a normal k-1 transitive subgroup
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I want to show that if G is k-transitive and N is a normal nontrivial subgroup then is K-1 transitive.
I know that I should use the fact G preserve the orbit of N's action of X. so for some Y on N's action, gY is an orbit of N's action on X.
Any help is appreciated.
group-theory
asked Nov 16 at 19:49
N3d4
102
add a comment |
up vote
0
down vote
favorite
I want to show that if G is k-transitive and N is a normal nontrivial subgroup then is K-1 transitive.
I know that I should use the fact G preserve the orbit of N's action of X. so for some Y on N's action, gY is an orbit of N's action on X.
Any help is appreciated.
group-theory
asked Nov 16 at 19:49
N3d4
102
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to show that if G is k-transitive and N is a normal nontrivial subgroup then is K-1 transitive.
I know that I should use the fact G preserve the orbit of N's action of X. so for some Y on N's action, gY is an orbit of N's action on X.
Any help is appreciated.
group-theory
asked Nov 16 at 19:49
N3d4
102
I want to show that if G is k-transitive and N is a normal nontrivial subgroup then is K-1 transitive.
I know that I should use the fact G preserve the orbit of N's action of X. so for some Y on N's action, gY is an orbit of N's action on X.
Any help is appreciated.
group-theory
group-theory
asked Nov 16 at 19:49
N3d4
102
asked Nov 16 at 19:49
N3d4
102
asked Nov 16 at 19:49
N3d4
102
asked Nov 16 at 19:49
N3d4
102
asked Nov 16 at 19:49
N3d4
102
102
add a comment |
add a comment |
1 Answer
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3
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You won't be able to show it, because it is not true. For example $S_4$ is $4$-transitive, but has a normal subgroup of order $4$, which is only $1$-transitive.
More generally, for all $n ge 2$, there is a group ${rm AGL}(n,2)$ of degree $2^n$ which is $3$-transitive but has a normal subgroup of order $2^n$ which is only $1$-transitive. (The case $n=2$ gives $S_4$.)
answered Nov 16 at 21:34
Derek Holt
52k53468
Thank you. But it is an exersise from book. So I think I can show it. I forgot to mention K>2.
– N3d4
Nov 17 at 5:51
As I said, $S_4$ is clearly a counterexample with $k=4$, so how can you possibly show it? Exercises in books can be wrong.
– Derek Holt
Nov 17 at 8:39
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You won't be able to show it, because it is not true. For example $S_4$ is $4$-transitive, but has a normal subgroup of order $4$, which is only $1$-transitive.
More generally, for all $n ge 2$, there is a group ${rm AGL}(n,2)$ of degree $2^n$ which is $3$-transitive but has a normal subgroup of order $2^n$ which is only $1$-transitive. (The case $n=2$ gives $S_4$.)
answered Nov 16 at 21:34
Derek Holt
52k53468
Thank you. But it is an exersise from book. So I think I can show it. I forgot to mention K>2.
– N3d4
Nov 17 at 5:51
As I said, $S_4$ is clearly a counterexample with $k=4$, so how can you possibly show it? Exercises in books can be wrong.
– Derek Holt
Nov 17 at 8:39
add a comment |
up vote
3
down vote
accepted
You won't be able to show it, because it is not true. For example $S_4$ is $4$-transitive, but has a normal subgroup of order $4$, which is only $1$-transitive.
More generally, for all $n ge 2$, there is a group ${rm AGL}(n,2)$ of degree $2^n$ which is $3$-transitive but has a normal subgroup of order $2^n$ which is only $1$-transitive. (The case $n=2$ gives $S_4$.)
answered Nov 16 at 21:34
Derek Holt
52k53468
Thank you. But it is an exersise from book. So I think I can show it. I forgot to mention K>2.
– N3d4
Nov 17 at 5:51
As I said, $S_4$ is clearly a counterexample with $k=4$, so how can you possibly show it? Exercises in books can be wrong.
– Derek Holt
Nov 17 at 8:39
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You won't be able to show it, because it is not true. For example $S_4$ is $4$-transitive, but has a normal subgroup of order $4$, which is only $1$-transitive.
More generally, for all $n ge 2$, there is a group ${rm AGL}(n,2)$ of degree $2^n$ which is $3$-transitive but has a normal subgroup of order $2^n$ which is only $1$-transitive. (The case $n=2$ gives $S_4$.)
answered Nov 16 at 21:34
Derek Holt
52k53468
You won't be able to show it, because it is not true. For example $S_4$ is $4$-transitive, but has a normal subgroup of order $4$, which is only $1$-transitive.
More generally, for all $n ge 2$, there is a group ${rm AGL}(n,2)$ of degree $2^n$ which is $3$-transitive but has a normal subgroup of order $2^n$ which is only $1$-transitive. (The case $n=2$ gives $S_4$.)
answered Nov 16 at 21:34
Derek Holt
52k53468
answered Nov 16 at 21:34
Derek Holt
52k53468
answered Nov 16 at 21:34
Derek Holt
52k53468
answered Nov 16 at 21:34
Derek Holt
52k53468
52k53468
Thank you. But it is an exersise from book. So I think I can show it. I forgot to mention K>2.
– N3d4
Nov 17 at 5:51
As I said, $S_4$ is clearly a counterexample with $k=4$, so how can you possibly show it? Exercises in books can be wrong.
– Derek Holt
Nov 17 at 8:39
add a comment |
Thank you. But it is an exersise from book. So I think I can show it. I forgot to mention K>2.
– N3d4
Nov 17 at 5:51
As I said, $S_4$ is clearly a counterexample with $k=4$, so how can you possibly show it? Exercises in books can be wrong.
– Derek Holt
Nov 17 at 8:39
Thank you. But it is an exersise from book. So I think I can show it. I forgot to mention K>2.
– N3d4
Nov 17 at 5:51
Thank you. But it is an exersise from book. So I think I can show it. I forgot to mention K>2.
– N3d4
Nov 17 at 5:51
As I said, $S_4$ is clearly a counterexample with $k=4$, so how can you possibly show it? Exercises in books can be wrong.
– Derek Holt
Nov 17 at 8:39
As I said, $S_4$ is clearly a counterexample with $k=4$, so how can you possibly show it? Exercises in books can be wrong.
– Derek Holt
Nov 17 at 8:39
add a comment |
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