Understanding rules on an Abelian group decomposition











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In Aluffi's book Algebra, just from possibility of writing an abelian group G isomorphic to $langle grangle oplus G/langle grangle$ it concludes that by induction $G cong mathbb{Z}/d_1mathbb{Z} oplus dots mathbb{Z}/d_nmathbb{Z}$ with specific rules on $d_i$'s!



1- So why not $G cong mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_1mathbb{Z} dots mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_2mathbb{Z} dots mathbb{Z}/p_2mathbb{Z} oplus dots dots dots mathbb{Z}/p_nmathbb{Z}$?



2- Why $d_i|d_{i+1}$? (a must/theorem or a choice/standard-rule?)



3- And if writing them as powers of primes so why following the below steps?
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    In Aluffi's book Algebra, just from possibility of writing an abelian group G isomorphic to $langle grangle oplus G/langle grangle$ it concludes that by induction $G cong mathbb{Z}/d_1mathbb{Z} oplus dots mathbb{Z}/d_nmathbb{Z}$ with specific rules on $d_i$'s!



    1- So why not $G cong mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_1mathbb{Z} dots mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_2mathbb{Z} dots mathbb{Z}/p_2mathbb{Z} oplus dots dots dots mathbb{Z}/p_nmathbb{Z}$?



    2- Why $d_i|d_{i+1}$? (a must/theorem or a choice/standard-rule?)



    3- And if writing them as powers of primes so why following the below steps?
    enter image description here










    share|cite|improve this question

















    This question has an open bounty worth +50
    reputation from 72D ending in 6 days.


    The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.


















      up vote
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      In Aluffi's book Algebra, just from possibility of writing an abelian group G isomorphic to $langle grangle oplus G/langle grangle$ it concludes that by induction $G cong mathbb{Z}/d_1mathbb{Z} oplus dots mathbb{Z}/d_nmathbb{Z}$ with specific rules on $d_i$'s!



      1- So why not $G cong mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_1mathbb{Z} dots mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_2mathbb{Z} dots mathbb{Z}/p_2mathbb{Z} oplus dots dots dots mathbb{Z}/p_nmathbb{Z}$?



      2- Why $d_i|d_{i+1}$? (a must/theorem or a choice/standard-rule?)



      3- And if writing them as powers of primes so why following the below steps?
      enter image description here










      share|cite|improve this question















      In Aluffi's book Algebra, just from possibility of writing an abelian group G isomorphic to $langle grangle oplus G/langle grangle$ it concludes that by induction $G cong mathbb{Z}/d_1mathbb{Z} oplus dots mathbb{Z}/d_nmathbb{Z}$ with specific rules on $d_i$'s!



      1- So why not $G cong mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_1mathbb{Z} dots mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_2mathbb{Z} dots mathbb{Z}/p_2mathbb{Z} oplus dots dots dots mathbb{Z}/p_nmathbb{Z}$?



      2- Why $d_i|d_{i+1}$? (a must/theorem or a choice/standard-rule?)



      3- And if writing them as powers of primes so why following the below steps?
      enter image description here







      abstract-algebra group-theory abelian-groups






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      edited 2 days ago









      Arturo Magidin

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      asked Nov 30 at 0:26









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      This question has an open bounty worth +50
      reputation from 72D ending in 6 days.


      The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.








      This question has an open bounty worth +50
      reputation from 72D ending in 6 days.


      The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
























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          I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $mathbb{Z}^r$ in the decomposition.



          1) We may not have necessarily have prime orders. Note $<g> cong mathbb{Z}/dmathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $mathbb{Z}/4mathbb{Z} notcong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$.



          2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds




          1. Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.






          share|cite|improve this answer























          • But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
            – André 3000
            17 hours ago












          • @André3000 Yes you're right. I'll edit.
            – Joel Pereira
            12 hours ago











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          1 Answer
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          1 Answer
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          active

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          up vote
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          I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $mathbb{Z}^r$ in the decomposition.



          1) We may not have necessarily have prime orders. Note $<g> cong mathbb{Z}/dmathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $mathbb{Z}/4mathbb{Z} notcong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$.



          2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds




          1. Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.






          share|cite|improve this answer























          • But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
            – André 3000
            17 hours ago












          • @André3000 Yes you're right. I'll edit.
            – Joel Pereira
            12 hours ago















          up vote
          0
          down vote













          I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $mathbb{Z}^r$ in the decomposition.



          1) We may not have necessarily have prime orders. Note $<g> cong mathbb{Z}/dmathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $mathbb{Z}/4mathbb{Z} notcong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$.



          2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds




          1. Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.






          share|cite|improve this answer























          • But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
            – André 3000
            17 hours ago












          • @André3000 Yes you're right. I'll edit.
            – Joel Pereira
            12 hours ago













          up vote
          0
          down vote










          up vote
          0
          down vote









          I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $mathbb{Z}^r$ in the decomposition.



          1) We may not have necessarily have prime orders. Note $<g> cong mathbb{Z}/dmathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $mathbb{Z}/4mathbb{Z} notcong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$.



          2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds




          1. Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.






          share|cite|improve this answer














          I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $mathbb{Z}^r$ in the decomposition.



          1) We may not have necessarily have prime orders. Note $<g> cong mathbb{Z}/dmathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $mathbb{Z}/4mathbb{Z} notcong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$.



          2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds




          1. Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 11 hours ago









          the_fox

          2,2791430




          2,2791430










          answered 20 hours ago









          Joel Pereira

          43618




          43618












          • But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
            – André 3000
            17 hours ago












          • @André3000 Yes you're right. I'll edit.
            – Joel Pereira
            12 hours ago


















          • But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
            – André 3000
            17 hours ago












          • @André3000 Yes you're right. I'll edit.
            – Joel Pereira
            12 hours ago
















          But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
          – André 3000
          17 hours ago






          But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
          – André 3000
          17 hours ago














          @André3000 Yes you're right. I'll edit.
          – Joel Pereira
          12 hours ago




          @André3000 Yes you're right. I'll edit.
          – Joel Pereira
          12 hours ago


















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