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A proof of the third Sylow theorem
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I'm trying to prove the third Sylow theorem, in particular that the number of Sylow $p$-subgroups conjugate to a Sylow $p$-subgroup $P$ (Denoted by $N_p$) is $N_p equiv 1 pmod p$. The book exercise is pointing at this method:
Show that the action of $P$ on the set of Sylow $p$-subgroups has $P$ as its only fixed point (easy to show). Then use this and the second Sylow theorem to prove the third.
My attempt:
Using the class formula, $$|P|=|Z| + sum_H [P:mathrmStab_P(H)]$$
where $Z$ is the set of fixed points of the action (which only has $1$ by the above) and the sum runs over the distinct orbits of the action. But from the second Sylow theorem, every Sylow $p$-subgroup is a conjugate of the other, so in particular the sum only has one term. If we denote the set $P$ is acting on by $S$, then by the orbit-stabilizer theorem, $Scong P / mathrmStab_P(H)$. But $|S|=[P:mathrmStab_P(H)]$ is precisely $N_p$, so I get by the class formula that $|P|=|Z| + N_p$ and taking $mod p$, $N_pequiv -1 pmod p$. What went wrong here?
group-theory proof-verification finite-groups sylow-theory
asked Jul 20 at 18:34
George
774715
 |Â
show 5 more comments
up vote
2
down vote
favorite
I'm trying to prove the third Sylow theorem, in particular that the number of Sylow $p$-subgroups conjugate to a Sylow $p$-subgroup $P$ (Denoted by $N_p$) is $N_p equiv 1 pmod p$. The book exercise is pointing at this method:
Show that the action of $P$ on the set of Sylow $p$-subgroups has $P$ as its only fixed point (easy to show). Then use this and the second Sylow theorem to prove the third.
My attempt:
Using the class formula, $$|P|=|Z| + sum_H [P:mathrmStab_P(H)]$$
where $Z$ is the set of fixed points of the action (which only has $1$ by the above) and the sum runs over the distinct orbits of the action. But from the second Sylow theorem, every Sylow $p$-subgroup is a conjugate of the other, so in particular the sum only has one term. If we denote the set $P$ is acting on by $S$, then by the orbit-stabilizer theorem, $Scong P / mathrmStab_P(H)$. But $|S|=[P:mathrmStab_P(H)]$ is precisely $N_p$, so I get by the class formula that $|P|=|Z| + N_p$ and taking $mod p$, $N_pequiv -1 pmod p$. What went wrong here?
group-theory proof-verification finite-groups sylow-theory
asked Jul 20 at 18:34
George
774715
1
Every two $p$-Sylow subgroups are conjugate via an element of the ambient group, not necessarily by an element of $P$.
– Gal Porat
Jul 20 at 18:38
But conjugating $P$ by an element of $P$ does not give me a new Sylow $p$-group. I guess if it is the case that conjugating any other Sylow subgroup by elements of $p$ cannot give me a distinct subgroup, then the class formula would give $|P|=|Z|+(p-1)N_p$ and the problem would be resolved?
– George
Jul 20 at 18:45
Actually I don't know where I got $p-1$ from
– George
Jul 20 at 18:51
I'm just pointing out a flaw in your argument for the sum having only one term.
– Gal Porat
Jul 20 at 18:52
If $cal S$ is the set of the p-Sylow subgroups, take one of them, say $Q$ and build the map $Qtimescal Stocal S$ via $qcdot P=qPq^-1$, this will be an action which gonna give you a class equation. The rest (on this fashion) depends on how much you know about her.
– janmarqz
Jul 20 at 19:16
 |Â
show 5 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to prove the third Sylow theorem, in particular that the number of Sylow $p$-subgroups conjugate to a Sylow $p$-subgroup $P$ (Denoted by $N_p$) is $N_p equiv 1 pmod p$. The book exercise is pointing at this method:
Show that the action of $P$ on the set of Sylow $p$-subgroups has $P$ as its only fixed point (easy to show). Then use this and the second Sylow theorem to prove the third.
My attempt:
Using the class formula, $$|P|=|Z| + sum_H [P:mathrmStab_P(H)]$$
where $Z$ is the set of fixed points of the action (which only has $1$ by the above) and the sum runs over the distinct orbits of the action. But from the second Sylow theorem, every Sylow $p$-subgroup is a conjugate of the other, so in particular the sum only has one term. If we denote the set $P$ is acting on by $S$, then by the orbit-stabilizer theorem, $Scong P / mathrmStab_P(H)$. But $|S|=[P:mathrmStab_P(H)]$ is precisely $N_p$, so I get by the class formula that $|P|=|Z| + N_p$ and taking $mod p$, $N_pequiv -1 pmod p$. What went wrong here?
group-theory proof-verification finite-groups sylow-theory
asked Jul 20 at 18:34
George
774715
I'm trying to prove the third Sylow theorem, in particular that the number of Sylow $p$-subgroups conjugate to a Sylow $p$-subgroup $P$ (Denoted by $N_p$) is $N_p equiv 1 pmod p$. The book exercise is pointing at this method:
Show that the action of $P$ on the set of Sylow $p$-subgroups has $P$ as its only fixed point (easy to show). Then use this and the second Sylow theorem to prove the third.
My attempt:
Using the class formula, $$|P|=|Z| + sum_H [P:mathrmStab_P(H)]$$
where $Z$ is the set of fixed points of the action (which only has $1$ by the above) and the sum runs over the distinct orbits of the action. But from the second Sylow theorem, every Sylow $p$-subgroup is a conjugate of the other, so in particular the sum only has one term. If we denote the set $P$ is acting on by $S$, then by the orbit-stabilizer theorem, $Scong P / mathrmStab_P(H)$. But $|S|=[P:mathrmStab_P(H)]$ is precisely $N_p$, so I get by the class formula that $|P|=|Z| + N_p$ and taking $mod p$, $N_pequiv -1 pmod p$. What went wrong here?
group-theory proof-verification finite-groups sylow-theory
asked Jul 20 at 18:34
George
774715
asked Jul 20 at 18:34
George
774715
asked Jul 20 at 18:34
George
774715
asked Jul 20 at 18:34
George
774715
774715
1
Every two $p$-Sylow subgroups are conjugate via an element of the ambient group, not necessarily by an element of $P$.
– Gal Porat
Jul 20 at 18:38
But conjugating $P$ by an element of $P$ does not give me a new Sylow $p$-group. I guess if it is the case that conjugating any other Sylow subgroup by elements of $p$ cannot give me a distinct subgroup, then the class formula would give $|P|=|Z|+(p-1)N_p$ and the problem would be resolved?
– George
Jul 20 at 18:45
Actually I don't know where I got $p-1$ from
– George
Jul 20 at 18:51
I'm just pointing out a flaw in your argument for the sum having only one term.
– Gal Porat
Jul 20 at 18:52
If $cal S$ is the set of the p-Sylow subgroups, take one of them, say $Q$ and build the map $Qtimescal Stocal S$ via $qcdot P=qPq^-1$, this will be an action which gonna give you a class equation. The rest (on this fashion) depends on how much you know about her.
– janmarqz
Jul 20 at 19:16
 |Â
show 5 more comments
1
Every two $p$-Sylow subgroups are conjugate via an element of the ambient group, not necessarily by an element of $P$.
– Gal Porat
Jul 20 at 18:38
But conjugating $P$ by an element of $P$ does not give me a new Sylow $p$-group. I guess if it is the case that conjugating any other Sylow subgroup by elements of $p$ cannot give me a distinct subgroup, then the class formula would give $|P|=|Z|+(p-1)N_p$ and the problem would be resolved?
– George
Jul 20 at 18:45
Actually I don't know where I got $p-1$ from
– George
Jul 20 at 18:51
I'm just pointing out a flaw in your argument for the sum having only one term.
– Gal Porat
Jul 20 at 18:52
If $cal S$ is the set of the p-Sylow subgroups, take one of them, say $Q$ and build the map $Qtimescal Stocal S$ via $qcdot P=qPq^-1$, this will be an action which gonna give you a class equation. The rest (on this fashion) depends on how much you know about her.
– janmarqz
Jul 20 at 19:16
1
1
Every two $p$-Sylow subgroups are conjugate via an element of the ambient group, not necessarily by an element of $P$.
– Gal Porat
Jul 20 at 18:38
Every two $p$-Sylow subgroups are conjugate via an element of the ambient group, not necessarily by an element of $P$.
– Gal Porat
Jul 20 at 18:38
But conjugating $P$ by an element of $P$ does not give me a new Sylow $p$-group. I guess if it is the case that conjugating any other Sylow subgroup by elements of $p$ cannot give me a distinct subgroup, then the class formula would give $|P|=|Z|+(p-1)N_p$ and the problem would be resolved?
– George
Jul 20 at 18:45
But conjugating $P$ by an element of $P$ does not give me a new Sylow $p$-group. I guess if it is the case that conjugating any other Sylow subgroup by elements of $p$ cannot give me a distinct subgroup, then the class formula would give $|P|=|Z|+(p-1)N_p$ and the problem would be resolved?
– George
Jul 20 at 18:45
Actually I don't know where I got $p-1$ from
– George
Jul 20 at 18:51
Actually I don't know where I got $p-1$ from
– George
Jul 20 at 18:51
I'm just pointing out a flaw in your argument for the sum having only one term.
– Gal Porat
Jul 20 at 18:52
I'm just pointing out a flaw in your argument for the sum having only one term.
– Gal Porat
Jul 20 at 18:52
If $cal S$ is the set of the p-Sylow subgroups, take one of them, say $Q$ and build the map $Qtimescal Stocal S$ via $qcdot P=qPq^-1$, this will be an action which gonna give you a class equation. The rest (on this fashion) depends on how much you know about her.
– janmarqz
Jul 20 at 19:16
If $cal S$ is the set of the p-Sylow subgroups, take one of them, say $Q$ and build the map $Qtimescal Stocal S$ via $qcdot P=qPq^-1$, this will be an action which gonna give you a class equation. The rest (on this fashion) depends on how much you know about her.
– janmarqz
Jul 20 at 19:16
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
Based on the comments of @janmarqz, I believe I have resolved the problem, so I will post for completeness (and others to check).
Let $S$ be the set of Sylow $p$-subgroups of the finite group $G$ and let $Pin S$ act on the set $S$ by conjugation.
First I'll show that $P$ is the only fixed point of the action:
Let $H$ be a Sylow $p$-subgroup that is fixed by the action of $P$. This is equivalent to saying that for every $gin P$, we have $gHg^-1=H$. In particular, $P$ is a subgroup of the normalizer $N_G(H)$ of $H$. Since $H$ is normal in its normalizer, then the subgroup $PH$ is normal in $N_G(H)$ by the second isomorphism theorem, and in particular, $|PH/H|=|P/Pcap H|$. The cardinality of the right hand side is a positive power of $p$, therefore $PH$ is a $p$-group. However, $H$ is a maximal $p$-group of $G$ and hence $PH=H$, which in turn implies that $P$ is a subgroup of $H$. By order considerations, $P=H$, hence $P$ is the only fixed point of the action.
Now, the class equation $$|S|=|S_P|+sum_H [P:mathrmStab_P(H)]$$
Clearly, $|S|=N_p$, and we know that $P$ is the only fixed point of the action, so $|Z|=1$, and necessarily $[P:mathrmStab_P(H)]>1$. Since $P$ is a $p$-group, then $p$ divides $[P:mathrmStab_P(H)]$, so taking the class equation $mod p$ gives the result.
edited Jul 20 at 22:54
janmarqz
6,02941629
answered Jul 20 at 20:48
George
774715
it would be nice to see how the only fixed point of the action is what you assert
– janmarqz
Jul 20 at 22:07
1
@janmarqz added
– George
Jul 20 at 22:47
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Based on the comments of @janmarqz, I believe I have resolved the problem, so I will post for completeness (and others to check).
Let $S$ be the set of Sylow $p$-subgroups of the finite group $G$ and let $Pin S$ act on the set $S$ by conjugation.
First I'll show that $P$ is the only fixed point of the action:
Let $H$ be a Sylow $p$-subgroup that is fixed by the action of $P$. This is equivalent to saying that for every $gin P$, we have $gHg^-1=H$. In particular, $P$ is a subgroup of the normalizer $N_G(H)$ of $H$. Since $H$ is normal in its normalizer, then the subgroup $PH$ is normal in $N_G(H)$ by the second isomorphism theorem, and in particular, $|PH/H|=|P/Pcap H|$. The cardinality of the right hand side is a positive power of $p$, therefore $PH$ is a $p$-group. However, $H$ is a maximal $p$-group of $G$ and hence $PH=H$, which in turn implies that $P$ is a subgroup of $H$. By order considerations, $P=H$, hence $P$ is the only fixed point of the action.
Now, the class equation $$|S|=|S_P|+sum_H [P:mathrmStab_P(H)]$$
Clearly, $|S|=N_p$, and we know that $P$ is the only fixed point of the action, so $|Z|=1$, and necessarily $[P:mathrmStab_P(H)]>1$. Since $P$ is a $p$-group, then $p$ divides $[P:mathrmStab_P(H)]$, so taking the class equation $mod p$ gives the result.
edited Jul 20 at 22:54
janmarqz
6,02941629
answered Jul 20 at 20:48
George
774715
it would be nice to see how the only fixed point of the action is what you assert
– janmarqz
Jul 20 at 22:07
1
@janmarqz added
– George
Jul 20 at 22:47
add a comment |Â
up vote
2
down vote
Based on the comments of @janmarqz, I believe I have resolved the problem, so I will post for completeness (and others to check).
Let $S$ be the set of Sylow $p$-subgroups of the finite group $G$ and let $Pin S$ act on the set $S$ by conjugation.
First I'll show that $P$ is the only fixed point of the action:
Let $H$ be a Sylow $p$-subgroup that is fixed by the action of $P$. This is equivalent to saying that for every $gin P$, we have $gHg^-1=H$. In particular, $P$ is a subgroup of the normalizer $N_G(H)$ of $H$. Since $H$ is normal in its normalizer, then the subgroup $PH$ is normal in $N_G(H)$ by the second isomorphism theorem, and in particular, $|PH/H|=|P/Pcap H|$. The cardinality of the right hand side is a positive power of $p$, therefore $PH$ is a $p$-group. However, $H$ is a maximal $p$-group of $G$ and hence $PH=H$, which in turn implies that $P$ is a subgroup of $H$. By order considerations, $P=H$, hence $P$ is the only fixed point of the action.
Now, the class equation $$|S|=|S_P|+sum_H [P:mathrmStab_P(H)]$$
Clearly, $|S|=N_p$, and we know that $P$ is the only fixed point of the action, so $|Z|=1$, and necessarily $[P:mathrmStab_P(H)]>1$. Since $P$ is a $p$-group, then $p$ divides $[P:mathrmStab_P(H)]$, so taking the class equation $mod p$ gives the result.
edited Jul 20 at 22:54
janmarqz
6,02941629
answered Jul 20 at 20:48
George
774715
it would be nice to see how the only fixed point of the action is what you assert
– janmarqz
Jul 20 at 22:07
1
@janmarqz added
– George
Jul 20 at 22:47
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Based on the comments of @janmarqz, I believe I have resolved the problem, so I will post for completeness (and others to check).
Let $S$ be the set of Sylow $p$-subgroups of the finite group $G$ and let $Pin S$ act on the set $S$ by conjugation.
First I'll show that $P$ is the only fixed point of the action:
Let $H$ be a Sylow $p$-subgroup that is fixed by the action of $P$. This is equivalent to saying that for every $gin P$, we have $gHg^-1=H$. In particular, $P$ is a subgroup of the normalizer $N_G(H)$ of $H$. Since $H$ is normal in its normalizer, then the subgroup $PH$ is normal in $N_G(H)$ by the second isomorphism theorem, and in particular, $|PH/H|=|P/Pcap H|$. The cardinality of the right hand side is a positive power of $p$, therefore $PH$ is a $p$-group. However, $H$ is a maximal $p$-group of $G$ and hence $PH=H$, which in turn implies that $P$ is a subgroup of $H$. By order considerations, $P=H$, hence $P$ is the only fixed point of the action.
Now, the class equation $$|S|=|S_P|+sum_H [P:mathrmStab_P(H)]$$
Clearly, $|S|=N_p$, and we know that $P$ is the only fixed point of the action, so $|Z|=1$, and necessarily $[P:mathrmStab_P(H)]>1$. Since $P$ is a $p$-group, then $p$ divides $[P:mathrmStab_P(H)]$, so taking the class equation $mod p$ gives the result.
edited Jul 20 at 22:54
janmarqz
6,02941629
answered Jul 20 at 20:48
George
774715
Based on the comments of @janmarqz, I believe I have resolved the problem, so I will post for completeness (and others to check).
Let $S$ be the set of Sylow $p$-subgroups of the finite group $G$ and let $Pin S$ act on the set $S$ by conjugation.
First I'll show that $P$ is the only fixed point of the action:
Let $H$ be a Sylow $p$-subgroup that is fixed by the action of $P$. This is equivalent to saying that for every $gin P$, we have $gHg^-1=H$. In particular, $P$ is a subgroup of the normalizer $N_G(H)$ of $H$. Since $H$ is normal in its normalizer, then the subgroup $PH$ is normal in $N_G(H)$ by the second isomorphism theorem, and in particular, $|PH/H|=|P/Pcap H|$. The cardinality of the right hand side is a positive power of $p$, therefore $PH$ is a $p$-group. However, $H$ is a maximal $p$-group of $G$ and hence $PH=H$, which in turn implies that $P$ is a subgroup of $H$. By order considerations, $P=H$, hence $P$ is the only fixed point of the action.
Now, the class equation $$|S|=|S_P|+sum_H [P:mathrmStab_P(H)]$$
Clearly, $|S|=N_p$, and we know that $P$ is the only fixed point of the action, so $|Z|=1$, and necessarily $[P:mathrmStab_P(H)]>1$. Since $P$ is a $p$-group, then $p$ divides $[P:mathrmStab_P(H)]$, so taking the class equation $mod p$ gives the result.
edited Jul 20 at 22:54
janmarqz
6,02941629
answered Jul 20 at 20:48
George
774715
edited Jul 20 at 22:54
janmarqz
6,02941629
edited Jul 20 at 22:54
janmarqz
6,02941629
edited Jul 20 at 22:54
janmarqz
6,02941629
6,02941629
answered Jul 20 at 20:48
George
774715
answered Jul 20 at 20:48
George
774715
answered Jul 20 at 20:48
George
774715
774715
it would be nice to see how the only fixed point of the action is what you assert
– janmarqz
Jul 20 at 22:07
1
@janmarqz added
– George
Jul 20 at 22:47
add a comment |Â
it would be nice to see how the only fixed point of the action is what you assert
– janmarqz
Jul 20 at 22:07
1
@janmarqz added
– George
Jul 20 at 22:47
it would be nice to see how the only fixed point of the action is what you assert
– janmarqz
Jul 20 at 22:07
it would be nice to see how the only fixed point of the action is what you assert
– janmarqz
Jul 20 at 22:07
1
1
@janmarqz added
– George
Jul 20 at 22:47
@janmarqz added
– George
Jul 20 at 22:47
add a comment |Â
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1
Every two $p$-Sylow subgroups are conjugate via an element of the ambient group, not necessarily by an element of $P$.
– Gal Porat
Jul 20 at 18:38
But conjugating $P$ by an element of $P$ does not give me a new Sylow $p$-group. I guess if it is the case that conjugating any other Sylow subgroup by elements of $p$ cannot give me a distinct subgroup, then the class formula would give $|P|=|Z|+(p-1)N_p$ and the problem would be resolved?
– George
Jul 20 at 18:45
Actually I don't know where I got $p-1$ from
– George
Jul 20 at 18:51
I'm just pointing out a flaw in your argument for the sum having only one term.
– Gal Porat
Jul 20 at 18:52
If $cal S$ is the set of the p-Sylow subgroups, take one of them, say $Q$ and build the map $Qtimescal Stocal S$ via $qcdot P=qPq^-1$, this will be an action which gonna give you a class equation. The rest (on this fashion) depends on how much you know about her.
– janmarqz
Jul 20 at 19:16