Why does group action by conjugation on sylow subgroups define a homomorphism into the symmetric group? The 2019 Stack Overflow Developer Survey Results Are InGroup Theory, group of order 55Finding the kernel of an action on conjugate subgroupsCheck:etermine the number of Sylow $2$-subgroups and Sylow $3$-subgroups that $G$ can have.On the number of Sylow subgroups in Symmetric GroupWhy the symmetric group $S_6$ has 10 Sylow 3-subgroups?Group action on Sylow subgroupsTransitive group action on Sylow subgroupsProof verification - the only group of order 24 without normal sylow subgroup is $S_4$.Prove that there is at most one non-abelian group of order $28$ all of whose Sylow $2$-subgroups are cyclicWhy does $A_5$ have $binom54$ Sylow 2-subgroups?Sylow Subgroups and Conjugation
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Why does group action by conjugation on sylow subgroups define a homomorphism into the symmetric group?
The 2019 Stack Overflow Developer Survey Results Are InGroup Theory, group of order 55Finding the kernel of an action on conjugate subgroupsCheck:etermine the number of Sylow $2$-subgroups and Sylow $3$-subgroups that $G$ can have.On the number of Sylow subgroups in Symmetric GroupWhy the symmetric group $S_6$ has 10 Sylow 3-subgroups?Group action on Sylow subgroupsTransitive group action on Sylow subgroupsProof verification - the only group of order 24 without normal sylow subgroup is $S_4$.Prove that there is at most one non-abelian group of order $28$ all of whose Sylow $2$-subgroups are cyclicWhy does $A_5$ have $binom54$ Sylow 2-subgroups?Sylow Subgroups and Conjugation
$begingroup$
Sylow theorems state that sylow p subgroups of a group G are conjugate. Often I see argumentation that if there are n sylow p subgroups in G then we can define a group action on it by conjugation and hence create a homomorphism from G into Symmetric group or order n. Please provide a proof that this homomorphism is legitimate and why conjugation by any element on a sylow p subgroup takes you to another sylow p subgroup?
group-theory sylow-theory
edited Apr 6 '15 at 11:12
Nicky Hekster
29.1k63456
asked Apr 6 '15 at 10:06
SundipSundip
314
$endgroup$
add a comment |
$begingroup$
Sylow theorems state that sylow p subgroups of a group G are conjugate. Often I see argumentation that if there are n sylow p subgroups in G then we can define a group action on it by conjugation and hence create a homomorphism from G into Symmetric group or order n. Please provide a proof that this homomorphism is legitimate and why conjugation by any element on a sylow p subgroup takes you to another sylow p subgroup?
group-theory sylow-theory
edited Apr 6 '15 at 11:12
Nicky Hekster
29.1k63456
asked Apr 6 '15 at 10:06
SundipSundip
314
$endgroup$
$begingroup$
This action is used in the proof that the Sylow $p$-subgroups are conjugate.
$endgroup$
– Derek Holt
Apr 6 '15 at 11:03
add a comment |
$begingroup$
Sylow theorems state that sylow p subgroups of a group G are conjugate. Often I see argumentation that if there are n sylow p subgroups in G then we can define a group action on it by conjugation and hence create a homomorphism from G into Symmetric group or order n. Please provide a proof that this homomorphism is legitimate and why conjugation by any element on a sylow p subgroup takes you to another sylow p subgroup?
group-theory sylow-theory
edited Apr 6 '15 at 11:12
Nicky Hekster
29.1k63456
asked Apr 6 '15 at 10:06
SundipSundip
314
$endgroup$
Sylow theorems state that sylow p subgroups of a group G are conjugate. Often I see argumentation that if there are n sylow p subgroups in G then we can define a group action on it by conjugation and hence create a homomorphism from G into Symmetric group or order n. Please provide a proof that this homomorphism is legitimate and why conjugation by any element on a sylow p subgroup takes you to another sylow p subgroup?
group-theory sylow-theory
group-theory sylow-theory
edited Apr 6 '15 at 11:12
Nicky Hekster
29.1k63456
asked Apr 6 '15 at 10:06
SundipSundip
314
edited Apr 6 '15 at 11:12
Nicky Hekster
29.1k63456
asked Apr 6 '15 at 10:06
SundipSundip
314
edited Apr 6 '15 at 11:12
Nicky Hekster
29.1k63456
edited Apr 6 '15 at 11:12
Nicky Hekster
29.1k63456
edited Apr 6 '15 at 11:12
Nicky Hekster
29.1k63456
29.1k63456
asked Apr 6 '15 at 10:06
SundipSundip
314
asked Apr 6 '15 at 10:06
SundipSundip
314
asked Apr 6 '15 at 10:06
SundipSundip
314
314
$begingroup$
This action is used in the proof that the Sylow $p$-subgroups are conjugate.
$endgroup$
– Derek Holt
Apr 6 '15 at 11:03
add a comment |
$begingroup$
This action is used in the proof that the Sylow $p$-subgroups are conjugate.
$endgroup$
– Derek Holt
Apr 6 '15 at 11:03
$begingroup$
This action is used in the proof that the Sylow $p$-subgroups are conjugate.
$endgroup$
– Derek Holt
Apr 6 '15 at 11:03
$begingroup$
This action is used in the proof that the Sylow $p$-subgroups are conjugate.
$endgroup$
– Derek Holt
Apr 6 '15 at 11:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In a group $G$ of cardinal $p^km$ with $gcd(p,m)=1$ a $p$-Sylow is defined to be any subgroup $S$ of $G$ whose cardinal is $p^k$. Suppose that $gin G$ then $gSg^-1$ is a subgroup of $G$ and its cardinal must be the same as the cardinal of $S$ because, in a group $G$ conjugation by an element is a one-to-one morphism (it is also onto but you don't need this here), so it sends subgroups to subgroups and the cardinal is constant. Finally $gSg^-1$ is a subgroup of cardinal $|S|=p^k$ that $gSg^-1$ is also a $p$-Sylow. This answers your last question.
For the first question. Suppose $X=S_1,...,S_n$ is the set of $p$-Sylows in $G$ (because of the first Sylow theorem, $ngeq 1$) with an arbitrary enumeration. It is easily seen by the remark above that $G$ acts on $X$ by conjugation. Namely if $S$ is a $p$-Sylow and $gin G$ then :
$$g.S:=gSg^-1in X$$
It is a group action because $1_G.S=S$ and $g.(h.S)=g(h.S)g^-1=ghSh^-1g^-1=(gh).S$. But you can see this action in a different way. Let $Bij(X)$ be the set of functions from $X$ to $X$ which are one to one and onto then define :
$$rho: G rightarrow Bij(X)$$
$$gmapsto [Smapsto gSg^-1] $$
The fact that $rho(g)$ is one to one and onto comes from the fact that $rho(g^-1)=rho(g)^-1$ furthermore from the rules of group action, $rho$ is actually a group morphism from $G$ to $Bij(X)$ (with the law of composition).
Now the last thing to use is that $X$ is in bijection with $1,...,n$. Let us set :
$$psi: 1,...,nrightarrow X $$
$$imapsto S_i $$
Then you have $psi^-1circrho(g)circpsi$ is a bijection of $1,...,n$ i.e. an element of $mathfrakS_n$ (a permutation). So the morphism you are looking for is explicitely :
$$rho_0:Grightarrow mathfrakS_n $$
$$gmapsto psi^-1circrho(g)circpsi $$
It should be remarked that this morphism is not canonical (it explicitely depends on your enumeration i.e. $psi$) nevertheless, it is canonical up to conjugation in $mathfrakS_n$.
answered Apr 6 '15 at 10:21
Clément GuérinClément Guérin
10k1936
$endgroup$
$begingroup$
Thanks. So if a group G has n sylow p subgroups then we can define homomorphism from G to Sn and if order of G is greater than n! then the kernel of the homomorphism is greater than 1 by the first isomorphism theorem and hence G cannot be simple. By sylow kernel cannot be entire group.
$endgroup$
– Sundip
Apr 6 '15 at 11:03
$begingroup$
Well, if $G$ is simple and $|G|>n!$ then the kernel of the morphism cannot be trivial but the kernel could be $G$ as well. Actually (because of the second Sylow's theorem) one can see that the morphism is trivial if and only if $n=1$ (i.e. the $p$-Sylow is unique). That is if $ngeq 2$ and $|G|>n!$ then $G$ cannot be simple.
$endgroup$
– Clément Guérin
Apr 6 '15 at 11:12
add a comment |
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$begingroup$
In a group $G$ of cardinal $p^km$ with $gcd(p,m)=1$ a $p$-Sylow is defined to be any subgroup $S$ of $G$ whose cardinal is $p^k$. Suppose that $gin G$ then $gSg^-1$ is a subgroup of $G$ and its cardinal must be the same as the cardinal of $S$ because, in a group $G$ conjugation by an element is a one-to-one morphism (it is also onto but you don't need this here), so it sends subgroups to subgroups and the cardinal is constant. Finally $gSg^-1$ is a subgroup of cardinal $|S|=p^k$ that $gSg^-1$ is also a $p$-Sylow. This answers your last question.
For the first question. Suppose $X=S_1,...,S_n$ is the set of $p$-Sylows in $G$ (because of the first Sylow theorem, $ngeq 1$) with an arbitrary enumeration. It is easily seen by the remark above that $G$ acts on $X$ by conjugation. Namely if $S$ is a $p$-Sylow and $gin G$ then :
$$g.S:=gSg^-1in X$$
It is a group action because $1_G.S=S$ and $g.(h.S)=g(h.S)g^-1=ghSh^-1g^-1=(gh).S$. But you can see this action in a different way. Let $Bij(X)$ be the set of functions from $X$ to $X$ which are one to one and onto then define :
$$rho: G rightarrow Bij(X)$$
$$gmapsto [Smapsto gSg^-1] $$
The fact that $rho(g)$ is one to one and onto comes from the fact that $rho(g^-1)=rho(g)^-1$ furthermore from the rules of group action, $rho$ is actually a group morphism from $G$ to $Bij(X)$ (with the law of composition).
Now the last thing to use is that $X$ is in bijection with $1,...,n$. Let us set :
$$psi: 1,...,nrightarrow X $$
$$imapsto S_i $$
Then you have $psi^-1circrho(g)circpsi$ is a bijection of $1,...,n$ i.e. an element of $mathfrakS_n$ (a permutation). So the morphism you are looking for is explicitely :
$$rho_0:Grightarrow mathfrakS_n $$
$$gmapsto psi^-1circrho(g)circpsi $$
It should be remarked that this morphism is not canonical (it explicitely depends on your enumeration i.e. $psi$) nevertheless, it is canonical up to conjugation in $mathfrakS_n$.
answered Apr 6 '15 at 10:21
Clément GuérinClément Guérin
10k1936
$endgroup$
$begingroup$
Thanks. So if a group G has n sylow p subgroups then we can define homomorphism from G to Sn and if order of G is greater than n! then the kernel of the homomorphism is greater than 1 by the first isomorphism theorem and hence G cannot be simple. By sylow kernel cannot be entire group.
$endgroup$
– Sundip
Apr 6 '15 at 11:03
$begingroup$
Well, if $G$ is simple and $|G|>n!$ then the kernel of the morphism cannot be trivial but the kernel could be $G$ as well. Actually (because of the second Sylow's theorem) one can see that the morphism is trivial if and only if $n=1$ (i.e. the $p$-Sylow is unique). That is if $ngeq 2$ and $|G|>n!$ then $G$ cannot be simple.
$endgroup$
– Clément Guérin
Apr 6 '15 at 11:12
add a comment |
$begingroup$
In a group $G$ of cardinal $p^km$ with $gcd(p,m)=1$ a $p$-Sylow is defined to be any subgroup $S$ of $G$ whose cardinal is $p^k$. Suppose that $gin G$ then $gSg^-1$ is a subgroup of $G$ and its cardinal must be the same as the cardinal of $S$ because, in a group $G$ conjugation by an element is a one-to-one morphism (it is also onto but you don't need this here), so it sends subgroups to subgroups and the cardinal is constant. Finally $gSg^-1$ is a subgroup of cardinal $|S|=p^k$ that $gSg^-1$ is also a $p$-Sylow. This answers your last question.
For the first question. Suppose $X=S_1,...,S_n$ is the set of $p$-Sylows in $G$ (because of the first Sylow theorem, $ngeq 1$) with an arbitrary enumeration. It is easily seen by the remark above that $G$ acts on $X$ by conjugation. Namely if $S$ is a $p$-Sylow and $gin G$ then :
$$g.S:=gSg^-1in X$$
It is a group action because $1_G.S=S$ and $g.(h.S)=g(h.S)g^-1=ghSh^-1g^-1=(gh).S$. But you can see this action in a different way. Let $Bij(X)$ be the set of functions from $X$ to $X$ which are one to one and onto then define :
$$rho: G rightarrow Bij(X)$$
$$gmapsto [Smapsto gSg^-1] $$
The fact that $rho(g)$ is one to one and onto comes from the fact that $rho(g^-1)=rho(g)^-1$ furthermore from the rules of group action, $rho$ is actually a group morphism from $G$ to $Bij(X)$ (with the law of composition).
Now the last thing to use is that $X$ is in bijection with $1,...,n$. Let us set :
$$psi: 1,...,nrightarrow X $$
$$imapsto S_i $$
Then you have $psi^-1circrho(g)circpsi$ is a bijection of $1,...,n$ i.e. an element of $mathfrakS_n$ (a permutation). So the morphism you are looking for is explicitely :
$$rho_0:Grightarrow mathfrakS_n $$
$$gmapsto psi^-1circrho(g)circpsi $$
It should be remarked that this morphism is not canonical (it explicitely depends on your enumeration i.e. $psi$) nevertheless, it is canonical up to conjugation in $mathfrakS_n$.
answered Apr 6 '15 at 10:21
Clément GuérinClément Guérin
10k1936
$endgroup$
$begingroup$
Thanks. So if a group G has n sylow p subgroups then we can define homomorphism from G to Sn and if order of G is greater than n! then the kernel of the homomorphism is greater than 1 by the first isomorphism theorem and hence G cannot be simple. By sylow kernel cannot be entire group.
$endgroup$
– Sundip
Apr 6 '15 at 11:03
$begingroup$
Well, if $G$ is simple and $|G|>n!$ then the kernel of the morphism cannot be trivial but the kernel could be $G$ as well. Actually (because of the second Sylow's theorem) one can see that the morphism is trivial if and only if $n=1$ (i.e. the $p$-Sylow is unique). That is if $ngeq 2$ and $|G|>n!$ then $G$ cannot be simple.
$endgroup$
– Clément Guérin
Apr 6 '15 at 11:12
add a comment |
$begingroup$
In a group $G$ of cardinal $p^km$ with $gcd(p,m)=1$ a $p$-Sylow is defined to be any subgroup $S$ of $G$ whose cardinal is $p^k$. Suppose that $gin G$ then $gSg^-1$ is a subgroup of $G$ and its cardinal must be the same as the cardinal of $S$ because, in a group $G$ conjugation by an element is a one-to-one morphism (it is also onto but you don't need this here), so it sends subgroups to subgroups and the cardinal is constant. Finally $gSg^-1$ is a subgroup of cardinal $|S|=p^k$ that $gSg^-1$ is also a $p$-Sylow. This answers your last question.
For the first question. Suppose $X=S_1,...,S_n$ is the set of $p$-Sylows in $G$ (because of the first Sylow theorem, $ngeq 1$) with an arbitrary enumeration. It is easily seen by the remark above that $G$ acts on $X$ by conjugation. Namely if $S$ is a $p$-Sylow and $gin G$ then :
$$g.S:=gSg^-1in X$$
It is a group action because $1_G.S=S$ and $g.(h.S)=g(h.S)g^-1=ghSh^-1g^-1=(gh).S$. But you can see this action in a different way. Let $Bij(X)$ be the set of functions from $X$ to $X$ which are one to one and onto then define :
$$rho: G rightarrow Bij(X)$$
$$gmapsto [Smapsto gSg^-1] $$
The fact that $rho(g)$ is one to one and onto comes from the fact that $rho(g^-1)=rho(g)^-1$ furthermore from the rules of group action, $rho$ is actually a group morphism from $G$ to $Bij(X)$ (with the law of composition).
Now the last thing to use is that $X$ is in bijection with $1,...,n$. Let us set :
$$psi: 1,...,nrightarrow X $$
$$imapsto S_i $$
Then you have $psi^-1circrho(g)circpsi$ is a bijection of $1,...,n$ i.e. an element of $mathfrakS_n$ (a permutation). So the morphism you are looking for is explicitely :
$$rho_0:Grightarrow mathfrakS_n $$
$$gmapsto psi^-1circrho(g)circpsi $$
It should be remarked that this morphism is not canonical (it explicitely depends on your enumeration i.e. $psi$) nevertheless, it is canonical up to conjugation in $mathfrakS_n$.
answered Apr 6 '15 at 10:21
Clément GuérinClément Guérin
10k1936
$endgroup$
In a group $G$ of cardinal $p^km$ with $gcd(p,m)=1$ a $p$-Sylow is defined to be any subgroup $S$ of $G$ whose cardinal is $p^k$. Suppose that $gin G$ then $gSg^-1$ is a subgroup of $G$ and its cardinal must be the same as the cardinal of $S$ because, in a group $G$ conjugation by an element is a one-to-one morphism (it is also onto but you don't need this here), so it sends subgroups to subgroups and the cardinal is constant. Finally $gSg^-1$ is a subgroup of cardinal $|S|=p^k$ that $gSg^-1$ is also a $p$-Sylow. This answers your last question.
For the first question. Suppose $X=S_1,...,S_n$ is the set of $p$-Sylows in $G$ (because of the first Sylow theorem, $ngeq 1$) with an arbitrary enumeration. It is easily seen by the remark above that $G$ acts on $X$ by conjugation. Namely if $S$ is a $p$-Sylow and $gin G$ then :
$$g.S:=gSg^-1in X$$
It is a group action because $1_G.S=S$ and $g.(h.S)=g(h.S)g^-1=ghSh^-1g^-1=(gh).S$. But you can see this action in a different way. Let $Bij(X)$ be the set of functions from $X$ to $X$ which are one to one and onto then define :
$$rho: G rightarrow Bij(X)$$
$$gmapsto [Smapsto gSg^-1] $$
The fact that $rho(g)$ is one to one and onto comes from the fact that $rho(g^-1)=rho(g)^-1$ furthermore from the rules of group action, $rho$ is actually a group morphism from $G$ to $Bij(X)$ (with the law of composition).
Now the last thing to use is that $X$ is in bijection with $1,...,n$. Let us set :
$$psi: 1,...,nrightarrow X $$
$$imapsto S_i $$
Then you have $psi^-1circrho(g)circpsi$ is a bijection of $1,...,n$ i.e. an element of $mathfrakS_n$ (a permutation). So the morphism you are looking for is explicitely :
$$rho_0:Grightarrow mathfrakS_n $$
$$gmapsto psi^-1circrho(g)circpsi $$
It should be remarked that this morphism is not canonical (it explicitely depends on your enumeration i.e. $psi$) nevertheless, it is canonical up to conjugation in $mathfrakS_n$.
answered Apr 6 '15 at 10:21
Clément GuérinClément Guérin
10k1936
answered Apr 6 '15 at 10:21
Clément GuérinClément Guérin
10k1936
answered Apr 6 '15 at 10:21
Clément GuérinClément Guérin
10k1936
answered Apr 6 '15 at 10:21
Clément GuérinClément Guérin
10k1936
10k1936
$begingroup$
Thanks. So if a group G has n sylow p subgroups then we can define homomorphism from G to Sn and if order of G is greater than n! then the kernel of the homomorphism is greater than 1 by the first isomorphism theorem and hence G cannot be simple. By sylow kernel cannot be entire group.
$endgroup$
– Sundip
Apr 6 '15 at 11:03
$begingroup$
Well, if $G$ is simple and $|G|>n!$ then the kernel of the morphism cannot be trivial but the kernel could be $G$ as well. Actually (because of the second Sylow's theorem) one can see that the morphism is trivial if and only if $n=1$ (i.e. the $p$-Sylow is unique). That is if $ngeq 2$ and $|G|>n!$ then $G$ cannot be simple.
$endgroup$
– Clément Guérin
Apr 6 '15 at 11:12
add a comment |
$begingroup$
Thanks. So if a group G has n sylow p subgroups then we can define homomorphism from G to Sn and if order of G is greater than n! then the kernel of the homomorphism is greater than 1 by the first isomorphism theorem and hence G cannot be simple. By sylow kernel cannot be entire group.
$endgroup$
– Sundip
Apr 6 '15 at 11:03
$begingroup$
Well, if $G$ is simple and $|G|>n!$ then the kernel of the morphism cannot be trivial but the kernel could be $G$ as well. Actually (because of the second Sylow's theorem) one can see that the morphism is trivial if and only if $n=1$ (i.e. the $p$-Sylow is unique). That is if $ngeq 2$ and $|G|>n!$ then $G$ cannot be simple.
$endgroup$
– Clément Guérin
Apr 6 '15 at 11:12
$begingroup$
Thanks. So if a group G has n sylow p subgroups then we can define homomorphism from G to Sn and if order of G is greater than n! then the kernel of the homomorphism is greater than 1 by the first isomorphism theorem and hence G cannot be simple. By sylow kernel cannot be entire group.
$endgroup$
– Sundip
Apr 6 '15 at 11:03
$begingroup$
Thanks. So if a group G has n sylow p subgroups then we can define homomorphism from G to Sn and if order of G is greater than n! then the kernel of the homomorphism is greater than 1 by the first isomorphism theorem and hence G cannot be simple. By sylow kernel cannot be entire group.
$endgroup$
– Sundip
Apr 6 '15 at 11:03
$begingroup$
Well, if $G$ is simple and $|G|>n!$ then the kernel of the morphism cannot be trivial but the kernel could be $G$ as well. Actually (because of the second Sylow's theorem) one can see that the morphism is trivial if and only if $n=1$ (i.e. the $p$-Sylow is unique). That is if $ngeq 2$ and $|G|>n!$ then $G$ cannot be simple.
$endgroup$
– Clément Guérin
Apr 6 '15 at 11:12
$begingroup$
Well, if $G$ is simple and $|G|>n!$ then the kernel of the morphism cannot be trivial but the kernel could be $G$ as well. Actually (because of the second Sylow's theorem) one can see that the morphism is trivial if and only if $n=1$ (i.e. the $p$-Sylow is unique). That is if $ngeq 2$ and $|G|>n!$ then $G$ cannot be simple.
$endgroup$
– Clément Guérin
Apr 6 '15 at 11:12
add a comment |
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$begingroup$
This action is used in the proof that the Sylow $p$-subgroups are conjugate.
$endgroup$
– Derek Holt
Apr 6 '15 at 11:03