If $G'=G$, and suppose $N_G(P)$ is solvable for $Psubseteq G$, show that $ G$ is simple.












1












$begingroup$


Given a finite nontrivial group $G$, Suppose $N_G(P)$ is solvable for every nontrivial $Psubseteq G$ of prime power order. If $G'=G$, (where $G'$ is the derived subgroup) show that $G$ is simple.



I understand the pieces of information given in the problem but how to put them together to make an argument is challenging to me. I will appreciate any help.



I am thinking that, if I start by assuming $G$ is not simple and that there exists a normal nontrivial subgroup $ Ntriangleleft G$.



And if I can show that the commutator subgroup $G'=1$ then since $G'=G$ I get a contradiction to the nontriviality assumption of the group.



Is this a good way to go? How should I view it?
Thanks










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Choose $P in {rm Syl}_p(N)$ for some prime $p$ dividing $N$. Then by the Frattini Argument $G = NN_G(P)$ and $G/N cong N_G(P)/N_N(P)$. Now use the facts that $N_G(P)$ is solvable and $G'=G$ to deduce that $G=N$.
    $endgroup$
    – Derek Holt
    Dec 4 '18 at 10:05










  • $begingroup$
    @DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
    $endgroup$
    – Cnine
    Dec 5 '18 at 23:08










  • $begingroup$
    That's the second isomorphism theorem
    $endgroup$
    – Derek Holt
    Dec 6 '18 at 8:08
















1












$begingroup$


Given a finite nontrivial group $G$, Suppose $N_G(P)$ is solvable for every nontrivial $Psubseteq G$ of prime power order. If $G'=G$, (where $G'$ is the derived subgroup) show that $G$ is simple.



I understand the pieces of information given in the problem but how to put them together to make an argument is challenging to me. I will appreciate any help.



I am thinking that, if I start by assuming $G$ is not simple and that there exists a normal nontrivial subgroup $ Ntriangleleft G$.



And if I can show that the commutator subgroup $G'=1$ then since $G'=G$ I get a contradiction to the nontriviality assumption of the group.



Is this a good way to go? How should I view it?
Thanks










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Choose $P in {rm Syl}_p(N)$ for some prime $p$ dividing $N$. Then by the Frattini Argument $G = NN_G(P)$ and $G/N cong N_G(P)/N_N(P)$. Now use the facts that $N_G(P)$ is solvable and $G'=G$ to deduce that $G=N$.
    $endgroup$
    – Derek Holt
    Dec 4 '18 at 10:05










  • $begingroup$
    @DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
    $endgroup$
    – Cnine
    Dec 5 '18 at 23:08










  • $begingroup$
    That's the second isomorphism theorem
    $endgroup$
    – Derek Holt
    Dec 6 '18 at 8:08














1












1








1





$begingroup$


Given a finite nontrivial group $G$, Suppose $N_G(P)$ is solvable for every nontrivial $Psubseteq G$ of prime power order. If $G'=G$, (where $G'$ is the derived subgroup) show that $G$ is simple.



I understand the pieces of information given in the problem but how to put them together to make an argument is challenging to me. I will appreciate any help.



I am thinking that, if I start by assuming $G$ is not simple and that there exists a normal nontrivial subgroup $ Ntriangleleft G$.



And if I can show that the commutator subgroup $G'=1$ then since $G'=G$ I get a contradiction to the nontriviality assumption of the group.



Is this a good way to go? How should I view it?
Thanks










share|cite|improve this question











$endgroup$




Given a finite nontrivial group $G$, Suppose $N_G(P)$ is solvable for every nontrivial $Psubseteq G$ of prime power order. If $G'=G$, (where $G'$ is the derived subgroup) show that $G$ is simple.



I understand the pieces of information given in the problem but how to put them together to make an argument is challenging to me. I will appreciate any help.



I am thinking that, if I start by assuming $G$ is not simple and that there exists a normal nontrivial subgroup $ Ntriangleleft G$.



And if I can show that the commutator subgroup $G'=1$ then since $G'=G$ I get a contradiction to the nontriviality assumption of the group.



Is this a good way to go? How should I view it?
Thanks







abstract-algebra group-theory finite-groups






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share|cite|improve this question













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edited Dec 4 '18 at 9:39









the_fox

2,58711533




2,58711533










asked Dec 4 '18 at 9:13









CnineCnine

895




895








  • 3




    $begingroup$
    Choose $P in {rm Syl}_p(N)$ for some prime $p$ dividing $N$. Then by the Frattini Argument $G = NN_G(P)$ and $G/N cong N_G(P)/N_N(P)$. Now use the facts that $N_G(P)$ is solvable and $G'=G$ to deduce that $G=N$.
    $endgroup$
    – Derek Holt
    Dec 4 '18 at 10:05










  • $begingroup$
    @DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
    $endgroup$
    – Cnine
    Dec 5 '18 at 23:08










  • $begingroup$
    That's the second isomorphism theorem
    $endgroup$
    – Derek Holt
    Dec 6 '18 at 8:08














  • 3




    $begingroup$
    Choose $P in {rm Syl}_p(N)$ for some prime $p$ dividing $N$. Then by the Frattini Argument $G = NN_G(P)$ and $G/N cong N_G(P)/N_N(P)$. Now use the facts that $N_G(P)$ is solvable and $G'=G$ to deduce that $G=N$.
    $endgroup$
    – Derek Holt
    Dec 4 '18 at 10:05










  • $begingroup$
    @DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
    $endgroup$
    – Cnine
    Dec 5 '18 at 23:08










  • $begingroup$
    That's the second isomorphism theorem
    $endgroup$
    – Derek Holt
    Dec 6 '18 at 8:08








3




3




$begingroup$
Choose $P in {rm Syl}_p(N)$ for some prime $p$ dividing $N$. Then by the Frattini Argument $G = NN_G(P)$ and $G/N cong N_G(P)/N_N(P)$. Now use the facts that $N_G(P)$ is solvable and $G'=G$ to deduce that $G=N$.
$endgroup$
– Derek Holt
Dec 4 '18 at 10:05




$begingroup$
Choose $P in {rm Syl}_p(N)$ for some prime $p$ dividing $N$. Then by the Frattini Argument $G = NN_G(P)$ and $G/N cong N_G(P)/N_N(P)$. Now use the facts that $N_G(P)$ is solvable and $G'=G$ to deduce that $G=N$.
$endgroup$
– Derek Holt
Dec 4 '18 at 10:05












$begingroup$
@DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
$endgroup$
– Cnine
Dec 5 '18 at 23:08




$begingroup$
@DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
$endgroup$
– Cnine
Dec 5 '18 at 23:08












$begingroup$
That's the second isomorphism theorem
$endgroup$
– Derek Holt
Dec 6 '18 at 8:08




$begingroup$
That's the second isomorphism theorem
$endgroup$
– Derek Holt
Dec 6 '18 at 8:08










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