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Let $A unlhd G$, then $n_p(G)=n_p(G/A)n_p(PA)$ for $Pin Syl_p(G)$ and $n_p(PA)=|A:N_A(P)|$
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$n_p(G)=n_p(G/A)n_p(PA)$ for $Pin Syl_p(G)$ and $n_p(PA)=|A:N_A(P)|$
This result is from the paper Guo, W., & Vdovin, E. P. (2018). Number of Sylow subgroups in finite groups. Journal of Group Theory, 21(4), 695-712.
The proof is as follows:
Consider a natural homomorphism $bar:Grightarrow overlineG=G/A$. Clearly, $n_p(PA)$ is the number of all Sylow $p$-subgroups $Q$ of $G$ such that $overlineP=overlineQ$ since $overlineP=overlineQ$ if and only if $Qleq PA$. By Sylow theorem, for every $Qin Syl_p(G)$, there exists $xin G$ such that $P^x=Q$, so $(PA)^x=QA$. This shows that $n_p(PA)$ does not depend on the choice of $P$, and so it is the same for any Sylow $p$-subgroups of $G$. Thus we obtain $n_p(G)=n_p(G/A)n_p(PA)$. By Sylow theorem, $$n_p(PA)=|PA:N_PA(P)|=|A:N_A(P)|$$ where the second identity follows from the Dedekind theorem.
Several questions/problems I have:
(i)If $Qleq PA$, I can only show that $overlineQleq overlineP$. I can't see why $overlinePleq overlineQ$ in this case.
(ii)I can't see why the previous argument would directly lead to the conclusion $n_p(G)=n_p(G/A)n_p(PA)$.
(iii)I search a lot about the Dedekind theorem mentioned in the text but the result I found is for continuity on real axis or for Galois group.
abstract-algebra group-theory sylow-theory
edited Aug 6 at 7:30
Nicky Hekster
26.9k53052
asked Aug 6 at 3:23
Alan Wang
4,431932
add a comment |Â
up vote
2
down vote
favorite
$n_p(G)=n_p(G/A)n_p(PA)$ for $Pin Syl_p(G)$ and $n_p(PA)=|A:N_A(P)|$
This result is from the paper Guo, W., & Vdovin, E. P. (2018). Number of Sylow subgroups in finite groups. Journal of Group Theory, 21(4), 695-712.
The proof is as follows:
Consider a natural homomorphism $bar:Grightarrow overlineG=G/A$. Clearly, $n_p(PA)$ is the number of all Sylow $p$-subgroups $Q$ of $G$ such that $overlineP=overlineQ$ since $overlineP=overlineQ$ if and only if $Qleq PA$. By Sylow theorem, for every $Qin Syl_p(G)$, there exists $xin G$ such that $P^x=Q$, so $(PA)^x=QA$. This shows that $n_p(PA)$ does not depend on the choice of $P$, and so it is the same for any Sylow $p$-subgroups of $G$. Thus we obtain $n_p(G)=n_p(G/A)n_p(PA)$. By Sylow theorem, $$n_p(PA)=|PA:N_PA(P)|=|A:N_A(P)|$$ where the second identity follows from the Dedekind theorem.
Several questions/problems I have:
(i)If $Qleq PA$, I can only show that $overlineQleq overlineP$. I can't see why $overlinePleq overlineQ$ in this case.
(ii)I can't see why the previous argument would directly lead to the conclusion $n_p(G)=n_p(G/A)n_p(PA)$.
(iii)I search a lot about the Dedekind theorem mentioned in the text but the result I found is for continuity on real axis or for Galois group.
abstract-algebra group-theory sylow-theory
edited Aug 6 at 7:30
Nicky Hekster
26.9k53052
asked Aug 6 at 3:23
Alan Wang
4,431932
Dedekind theorem is if $Hle K$ and $L$ are subgroups, then $HLcap K=H(Lcap K)$ as sets.
– Steve D
Aug 6 at 3:29
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$n_p(G)=n_p(G/A)n_p(PA)$ for $Pin Syl_p(G)$ and $n_p(PA)=|A:N_A(P)|$
This result is from the paper Guo, W., & Vdovin, E. P. (2018). Number of Sylow subgroups in finite groups. Journal of Group Theory, 21(4), 695-712.
The proof is as follows:
Consider a natural homomorphism $bar:Grightarrow overlineG=G/A$. Clearly, $n_p(PA)$ is the number of all Sylow $p$-subgroups $Q$ of $G$ such that $overlineP=overlineQ$ since $overlineP=overlineQ$ if and only if $Qleq PA$. By Sylow theorem, for every $Qin Syl_p(G)$, there exists $xin G$ such that $P^x=Q$, so $(PA)^x=QA$. This shows that $n_p(PA)$ does not depend on the choice of $P$, and so it is the same for any Sylow $p$-subgroups of $G$. Thus we obtain $n_p(G)=n_p(G/A)n_p(PA)$. By Sylow theorem, $$n_p(PA)=|PA:N_PA(P)|=|A:N_A(P)|$$ where the second identity follows from the Dedekind theorem.
Several questions/problems I have:
(i)If $Qleq PA$, I can only show that $overlineQleq overlineP$. I can't see why $overlinePleq overlineQ$ in this case.
(ii)I can't see why the previous argument would directly lead to the conclusion $n_p(G)=n_p(G/A)n_p(PA)$.
(iii)I search a lot about the Dedekind theorem mentioned in the text but the result I found is for continuity on real axis or for Galois group.
abstract-algebra group-theory sylow-theory
edited Aug 6 at 7:30
Nicky Hekster
26.9k53052
asked Aug 6 at 3:23
Alan Wang
4,431932
$n_p(G)=n_p(G/A)n_p(PA)$ for $Pin Syl_p(G)$ and $n_p(PA)=|A:N_A(P)|$
This result is from the paper Guo, W., & Vdovin, E. P. (2018). Number of Sylow subgroups in finite groups. Journal of Group Theory, 21(4), 695-712.
The proof is as follows:
Consider a natural homomorphism $bar:Grightarrow overlineG=G/A$. Clearly, $n_p(PA)$ is the number of all Sylow $p$-subgroups $Q$ of $G$ such that $overlineP=overlineQ$ since $overlineP=overlineQ$ if and only if $Qleq PA$. By Sylow theorem, for every $Qin Syl_p(G)$, there exists $xin G$ such that $P^x=Q$, so $(PA)^x=QA$. This shows that $n_p(PA)$ does not depend on the choice of $P$, and so it is the same for any Sylow $p$-subgroups of $G$. Thus we obtain $n_p(G)=n_p(G/A)n_p(PA)$. By Sylow theorem, $$n_p(PA)=|PA:N_PA(P)|=|A:N_A(P)|$$ where the second identity follows from the Dedekind theorem.
Several questions/problems I have:
(i)If $Qleq PA$, I can only show that $overlineQleq overlineP$. I can't see why $overlinePleq overlineQ$ in this case.
(ii)I can't see why the previous argument would directly lead to the conclusion $n_p(G)=n_p(G/A)n_p(PA)$.
(iii)I search a lot about the Dedekind theorem mentioned in the text but the result I found is for continuity on real axis or for Galois group.
abstract-algebra group-theory sylow-theory
edited Aug 6 at 7:30
Nicky Hekster
26.9k53052
asked Aug 6 at 3:23
Alan Wang
4,431932
edited Aug 6 at 7:30
Nicky Hekster
26.9k53052
edited Aug 6 at 7:30
Nicky Hekster
26.9k53052
edited Aug 6 at 7:30
Nicky Hekster
26.9k53052
26.9k53052
asked Aug 6 at 3:23
Alan Wang
4,431932
asked Aug 6 at 3:23
Alan Wang
4,431932
asked Aug 6 at 3:23
Alan Wang
4,431932
4,431932
Dedekind theorem is if $Hle K$ and $L$ are subgroups, then $HLcap K=H(Lcap K)$ as sets.
– Steve D
Aug 6 at 3:29
add a comment |Â
Dedekind theorem is if $Hle K$ and $L$ are subgroups, then $HLcap K=H(Lcap K)$ as sets.
– Steve D
Aug 6 at 3:29
Dedekind theorem is if $Hle K$ and $L$ are subgroups, then $HLcap K=H(Lcap K)$ as sets.
– Steve D
Aug 6 at 3:29
Dedekind theorem is if $Hle K$ and $L$ are subgroups, then $HLcap K=H(Lcap K)$ as sets.
– Steve D
Aug 6 at 3:29
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Let me offer a slightly different proof, which might illuminate what's going on.
The Sylow subgroups of $G/A$ look like $PA/A$, where $P$ is a Sylow subgroup of $G$. Thus $G$ acts on the Sylow subgroups of $G/A$, and it acts transitively (since it acts transitively on its own Sylow subgroups). This action is just the conjugation action on $PA$ up in $G$. By the orbit-stabilizer theorem, then,
$$ n_p(G/A) = [G:N_G(PA)] $$
We also have $n_p(G)=[G:N_G(P)]$. Now if $gin N_G(P)$, then $gin N_G(PA)$, since $A$ is normal. So we can write
$$ n_p(G) = n_p(G/A)cdot[N_G(PA):N_G(P)]$$
That last number is just $n_p(N_G(PA))$. But $Ple PAlhd N_G(PA)$, so every Sylow subgroup of $N_G(PA)$ is one of $PA$; thus that last number is also $n_p(PA)$, which gives you the first claim.
For your second claim, let $A$ act on the Sylow subgroups of $PA$. If $Q$ is a Sylow subgroup of $PA$, then it is conjugate to $P$. This there is some $pain PA$ such that
$$ Q = P^pa=P^a $$
Thus $A$ acts transitively on the Sylow subgroups of $PA$, and by the orbit-stabilizer theorem again, $n_p(PA)=[A:N_A(P)]$.
Edit: below is my original argument for the second part.
For your second claim, note that $N_G(PA)=N_G(P)A$. This follows, for example, from Frattini's argument applies to $PAlhd N_G(PA)$. But then we have
beginalign*
|N_G(PA)| &= |N_G(P)A|\
&= fracAN_G(P)cap A
endalign*
Note that the subgroup in the denominator there is just $N_A(P)$. So we then have from above
beginalign*
n_p(PA) &= [N_G(PA):N_G(P)]\
&= frac\
&= fracA\
&= frac\
&= [A:N_A(P)]
endalign*
answered Aug 6 at 4:16
Steve D
2,097419
May I know why Sylow subgroups of $G/A$ look like $PA/A$? I can understand your answer except this part.
– Alan Wang
Aug 6 at 5:00
@AlanWang: subgroups of $G/A$ are just subgroups of $G$ containing $A$ ( this is the "fourth isomorphism theorem"). $PA/Acong P/(Pcap A)$ is a p-group, and $[G:PA]=[G/A:PA/A]$ is not divisible by $p$.
– Steve D
Aug 6 at 13:15
From $PA/Acong P/(Pcap A)$ and $PA/A$ is a $p$-group, I can only manage to obtain that $p$ divides $|P|$. I still can't see why $P$ is a Sylow $p$-subgroup of $G$.
– Alan Wang
Aug 6 at 13:22
What I tried is : Write $|G|=p^nk$ and $|A|=p^ml$ where $k,l$ and $p$ are relatively prime. So $|PA/A|=p^n-m$.
– Alan Wang
Aug 6 at 13:31
@AlanWang: I'm assuming $P$ is a Sylow subgroup of $G$, then showing $PA/A$ is a Sylow subgroup of $G/A$.
– Steve D
Aug 6 at 13:35
 |Â
show 1 more comment
up vote
2
down vote
Additional remark and further refinement, a Theorem of Marshall Hall:
Theorem (M. Hall, 1967) Let $A unlhd G$, $P in Syl_p(G)$, then $n_p(G)=a_pb_pc_p$, where
$a_p = #Syl_p(G/A)$
$b_p=#Syl_p(A)$ and
$c_p=#Syl_p(N_PA(P cap A)/(P cap A))$.
answered Aug 6 at 13:23
Nicky Hekster
26.9k53052
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let me offer a slightly different proof, which might illuminate what's going on.
The Sylow subgroups of $G/A$ look like $PA/A$, where $P$ is a Sylow subgroup of $G$. Thus $G$ acts on the Sylow subgroups of $G/A$, and it acts transitively (since it acts transitively on its own Sylow subgroups). This action is just the conjugation action on $PA$ up in $G$. By the orbit-stabilizer theorem, then,
$$ n_p(G/A) = [G:N_G(PA)] $$
We also have $n_p(G)=[G:N_G(P)]$. Now if $gin N_G(P)$, then $gin N_G(PA)$, since $A$ is normal. So we can write
$$ n_p(G) = n_p(G/A)cdot[N_G(PA):N_G(P)]$$
That last number is just $n_p(N_G(PA))$. But $Ple PAlhd N_G(PA)$, so every Sylow subgroup of $N_G(PA)$ is one of $PA$; thus that last number is also $n_p(PA)$, which gives you the first claim.
For your second claim, let $A$ act on the Sylow subgroups of $PA$. If $Q$ is a Sylow subgroup of $PA$, then it is conjugate to $P$. This there is some $pain PA$ such that
$$ Q = P^pa=P^a $$
Thus $A$ acts transitively on the Sylow subgroups of $PA$, and by the orbit-stabilizer theorem again, $n_p(PA)=[A:N_A(P)]$.
Edit: below is my original argument for the second part.
For your second claim, note that $N_G(PA)=N_G(P)A$. This follows, for example, from Frattini's argument applies to $PAlhd N_G(PA)$. But then we have
beginalign*
|N_G(PA)| &= |N_G(P)A|\
&= fracAN_G(P)cap A
endalign*
Note that the subgroup in the denominator there is just $N_A(P)$. So we then have from above
beginalign*
n_p(PA) &= [N_G(PA):N_G(P)]\
&= frac\
&= fracA\
&= frac\
&= [A:N_A(P)]
endalign*
answered Aug 6 at 4:16
Steve D
2,097419
May I know why Sylow subgroups of $G/A$ look like $PA/A$? I can understand your answer except this part.
– Alan Wang
Aug 6 at 5:00
@AlanWang: subgroups of $G/A$ are just subgroups of $G$ containing $A$ ( this is the "fourth isomorphism theorem"). $PA/Acong P/(Pcap A)$ is a p-group, and $[G:PA]=[G/A:PA/A]$ is not divisible by $p$.
– Steve D
Aug 6 at 13:15
From $PA/Acong P/(Pcap A)$ and $PA/A$ is a $p$-group, I can only manage to obtain that $p$ divides $|P|$. I still can't see why $P$ is a Sylow $p$-subgroup of $G$.
– Alan Wang
Aug 6 at 13:22
What I tried is : Write $|G|=p^nk$ and $|A|=p^ml$ where $k,l$ and $p$ are relatively prime. So $|PA/A|=p^n-m$.
– Alan Wang
Aug 6 at 13:31
@AlanWang: I'm assuming $P$ is a Sylow subgroup of $G$, then showing $PA/A$ is a Sylow subgroup of $G/A$.
– Steve D
Aug 6 at 13:35
 |Â
show 1 more comment
up vote
4
down vote
accepted
Let me offer a slightly different proof, which might illuminate what's going on.
The Sylow subgroups of $G/A$ look like $PA/A$, where $P$ is a Sylow subgroup of $G$. Thus $G$ acts on the Sylow subgroups of $G/A$, and it acts transitively (since it acts transitively on its own Sylow subgroups). This action is just the conjugation action on $PA$ up in $G$. By the orbit-stabilizer theorem, then,
$$ n_p(G/A) = [G:N_G(PA)] $$
We also have $n_p(G)=[G:N_G(P)]$. Now if $gin N_G(P)$, then $gin N_G(PA)$, since $A$ is normal. So we can write
$$ n_p(G) = n_p(G/A)cdot[N_G(PA):N_G(P)]$$
That last number is just $n_p(N_G(PA))$. But $Ple PAlhd N_G(PA)$, so every Sylow subgroup of $N_G(PA)$ is one of $PA$; thus that last number is also $n_p(PA)$, which gives you the first claim.
For your second claim, let $A$ act on the Sylow subgroups of $PA$. If $Q$ is a Sylow subgroup of $PA$, then it is conjugate to $P$. This there is some $pain PA$ such that
$$ Q = P^pa=P^a $$
Thus $A$ acts transitively on the Sylow subgroups of $PA$, and by the orbit-stabilizer theorem again, $n_p(PA)=[A:N_A(P)]$.
Edit: below is my original argument for the second part.
For your second claim, note that $N_G(PA)=N_G(P)A$. This follows, for example, from Frattini's argument applies to $PAlhd N_G(PA)$. But then we have
beginalign*
|N_G(PA)| &= |N_G(P)A|\
&= fracAN_G(P)cap A
endalign*
Note that the subgroup in the denominator there is just $N_A(P)$. So we then have from above
beginalign*
n_p(PA) &= [N_G(PA):N_G(P)]\
&= frac\
&= fracA\
&= frac\
&= [A:N_A(P)]
endalign*
answered Aug 6 at 4:16
Steve D
2,097419
May I know why Sylow subgroups of $G/A$ look like $PA/A$? I can understand your answer except this part.
– Alan Wang
Aug 6 at 5:00
@AlanWang: subgroups of $G/A$ are just subgroups of $G$ containing $A$ ( this is the "fourth isomorphism theorem"). $PA/Acong P/(Pcap A)$ is a p-group, and $[G:PA]=[G/A:PA/A]$ is not divisible by $p$.
– Steve D
Aug 6 at 13:15
From $PA/Acong P/(Pcap A)$ and $PA/A$ is a $p$-group, I can only manage to obtain that $p$ divides $|P|$. I still can't see why $P$ is a Sylow $p$-subgroup of $G$.
– Alan Wang
Aug 6 at 13:22
What I tried is : Write $|G|=p^nk$ and $|A|=p^ml$ where $k,l$ and $p$ are relatively prime. So $|PA/A|=p^n-m$.
– Alan Wang
Aug 6 at 13:31
@AlanWang: I'm assuming $P$ is a Sylow subgroup of $G$, then showing $PA/A$ is a Sylow subgroup of $G/A$.
– Steve D
Aug 6 at 13:35
 |Â
show 1 more comment
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let me offer a slightly different proof, which might illuminate what's going on.
The Sylow subgroups of $G/A$ look like $PA/A$, where $P$ is a Sylow subgroup of $G$. Thus $G$ acts on the Sylow subgroups of $G/A$, and it acts transitively (since it acts transitively on its own Sylow subgroups). This action is just the conjugation action on $PA$ up in $G$. By the orbit-stabilizer theorem, then,
$$ n_p(G/A) = [G:N_G(PA)] $$
We also have $n_p(G)=[G:N_G(P)]$. Now if $gin N_G(P)$, then $gin N_G(PA)$, since $A$ is normal. So we can write
$$ n_p(G) = n_p(G/A)cdot[N_G(PA):N_G(P)]$$
That last number is just $n_p(N_G(PA))$. But $Ple PAlhd N_G(PA)$, so every Sylow subgroup of $N_G(PA)$ is one of $PA$; thus that last number is also $n_p(PA)$, which gives you the first claim.
For your second claim, let $A$ act on the Sylow subgroups of $PA$. If $Q$ is a Sylow subgroup of $PA$, then it is conjugate to $P$. This there is some $pain PA$ such that
$$ Q = P^pa=P^a $$
Thus $A$ acts transitively on the Sylow subgroups of $PA$, and by the orbit-stabilizer theorem again, $n_p(PA)=[A:N_A(P)]$.
Edit: below is my original argument for the second part.
For your second claim, note that $N_G(PA)=N_G(P)A$. This follows, for example, from Frattini's argument applies to $PAlhd N_G(PA)$. But then we have
beginalign*
|N_G(PA)| &= |N_G(P)A|\
&= fracAN_G(P)cap A
endalign*
Note that the subgroup in the denominator there is just $N_A(P)$. So we then have from above
beginalign*
n_p(PA) &= [N_G(PA):N_G(P)]\
&= frac\
&= fracA\
&= frac\
&= [A:N_A(P)]
endalign*
answered Aug 6 at 4:16
Steve D
2,097419
Let me offer a slightly different proof, which might illuminate what's going on.
The Sylow subgroups of $G/A$ look like $PA/A$, where $P$ is a Sylow subgroup of $G$. Thus $G$ acts on the Sylow subgroups of $G/A$, and it acts transitively (since it acts transitively on its own Sylow subgroups). This action is just the conjugation action on $PA$ up in $G$. By the orbit-stabilizer theorem, then,
$$ n_p(G/A) = [G:N_G(PA)] $$
We also have $n_p(G)=[G:N_G(P)]$. Now if $gin N_G(P)$, then $gin N_G(PA)$, since $A$ is normal. So we can write
$$ n_p(G) = n_p(G/A)cdot[N_G(PA):N_G(P)]$$
That last number is just $n_p(N_G(PA))$. But $Ple PAlhd N_G(PA)$, so every Sylow subgroup of $N_G(PA)$ is one of $PA$; thus that last number is also $n_p(PA)$, which gives you the first claim.
For your second claim, let $A$ act on the Sylow subgroups of $PA$. If $Q$ is a Sylow subgroup of $PA$, then it is conjugate to $P$. This there is some $pain PA$ such that
$$ Q = P^pa=P^a $$
Thus $A$ acts transitively on the Sylow subgroups of $PA$, and by the orbit-stabilizer theorem again, $n_p(PA)=[A:N_A(P)]$.
Edit: below is my original argument for the second part.
For your second claim, note that $N_G(PA)=N_G(P)A$. This follows, for example, from Frattini's argument applies to $PAlhd N_G(PA)$. But then we have
beginalign*
|N_G(PA)| &= |N_G(P)A|\
&= fracAN_G(P)cap A
endalign*
Note that the subgroup in the denominator there is just $N_A(P)$. So we then have from above
beginalign*
n_p(PA) &= [N_G(PA):N_G(P)]\
&= frac\
&= fracA\
&= frac\
&= [A:N_A(P)]
endalign*
answered Aug 6 at 4:16
Steve D
2,097419
edited Aug 6 at 4:37
answered Aug 6 at 4:16
Steve D
2,097419
answered Aug 6 at 4:16
Steve D
2,097419
answered Aug 6 at 4:16
Steve D
2,097419
2,097419
May I know why Sylow subgroups of $G/A$ look like $PA/A$? I can understand your answer except this part.
– Alan Wang
Aug 6 at 5:00
@AlanWang: subgroups of $G/A$ are just subgroups of $G$ containing $A$ ( this is the "fourth isomorphism theorem"). $PA/Acong P/(Pcap A)$ is a p-group, and $[G:PA]=[G/A:PA/A]$ is not divisible by $p$.
– Steve D
Aug 6 at 13:15
From $PA/Acong P/(Pcap A)$ and $PA/A$ is a $p$-group, I can only manage to obtain that $p$ divides $|P|$. I still can't see why $P$ is a Sylow $p$-subgroup of $G$.
– Alan Wang
Aug 6 at 13:22
What I tried is : Write $|G|=p^nk$ and $|A|=p^ml$ where $k,l$ and $p$ are relatively prime. So $|PA/A|=p^n-m$.
– Alan Wang
Aug 6 at 13:31
@AlanWang: I'm assuming $P$ is a Sylow subgroup of $G$, then showing $PA/A$ is a Sylow subgroup of $G/A$.
– Steve D
Aug 6 at 13:35
 |Â
show 1 more comment
May I know why Sylow subgroups of $G/A$ look like $PA/A$? I can understand your answer except this part.
– Alan Wang
Aug 6 at 5:00
@AlanWang: subgroups of $G/A$ are just subgroups of $G$ containing $A$ ( this is the "fourth isomorphism theorem"). $PA/Acong P/(Pcap A)$ is a p-group, and $[G:PA]=[G/A:PA/A]$ is not divisible by $p$.
– Steve D
Aug 6 at 13:15
From $PA/Acong P/(Pcap A)$ and $PA/A$ is a $p$-group, I can only manage to obtain that $p$ divides $|P|$. I still can't see why $P$ is a Sylow $p$-subgroup of $G$.
– Alan Wang
Aug 6 at 13:22
What I tried is : Write $|G|=p^nk$ and $|A|=p^ml$ where $k,l$ and $p$ are relatively prime. So $|PA/A|=p^n-m$.
– Alan Wang
Aug 6 at 13:31
@AlanWang: I'm assuming $P$ is a Sylow subgroup of $G$, then showing $PA/A$ is a Sylow subgroup of $G/A$.
– Steve D
Aug 6 at 13:35
May I know why Sylow subgroups of $G/A$ look like $PA/A$? I can understand your answer except this part.
– Alan Wang
Aug 6 at 5:00
May I know why Sylow subgroups of $G/A$ look like $PA/A$? I can understand your answer except this part.
– Alan Wang
Aug 6 at 5:00
@AlanWang: subgroups of $G/A$ are just subgroups of $G$ containing $A$ ( this is the "fourth isomorphism theorem"). $PA/Acong P/(Pcap A)$ is a p-group, and $[G:PA]=[G/A:PA/A]$ is not divisible by $p$.
– Steve D
Aug 6 at 13:15
@AlanWang: subgroups of $G/A$ are just subgroups of $G$ containing $A$ ( this is the "fourth isomorphism theorem"). $PA/Acong P/(Pcap A)$ is a p-group, and $[G:PA]=[G/A:PA/A]$ is not divisible by $p$.
– Steve D
Aug 6 at 13:15
From $PA/Acong P/(Pcap A)$ and $PA/A$ is a $p$-group, I can only manage to obtain that $p$ divides $|P|$. I still can't see why $P$ is a Sylow $p$-subgroup of $G$.
– Alan Wang
Aug 6 at 13:22
From $PA/Acong P/(Pcap A)$ and $PA/A$ is a $p$-group, I can only manage to obtain that $p$ divides $|P|$. I still can't see why $P$ is a Sylow $p$-subgroup of $G$.
– Alan Wang
Aug 6 at 13:22
What I tried is : Write $|G|=p^nk$ and $|A|=p^ml$ where $k,l$ and $p$ are relatively prime. So $|PA/A|=p^n-m$.
– Alan Wang
Aug 6 at 13:31
What I tried is : Write $|G|=p^nk$ and $|A|=p^ml$ where $k,l$ and $p$ are relatively prime. So $|PA/A|=p^n-m$.
– Alan Wang
Aug 6 at 13:31
@AlanWang: I'm assuming $P$ is a Sylow subgroup of $G$, then showing $PA/A$ is a Sylow subgroup of $G/A$.
– Steve D
Aug 6 at 13:35
@AlanWang: I'm assuming $P$ is a Sylow subgroup of $G$, then showing $PA/A$ is a Sylow subgroup of $G/A$.
– Steve D
Aug 6 at 13:35
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up vote
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Additional remark and further refinement, a Theorem of Marshall Hall:
Theorem (M. Hall, 1967) Let $A unlhd G$, $P in Syl_p(G)$, then $n_p(G)=a_pb_pc_p$, where
$a_p = #Syl_p(G/A)$
$b_p=#Syl_p(A)$ and
$c_p=#Syl_p(N_PA(P cap A)/(P cap A))$.
answered Aug 6 at 13:23
Nicky Hekster
26.9k53052
add a comment |Â
up vote
2
down vote
Additional remark and further refinement, a Theorem of Marshall Hall:
Theorem (M. Hall, 1967) Let $A unlhd G$, $P in Syl_p(G)$, then $n_p(G)=a_pb_pc_p$, where
$a_p = #Syl_p(G/A)$
$b_p=#Syl_p(A)$ and
$c_p=#Syl_p(N_PA(P cap A)/(P cap A))$.
answered Aug 6 at 13:23
Nicky Hekster
26.9k53052
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Additional remark and further refinement, a Theorem of Marshall Hall:
Theorem (M. Hall, 1967) Let $A unlhd G$, $P in Syl_p(G)$, then $n_p(G)=a_pb_pc_p$, where
$a_p = #Syl_p(G/A)$
$b_p=#Syl_p(A)$ and
$c_p=#Syl_p(N_PA(P cap A)/(P cap A))$.
answered Aug 6 at 13:23
Nicky Hekster
26.9k53052
Additional remark and further refinement, a Theorem of Marshall Hall:
Theorem (M. Hall, 1967) Let $A unlhd G$, $P in Syl_p(G)$, then $n_p(G)=a_pb_pc_p$, where
$a_p = #Syl_p(G/A)$
$b_p=#Syl_p(A)$ and
$c_p=#Syl_p(N_PA(P cap A)/(P cap A))$.
answered Aug 6 at 13:23
Nicky Hekster
26.9k53052
edited Aug 10 at 12:28
answered Aug 6 at 13:23
Nicky Hekster
26.9k53052
answered Aug 6 at 13:23
Nicky Hekster
26.9k53052
answered Aug 6 at 13:23
Nicky Hekster
26.9k53052
26.9k53052
add a comment |Â
add a comment |Â
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Dedekind theorem is if $Hle K$ and $L$ are subgroups, then $HLcap K=H(Lcap K)$ as sets.
– Steve D
Aug 6 at 3:29