Question

prove that G={1,-1,i,-i} forms an abelian group with respect to multiplication​

Answer 1

Answer:

In the attachment, I have formed cayley's table to  verify the following axioms.

1.Closure axiom:

From the cayley's table, it is clear that

all the elements in ther cayleys table are

elements of G.

This implies G is closed under multiplication.

Hence closure axiom is true.

2.Associative axiom:

Multiplication of complex numbers is always associative.

Hence associative axiom is true .

3. Identity axiom:

Clearly the identity element is 1 and

1 ∈ G

4. Inverse axiom:

From the table,

Inverse of 1 is 1

Inverse of -1 is -1

Inverse of i is -i

Inverse of -i is i

since each element of G has unique inverse, Inverse axiom is true.

5. Commutative axiom:

Clearly multiplication of complex number is always commutative.

Hence commutative axiom is true.

Therefore, G is an abelian group under multiplication.

Answer 2

Answer:

prove that G={1,-1,i,-i} forms an abelian group with respect to multiplication