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.










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  • 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

















up vote
0
down vote

favorite
1












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.










share|cite|improve this question
























  • 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















up vote
0
down vote

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down vote

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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.










share|cite|improve this question















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






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share|cite|improve this question













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share|cite|improve this question








edited 17 hours ago

























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




















  • 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

















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