Proving the Existence of One Cyclic Subgroup in a Cyclic Group of Order n

murmillo
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Homework Statement


Let G be a cyclic group of order n, and let r be an integer dividing n. Prove that G contains exactly one subgroup of order r.


Homework Equations



cyclic group, subgroup

The Attempt at a Solution


Say the group G is {x^0, x^1, ..., x^(n-1)}
If there is a subgroup H of order r, it must be cyclic, because: why? I can't figure it out, but I have a feeling that it must be cyclic.

H is generated by some element, call it b=x^m. Since x^r = 0, we have (x^m)(x^m)... (r times) = 0. Thus mr=n and H must be the cyclic group generated by x^(n/r).

I have a feeling that I have the right idea but I don't know how to show that a group is cyclic. Could someone help?
 
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Oh wait, I can use the fact that the cyclic subgroup G is isomorphic to Z/nZ with additive law of composition. Then can I say that any subgroup of Z/nZ must be in the form Z/mZ and then use the isomorphism idea to get back to H being a cyclic subgroup?
 
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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