Are Abelian Groups of Relatively Prime Orders Isomorphic?

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Homework Statement



Let m and n be relatively prime positive integers. Show that if there are, up to isomorphism, r abelian groups of order m and s of order n, then there are rs abelian groups of order mn.

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The Attempt at a Solution



I'm not sure how to go about this. I was thinking of saying that since m and n are relatively prime, the gcd(m,n)=1; wouldn't this then imply that the group order would be mn? Because mn is the lcm of m and n?

Any help is appreciated.
 
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\hbox{gcd}(m,n)=1 \Rightarrow C_{mn}\cong C_{m} \times C_{n}.

HINT:1. If C_m=<a>, C_n=<b> then prove that C_{m} \times C_{n}=<d> where d=(a,b)
2. Cyclic groups same orders are isomorphic.
 
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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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