Is the set of three cube roots of 1 a multiplicative group? Prove

Expert Answers
embizze eNotes educator| Certified Educator

The cube roots of 1 are `{1,-1/2+3/2i,-1/2-3/2i}`

In order for this set to be a group under multiplication it mustobey the following axioms; closure, associativity, the existence of a unique identity, and every element must have a unique inverse.

The given set is not a group as it is not closed under multiplication.

`(-1/2+3/2i)(-1/2-3/2i)=5/2` which is not an element of the set.

pramodpandey | Student

It is groupĀ  under ordinary multplication ( you may ref Topics in Algebra by I. N. Herstein ,Sarlang or Jacobson Nathan)

`w^3=1`

`w^3-1=0`

`(w-1)(w^2+w+1)=0`

`w=1 ,(-1+sqrt(3)i)/2 ,(-1-sqrt(3)i)/2`

`say`

`w=omega,omega^2,1`

`where`

`omega=(-1+sqrt(3)i)/2`

`omega^2=(-1-sqrt(3)i)/2`

` `

`(i) closed`

(ii) Associative

(iii) existance of identity=1

(iv) inverse exist for every element as `omega and omega^2` are inverse of each other.

Moreover it is an abelian group.