What is the order of an element in a finite abelian group of odd order?

arshavin
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Homework Statement



Let G be a finite abelian group of odd order. Prove that the product of all the elements
of G is the identity.




The Attempt at a Solution



easy to see the case when each element has inverse which is not itself.
 
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Indeed, if every element has an inverse which is not itself, then this is true. So, what you need to do is actually show that no element has itself as inverse. So, assume that g has itself as inverse, then what is the order of g?
 
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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