Number of generators of subgroup Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Torsion subgroupOn the minimal number of generators of a finite groupBound number of generators of a subgroup of a nilpotent group?Minimal number of generators for a finitely generated abelian $p$-groupA question on finitely generated Abelian groups with a minimal number of generatorsFactoring an Abelian groupThe number of internal direct summands of a finitely generated abelian groupFree group generated by two generators is isomorphic to product of two infinite cyclic groupsAlternative proof of the Fundamental Theorem of Abelian Groups??Hungerford Chapter 2 Section 2 Problem 2 WITHOUT using the structure theorem of finite abelian groups

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Number of generators of subgroup



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Torsion subgroupOn the minimal number of generators of a finite groupBound number of generators of a subgroup of a nilpotent group?Minimal number of generators for a finitely generated abelian $p$-groupA question on finitely generated Abelian groups with a minimal number of generatorsFactoring an Abelian groupThe number of internal direct summands of a finitely generated abelian groupFree group generated by two generators is isomorphic to product of two infinite cyclic groupsAlternative proof of the Fundamental Theorem of Abelian Groups??Hungerford Chapter 2 Section 2 Problem 2 WITHOUT using the structure theorem of finite abelian groups










1












$begingroup$


I am trying to prove the following.



let $G$ be a finitely generated abelian group, and $H<G$ a subgroup such that there exists a subgroup $K<G$ and we can write $G=H oplus K$. Is it true that the minimal number of generators of H is strictly smaller than the minimal number of generators of $G$?



Clearly if G can not be written as a direct summand of $H$ then this is not true, just consider $G= mathbbZ$ and $H=2mathbbZ$.



I would like to prove it because I believe it can provide a simpler proof for the characterization of finitely generated abelian groups.










share|cite|improve this question











$endgroup$











  • $begingroup$
    $mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
    $endgroup$
    – lulu
    5 hours ago






  • 2




    $begingroup$
    Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
    $endgroup$
    – lulu
    5 hours ago










  • $begingroup$
    Thank you for pointing that out. I will edit to correct it.
    $endgroup$
    – Charles
    5 hours ago















1












$begingroup$


I am trying to prove the following.



let $G$ be a finitely generated abelian group, and $H<G$ a subgroup such that there exists a subgroup $K<G$ and we can write $G=H oplus K$. Is it true that the minimal number of generators of H is strictly smaller than the minimal number of generators of $G$?



Clearly if G can not be written as a direct summand of $H$ then this is not true, just consider $G= mathbbZ$ and $H=2mathbbZ$.



I would like to prove it because I believe it can provide a simpler proof for the characterization of finitely generated abelian groups.










share|cite|improve this question











$endgroup$











  • $begingroup$
    $mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
    $endgroup$
    – lulu
    5 hours ago






  • 2




    $begingroup$
    Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
    $endgroup$
    – lulu
    5 hours ago










  • $begingroup$
    Thank you for pointing that out. I will edit to correct it.
    $endgroup$
    – Charles
    5 hours ago













1












1








1





$begingroup$


I am trying to prove the following.



let $G$ be a finitely generated abelian group, and $H<G$ a subgroup such that there exists a subgroup $K<G$ and we can write $G=H oplus K$. Is it true that the minimal number of generators of H is strictly smaller than the minimal number of generators of $G$?



Clearly if G can not be written as a direct summand of $H$ then this is not true, just consider $G= mathbbZ$ and $H=2mathbbZ$.



I would like to prove it because I believe it can provide a simpler proof for the characterization of finitely generated abelian groups.










share|cite|improve this question











$endgroup$




I am trying to prove the following.



let $G$ be a finitely generated abelian group, and $H<G$ a subgroup such that there exists a subgroup $K<G$ and we can write $G=H oplus K$. Is it true that the minimal number of generators of H is strictly smaller than the minimal number of generators of $G$?



Clearly if G can not be written as a direct summand of $H$ then this is not true, just consider $G= mathbbZ$ and $H=2mathbbZ$.



I would like to prove it because I believe it can provide a simpler proof for the characterization of finitely generated abelian groups.







group-theory abelian-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago







Charles

















asked 5 hours ago









CharlesCharles

582420




582420











  • $begingroup$
    $mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
    $endgroup$
    – lulu
    5 hours ago






  • 2




    $begingroup$
    Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
    $endgroup$
    – lulu
    5 hours ago










  • $begingroup$
    Thank you for pointing that out. I will edit to correct it.
    $endgroup$
    – Charles
    5 hours ago
















  • $begingroup$
    $mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
    $endgroup$
    – lulu
    5 hours ago






  • 2




    $begingroup$
    Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
    $endgroup$
    – lulu
    5 hours ago










  • $begingroup$
    Thank you for pointing that out. I will edit to correct it.
    $endgroup$
    – Charles
    5 hours ago















$begingroup$
$mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
$endgroup$
– lulu
5 hours ago




$begingroup$
$mathbb Zbig / 2mathbb Z oplus mathbb Zbig / 3mathbb Z $ is cyclic.
$endgroup$
– lulu
5 hours ago




2




2




$begingroup$
Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
$endgroup$
– lulu
5 hours ago




$begingroup$
Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so,
$endgroup$
– lulu
5 hours ago












$begingroup$
Thank you for pointing that out. I will edit to correct it.
$endgroup$
– Charles
5 hours ago




$begingroup$
Thank you for pointing that out. I will edit to correct it.
$endgroup$
– Charles
5 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

No, it is not true. Consider $mathbbZ_2oplusmathbbZ_3$. This has a generator $(1,1)$. Note that
$$0oplusmathbbZ_3<mathbbZ_2oplusmathbbZ_3 ,$$
and
$$(mathbbZ_2oplus 0)oplus(0oplusmathbbZ_3)=mathbbZ_2oplusmathbbZ_3.$$
However, $0oplusmathbbZ_3$ is generated by $(0,1).$






share|cite|improve this answer









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    1 Answer
    1






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    active

    oldest

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    4












    $begingroup$

    No, it is not true. Consider $mathbbZ_2oplusmathbbZ_3$. This has a generator $(1,1)$. Note that
    $$0oplusmathbbZ_3<mathbbZ_2oplusmathbbZ_3 ,$$
    and
    $$(mathbbZ_2oplus 0)oplus(0oplusmathbbZ_3)=mathbbZ_2oplusmathbbZ_3.$$
    However, $0oplusmathbbZ_3$ is generated by $(0,1).$






    share|cite|improve this answer









    $endgroup$

















      4












      $begingroup$

      No, it is not true. Consider $mathbbZ_2oplusmathbbZ_3$. This has a generator $(1,1)$. Note that
      $$0oplusmathbbZ_3<mathbbZ_2oplusmathbbZ_3 ,$$
      and
      $$(mathbbZ_2oplus 0)oplus(0oplusmathbbZ_3)=mathbbZ_2oplusmathbbZ_3.$$
      However, $0oplusmathbbZ_3$ is generated by $(0,1).$






      share|cite|improve this answer









      $endgroup$















        4












        4








        4





        $begingroup$

        No, it is not true. Consider $mathbbZ_2oplusmathbbZ_3$. This has a generator $(1,1)$. Note that
        $$0oplusmathbbZ_3<mathbbZ_2oplusmathbbZ_3 ,$$
        and
        $$(mathbbZ_2oplus 0)oplus(0oplusmathbbZ_3)=mathbbZ_2oplusmathbbZ_3.$$
        However, $0oplusmathbbZ_3$ is generated by $(0,1).$






        share|cite|improve this answer









        $endgroup$



        No, it is not true. Consider $mathbbZ_2oplusmathbbZ_3$. This has a generator $(1,1)$. Note that
        $$0oplusmathbbZ_3<mathbbZ_2oplusmathbbZ_3 ,$$
        and
        $$(mathbbZ_2oplus 0)oplus(0oplusmathbbZ_3)=mathbbZ_2oplusmathbbZ_3.$$
        However, $0oplusmathbbZ_3$ is generated by $(0,1).$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 5 hours ago









        MelodyMelody

        1,42212




        1,42212



























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