Hall subgroup that has normal complement
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Let $H$ be a hall subgroup of finite group $G$. If $Hleq Z(N_{G}(H))$ then $H$ has a normal complement in $G$.
I want to know if we assume $|H|=n$ , $H$ is the only subgroup of $ N_{G}(H)$ with order $n$ ?
Or is the below statment true for hall subgroups?
" Let $H$ be a hall subgroup of finite group $G$. Also assume that $a,b in C_{G}(H)$.If $a,b$ are conjugate in $G$ then they are conjugate in $ N_{G}(H)$".
If one of the two is true I can reach the answer of the main question.
group-theory finite-groups
asked 20 hours ago
Yasmin
1009
|
show 3 more comments
up vote
0
down vote
favorite
Let $H$ be a hall subgroup of finite group $G$. If $Hleq Z(N_{G}(H))$ then $H$ has a normal complement in $G$.
I want to know if we assume $|H|=n$ , $H$ is the only subgroup of $ N_{G}(H)$ with order $n$ ?
Or is the below statment true for hall subgroups?
" Let $H$ be a hall subgroup of finite group $G$. Also assume that $a,b in C_{G}(H)$.If $a,b$ are conjugate in $G$ then they are conjugate in $ N_{G}(H)$".
If one of the two is true I can reach the answer of the main question.
group-theory finite-groups
asked 20 hours ago
Yasmin
1009
Any two Hall $pi$-subgroups are conjugate, and you know $H$ is normal in $N_G(H)$.
– user10354138
19 hours ago
@user10354138 Don't you need to assume the group is solvable to have all Hall $pi$-subgroups be conjugate? Or am I misremembering which parts of those theorems require solvability?
– Tobias Kildetoft
19 hours ago
@user10354138 in the way you say we need $G$ be solvable.
– Yasmin
19 hours ago
oops, sorry, forgotten about that part.
– user10354138
19 hours ago
As I commented on the (now deleted) MO question, if $H$ is a normal Hall subgroup and $K$ is a subgroup such that $|K|$ divides $|H|$ then $Kleq H$ (just consider the subgroup $HK$).
– Tobias Kildetoft
19 hours ago
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $H$ be a hall subgroup of finite group $G$. If $Hleq Z(N_{G}(H))$ then $H$ has a normal complement in $G$.
I want to know if we assume $|H|=n$ , $H$ is the only subgroup of $ N_{G}(H)$ with order $n$ ?
Or is the below statment true for hall subgroups?
" Let $H$ be a hall subgroup of finite group $G$. Also assume that $a,b in C_{G}(H)$.If $a,b$ are conjugate in $G$ then they are conjugate in $ N_{G}(H)$".
If one of the two is true I can reach the answer of the main question.
group-theory finite-groups
asked 20 hours ago
Yasmin
1009
Let $H$ be a hall subgroup of finite group $G$. If $Hleq Z(N_{G}(H))$ then $H$ has a normal complement in $G$.
I want to know if we assume $|H|=n$ , $H$ is the only subgroup of $ N_{G}(H)$ with order $n$ ?
Or is the below statment true for hall subgroups?
" Let $H$ be a hall subgroup of finite group $G$. Also assume that $a,b in C_{G}(H)$.If $a,b$ are conjugate in $G$ then they are conjugate in $ N_{G}(H)$".
If one of the two is true I can reach the answer of the main question.
group-theory finite-groups
group-theory finite-groups
asked 20 hours ago
Yasmin
1009
asked 20 hours ago
Yasmin
1009
edited 17 hours ago
asked 20 hours ago
Yasmin
1009
asked 20 hours ago
Yasmin
1009
asked 20 hours ago
Yasmin
1009
1009
Any two Hall $pi$-subgroups are conjugate, and you know $H$ is normal in $N_G(H)$.
– user10354138
19 hours ago
@user10354138 Don't you need to assume the group is solvable to have all Hall $pi$-subgroups be conjugate? Or am I misremembering which parts of those theorems require solvability?
– Tobias Kildetoft
19 hours ago
@user10354138 in the way you say we need $G$ be solvable.
– Yasmin
19 hours ago
oops, sorry, forgotten about that part.
– user10354138
19 hours ago
As I commented on the (now deleted) MO question, if $H$ is a normal Hall subgroup and $K$ is a subgroup such that $|K|$ divides $|H|$ then $Kleq H$ (just consider the subgroup $HK$).
– Tobias Kildetoft
19 hours ago
|
show 3 more comments
Any two Hall $pi$-subgroups are conjugate, and you know $H$ is normal in $N_G(H)$.
– user10354138
19 hours ago
@user10354138 Don't you need to assume the group is solvable to have all Hall $pi$-subgroups be conjugate? Or am I misremembering which parts of those theorems require solvability?
– Tobias Kildetoft
19 hours ago
@user10354138 in the way you say we need $G$ be solvable.
– Yasmin
19 hours ago
oops, sorry, forgotten about that part.
– user10354138
19 hours ago
As I commented on the (now deleted) MO question, if $H$ is a normal Hall subgroup and $K$ is a subgroup such that $|K|$ divides $|H|$ then $Kleq H$ (just consider the subgroup $HK$).
– Tobias Kildetoft
19 hours ago
Any two Hall $pi$-subgroups are conjugate, and you know $H$ is normal in $N_G(H)$.
– user10354138
19 hours ago
Any two Hall $pi$-subgroups are conjugate, and you know $H$ is normal in $N_G(H)$.
– user10354138
19 hours ago
@user10354138 Don't you need to assume the group is solvable to have all Hall $pi$-subgroups be conjugate? Or am I misremembering which parts of those theorems require solvability?
– Tobias Kildetoft
19 hours ago
@user10354138 Don't you need to assume the group is solvable to have all Hall $pi$-subgroups be conjugate? Or am I misremembering which parts of those theorems require solvability?
– Tobias Kildetoft
19 hours ago
@user10354138 in the way you say we need $G$ be solvable.
– Yasmin
19 hours ago
@user10354138 in the way you say we need $G$ be solvable.
– Yasmin
19 hours ago
oops, sorry, forgotten about that part.
– user10354138
19 hours ago
oops, sorry, forgotten about that part.
– user10354138
19 hours ago
As I commented on the (now deleted) MO question, if $H$ is a normal Hall subgroup and $K$ is a subgroup such that $|K|$ divides $|H|$ then $Kleq H$ (just consider the subgroup $HK$).
– Tobias Kildetoft
19 hours ago
As I commented on the (now deleted) MO question, if $H$ is a normal Hall subgroup and $K$ is a subgroup such that $|K|$ divides $|H|$ then $Kleq H$ (just consider the subgroup $HK$).
– Tobias Kildetoft
19 hours ago
|
show 3 more comments
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4MqJgSyEtLOtXPnF2nQln0i,0gi
Any two Hall $pi$-subgroups are conjugate, and you know $H$ is normal in $N_G(H)$.
– user10354138
19 hours ago
@user10354138 Don't you need to assume the group is solvable to have all Hall $pi$-subgroups be conjugate? Or am I misremembering which parts of those theorems require solvability?
– Tobias Kildetoft
19 hours ago
@user10354138 in the way you say we need $G$ be solvable.
– Yasmin
19 hours ago
oops, sorry, forgotten about that part.
– user10354138
19 hours ago
As I commented on the (now deleted) MO question, if $H$ is a normal Hall subgroup and $K$ is a subgroup such that $|K|$ divides $|H|$ then $Kleq H$ (just consider the subgroup $HK$).
– Tobias Kildetoft
19 hours ago