Prove G is Cyclic: Prime p Order Group

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1. Let p be a prime and G a group whose order is p. Prove that G is cyclic.



2. I know that if p is prime, then the only possible subgroups of G are {e} and G itself. But, how to use this fact to show that G is cyclic?
 
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Pick an element g of G that is not e and consider the subgroup generated by g.
 
HINT: Lagrange
 
Last edited:
Dick said:
Pick an element g of G that is not e and consider the subgroup generated by g.

Ok, this may sound stupid, but, how do I know that <g>=G?
 
catherinenanc said:
Ok, this may sound stupid, but, how do I know that <g>=G?

Oh, wait, there are only two subgroups,so it has to be.

Thanks!
 
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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