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I know that the three cube roots of one are 1, `[1+sqrt(3)i]/(2)` `and...

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jolas | eNoter

Posted June 14, 2013 at 12:24 PM via web

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I know that the three cube roots of one are 1, `[1+sqrt(3)i]/(2)`

`and [1-sqrt(3)i]/(2)`

but how can i show whether they are multiplicative group? Please show me how they obey the four group axioms such as closure, associativity, identity and inverse.

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted June 14, 2013 at 2:48 PM (Answer #1)

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Show that the three roots of unity form a multiplicative group:

We have the set `{1,1/2+sqrt(3)/2 i,1/2-sqrt(3)/2 i}` and the binary operation of multiplication in the complex numbers.

(a) Closure: Since 1 is the identity for multiplication in the complex numbers (or 1 + 0i if you prefer), and multiplication in the complex numbers is commutative, it suffices to show that `(1/2+sqrt(3)/2 i)(1/2 - sqrt(3)/2 i)=1/4 -(-3/4)=1` is in the set. Thus the set is closed under multiplication.

** Alternatively you could create a Cayley table and show that all six products are in the group. **

(b) Multiplication is associative in the complex numbers.

** If you label the set {a,b,c} you could show that for all six possible permutations of a,b,c that `(a*b)*c=a*(b*c)`

e.g. `(a*b)*c=a*(b*c)`

`(b*a)*c=b*(a*c)` etc...

(c) The identity for complex multiplication is 1 which is an element of the set.

(d) Inverses: 1 is its own inverse. We need to show that `(1/2+sqrt(3)/2 i)(1/2-sqrt(3)/2 i)=(1/2-sqrt(3)/2 i)(1/2+sqrt(3)/2 i)=1`

Thus every element has an inverse. `(1/2+sqrt(3)/2 i)` is the multiplicative inverse of `(1/2 - sqrt(3)/2 i)` and vice versa.

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