Is H a Free Commutative Group of Rank n in Z^n?

charlamov
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show that H is subgroup of finite index in Z^n exactly when H is free comutative group of rank n
 
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charlamov said:
show that H is subgroup of finite index in Z^n exactly when H is free comutative group of rank n


What've you done? Do you know the elementary divisors theorem for fin. gen, abelian groups?

DonAntonio
 
charlamov said:
show that H is subgroup of finite index in Z^n exactly when H is free comutative group of rank n



THis question doesn't belong at all in science education, but in "Abstract Algebra" in the math department
 

Thread 'Finding the nth roots of a complex number'

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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