Question about the proof of Second Isomorphism Theorem












4












$begingroup$


The Second Isomorphism Theorem:
Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
$$H/(Hcap N)cong(HN)/N$$



There is the proof of Abstract Algebra Thomas by W. Judson:




Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
$$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
$$HN/N=phi(H)cong H/kerphi$$
Since
$$kerphi={hin H:hin N}=Hcap N$$
$HN/N=phi(H)cong H/Hcap N$




My question:



Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.



Thank you.






abstract-algebra group-theory group-isomorphism group-homomorphism







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    4












    $begingroup$


    The Second Isomorphism Theorem:
    Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
    $$H/(Hcap N)cong(HN)/N$$



    There is the proof of Abstract Algebra Thomas by W. Judson:




    Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
    $$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
    By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
    $$HN/N=phi(H)cong H/kerphi$$
    Since
    $$kerphi={hin H:hin N}=Hcap N$$
    $HN/N=phi(H)cong H/Hcap N$




    My question:



    Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.



    Thank you.






    abstract-algebra group-theory group-isomorphism group-homomorphism







    share|cite|improve this question









    New contributor




    NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4





      $begingroup$


      The Second Isomorphism Theorem:
      Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
      $$H/(Hcap N)cong(HN)/N$$



      There is the proof of Abstract Algebra Thomas by W. Judson:




      Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
      $$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
      By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
      $$HN/N=phi(H)cong H/kerphi$$
      Since
      $$kerphi={hin H:hin N}=Hcap N$$
      $HN/N=phi(H)cong H/Hcap N$




      My question:



      Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.



      Thank you.






      abstract-algebra group-theory group-isomorphism group-homomorphism







      share|cite|improve this question









      New contributor




      NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      The Second Isomorphism Theorem:
      Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
      $$H/(Hcap N)cong(HN)/N$$



      There is the proof of Abstract Algebra Thomas by W. Judson:




      Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
      $$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
      By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
      $$HN/N=phi(H)cong H/kerphi$$
      Since
      $$kerphi={hin H:hin N}=Hcap N$$
      $HN/N=phi(H)cong H/Hcap N$




      My question:



      Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.



      Thank you.






      abstract-algebra group-theory group-isomorphism group-homomorphism




      abstract-algebra group-theory group-isomorphism group-homomorphism






      share|cite|improve this question









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      NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











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      edited 3 hours ago









      Andrews

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      asked 4 hours ago









      NiaBieNiaBie

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          $begingroup$

          The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.






          share|cite|improve this answer









          $endgroup$













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            active

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            $begingroup$

            The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.






                share|cite|improve this answer









                $endgroup$



                The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                Joshua MundingerJoshua Mundinger

                2,7621028




                2,7621028






















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