a few questions about what's going on in this proof of sylow's theorem I found
$begingroup$
Note: If someone wants to even just answer my first question in the comments until someone else decides to give a full answer I'd be pretty happy. I just want to know there's no mistakes in it before I start trying to understand it fully and remember it.
I want to prove the claim that the number of P-sylow subgroups is given by $n_p=1+kp$ and in particular there is at least one P-sylow subgroup.
I found the following proof
Let P be any Sylow p-subgroup. If g ∈ G be a p-element and $gPg^{−1} = P$ * then g ∈ P. To see this, consider the subgroup R generated by g and P. By assumption, $g ∈ N_G(P)$, so$ R ≤ N_G(P)$. Hence, P is a normal subgroup of R. We find |R| = |R/P|·|P|. But |R/P| is a cyclic group generated by the coset gP. Then gP is a p-element since g is. Hence |R| is a power of p since all its elements are p-elements. Let Sp be the set of all Sylow p-subgroups of G. Then G acts on this set by conjugation. Let P,Q ∈ Sp be two distinct subgroups. Then Q cannot be fixed under conjugation by all the elements of P because of *
Let O be the P-orbit of Q under conjugation. Then the size of the orbit must be divisible by p because of the order-stabilizer equation:
|O| =
$|P| /|Stab_P(Q)|$
.
Since |P| is a power of p, the size of any orbit must be a power of p. The case |O| = $p^0$ = 1 is ruled out since Q cannot be fixed by all the elements of P. We find that the set of all Sylow p-subgroups is the union of P-orbits. There is only one orbit of order one, {P}, while the other orbits must have orders a positive power of p. We conclude np = |Sp| ≡ 1 mod p.
My questions :
1) ( most importantly) is this proof 100% valid. My lecturer gave us another proof but I find it much harder to follow.
2) In my lecturers proof he let G act on S by right multiplication. Why decide one or the other when it comes to action through conjugation or multiplication ?
3) at the very end of the proof
where it says "the others must have prime power" then why is it mod $p$ instead of mod $p^alpha$
group-theory proof-explanation sylow-theory
asked Dec 22 '18 at 23:38
can'tcauchycan'tcauchy
989417
$endgroup$
add a comment |
$begingroup$
Note: If someone wants to even just answer my first question in the comments until someone else decides to give a full answer I'd be pretty happy. I just want to know there's no mistakes in it before I start trying to understand it fully and remember it.
I want to prove the claim that the number of P-sylow subgroups is given by $n_p=1+kp$ and in particular there is at least one P-sylow subgroup.
I found the following proof
Let P be any Sylow p-subgroup. If g ∈ G be a p-element and $gPg^{−1} = P$ * then g ∈ P. To see this, consider the subgroup R generated by g and P. By assumption, $g ∈ N_G(P)$, so$ R ≤ N_G(P)$. Hence, P is a normal subgroup of R. We find |R| = |R/P|·|P|. But |R/P| is a cyclic group generated by the coset gP. Then gP is a p-element since g is. Hence |R| is a power of p since all its elements are p-elements. Let Sp be the set of all Sylow p-subgroups of G. Then G acts on this set by conjugation. Let P,Q ∈ Sp be two distinct subgroups. Then Q cannot be fixed under conjugation by all the elements of P because of *
Let O be the P-orbit of Q under conjugation. Then the size of the orbit must be divisible by p because of the order-stabilizer equation:
|O| =
$|P| /|Stab_P(Q)|$
.
Since |P| is a power of p, the size of any orbit must be a power of p. The case |O| = $p^0$ = 1 is ruled out since Q cannot be fixed by all the elements of P. We find that the set of all Sylow p-subgroups is the union of P-orbits. There is only one orbit of order one, {P}, while the other orbits must have orders a positive power of p. We conclude np = |Sp| ≡ 1 mod p.
My questions :
1) ( most importantly) is this proof 100% valid. My lecturer gave us another proof but I find it much harder to follow.
2) In my lecturers proof he let G act on S by right multiplication. Why decide one or the other when it comes to action through conjugation or multiplication ?
3) at the very end of the proof
where it says "the others must have prime power" then why is it mod $p$ instead of mod $p^alpha$
group-theory proof-explanation sylow-theory
asked Dec 22 '18 at 23:38
can'tcauchycan'tcauchy
989417
$endgroup$
add a comment |
$begingroup$
Note: If someone wants to even just answer my first question in the comments until someone else decides to give a full answer I'd be pretty happy. I just want to know there's no mistakes in it before I start trying to understand it fully and remember it.
I want to prove the claim that the number of P-sylow subgroups is given by $n_p=1+kp$ and in particular there is at least one P-sylow subgroup.
I found the following proof
Let P be any Sylow p-subgroup. If g ∈ G be a p-element and $gPg^{−1} = P$ * then g ∈ P. To see this, consider the subgroup R generated by g and P. By assumption, $g ∈ N_G(P)$, so$ R ≤ N_G(P)$. Hence, P is a normal subgroup of R. We find |R| = |R/P|·|P|. But |R/P| is a cyclic group generated by the coset gP. Then gP is a p-element since g is. Hence |R| is a power of p since all its elements are p-elements. Let Sp be the set of all Sylow p-subgroups of G. Then G acts on this set by conjugation. Let P,Q ∈ Sp be two distinct subgroups. Then Q cannot be fixed under conjugation by all the elements of P because of *
Let O be the P-orbit of Q under conjugation. Then the size of the orbit must be divisible by p because of the order-stabilizer equation:
|O| =
$|P| /|Stab_P(Q)|$
.
Since |P| is a power of p, the size of any orbit must be a power of p. The case |O| = $p^0$ = 1 is ruled out since Q cannot be fixed by all the elements of P. We find that the set of all Sylow p-subgroups is the union of P-orbits. There is only one orbit of order one, {P}, while the other orbits must have orders a positive power of p. We conclude np = |Sp| ≡ 1 mod p.
My questions :
1) ( most importantly) is this proof 100% valid. My lecturer gave us another proof but I find it much harder to follow.
2) In my lecturers proof he let G act on S by right multiplication. Why decide one or the other when it comes to action through conjugation or multiplication ?
3) at the very end of the proof
where it says "the others must have prime power" then why is it mod $p$ instead of mod $p^alpha$
group-theory proof-explanation sylow-theory
asked Dec 22 '18 at 23:38
can'tcauchycan'tcauchy
989417
$endgroup$
Note: If someone wants to even just answer my first question in the comments until someone else decides to give a full answer I'd be pretty happy. I just want to know there's no mistakes in it before I start trying to understand it fully and remember it.
I want to prove the claim that the number of P-sylow subgroups is given by $n_p=1+kp$ and in particular there is at least one P-sylow subgroup.
I found the following proof
Let P be any Sylow p-subgroup. If g ∈ G be a p-element and $gPg^{−1} = P$ * then g ∈ P. To see this, consider the subgroup R generated by g and P. By assumption, $g ∈ N_G(P)$, so$ R ≤ N_G(P)$. Hence, P is a normal subgroup of R. We find |R| = |R/P|·|P|. But |R/P| is a cyclic group generated by the coset gP. Then gP is a p-element since g is. Hence |R| is a power of p since all its elements are p-elements. Let Sp be the set of all Sylow p-subgroups of G. Then G acts on this set by conjugation. Let P,Q ∈ Sp be two distinct subgroups. Then Q cannot be fixed under conjugation by all the elements of P because of *
Let O be the P-orbit of Q under conjugation. Then the size of the orbit must be divisible by p because of the order-stabilizer equation:
|O| =
$|P| /|Stab_P(Q)|$
.
Since |P| is a power of p, the size of any orbit must be a power of p. The case |O| = $p^0$ = 1 is ruled out since Q cannot be fixed by all the elements of P. We find that the set of all Sylow p-subgroups is the union of P-orbits. There is only one orbit of order one, {P}, while the other orbits must have orders a positive power of p. We conclude np = |Sp| ≡ 1 mod p.
My questions :
1) ( most importantly) is this proof 100% valid. My lecturer gave us another proof but I find it much harder to follow.
2) In my lecturers proof he let G act on S by right multiplication. Why decide one or the other when it comes to action through conjugation or multiplication ?
3) at the very end of the proof
where it says "the others must have prime power" then why is it mod $p$ instead of mod $p^alpha$
group-theory proof-explanation sylow-theory
group-theory proof-explanation sylow-theory
asked Dec 22 '18 at 23:38
can'tcauchycan'tcauchy
989417
asked Dec 22 '18 at 23:38
can'tcauchycan'tcauchy
989417
edited Dec 23 '18 at 1:16
can'tcauchy
asked Dec 22 '18 at 23:38
can'tcauchycan'tcauchy
989417
asked Dec 22 '18 at 23:38
can'tcauchycan'tcauchy
989417
asked Dec 22 '18 at 23:38
can'tcauchycan'tcauchy
989417
989417
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