Let $N$ be an Abelian normal subgroup of $G$, if $G/N$ is perfect, then also $G'$ is perfect.












1












$begingroup$


I'm reading Kurzweil & Stellmacher's "The Theory of Finite Groups", its 1.5.3 says:




Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect,
then also $G'$ is perfect.



Proof. From 1.5.1, applied to the natural epimorphism, we obtain
$$G/N = (G/N)' = G'N/N$$



and thus $G = G'N$. Since also $G'/N cap G'$ ($cong G/N$) is perfect, the same
argument gives $G' = G''(N cap G')$. It follows that $G = G''N$ and $G/G'' cong
N/N cap G''$. Now 1.5.2 implies $G' = G''$ since $N$ is Abelian. $square$




I'm a bit lost here: why $G'/N cap G'$ is perfect and $G'/N cap G' cong G/N$?






group-theory abelian-groups normal-subgroups group-homomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is sufficient to show that $G'/Ncap G'cong G/N$, since $G/N$ is perfect by hypothesis.
    $endgroup$
    – Shaun
    Dec 29 '18 at 10:48






  • 1




    $begingroup$
    This looks like a job for an isomorphism theorem.
    $endgroup$
    – Shaun
    Dec 29 '18 at 10:50


















1












$begingroup$


I'm reading Kurzweil & Stellmacher's "The Theory of Finite Groups", its 1.5.3 says:




Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect,
then also $G'$ is perfect.



Proof. From 1.5.1, applied to the natural epimorphism, we obtain
$$G/N = (G/N)' = G'N/N$$



and thus $G = G'N$. Since also $G'/N cap G'$ ($cong G/N$) is perfect, the same
argument gives $G' = G''(N cap G')$. It follows that $G = G''N$ and $G/G'' cong
N/N cap G''$. Now 1.5.2 implies $G' = G''$ since $N$ is Abelian. $square$




I'm a bit lost here: why $G'/N cap G'$ is perfect and $G'/N cap G' cong G/N$?






group-theory abelian-groups normal-subgroups group-homomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is sufficient to show that $G'/Ncap G'cong G/N$, since $G/N$ is perfect by hypothesis.
    $endgroup$
    – Shaun
    Dec 29 '18 at 10:48






  • 1




    $begingroup$
    This looks like a job for an isomorphism theorem.
    $endgroup$
    – Shaun
    Dec 29 '18 at 10:50
















1












1








1





$begingroup$


I'm reading Kurzweil & Stellmacher's "The Theory of Finite Groups", its 1.5.3 says:




Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect,
then also $G'$ is perfect.



Proof. From 1.5.1, applied to the natural epimorphism, we obtain
$$G/N = (G/N)' = G'N/N$$



and thus $G = G'N$. Since also $G'/N cap G'$ ($cong G/N$) is perfect, the same
argument gives $G' = G''(N cap G')$. It follows that $G = G''N$ and $G/G'' cong
N/N cap G''$. Now 1.5.2 implies $G' = G''$ since $N$ is Abelian. $square$




I'm a bit lost here: why $G'/N cap G'$ is perfect and $G'/N cap G' cong G/N$?






group-theory abelian-groups normal-subgroups group-homomorphism







share|cite|improve this question











$endgroup$




I'm reading Kurzweil & Stellmacher's "The Theory of Finite Groups", its 1.5.3 says:




Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect,
then also $G'$ is perfect.



Proof. From 1.5.1, applied to the natural epimorphism, we obtain
$$G/N = (G/N)' = G'N/N$$



and thus $G = G'N$. Since also $G'/N cap G'$ ($cong G/N$) is perfect, the same
argument gives $G' = G''(N cap G')$. It follows that $G = G''N$ and $G/G'' cong
N/N cap G''$. Now 1.5.2 implies $G' = G''$ since $N$ is Abelian. $square$




I'm a bit lost here: why $G'/N cap G'$ is perfect and $G'/N cap G' cong G/N$?






group-theory abelian-groups normal-subgroups group-homomorphism




group-theory abelian-groups normal-subgroups group-homomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 10:50









Shaun

9,268113684




9,268113684










asked Dec 29 '18 at 10:34









athosathos

91511340




91511340












  • $begingroup$
    It is sufficient to show that $G'/Ncap G'cong G/N$, since $G/N$ is perfect by hypothesis.
    $endgroup$
    – Shaun
    Dec 29 '18 at 10:48






  • 1




    $begingroup$
    This looks like a job for an isomorphism theorem.
    $endgroup$
    – Shaun
    Dec 29 '18 at 10:50




















  • $begingroup$
    It is sufficient to show that $G'/Ncap G'cong G/N$, since $G/N$ is perfect by hypothesis.
    $endgroup$
    – Shaun
    Dec 29 '18 at 10:48






  • 1




    $begingroup$
    This looks like a job for an isomorphism theorem.
    $endgroup$
    – Shaun
    Dec 29 '18 at 10:50


















$begingroup$
It is sufficient to show that $G'/Ncap G'cong G/N$, since $G/N$ is perfect by hypothesis.
$endgroup$
– Shaun
Dec 29 '18 at 10:48




$begingroup$
It is sufficient to show that $G'/Ncap G'cong G/N$, since $G/N$ is perfect by hypothesis.
$endgroup$
– Shaun
Dec 29 '18 at 10:48




1




1




$begingroup$
This looks like a job for an isomorphism theorem.
$endgroup$
– Shaun
Dec 29 '18 at 10:50






$begingroup$
This looks like a job for an isomorphism theorem.
$endgroup$
– Shaun
Dec 29 '18 at 10:50












1 Answer
1






active

oldest

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4












$begingroup$

It's the second isomorphism theorem applied here:
$$G'/(Ncap G') cong G'N/N$$
and we already know $G'N/N=G/N$ which is perfect by hypothesis.



Consequently, by the same reasons as in the first part of the proof, we get $G'=G''N$, and thus $G=G'N=G''NN=G''N$.






share|cite|improve this answer









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    $begingroup$

    It's the second isomorphism theorem applied here:
    $$G'/(Ncap G') cong G'N/N$$
    and we already know $G'N/N=G/N$ which is perfect by hypothesis.



    Consequently, by the same reasons as in the first part of the proof, we get $G'=G''N$, and thus $G=G'N=G''NN=G''N$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      It's the second isomorphism theorem applied here:
      $$G'/(Ncap G') cong G'N/N$$
      and we already know $G'N/N=G/N$ which is perfect by hypothesis.



      Consequently, by the same reasons as in the first part of the proof, we get $G'=G''N$, and thus $G=G'N=G''NN=G''N$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        It's the second isomorphism theorem applied here:
        $$G'/(Ncap G') cong G'N/N$$
        and we already know $G'N/N=G/N$ which is perfect by hypothesis.



        Consequently, by the same reasons as in the first part of the proof, we get $G'=G''N$, and thus $G=G'N=G''NN=G''N$.






        share|cite|improve this answer









        $endgroup$



        It's the second isomorphism theorem applied here:
        $$G'/(Ncap G') cong G'N/N$$
        and we already know $G'N/N=G/N$ which is perfect by hypothesis.



        Consequently, by the same reasons as in the first part of the proof, we get $G'=G''N$, and thus $G=G'N=G''NN=G''N$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 11:25









        BerciBerci

        61.1k23674




        61.1k23674






























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