Abstract Algebra and cyclic subgroups

brydustin
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Homework Statement


from Algebra by Michael Artin, chapter 2, question 5 under section 2(subgroups)

An nth root of unity is a complex number z such that z^n =1. Prove that the nth roots of unity form a cyclic subgroup of C^(x) (the complex numbers under multiplication) of order n.


Homework Equations



cyclic subgroup = {z,z^2, ... , z^(n-1), z^n = 1}

Closed under multiplication, is associative, has an identity and is cyclic (z^(n+b) = z^b becausse z^n = 1).

The Attempt at a Solution



What's the point of this problem? Have I "proven" it, I'm trying to prepare for GRE and I am reviewing algebra (I'm a graduate, this isn't actually homework), but I don't see how its not immediately obvious that its a cyclic subgroup (put another way: What is Artin trying to show his readers?)
Thanks
 
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You left out the requirement that every element has an inverse. You also have the definition of cyclic wrong (it's that every element can be written as g^i for some element g that you haven't identified).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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