For which abelian groups $G$ is there a short exact sequence $0 rightarrow mathbb{Z}/p^2 rightarrow G...












4












$begingroup$


I am trying to find for which abelian groups $G$ is there a short exact sequence.



$0 rightarrow mathbb{Z}/p^2 rightarrow G rightarrow mathbb{Z}/p^2 rightarrow 0$?



I have reasoned as follows: consider the sequence with the maps as follows $$0 xrightarrow{phi} mathbb{Z}/p^2 xrightarrow{f} G xrightarrow{g} mathbb{Z}/p^2 xrightarrow{psi} 0$$



Since we want the sequence to be exact, we need that $ker f= im(phi) = 0$ (so $f$ is injective) and we need that $im(g) = ker (psi) = mathbb{Z}/p^2$ (so $g$ is surjective). On top of that, we know that $im( f) = ker g$. Then, by the first isomorphism theorem, we know that $mathbb{Z}/p^2 / ker f = mathbb{Z}/p^2 cong im(f) = ker g$. So we know $ker g = mathbb{Z}/p^2$ and $im(g) = mathbb{Z}/p^2$. This leads me to think that one possibility for $G$ is $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$. However, the answers say that other possibilities could be $mathbb{Z}/p^4$ and $mathbb{Z}/p^3 oplus mathbb{Z}/p$ and I don't see how they came with this possibility, can anyone explain this solution? Also, is my argument for why $G$ can be $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$ correct? Thanks for your help!






abstract-algebra abelian-groups exact-sequence







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  • 1




    $begingroup$
    Use the classification of f.g. abelian groups.
    $endgroup$
    – anomaly
    Dec 27 '18 at 1:51
















4












$begingroup$


I am trying to find for which abelian groups $G$ is there a short exact sequence.



$0 rightarrow mathbb{Z}/p^2 rightarrow G rightarrow mathbb{Z}/p^2 rightarrow 0$?



I have reasoned as follows: consider the sequence with the maps as follows $$0 xrightarrow{phi} mathbb{Z}/p^2 xrightarrow{f} G xrightarrow{g} mathbb{Z}/p^2 xrightarrow{psi} 0$$



Since we want the sequence to be exact, we need that $ker f= im(phi) = 0$ (so $f$ is injective) and we need that $im(g) = ker (psi) = mathbb{Z}/p^2$ (so $g$ is surjective). On top of that, we know that $im( f) = ker g$. Then, by the first isomorphism theorem, we know that $mathbb{Z}/p^2 / ker f = mathbb{Z}/p^2 cong im(f) = ker g$. So we know $ker g = mathbb{Z}/p^2$ and $im(g) = mathbb{Z}/p^2$. This leads me to think that one possibility for $G$ is $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$. However, the answers say that other possibilities could be $mathbb{Z}/p^4$ and $mathbb{Z}/p^3 oplus mathbb{Z}/p$ and I don't see how they came with this possibility, can anyone explain this solution? Also, is my argument for why $G$ can be $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$ correct? Thanks for your help!






abstract-algebra abelian-groups exact-sequence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Use the classification of f.g. abelian groups.
    $endgroup$
    – anomaly
    Dec 27 '18 at 1:51














4












4








4


1



$begingroup$


I am trying to find for which abelian groups $G$ is there a short exact sequence.



$0 rightarrow mathbb{Z}/p^2 rightarrow G rightarrow mathbb{Z}/p^2 rightarrow 0$?



I have reasoned as follows: consider the sequence with the maps as follows $$0 xrightarrow{phi} mathbb{Z}/p^2 xrightarrow{f} G xrightarrow{g} mathbb{Z}/p^2 xrightarrow{psi} 0$$



Since we want the sequence to be exact, we need that $ker f= im(phi) = 0$ (so $f$ is injective) and we need that $im(g) = ker (psi) = mathbb{Z}/p^2$ (so $g$ is surjective). On top of that, we know that $im( f) = ker g$. Then, by the first isomorphism theorem, we know that $mathbb{Z}/p^2 / ker f = mathbb{Z}/p^2 cong im(f) = ker g$. So we know $ker g = mathbb{Z}/p^2$ and $im(g) = mathbb{Z}/p^2$. This leads me to think that one possibility for $G$ is $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$. However, the answers say that other possibilities could be $mathbb{Z}/p^4$ and $mathbb{Z}/p^3 oplus mathbb{Z}/p$ and I don't see how they came with this possibility, can anyone explain this solution? Also, is my argument for why $G$ can be $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$ correct? Thanks for your help!






abstract-algebra abelian-groups exact-sequence







share|cite|improve this question









$endgroup$




I am trying to find for which abelian groups $G$ is there a short exact sequence.



$0 rightarrow mathbb{Z}/p^2 rightarrow G rightarrow mathbb{Z}/p^2 rightarrow 0$?



I have reasoned as follows: consider the sequence with the maps as follows $$0 xrightarrow{phi} mathbb{Z}/p^2 xrightarrow{f} G xrightarrow{g} mathbb{Z}/p^2 xrightarrow{psi} 0$$



Since we want the sequence to be exact, we need that $ker f= im(phi) = 0$ (so $f$ is injective) and we need that $im(g) = ker (psi) = mathbb{Z}/p^2$ (so $g$ is surjective). On top of that, we know that $im( f) = ker g$. Then, by the first isomorphism theorem, we know that $mathbb{Z}/p^2 / ker f = mathbb{Z}/p^2 cong im(f) = ker g$. So we know $ker g = mathbb{Z}/p^2$ and $im(g) = mathbb{Z}/p^2$. This leads me to think that one possibility for $G$ is $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$. However, the answers say that other possibilities could be $mathbb{Z}/p^4$ and $mathbb{Z}/p^3 oplus mathbb{Z}/p$ and I don't see how they came with this possibility, can anyone explain this solution? Also, is my argument for why $G$ can be $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$ correct? Thanks for your help!






abstract-algebra abelian-groups exact-sequence




abstract-algebra abelian-groups exact-sequence






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asked Dec 27 '18 at 1:21









BOlivianoperuano84BOlivianoperuano84

1778




1778








  • 1




    $begingroup$
    Use the classification of f.g. abelian groups.
    $endgroup$
    – anomaly
    Dec 27 '18 at 1:51














  • 1




    $begingroup$
    Use the classification of f.g. abelian groups.
    $endgroup$
    – anomaly
    Dec 27 '18 at 1:51








1




1




$begingroup$
Use the classification of f.g. abelian groups.
$endgroup$
– anomaly
Dec 27 '18 at 1:51




$begingroup$
Use the classification of f.g. abelian groups.
$endgroup$
– anomaly
Dec 27 '18 at 1:51










1 Answer
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$begingroup$

This short exact sequence is a fancy way of saying that $G$ has a subgroup $H$ such that $H$ and $G / H$ are both isomorphic to $mathbb Z / p^2$.



[Think of $f$ as the embedding of $H$ into $G$, and think of $g$ as the projection of $G$ onto $G/H$. It's clear that $f$ is injective, and $g$ is surjective, and ${rm im}(f) = {rm ker}(g)$, which is what your short exact sequence says.]



For example:




  • $G = mathbb Z / p^2 oplus mathbb Z / p^2$, with $H$ generated by $(1, 0)$.


  • $G = mathbb Z / p^4$, with $H$ generated by $p^2$.


  • $G = mathbb Z / p^3 oplus mathbb Z / p$, with $H$ generated by $(p, 1)$.







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    $begingroup$

    This short exact sequence is a fancy way of saying that $G$ has a subgroup $H$ such that $H$ and $G / H$ are both isomorphic to $mathbb Z / p^2$.



    [Think of $f$ as the embedding of $H$ into $G$, and think of $g$ as the projection of $G$ onto $G/H$. It's clear that $f$ is injective, and $g$ is surjective, and ${rm im}(f) = {rm ker}(g)$, which is what your short exact sequence says.]



    For example:




    • $G = mathbb Z / p^2 oplus mathbb Z / p^2$, with $H$ generated by $(1, 0)$.


    • $G = mathbb Z / p^4$, with $H$ generated by $p^2$.


    • $G = mathbb Z / p^3 oplus mathbb Z / p$, with $H$ generated by $(p, 1)$.







    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      This short exact sequence is a fancy way of saying that $G$ has a subgroup $H$ such that $H$ and $G / H$ are both isomorphic to $mathbb Z / p^2$.



      [Think of $f$ as the embedding of $H$ into $G$, and think of $g$ as the projection of $G$ onto $G/H$. It's clear that $f$ is injective, and $g$ is surjective, and ${rm im}(f) = {rm ker}(g)$, which is what your short exact sequence says.]



      For example:




      • $G = mathbb Z / p^2 oplus mathbb Z / p^2$, with $H$ generated by $(1, 0)$.


      • $G = mathbb Z / p^4$, with $H$ generated by $p^2$.


      • $G = mathbb Z / p^3 oplus mathbb Z / p$, with $H$ generated by $(p, 1)$.







      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        This short exact sequence is a fancy way of saying that $G$ has a subgroup $H$ such that $H$ and $G / H$ are both isomorphic to $mathbb Z / p^2$.



        [Think of $f$ as the embedding of $H$ into $G$, and think of $g$ as the projection of $G$ onto $G/H$. It's clear that $f$ is injective, and $g$ is surjective, and ${rm im}(f) = {rm ker}(g)$, which is what your short exact sequence says.]



        For example:




        • $G = mathbb Z / p^2 oplus mathbb Z / p^2$, with $H$ generated by $(1, 0)$.


        • $G = mathbb Z / p^4$, with $H$ generated by $p^2$.


        • $G = mathbb Z / p^3 oplus mathbb Z / p$, with $H$ generated by $(p, 1)$.







        share|cite|improve this answer









        $endgroup$



        This short exact sequence is a fancy way of saying that $G$ has a subgroup $H$ such that $H$ and $G / H$ are both isomorphic to $mathbb Z / p^2$.



        [Think of $f$ as the embedding of $H$ into $G$, and think of $g$ as the projection of $G$ onto $G/H$. It's clear that $f$ is injective, and $g$ is surjective, and ${rm im}(f) = {rm ker}(g)$, which is what your short exact sequence says.]



        For example:




        • $G = mathbb Z / p^2 oplus mathbb Z / p^2$, with $H$ generated by $(1, 0)$.


        • $G = mathbb Z / p^4$, with $H$ generated by $p^2$.


        • $G = mathbb Z / p^3 oplus mathbb Z / p$, with $H$ generated by $(p, 1)$.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 1:39









        Kenny WongKenny Wong

        19.1k21441




        19.1k21441






























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