# Cyclic group/ subgroup proof

1. Aug 19, 2010

### murmillo

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

cyclic group, subgroup

3. 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?

2. Aug 19, 2010

### murmillo

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?