Isomorphism to C_n with n prime

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Homework Statement



Prove taht if the order n of a group G is a prime number, then G must be isomorphic to the cyclic group fo order n, C_n.

The Attempt at a Solution



We have previously proven that a group can can be written as S = \{A,A^2,A^3,A^4...,A^n = E\} where E is the identity and the group is of order n. We also have Lagrange that tells us, in this case, the order of every element in S is n if the order is prime.

So let's say we have the group G = \{A, A^2, A^3,..,A^g\} where G is of order g which is prime and the cyclic group C_n = \{C, C^2, C^3,...,C^n\} where n is again the prime order of the group. By this we know that A^m \ne E , m<g and similarly C^m \ne E ,m<n.

Now it seems like you can make an absolutely arbitrary 1 to 1 mapping from G -> C_n, so is my best bet to try to prove that it's possible to make a non-1to1 mapping and show that it must not work?
 
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How much data about a homomorphism \varphi: G \to C_n do you need to completely determine \varphi? It isn't much.
 
I suppose you just need to map one element and then by construction, the rest is figured out. That is, if I pick some A_1 \epsilon G and map it to some C_1 \epsilon C_n, then I should be able to say A_1^2 = A_2 = C_1^2=C_2 which gives me a unique mapping I would think.
 
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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