Number of generators of subgroup












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 $G= mathbb{Z}$ and $H=2mathbb{Z}$.



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











$endgroup$












  • $begingroup$
    $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 $G= mathbb{Z}$ and $H=2mathbb{Z}$.



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











$endgroup$












  • $begingroup$
    $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 $G= mathbb{Z}$ and $H=2mathbb{Z}$.



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











$endgroup$




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 $G= mathbb{Z}$ and $H=2mathbb{Z}$.



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




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
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$begingroup$

No, it is not true. Consider $mathbb{Z}_2oplusmathbb{Z}_3$. This has a generator $(1,1)$. Note that
$$0oplusmathbb{Z}_3<mathbb{Z}_2oplusmathbb{Z}_3 ,$$
and
$$(mathbb{Z}_2oplus 0)oplus(0oplusmathbb{Z_3})=mathbb{Z}_2oplusmathbb{Z}_3.$$
However, $0oplusmathbb{Z}_3$ is generated by $(0,1).$






share|cite|improve this answer









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    $begingroup$

    No, it is not true. Consider $mathbb{Z}_2oplusmathbb{Z}_3$. This has a generator $(1,1)$. Note that
    $$0oplusmathbb{Z}_3<mathbb{Z}_2oplusmathbb{Z}_3 ,$$
    and
    $$(mathbb{Z}_2oplus 0)oplus(0oplusmathbb{Z_3})=mathbb{Z}_2oplusmathbb{Z}_3.$$
    However, $0oplusmathbb{Z}_3$ is generated by $(0,1).$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      No, it is not true. Consider $mathbb{Z}_2oplusmathbb{Z}_3$. This has a generator $(1,1)$. Note that
      $$0oplusmathbb{Z}_3<mathbb{Z}_2oplusmathbb{Z}_3 ,$$
      and
      $$(mathbb{Z}_2oplus 0)oplus(0oplusmathbb{Z_3})=mathbb{Z}_2oplusmathbb{Z}_3.$$
      However, $0oplusmathbb{Z}_3$ is generated by $(0,1).$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        No, it is not true. Consider $mathbb{Z}_2oplusmathbb{Z}_3$. This has a generator $(1,1)$. Note that
        $$0oplusmathbb{Z}_3<mathbb{Z}_2oplusmathbb{Z}_3 ,$$
        and
        $$(mathbb{Z}_2oplus 0)oplus(0oplusmathbb{Z_3})=mathbb{Z}_2oplusmathbb{Z}_3.$$
        However, $0oplusmathbb{Z}_3$ is generated by $(0,1).$






        share|cite|improve this answer









        $endgroup$



        No, it is not true. Consider $mathbb{Z}_2oplusmathbb{Z}_3$. This has a generator $(1,1)$. Note that
        $$0oplusmathbb{Z}_3<mathbb{Z}_2oplusmathbb{Z}_3 ,$$
        and
        $$(mathbb{Z}_2oplus 0)oplus(0oplusmathbb{Z_3})=mathbb{Z}_2oplusmathbb{Z}_3.$$
        However, $0oplusmathbb{Z}_3$ 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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