Question

Prove that the set g = (1, 2, 3, 4, 5, 6) is an abelian group under addition modulo 7.

Answer 1

1+2=3,4is taken common 5+6=11,11-4=7

Answer 2

Note :

• Group : An algebraic system (G,*) is said to be a group if the following condition are satisfied :

1. G is closed under *
2. G is associative under *
3. G has a unique identity element
4. Every element of G has a unique inverse in G

• Moreover , if a group (G,*) also holds commutative property , then it is called commutative group or abelian group .

Solution :

Given :

Z₇ = { 0 , 1 , 2 , 3 , 4 , 5 , 6 }

To prove :

Z₇ is an abelian group .

Proof :

For Cayley's table (composition table) please refer to the attachment .

1) Closure property :

All the elements of the composition table are the elements of Z₇ . ie. a +₇ b ∈ Z₇ ∀ a , b ∈ Z₇ .

2) Associative property :

We know that , the addition of integers is associative , thus a +₇ (b +₇ c) = (a +₇ b) +₇ c ∀ a , b , c ∈ Z₇ .

3) Existence of identity :

We have 0 ∈ Z₇ such that 0 +₇ a = a +₇ 0 = a ∀ a ∈ Z₇ .

Thus , 0 is the identity element in Z₇ .

4) Existence of inverse element :

∀ a ∈ Z₇ , there exists a⁻¹ ∈ Z₇ such that a +₇ a⁻¹ = a⁻¹ +₇ a = 0 where a⁻¹ is called the inverse of a .

Here ,

0⁻¹ = 0

1⁻¹ = 6

2⁻¹ = 5

3⁻¹ = 4

4⁻¹ = 3

5⁻¹ = 2

6⁻¹ = 1

5) Commutative property :

The Cayley's table is symmetrical about the principal diagonal , thus Z₇ is commutative , ie. a +₇ b = a +₇ b ∀ a , b ∈ Z₇ .

Hence , Z₇ is an abelian group .