Cyclic subgroups of an Abelian group

dancavallaro
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Homework Statement


If G is an Abelian group and contains cyclic subgroups of orders 4 and 6, what other sizes of cyclic subgroups must G contain?


Homework Equations


A cyclic group of order n has cyclic subgroups with orders corresponding to all of n's divisors.


The Attempt at a Solution


I know that a cyclic group of order n has cyclic subgroups with orders corresponding to all of n's divisors, so I'm inclined to say that G must have cyclic subgroups with orders 1, 2, and 3. But I also have a hunch that the subgroups of orders 4 and 6 combine in some way, so maybe there's also a cyclic subgroup of order lcm(4,6) = 12? Am I on the right track here?
 
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Yes you are. Since 4 and 6 divide n, so must 12. (Think of the prime factorization of n).
 
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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