If $G'=G$, and suppose $N_G(P)$ is solvable for $Psubseteq G$, show that $ G$ is simple.
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
edited Dec 4 '18 at 9:39
the_fox
2,47211431
asked Dec 4 '18 at 9:13
CnineCnine
895
add a comment |
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
edited Dec 4 '18 at 9:39
the_fox
2,47211431
asked Dec 4 '18 at 9:13
CnineCnine
895
3
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 $G=N$.
– Derek Holt
Dec 4 '18 at 10:05
@DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
– Cnine
Dec 5 '18 at 23:08
That's the second isomorphism theorem
– Derek Holt
Dec 6 '18 at 8:08
add a comment |
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
edited Dec 4 '18 at 9:39
the_fox
2,47211431
asked Dec 4 '18 at 9:13
CnineCnine
895
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
abstract-algebra group-theory finite-groups
edited Dec 4 '18 at 9:39
the_fox
2,47211431
asked Dec 4 '18 at 9:13
CnineCnine
895
edited Dec 4 '18 at 9:39
the_fox
2,47211431
asked Dec 4 '18 at 9:13
CnineCnine
895
edited Dec 4 '18 at 9:39
the_fox
2,47211431
edited Dec 4 '18 at 9:39
the_fox
2,47211431
edited Dec 4 '18 at 9:39
the_fox
2,47211431
2,47211431
asked Dec 4 '18 at 9:13
CnineCnine
895
asked Dec 4 '18 at 9:13
CnineCnine
895
asked Dec 4 '18 at 9:13
CnineCnine
895
895
3
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 $G=N$.
– Derek Holt
Dec 4 '18 at 10:05
@DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
– Cnine
Dec 5 '18 at 23:08
That's the second isomorphism theorem
– Derek Holt
Dec 6 '18 at 8:08
add a comment |
3
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 $G=N$.
– Derek Holt
Dec 4 '18 at 10:05
@DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
– Cnine
Dec 5 '18 at 23:08
That's the second isomorphism theorem
– Derek Holt
Dec 6 '18 at 8:08
3
3
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 $G=N$.
– Derek Holt
Dec 4 '18 at 10:05
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 $G=N$.
– Derek Holt
Dec 4 '18 at 10:05
@DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
– Cnine
Dec 5 '18 at 23:08
@DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
– Cnine
Dec 5 '18 at 23:08
That's the second isomorphism theorem
– Derek Holt
Dec 6 '18 at 8:08
That's the second isomorphism theorem
– Derek Holt
Dec 6 '18 at 8:08
add a comment |
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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 $G=N$.
– Derek Holt
Dec 4 '18 at 10:05
@DerekHolt, thank you. but why is $G/N=N_G(P)/N_N(P)$.?
– Cnine
Dec 5 '18 at 23:08
That's the second isomorphism theorem
– Derek Holt
Dec 6 '18 at 8:08