Classify 3-transitive permutation group of degree 6

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This problem is from the book "Introduction to Group Theory" by Derek J.S Robinson. So, the notations are according to the book.

Prove that the only 3-transitive permutation groups of degree $6$ are $S_6$, $A_6$, and $L(5)$.

First I elaborate a little bit on $L(5)$. Let $F_5$ be the finite field of order 5. Consider $F=F_5cup infty $. We call $alpha:Fto F$ defined by $alpha(x)=fracax+bcx+d$, where $ad-bcneq 0$ to be a fractional linear transformation. Then $L(5)$ is the set of all fractional linear transformations. Clearly $L(5)$ is a group under composition of maps. We have $|L(5)|=120$ and $L(5)$ acts on $F$ by evalutation. We have $|F|=6$. It can be shown that kernel of this action is trivial and it is sharply $3-$ transitive. So $L(5)$ is a 3-transitive permutation group of degree $6$.

Now I write what I have tried: Clearly as mentioned $S_6, A_6, L(5)$ are 3-transitive permutation groups of degree 6. Now, if $G$ is a 3-transitive group of degree 6 , we know that $120||G|$. So $|G|=120, 240, 360, 720$. If $|G|=720$ or $360$, then cleary $S_6$ and $A_6$ are only choices. Now at this point I am stuck. I have to contradict the fact $|G|=240$. And then I have to prove that if $|G|=120$, then $Gcong L(5)$. But I don't have a clue about how to do it!

Thanks in advance for any kind of help!

abstract-algebra group-theory permutations

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asked Jul 25 at 12:02

Riju

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  Can you classify $2$-transitive subgroups of $S_5$?
  – Lord Shark the Unknown
  Jul 25 at 12:05
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

This problem is from the book "Introduction to Group Theory" by Derek J.S Robinson. So, the notations are according to the book.

Prove that the only 3-transitive permutation groups of degree $6$ are $S_6$, $A_6$, and $L(5)$.

First I elaborate a little bit on $L(5)$. Let $F_5$ be the finite field of order 5. Consider $F=F_5cup infty $. We call $alpha:Fto F$ defined by $alpha(x)=fracax+bcx+d$, where $ad-bcneq 0$ to be a fractional linear transformation. Then $L(5)$ is the set of all fractional linear transformations. Clearly $L(5)$ is a group under composition of maps. We have $|L(5)|=120$ and $L(5)$ acts on $F$ by evalutation. We have $|F|=6$. It can be shown that kernel of this action is trivial and it is sharply $3-$ transitive. So $L(5)$ is a 3-transitive permutation group of degree $6$.

Now I write what I have tried: Clearly as mentioned $S_6, A_6, L(5)$ are 3-transitive permutation groups of degree 6. Now, if $G$ is a 3-transitive group of degree 6 , we know that $120||G|$. So $|G|=120, 240, 360, 720$. If $|G|=720$ or $360$, then cleary $S_6$ and $A_6$ are only choices. Now at this point I am stuck. I have to contradict the fact $|G|=240$. And then I have to prove that if $|G|=120$, then $Gcong L(5)$. But I don't have a clue about how to do it!

Thanks in advance for any kind of help!

abstract-algebra group-theory permutations

share|cite|improve this question

asked Jul 25 at 12:02

Riju

2,473314

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Can you classify $2$-transitive subgroups of $S_5$?
  – Lord Shark the Unknown
  Jul 25 at 12:05
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

This problem is from the book "Introduction to Group Theory" by Derek J.S Robinson. So, the notations are according to the book.

Prove that the only 3-transitive permutation groups of degree $6$ are $S_6$, $A_6$, and $L(5)$.

First I elaborate a little bit on $L(5)$. Let $F_5$ be the finite field of order 5. Consider $F=F_5cup infty $. We call $alpha:Fto F$ defined by $alpha(x)=fracax+bcx+d$, where $ad-bcneq 0$ to be a fractional linear transformation. Then $L(5)$ is the set of all fractional linear transformations. Clearly $L(5)$ is a group under composition of maps. We have $|L(5)|=120$ and $L(5)$ acts on $F$ by evalutation. We have $|F|=6$. It can be shown that kernel of this action is trivial and it is sharply $3-$ transitive. So $L(5)$ is a 3-transitive permutation group of degree $6$.

Now I write what I have tried: Clearly as mentioned $S_6, A_6, L(5)$ are 3-transitive permutation groups of degree 6. Now, if $G$ is a 3-transitive group of degree 6 , we know that $120||G|$. So $|G|=120, 240, 360, 720$. If $|G|=720$ or $360$, then cleary $S_6$ and $A_6$ are only choices. Now at this point I am stuck. I have to contradict the fact $|G|=240$. And then I have to prove that if $|G|=120$, then $Gcong L(5)$. But I don't have a clue about how to do it!

Thanks in advance for any kind of help!

abstract-algebra group-theory permutations

share|cite|improve this question

asked Jul 25 at 12:02

Riju

2,473314

This problem is from the book "Introduction to Group Theory" by Derek J.S Robinson. So, the notations are according to the book.

Prove that the only 3-transitive permutation groups of degree $6$ are $S_6$, $A_6$, and $L(5)$.

First I elaborate a little bit on $L(5)$. Let $F_5$ be the finite field of order 5. Consider $F=F_5cup infty $. We call $alpha:Fto F$ defined by $alpha(x)=fracax+bcx+d$, where $ad-bcneq 0$ to be a fractional linear transformation. Then $L(5)$ is the set of all fractional linear transformations. Clearly $L(5)$ is a group under composition of maps. We have $|L(5)|=120$ and $L(5)$ acts on $F$ by evalutation. We have $|F|=6$. It can be shown that kernel of this action is trivial and it is sharply $3-$ transitive. So $L(5)$ is a 3-transitive permutation group of degree $6$.

Now I write what I have tried: Clearly as mentioned $S_6, A_6, L(5)$ are 3-transitive permutation groups of degree 6. Now, if $G$ is a 3-transitive group of degree 6 , we know that $120||G|$. So $|G|=120, 240, 360, 720$. If $|G|=720$ or $360$, then cleary $S_6$ and $A_6$ are only choices. Now at this point I am stuck. I have to contradict the fact $|G|=240$. And then I have to prove that if $|G|=120$, then $Gcong L(5)$. But I don't have a clue about how to do it!

Thanks in advance for any kind of help!

abstract-algebra group-theory permutations

share|cite|improve this question

asked Jul 25 at 12:02

Riju

2,473314

share|cite|improve this question

share|cite|improve this question

asked Jul 25 at 12:02

Riju

2,473314

asked Jul 25 at 12:02

Riju

2,473314

asked Jul 25 at 12:02

Riju

2,473314

2,473314

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Can you classify $2$-transitive subgroups of $S_5$?
  – Lord Shark the Unknown
  Jul 25 at 12:05
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Can you classify $2$-transitive subgroups of $S_5$?
  – Lord Shark the Unknown
  Jul 25 at 12:05
  

  
  
  
  

1

1

Can you classify $2$-transitive subgroups of $S_5$?
– Lord Shark the Unknown
Jul 25 at 12:05

Can you classify $2$-transitive subgroups of $S_5$?
– Lord Shark the Unknown
Jul 25 at 12:05

add a comment | 

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To rule out $|G|=240$, observe that $|S_6:G|=3$, so $S_6$ acts transitively
on the three-element set of left cosets of $G$. Then $S_6$
will have a normal subgroup of index $3$ or $6$. This would contradict
the simplicity of $A_6$.

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answered Jul 25 at 12:09

Lord Shark the Unknown

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up vote
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To rule out $|G|=240$, observe that $|S_6:G|=3$, so $S_6$ acts transitively
on the three-element set of left cosets of $G$. Then $S_6$
will have a normal subgroup of index $3$ or $6$. This would contradict
the simplicity of $A_6$.

share|cite|improve this answer

answered Jul 25 at 12:09

Lord Shark the Unknown

85k950111

add a comment | 

up vote
2
down vote

To rule out $|G|=240$, observe that $|S_6:G|=3$, so $S_6$ acts transitively
on the three-element set of left cosets of $G$. Then $S_6$
will have a normal subgroup of index $3$ or $6$. This would contradict
the simplicity of $A_6$.

share|cite|improve this answer

answered Jul 25 at 12:09

Lord Shark the Unknown

85k950111

add a comment | 

up vote
2
down vote

up vote
2
down vote

To rule out $|G|=240$, observe that $|S_6:G|=3$, so $S_6$ acts transitively
on the three-element set of left cosets of $G$. Then $S_6$
will have a normal subgroup of index $3$ or $6$. This would contradict
the simplicity of $A_6$.

share|cite|improve this answer

answered Jul 25 at 12:09

Lord Shark the Unknown

85k950111

To rule out $|G|=240$, observe that $|S_6:G|=3$, so $S_6$ acts transitively
on the three-element set of left cosets of $G$. Then $S_6$
will have a normal subgroup of index $3$ or $6$. This would contradict
the simplicity of $A_6$.

share|cite|improve this answer

answered Jul 25 at 12:09

Lord Shark the Unknown

85k950111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 12:09

Lord Shark the Unknown

85k950111

answered Jul 25 at 12:09

Lord Shark the Unknown

85k950111

answered Jul 25 at 12:09

Lord Shark the Unknown

85k950111

85k950111

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