Theorem of Finitely Generated Abelian Groups

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Homework Statement


Are the groups ##\mathbb{Z}_8 \times \mathbb{Z}_{10} \times \mathbb{Z}_{24}## and ##\mathbb{Z}_4 \times \mathbb{Z}_{12} \times \mathbb{Z}_{40}## isomorphic? Why or why not?

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The Attempt at a Solution


I think I am misunderstanding the Theorem of Finitely Generated Abelian Groups, because to me it seems that we can just decompose each direct product into ##\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5##, and so they are isomorphic. Why can't this be done, and why are they not isomorphic?
 
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Start small and simple. Draw up an addition table for ##\mathbb Z_4## and another for ##\mathbb Z_2\oplus\mathbb Z_2## and observe how they differ, preventing isomorphism.
 
andrewkirk said:
Start small and simple. Draw up an addition table for ##\mathbb Z_4## and another for ##\mathbb Z_2\oplus\mathbb Z_2## and observe how they differ, preventing isomorphism.
Oh right, because ##\mathbb{Z}_2 \times \mathbb{Z}_2## is isomorphic to the Klein-4 group, and the not ##\mathbb{Z}_4##.

So then how can I answer whether the two groups that I have are isomorphic in general?
 
One way to prove non-isomorphism is to look for an algebraic feature of one group that the other group does not have.
One way to prove isomorphism is to construct an isomorphism between the two.

You might try looking at the number of elements of order 8 in each of the two groups.
 
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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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