$x,y in Z_G(P)$ such that $gxg^{-1}=y$, show that there exists $nin N_G(P)$ such that $nxn^{-1}=y$.
up vote
2
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Past year paper question:
Let $P$ be a Sylow p-subgroup of a finite group $G$.
Define $Z_G(P)= { g in G mid pg=gp quad forall p in P }$
Suppose $x,y in Z_G(P)$ such that $gxg^{-1}=y$ for some $g in G$. Show that there exists $nin N_G(P)$ such that $nxn^{-1}=y$.
I'm guessing I need to to use the fact that $gxg^{-1}=y$, and maybe find some $z$ such that $zgxg^{-1}z^{-1}=y$ and $(zg) in N_G(P)$. Then $zg$ would be the $n$ that I need to find.
$zgxg^{-1}z^{-1}=y$ would mean that $zyz^{-1}=y$ so $z$ must commute with $y$. Also, $(zg) in N_G(P)$ would mean that $zgPg^{-1}z^{-1}=P$. But how does one proceed from here?
Some other things:
I get that $yp=py implies (gxg^{-1})p = (gxg^{-1})p$
I get some easy facts that $P triangleleft N_G(P), Z_G(P) leq N_G(P)$, and hence $PZ_G(P)$ is a subgroup of $N_G(P)$. And maybe I need to define some clever group action but I think of one that leads somewhere.
abstract-algebra group-theory
asked Nov 22 at 14:02
eatfood
1827
|
show 1 more comment
up vote
2
down vote
favorite
Past year paper question:
Let $P$ be a Sylow p-subgroup of a finite group $G$.
Define $Z_G(P)= { g in G mid pg=gp quad forall p in P }$
Suppose $x,y in Z_G(P)$ such that $gxg^{-1}=y$ for some $g in G$. Show that there exists $nin N_G(P)$ such that $nxn^{-1}=y$.
I'm guessing I need to to use the fact that $gxg^{-1}=y$, and maybe find some $z$ such that $zgxg^{-1}z^{-1}=y$ and $(zg) in N_G(P)$. Then $zg$ would be the $n$ that I need to find.
$zgxg^{-1}z^{-1}=y$ would mean that $zyz^{-1}=y$ so $z$ must commute with $y$. Also, $(zg) in N_G(P)$ would mean that $zgPg^{-1}z^{-1}=P$. But how does one proceed from here?
Some other things:
I get that $yp=py implies (gxg^{-1})p = (gxg^{-1})p$
I get some easy facts that $P triangleleft N_G(P), Z_G(P) leq N_G(P)$, and hence $PZ_G(P)$ is a subgroup of $N_G(P)$. And maybe I need to define some clever group action but I think of one that leads somewhere.
abstract-algebra group-theory
asked Nov 22 at 14:02
eatfood
1827
What would $n$ be? I think $g$ does not need to be inside $N_G(P)$.
– eatfood
Nov 22 at 14:14
2
Here is a hint. $y=gxg^{-1} in Z_G(gPg^{-1})$, so $P$ and $gPg^{-1}$ are both Sylow $p$-subgroups of $Z_G(y)$, and hence they are conjugate in $Z_G(y)$.
– Derek Holt
Nov 22 at 14:18
Indeed I did, @DerekHolt. Thanks.
– DonAntonio
Nov 22 at 15:17
@DerekHolt Ok, so $Z_G(y)$ contains $P$, so $P$ is Sylow p-subgroup of $Z_G(y)$. $y in Z_G(gPg^{-1})$ means $y$ commutes with $gpg^{-1}$ for all $p$. Which means that $gpg^{-1} in Z_G(y)$ for all $p$, so $gPg^{-1}$ is also a Sylow p-subgroup of $Z_G(y)$. Then Sylow p-subgroups are conjugate implies there exist a $z$ such that $ zgPg^{-1}z^{-1} = P$. And then $n=zg$. Thank you so much. One more question: how do you know to consider $Z_G(y)$? I always get stuck when I need to think creatively...
– eatfood
Nov 22 at 15:31
My solution is the same as the_foxes. I don't really know why one would think of considering $Z_G(y)$, but I knew from experience that you would need to use the conjugacy part of Sylow' Theorem somewhere. This type of argument is generally call the Frattini Argument. I guess $x in C_G(P)$ immediately implies $x^g in Z_G(P^g)$, but we already know that $y in Z_G(P)$, so both $P$ and $P^g$ centralize $y$.
– Derek Holt
Nov 22 at 18:44
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Past year paper question:
Let $P$ be a Sylow p-subgroup of a finite group $G$.
Define $Z_G(P)= { g in G mid pg=gp quad forall p in P }$
Suppose $x,y in Z_G(P)$ such that $gxg^{-1}=y$ for some $g in G$. Show that there exists $nin N_G(P)$ such that $nxn^{-1}=y$.
I'm guessing I need to to use the fact that $gxg^{-1}=y$, and maybe find some $z$ such that $zgxg^{-1}z^{-1}=y$ and $(zg) in N_G(P)$. Then $zg$ would be the $n$ that I need to find.
$zgxg^{-1}z^{-1}=y$ would mean that $zyz^{-1}=y$ so $z$ must commute with $y$. Also, $(zg) in N_G(P)$ would mean that $zgPg^{-1}z^{-1}=P$. But how does one proceed from here?
Some other things:
I get that $yp=py implies (gxg^{-1})p = (gxg^{-1})p$
I get some easy facts that $P triangleleft N_G(P), Z_G(P) leq N_G(P)$, and hence $PZ_G(P)$ is a subgroup of $N_G(P)$. And maybe I need to define some clever group action but I think of one that leads somewhere.
abstract-algebra group-theory
asked Nov 22 at 14:02
eatfood
1827
Past year paper question:
Let $P$ be a Sylow p-subgroup of a finite group $G$.
Define $Z_G(P)= { g in G mid pg=gp quad forall p in P }$
Suppose $x,y in Z_G(P)$ such that $gxg^{-1}=y$ for some $g in G$. Show that there exists $nin N_G(P)$ such that $nxn^{-1}=y$.
I'm guessing I need to to use the fact that $gxg^{-1}=y$, and maybe find some $z$ such that $zgxg^{-1}z^{-1}=y$ and $(zg) in N_G(P)$. Then $zg$ would be the $n$ that I need to find.
$zgxg^{-1}z^{-1}=y$ would mean that $zyz^{-1}=y$ so $z$ must commute with $y$. Also, $(zg) in N_G(P)$ would mean that $zgPg^{-1}z^{-1}=P$. But how does one proceed from here?
Some other things:
I get that $yp=py implies (gxg^{-1})p = (gxg^{-1})p$
I get some easy facts that $P triangleleft N_G(P), Z_G(P) leq N_G(P)$, and hence $PZ_G(P)$ is a subgroup of $N_G(P)$. And maybe I need to define some clever group action but I think of one that leads somewhere.
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 22 at 14:02
eatfood
1827
asked Nov 22 at 14:02
eatfood
1827
edited Nov 22 at 14:12
asked Nov 22 at 14:02
eatfood
1827
asked Nov 22 at 14:02
eatfood
1827
asked Nov 22 at 14:02
eatfood
1827
1827
What would $n$ be? I think $g$ does not need to be inside $N_G(P)$.
– eatfood
Nov 22 at 14:14
2
Here is a hint. $y=gxg^{-1} in Z_G(gPg^{-1})$, so $P$ and $gPg^{-1}$ are both Sylow $p$-subgroups of $Z_G(y)$, and hence they are conjugate in $Z_G(y)$.
– Derek Holt
Nov 22 at 14:18
Indeed I did, @DerekHolt. Thanks.
– DonAntonio
Nov 22 at 15:17
@DerekHolt Ok, so $Z_G(y)$ contains $P$, so $P$ is Sylow p-subgroup of $Z_G(y)$. $y in Z_G(gPg^{-1})$ means $y$ commutes with $gpg^{-1}$ for all $p$. Which means that $gpg^{-1} in Z_G(y)$ for all $p$, so $gPg^{-1}$ is also a Sylow p-subgroup of $Z_G(y)$. Then Sylow p-subgroups are conjugate implies there exist a $z$ such that $ zgPg^{-1}z^{-1} = P$. And then $n=zg$. Thank you so much. One more question: how do you know to consider $Z_G(y)$? I always get stuck when I need to think creatively...
– eatfood
Nov 22 at 15:31
My solution is the same as the_foxes. I don't really know why one would think of considering $Z_G(y)$, but I knew from experience that you would need to use the conjugacy part of Sylow' Theorem somewhere. This type of argument is generally call the Frattini Argument. I guess $x in C_G(P)$ immediately implies $x^g in Z_G(P^g)$, but we already know that $y in Z_G(P)$, so both $P$ and $P^g$ centralize $y$.
– Derek Holt
Nov 22 at 18:44
|
show 1 more comment
What would $n$ be? I think $g$ does not need to be inside $N_G(P)$.
– eatfood
Nov 22 at 14:14
2
Here is a hint. $y=gxg^{-1} in Z_G(gPg^{-1})$, so $P$ and $gPg^{-1}$ are both Sylow $p$-subgroups of $Z_G(y)$, and hence they are conjugate in $Z_G(y)$.
– Derek Holt
Nov 22 at 14:18
Indeed I did, @DerekHolt. Thanks.
– DonAntonio
Nov 22 at 15:17
@DerekHolt Ok, so $Z_G(y)$ contains $P$, so $P$ is Sylow p-subgroup of $Z_G(y)$. $y in Z_G(gPg^{-1})$ means $y$ commutes with $gpg^{-1}$ for all $p$. Which means that $gpg^{-1} in Z_G(y)$ for all $p$, so $gPg^{-1}$ is also a Sylow p-subgroup of $Z_G(y)$. Then Sylow p-subgroups are conjugate implies there exist a $z$ such that $ zgPg^{-1}z^{-1} = P$. And then $n=zg$. Thank you so much. One more question: how do you know to consider $Z_G(y)$? I always get stuck when I need to think creatively...
– eatfood
Nov 22 at 15:31
My solution is the same as the_foxes. I don't really know why one would think of considering $Z_G(y)$, but I knew from experience that you would need to use the conjugacy part of Sylow' Theorem somewhere. This type of argument is generally call the Frattini Argument. I guess $x in C_G(P)$ immediately implies $x^g in Z_G(P^g)$, but we already know that $y in Z_G(P)$, so both $P$ and $P^g$ centralize $y$.
– Derek Holt
Nov 22 at 18:44
What would $n$ be? I think $g$ does not need to be inside $N_G(P)$.
– eatfood
Nov 22 at 14:14
What would $n$ be? I think $g$ does not need to be inside $N_G(P)$.
– eatfood
Nov 22 at 14:14
2
2
Here is a hint. $y=gxg^{-1} in Z_G(gPg^{-1})$, so $P$ and $gPg^{-1}$ are both Sylow $p$-subgroups of $Z_G(y)$, and hence they are conjugate in $Z_G(y)$.
– Derek Holt
Nov 22 at 14:18
Here is a hint. $y=gxg^{-1} in Z_G(gPg^{-1})$, so $P$ and $gPg^{-1}$ are both Sylow $p$-subgroups of $Z_G(y)$, and hence they are conjugate in $Z_G(y)$.
– Derek Holt
Nov 22 at 14:18
Indeed I did, @DerekHolt. Thanks.
– DonAntonio
Nov 22 at 15:17
Indeed I did, @DerekHolt. Thanks.
– DonAntonio
Nov 22 at 15:17
@DerekHolt Ok, so $Z_G(y)$ contains $P$, so $P$ is Sylow p-subgroup of $Z_G(y)$. $y in Z_G(gPg^{-1})$ means $y$ commutes with $gpg^{-1}$ for all $p$. Which means that $gpg^{-1} in Z_G(y)$ for all $p$, so $gPg^{-1}$ is also a Sylow p-subgroup of $Z_G(y)$. Then Sylow p-subgroups are conjugate implies there exist a $z$ such that $ zgPg^{-1}z^{-1} = P$. And then $n=zg$. Thank you so much. One more question: how do you know to consider $Z_G(y)$? I always get stuck when I need to think creatively...
– eatfood
Nov 22 at 15:31
@DerekHolt Ok, so $Z_G(y)$ contains $P$, so $P$ is Sylow p-subgroup of $Z_G(y)$. $y in Z_G(gPg^{-1})$ means $y$ commutes with $gpg^{-1}$ for all $p$. Which means that $gpg^{-1} in Z_G(y)$ for all $p$, so $gPg^{-1}$ is also a Sylow p-subgroup of $Z_G(y)$. Then Sylow p-subgroups are conjugate implies there exist a $z$ such that $ zgPg^{-1}z^{-1} = P$. And then $n=zg$. Thank you so much. One more question: how do you know to consider $Z_G(y)$? I always get stuck when I need to think creatively...
– eatfood
Nov 22 at 15:31
My solution is the same as the_foxes. I don't really know why one would think of considering $Z_G(y)$, but I knew from experience that you would need to use the conjugacy part of Sylow' Theorem somewhere. This type of argument is generally call the Frattini Argument. I guess $x in C_G(P)$ immediately implies $x^g in Z_G(P^g)$, but we already know that $y in Z_G(P)$, so both $P$ and $P^g$ centralize $y$.
– Derek Holt
Nov 22 at 18:44
My solution is the same as the_foxes. I don't really know why one would think of considering $Z_G(y)$, but I knew from experience that you would need to use the conjugacy part of Sylow' Theorem somewhere. This type of argument is generally call the Frattini Argument. I guess $x in C_G(P)$ immediately implies $x^g in Z_G(P^g)$, but we already know that $y in Z_G(P)$, so both $P$ and $P^g$ centralize $y$.
– Derek Holt
Nov 22 at 18:44
|
show 1 more comment
1 Answer
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I'll write $C_G(P)$ for the centraliser of $P$ in $G$, which is what you have defined as $Z_G(P)$. We are given $x, y in C_G(P)$ which are conjugate in $G$ and we want to show that they are in fact conjugate in $N_G(P)$.
So let $y=x^g$ for some $g in G$ and note that since $x in C_G(P)$ we have $y=x^g in C_G(P)^g = C_G(P^g)$. Then $P^g leq C_G(y)$, thus $P^g$ is a Sylow $p$-subgroup of $C_G(y)$. We also have that $y in C_G(P)$ implies $P leq C_G(y)$ so $P$ is also a Sylow $p$-subgroup of $C_G(y)$.
We deduce that $P$ and $P^g$ are conjugate in $C_G(y)$ by Sylow's theorem, so $(P^g)^c = P$ for some $c in C_G(y)$. Since $(P^g)^c = P^{gc}$ we see that $gc$ normalises $P$, thus $gc in N_G(P)$ and also $x^{gc}=y^c=y$, so $x, y$ are conjugate in $N_G(P)$, which is what we wanted to prove.
answered Nov 22 at 14:18
the_fox
2,3191430
add a comment |
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I'll write $C_G(P)$ for the centraliser of $P$ in $G$, which is what you have defined as $Z_G(P)$. We are given $x, y in C_G(P)$ which are conjugate in $G$ and we want to show that they are in fact conjugate in $N_G(P)$.
So let $y=x^g$ for some $g in G$ and note that since $x in C_G(P)$ we have $y=x^g in C_G(P)^g = C_G(P^g)$. Then $P^g leq C_G(y)$, thus $P^g$ is a Sylow $p$-subgroup of $C_G(y)$. We also have that $y in C_G(P)$ implies $P leq C_G(y)$ so $P$ is also a Sylow $p$-subgroup of $C_G(y)$.
We deduce that $P$ and $P^g$ are conjugate in $C_G(y)$ by Sylow's theorem, so $(P^g)^c = P$ for some $c in C_G(y)$. Since $(P^g)^c = P^{gc}$ we see that $gc$ normalises $P$, thus $gc in N_G(P)$ and also $x^{gc}=y^c=y$, so $x, y$ are conjugate in $N_G(P)$, which is what we wanted to prove.
answered Nov 22 at 14:18
the_fox
2,3191430
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up vote
3
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accepted
I'll write $C_G(P)$ for the centraliser of $P$ in $G$, which is what you have defined as $Z_G(P)$. We are given $x, y in C_G(P)$ which are conjugate in $G$ and we want to show that they are in fact conjugate in $N_G(P)$.
So let $y=x^g$ for some $g in G$ and note that since $x in C_G(P)$ we have $y=x^g in C_G(P)^g = C_G(P^g)$. Then $P^g leq C_G(y)$, thus $P^g$ is a Sylow $p$-subgroup of $C_G(y)$. We also have that $y in C_G(P)$ implies $P leq C_G(y)$ so $P$ is also a Sylow $p$-subgroup of $C_G(y)$.
We deduce that $P$ and $P^g$ are conjugate in $C_G(y)$ by Sylow's theorem, so $(P^g)^c = P$ for some $c in C_G(y)$. Since $(P^g)^c = P^{gc}$ we see that $gc$ normalises $P$, thus $gc in N_G(P)$ and also $x^{gc}=y^c=y$, so $x, y$ are conjugate in $N_G(P)$, which is what we wanted to prove.
answered Nov 22 at 14:18
the_fox
2,3191430
add a comment |
up vote
3
down vote
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up vote
3
down vote
accepted
I'll write $C_G(P)$ for the centraliser of $P$ in $G$, which is what you have defined as $Z_G(P)$. We are given $x, y in C_G(P)$ which are conjugate in $G$ and we want to show that they are in fact conjugate in $N_G(P)$.
So let $y=x^g$ for some $g in G$ and note that since $x in C_G(P)$ we have $y=x^g in C_G(P)^g = C_G(P^g)$. Then $P^g leq C_G(y)$, thus $P^g$ is a Sylow $p$-subgroup of $C_G(y)$. We also have that $y in C_G(P)$ implies $P leq C_G(y)$ so $P$ is also a Sylow $p$-subgroup of $C_G(y)$.
We deduce that $P$ and $P^g$ are conjugate in $C_G(y)$ by Sylow's theorem, so $(P^g)^c = P$ for some $c in C_G(y)$. Since $(P^g)^c = P^{gc}$ we see that $gc$ normalises $P$, thus $gc in N_G(P)$ and also $x^{gc}=y^c=y$, so $x, y$ are conjugate in $N_G(P)$, which is what we wanted to prove.
answered Nov 22 at 14:18
the_fox
2,3191430
I'll write $C_G(P)$ for the centraliser of $P$ in $G$, which is what you have defined as $Z_G(P)$. We are given $x, y in C_G(P)$ which are conjugate in $G$ and we want to show that they are in fact conjugate in $N_G(P)$.
So let $y=x^g$ for some $g in G$ and note that since $x in C_G(P)$ we have $y=x^g in C_G(P)^g = C_G(P^g)$. Then $P^g leq C_G(y)$, thus $P^g$ is a Sylow $p$-subgroup of $C_G(y)$. We also have that $y in C_G(P)$ implies $P leq C_G(y)$ so $P$ is also a Sylow $p$-subgroup of $C_G(y)$.
We deduce that $P$ and $P^g$ are conjugate in $C_G(y)$ by Sylow's theorem, so $(P^g)^c = P$ for some $c in C_G(y)$. Since $(P^g)^c = P^{gc}$ we see that $gc$ normalises $P$, thus $gc in N_G(P)$ and also $x^{gc}=y^c=y$, so $x, y$ are conjugate in $N_G(P)$, which is what we wanted to prove.
answered Nov 22 at 14:18
the_fox
2,3191430
answered Nov 22 at 14:18
the_fox
2,3191430
answered Nov 22 at 14:18
the_fox
2,3191430
answered Nov 22 at 14:18
the_fox
2,3191430
2,3191430
add a comment |
add a comment |
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What would $n$ be? I think $g$ does not need to be inside $N_G(P)$.
– eatfood
Nov 22 at 14:14
2
Here is a hint. $y=gxg^{-1} in Z_G(gPg^{-1})$, so $P$ and $gPg^{-1}$ are both Sylow $p$-subgroups of $Z_G(y)$, and hence they are conjugate in $Z_G(y)$.
– Derek Holt
Nov 22 at 14:18
Indeed I did, @DerekHolt. Thanks.
– DonAntonio
Nov 22 at 15:17
@DerekHolt Ok, so $Z_G(y)$ contains $P$, so $P$ is Sylow p-subgroup of $Z_G(y)$. $y in Z_G(gPg^{-1})$ means $y$ commutes with $gpg^{-1}$ for all $p$. Which means that $gpg^{-1} in Z_G(y)$ for all $p$, so $gPg^{-1}$ is also a Sylow p-subgroup of $Z_G(y)$. Then Sylow p-subgroups are conjugate implies there exist a $z$ such that $ zgPg^{-1}z^{-1} = P$. And then $n=zg$. Thank you so much. One more question: how do you know to consider $Z_G(y)$? I always get stuck when I need to think creatively...
– eatfood
Nov 22 at 15:31
My solution is the same as the_foxes. I don't really know why one would think of considering $Z_G(y)$, but I knew from experience that you would need to use the conjugacy part of Sylow' Theorem somewhere. This type of argument is generally call the Frattini Argument. I guess $x in C_G(P)$ immediately implies $x^g in Z_G(P^g)$, but we already know that $y in Z_G(P)$, so both $P$ and $P^g$ centralize $y$.
– Derek Holt
Nov 22 at 18:44