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

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### 2 Answers

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.

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I think sir the cube roots of 1 are 1, -1/2 + `(sqrt(3)/2)i and -1/2-(sqrt(3)/2)i`

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Thank you so much sir, I am taking home study program and my professor only minds on sending me problem sets. Sir, this is one of the 10 item-problems my professor gave me. Can you show if this question is a multiplicative group under different axioms you have mentioned. Please sir, i'm going to consider this as a pattern to answer the remaining questions.

Thanks.

It is group 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.**

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thanks sir, now i know i am correct with 3 cube roots of 1. I am a little bit confused when i saw different answer... I know i am on the right track. God bless!