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Math Help - Uniqueness in the Fundamental Theorem of Finitely Generated Abelian Groups

  1. #1
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    Uniqueness in the Fundamental Theorem of Finitely Generated Abelian Groups

    The name of the theorem is almost as long as the proof!

    I am curious, does anyone has a good link, (I tried searching) that shows uniqueness. Meaning, I know the proof to show any non-trivial finitely generated abelian group is isomorphic to,
    \mathbb{Z}_{p_1^{a_1}}\times ... \times \mathbb{Z}_{p_n^{a_n}}\times \mathbb{Z}\times ... \times \mathbb{Z}.
    But, I am interested in knowing uniqueness.
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  2. #2
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    I don’t know of any links. But there is a good discussion of uniqueness is section 20 of THEORY of GROUPS by Kurosh. That is a very old book. You will need a fairly good mathematics library to find it.
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    I have made some progress on this.
    Thus, far I understand why the "Betti Number" (the number of full integer groups) is unique.
    It really is an elegant explanation.
    It turns out (and this is simple to show) that the torsion subgroup of G (finitely generated) is the product of all those primely factored groups. Thus, G/T(G)\simeq \mathbb{Z}\times ... \times \mathbb{Z}, which is free abelian. Thus, all ranks have the same cardinality and hence the number of Z's (Betti number) is fixed. Thus, showing uniqueness. Thus, the problem reduces to showing the uniquess of the factorization of the torsion subgroup.
    \mathbb{Z}_{p_1^{n_1}}\times ... \times \mathbb{Z}_{p_k^{n_k}}
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    I have made further progress.
    In the previous post it was shown that the number of the infinite groups was unique and the the only step now was to show uniquness of the torsian subgroup. The problem reduces to showing (all in the same prime) that if,
    \mathbb{Z}_{p^{a_1}}\times ... \times \mathbb{Z}_{p^{a_n}}\simeq \mathbb{Z}_{p^{b_1}}\times ... \times \mathbb{Z}_{p^{b_m}}
    And they are written in ascending power order then n=m and a_i=b_i. That is the representation is unique.
    I can show n=m but not the second part.
    Thus, far I see that if I can show that a_1=b_1 then I can "cancel" the cyclic group factor and apply the argument again.

    Plato you have any ideas?
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    TPH, I really wish that I could help you. It is clear to me that at this point you know far more about this then I do now. You see, I have not thought about this for years and years: not since I finish prelims in algebra. But I did go back and reread the section in Kurosh, the group theory text we used. It appears you have it correct. I will quote his summation.

    “If a finitely generated Abelian group is decomposed into the direct sum of indecomposable summands, then the number of infinite cyclic summands and the totality of the orders of the primary cyclic summands is independent of the decomposition, that is, of the choice of a basis.
    In other words, any two decompositions of a finitely generated Abelian into the direct sum of indecomposable cyclic groups are isomorphic.”
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    Quote Originally Posted by Plato View Post
    TPH, I really wish that I could help you. It is clear to me that at this point you know far more about this then I do now. You see, I have not thought about this for years and years: not since I finish prelims in algebra. But I did go back and reread the section in Kurosh, the group theory text we used. It appears you have it correct. I will quote his summation.

    “If a finitely generated Abelian group is decomposed into the direct sum of indecomposable summands, then the number of infinite cyclic summands and the totality of the orders of the primary cyclic summands is independent of the decomposition, that is, of the choice of a basis.
    In other words, any two decompositions of a finitely generated Abelian into the direct sum of indecomposable cyclic groups are isomorphic.”
    I finished the proof finally. I can post it (the uniqueness part) if you thus wish. My author left the proof as an excerise, had it when authors do that, especially when it is poorly explained.
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