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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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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.

Posted by jolas on June 14, 2013 at 12:24 PM via web and tagged with abstract algebra, algebra, group, math, modern math

High School Teacher

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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.

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