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Let G be a finite group. $P$ is $p$-sylow of G, $Nlhd G$ so $Pcap N$ is P sylow of N.
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Let G be a finite group. $P$ is $p$-sylow of G, $Nlhd G$ so $Pcap N$ is P sylow of N.
I need to prove or give Counterexample.
If I can assume that $p| |N|$ because otherwise there is not $p$-group sylow to $N$. Then I know how to prove this statement but the question if I can assume this or not?
If this statement is false can you help me find Counterexample?
It is prove is ok? and you can answer the questions in the parenthesis?
If $pmid |N|$ then the order of any element of $Pcap N$ is power of p besides the unit element because the order of any element in $Pcap N$ dividing the order of $P$. Thus $Pcap N$ is p-group.
I will show that every $p$-group which is subgroup of H contained in the conjugacy class of $Pcap H$. (but why do I need to prove it, what to I get from it?)
Meaning in a group $g(Pcap N)g^-1cong Pcap N$ for $gin G$.
from here we will get that every subgroup $Pcap N$ is max p- subgroup of N (why?)
and that $Pcap N$ os p-sylow subgroup of N (why?).
for that let $M$ be $p$-subgroup of $N$ (some $N$) then she is p-subgroup of $G$. Now, P p-subgroup of $G$ then exists $gin G$ such as $Nsubseteq gPg^-1$ (why?).
On the other hand, $Nlhd G$ so $Msubseteq N=gNg^-1$ (why do we know that $Msubseteq N$?).
then we get that: $Msubseteq gNg^-1cap gPg^-1=g(Ncap P)g^-1$ (why the last equality ?)
finite-groups sylow-theory
asked 2 days ago
Rimon
33
add a comment |Â
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0
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Let G be a finite group. $P$ is $p$-sylow of G, $Nlhd G$ so $Pcap N$ is P sylow of N.
I need to prove or give Counterexample.
If I can assume that $p| |N|$ because otherwise there is not $p$-group sylow to $N$. Then I know how to prove this statement but the question if I can assume this or not?
If this statement is false can you help me find Counterexample?
It is prove is ok? and you can answer the questions in the parenthesis?
If $pmid |N|$ then the order of any element of $Pcap N$ is power of p besides the unit element because the order of any element in $Pcap N$ dividing the order of $P$. Thus $Pcap N$ is p-group.
I will show that every $p$-group which is subgroup of H contained in the conjugacy class of $Pcap H$. (but why do I need to prove it, what to I get from it?)
Meaning in a group $g(Pcap N)g^-1cong Pcap N$ for $gin G$.
from here we will get that every subgroup $Pcap N$ is max p- subgroup of N (why?)
and that $Pcap N$ os p-sylow subgroup of N (why?).
for that let $M$ be $p$-subgroup of $N$ (some $N$) then she is p-subgroup of $G$. Now, P p-subgroup of $G$ then exists $gin G$ such as $Nsubseteq gPg^-1$ (why?).
On the other hand, $Nlhd G$ so $Msubseteq N=gNg^-1$ (why do we know that $Msubseteq N$?).
then we get that: $Msubseteq gNg^-1cap gPg^-1=g(Ncap P)g^-1$ (why the last equality ?)
finite-groups sylow-theory
asked 2 days ago
Rimon
33
Since $P$ is $p$-sylow, $Pcap N$ has order a power of $p$, so if $p$ does not divide the order of $N$, then $Pcap N$ is trivial and a $p$-sylow. So you can assume $p$ divides the order of $N$.
– xarles
2 days ago
add a comment |Â
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up vote
0
down vote
favorite
Let G be a finite group. $P$ is $p$-sylow of G, $Nlhd G$ so $Pcap N$ is P sylow of N.
I need to prove or give Counterexample.
If I can assume that $p| |N|$ because otherwise there is not $p$-group sylow to $N$. Then I know how to prove this statement but the question if I can assume this or not?
If this statement is false can you help me find Counterexample?
It is prove is ok? and you can answer the questions in the parenthesis?
If $pmid |N|$ then the order of any element of $Pcap N$ is power of p besides the unit element because the order of any element in $Pcap N$ dividing the order of $P$. Thus $Pcap N$ is p-group.
I will show that every $p$-group which is subgroup of H contained in the conjugacy class of $Pcap H$. (but why do I need to prove it, what to I get from it?)
Meaning in a group $g(Pcap N)g^-1cong Pcap N$ for $gin G$.
from here we will get that every subgroup $Pcap N$ is max p- subgroup of N (why?)
and that $Pcap N$ os p-sylow subgroup of N (why?).
for that let $M$ be $p$-subgroup of $N$ (some $N$) then she is p-subgroup of $G$. Now, P p-subgroup of $G$ then exists $gin G$ such as $Nsubseteq gPg^-1$ (why?).
On the other hand, $Nlhd G$ so $Msubseteq N=gNg^-1$ (why do we know that $Msubseteq N$?).
then we get that: $Msubseteq gNg^-1cap gPg^-1=g(Ncap P)g^-1$ (why the last equality ?)
finite-groups sylow-theory
asked 2 days ago
Rimon
33
Let G be a finite group. $P$ is $p$-sylow of G, $Nlhd G$ so $Pcap N$ is P sylow of N.
I need to prove or give Counterexample.
If I can assume that $p| |N|$ because otherwise there is not $p$-group sylow to $N$. Then I know how to prove this statement but the question if I can assume this or not?
If this statement is false can you help me find Counterexample?
It is prove is ok? and you can answer the questions in the parenthesis?
If $pmid |N|$ then the order of any element of $Pcap N$ is power of p besides the unit element because the order of any element in $Pcap N$ dividing the order of $P$. Thus $Pcap N$ is p-group.
I will show that every $p$-group which is subgroup of H contained in the conjugacy class of $Pcap H$. (but why do I need to prove it, what to I get from it?)
Meaning in a group $g(Pcap N)g^-1cong Pcap N$ for $gin G$.
from here we will get that every subgroup $Pcap N$ is max p- subgroup of N (why?)
and that $Pcap N$ os p-sylow subgroup of N (why?).
for that let $M$ be $p$-subgroup of $N$ (some $N$) then she is p-subgroup of $G$. Now, P p-subgroup of $G$ then exists $gin G$ such as $Nsubseteq gPg^-1$ (why?).
On the other hand, $Nlhd G$ so $Msubseteq N=gNg^-1$ (why do we know that $Msubseteq N$?).
then we get that: $Msubseteq gNg^-1cap gPg^-1=g(Ncap P)g^-1$ (why the last equality ?)
finite-groups sylow-theory
asked 2 days ago
Rimon
33
edited 2 days ago
asked 2 days ago
Rimon
33
asked 2 days ago
Rimon
33
asked 2 days ago
Rimon
33
33
Since $P$ is $p$-sylow, $Pcap N$ has order a power of $p$, so if $p$ does not divide the order of $N$, then $Pcap N$ is trivial and a $p$-sylow. So you can assume $p$ divides the order of $N$.
– xarles
2 days ago
add a comment |Â
Since $P$ is $p$-sylow, $Pcap N$ has order a power of $p$, so if $p$ does not divide the order of $N$, then $Pcap N$ is trivial and a $p$-sylow. So you can assume $p$ divides the order of $N$.
– xarles
2 days ago
Since $P$ is $p$-sylow, $Pcap N$ has order a power of $p$, so if $p$ does not divide the order of $N$, then $Pcap N$ is trivial and a $p$-sylow. So you can assume $p$ divides the order of $N$.
– xarles
2 days ago
Since $P$ is $p$-sylow, $Pcap N$ has order a power of $p$, so if $p$ does not divide the order of $N$, then $Pcap N$ is trivial and a $p$-sylow. So you can assume $p$ divides the order of $N$.
– xarles
2 days ago
add a comment |Â
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Since $P$ is $p$-sylow, $Pcap N$ has order a power of $p$, so if $p$ does not divide the order of $N$, then $Pcap N$ is trivial and a $p$-sylow. So you can assume $p$ divides the order of $N$.
– xarles
2 days ago