup of G if only if xy^-1 ϵH, for all x, y ϵH

Theorem: Let G be a group. A non-empty subset H of G is a subgroup of G if only if xy-1 ϵH, for all x, y ϵH

Proof:

(⇒) Let H be a subgroup of G and H is  non-empty subset.

Let x, y ϵH

Then y-1ϵH (since H is a subgroup of G)

Since x ϵH,  y-1 ϵH

Then xy-1ϵH (Since H is a subgroup of G)

 

(⇐) Assume that xy-1 ϵH, for all x, y ϵH

Let x = y then, we have, xx-1 ϵ H

⇒eϵ H

Therefore, identity law holds

Let x ϵH and eϵH

Then, x.e. ϵH

x-1 ϵH (Since x.x-1 = e)

Therefore, inverse law holds.

Let x, y ϵH

then y-1ϵH and (y-1)-1 ϵ H

⇒ xy ϵ H

Therefore, closure law holds.

Therefore, H is a subgroup of G.

Subgroup

A non empty subset H of a group (G, *) is called a subgroup  of G if it is has following properties:

  1. Closure law: If a ϵ H and b ϵ H then a*b ϵ H
  2. Identity law: e ϵ H
  3. Inverse law: If a ϵ H then a-1 ϵ H

Let G be a group and a, b ϵ G with O(a) = 5, a^3b = ba^3 .Prove that G is an abelian group.

Let G be a group and a, b ϵ G with O(a) = 5, a3b = ba3 .Prove that G is an abelian group.

Proof:

Given a3b =  ba3

Multiply left side by a2,

a2.a3.b = a2.b.a3

a5.b = a2.b.a3

1.b = a2.b.a3 [since O(a) = 5 i.e., a5 = 1]

Multiply right side by a2,

ba2 = a2ba3a2

ba2 = a2ba5 [since O(a) = 5 i.e., a5 = 1]

ba2 = a2b

Multiply right side by a

 

ba3 = a2ba

a3b = a2ba [since a3b = ba3]

a2.a.b = a2b.a

ab = ba [by left cancellation law]

Therefore, G is abelian group.

Order of Group

Let G be a finite group. Then the number of elements in G is called the order of the group G and is denoted by O(G) or |G|

Definition:

Let G be group and a ϵ G. IF there exists a positive integer n such that an = e, then the least such positive integer n is called the order of a.

If no such positive integer exists then we say that a is of infinite order.

Example:

G = { 1, -1, i, -I}

O(G) = 4


Let G be a group and a, b ϵ G with O(a) = 5, a^3b = ba^3 .Prove that G is an abelian group.


 

If G is a group that (ab)^n = a^n.b^n , for all a, b ϵG and for three consecutive integers n. Then prove that G is an abelian group.

If  G is a group that (ab)n = an.bn , for all a, b ϵG and for three consecutive integers n. Then prove that G is an abelian group.

Proof:

Let a and b be any two elements of G.

Suppose n, n+1 , n + e are three consecutive integers such that

(ab)n = an.bn …………….(1)

(ab)n+1 = an+1.bn+1 ………..(2)

(ab)n+2 = an+2. bn+2……………(3)

Equation (2) can also be written as,

(ab)(ab)n = a.an.b.bn

a.b.an.bn = a.an.b.bn

b.an = an.b [by cancellation law]

Equation (3) can also be written as,

(ab)(ab)n+1 = a.an+1.b.bn+1

a.b.an+1.bn+1 = a.an+1.b.bn+1

b.an+1 = an+1.b [by cancellation law]

b.an.a = an.a.b

an.b.a = an.a.b

⇒ ba = ab [by left cancellation law]

Therefore, G is an abelian group.

Prove that if (ab)^-1 = a^-1b^-1, for all a, b ϵG then G is abelian

Prove that if (ab)-1 = a-1b-1, for all a, b ϵG then G is abelian

Proof:

Given (ab)-1 = a-1.b-1…………………..(1), for all a, b ϵG

Since G is a group, we have (ab)-1 = b-1.a-1 ………….(2), for all a, b ϵ G

From (1) and (2), b-1a-1 = a-1b-1

Taking the inverse

(b-1a-1)-1 = (a-1 b-1)-1

(b-1)-1(a-1)-1 = (a‑1)-1(b-1)-1 [since (a-1)-1 = a, for all a ϵG]

⇒ba = ab

⸫ G is an abelian group.