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simplicity of the Janko group $J_1$

6

2

$begingroup$

In the paper, Janko shows the simplicity of Janko group $J_1$ at the Lemma 2.1. In this proof it says "By a transfer theorem all involutions are conjugate in $G$", but I cannot understand.

1. Some propositions are named "transfer theorem", such as Burnside's transfer theorem, Thompson's transfer theorem, etc... . I want to know which proposition is used.


2. I found an Additional description for the proof, but this part of proof is not correct (because the author uses the result of the Janko's paper, and the original proof doesn't use any specific calculations). I want to know theoretic approach of the problem.

group-theory finite-groups simple-groups

share|cite|improve this question

asked Mar 27 at 17:26

ritosonnritosonn

435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Strictly speaking, the statement is as follows: a finite group $G$ with three properties (a) all Sylow 2-group of $G$ is abelian (b) $G$ doesn't have subgroups index 2 (c) $G$ has an involution $t$ and $C_G(t)cong mathbbZ_2times mathfrakA_5$, then $G$ is simple.
  $endgroup$
  – ritosonn
  Mar 28 at 4:07
  

  
  
  
  

add a comment | 

6

2

$begingroup$

In the paper, Janko shows the simplicity of Janko group $J_1$ at the Lemma 2.1. In this proof it says "By a transfer theorem all involutions are conjugate in $G$", but I cannot understand.

1. Some propositions are named "transfer theorem", such as Burnside's transfer theorem, Thompson's transfer theorem, etc... . I want to know which proposition is used.


2. I found an Additional description for the proof, but this part of proof is not correct (because the author uses the result of the Janko's paper, and the original proof doesn't use any specific calculations). I want to know theoretic approach of the problem.

group-theory finite-groups simple-groups

share|cite|improve this question

asked Mar 27 at 17:26

ritosonnritosonn

435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Strictly speaking, the statement is as follows: a finite group $G$ with three properties (a) all Sylow 2-group of $G$ is abelian (b) $G$ doesn't have subgroups index 2 (c) $G$ has an involution $t$ and $C_G(t)cong mathbbZ_2times mathfrakA_5$, then $G$ is simple.
  $endgroup$
  – ritosonn
  Mar 28 at 4:07
  

  
  
  
  

add a comment | 

$begingroup$

In the paper, Janko shows the simplicity of Janko group $J_1$ at the Lemma 2.1. In this proof it says "By a transfer theorem all involutions are conjugate in $G$", but I cannot understand.

1. Some propositions are named "transfer theorem", such as Burnside's transfer theorem, Thompson's transfer theorem, etc... . I want to know which proposition is used.


2. I found an Additional description for the proof, but this part of proof is not correct (because the author uses the result of the Janko's paper, and the original proof doesn't use any specific calculations). I want to know theoretic approach of the problem.

group-theory finite-groups simple-groups

share|cite|improve this question

asked Mar 27 at 17:26

ritosonnritosonn

435

$endgroup$

share|cite|improve this question

asked Mar 27 at 17:26

ritosonnritosonn

435

share|cite|improve this question

asked Mar 27 at 17:26

ritosonnritosonn

435

asked Mar 27 at 17:26

ritosonnritosonn

435

asked Mar 27 at 17:26

ritosonnritosonn

435

1 Answer
1

active

oldest

votes

6

$begingroup$

The group in question is simple and has elementary abelian Sylow 2-subgroups of order 8.

A crucial fact that you need to know is that, if finite group $G$ has an abelian Sylow $p$-subgroup $P$, then $p$-transfer is controlled by $N_G(P)$ - see here for example.

This means that the largest $2$-quotient of $G$ (which is trivial by assumption) is isomorphic to the largest $2$-quotient of $N_G(P)$.

Now $N_G(P)/C_G(P) le rm Aut(P) cong rm GL(3,2)$. If $|N_G(P)/C_G(P)|$ is divisible by $7$, then the involutions in $P$ are all conjugate under an element of order $7$. That is what we are trying to prove.

Now $|rm GL(3,2)| = 8 times 3 times 7$, and $|N_G(P)/C_G(P)|$ is odd, so the only other options are $|N_G(P)/C_G(P)| = 1$ or $3$. In both cases, the action is reducible with at least one element of order $2$ centralizex by $N_G(P)$, and then $N_G(P)$ has a quotient group of order $2$. Hence so does $G$, contradicting the assumption.

share|cite|improve this answer

edited Mar 28 at 4:03

answered Mar 28 at 0:54

Derek HoltDerek Holt

54.7k53574

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. I doubt this answer uses the simplicity of $J_1$, is this a circular reasoning?
  $endgroup$
  – ritosonn
  Mar 28 at 3:53
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes you are right. But we are assuming that $G$ has no subgroup of index $2$, which is enough for this argument to work.
  $endgroup$
  – Derek Holt
  Mar 28 at 4:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have another question. In the latter of the proof, you use the result "the Sylow $2$-group of $J_1$ is $(mathbbZ_2)^3$". Is it necessary of the proof (or is the result easy to show) ?
  $endgroup$
  – ritosonn
  Mar 29 at 17:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This follows from the given assumptions that the Sylow $2$-subgroups are abelian, and that the centralizer of an element $t$ of order $2$ has the structure $C_G(t) = C_2 times A_5$.
  $endgroup$
  – Derek Holt
  Mar 29 at 20:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had misunderstood, but we could calculate the concrete structure of Sylow 2-subgroup... . As you say we can take the subgroup of $C_G(t)$ isomorphic to the $(mathbbZ_2)^3$, and $|C_G(t)|=120=2^3times 15$, so it's the maximal 2-subgroup of $C_G(t)$. In addition to that, if we take the Sylow 2-group $P$ which includes $langle trangle cong mathbbZ_2$, then $P$ must be the subgroup of $C_G(t)$ because $P$ is abelian from the assumption (a). So the Sylow 2-group of $J_1$ is $(mathbbZ_2)^3$.
  $endgroup$
  – ritosonn
  Mar 29 at 21:39
  
  

  
  
  
  

add a comment | 

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6

$begingroup$

The group in question is simple and has elementary abelian Sylow 2-subgroups of order 8.

A crucial fact that you need to know is that, if finite group $G$ has an abelian Sylow $p$-subgroup $P$, then $p$-transfer is controlled by $N_G(P)$ - see here for example.

This means that the largest $2$-quotient of $G$ (which is trivial by assumption) is isomorphic to the largest $2$-quotient of $N_G(P)$.

Now $N_G(P)/C_G(P) le rm Aut(P) cong rm GL(3,2)$. If $|N_G(P)/C_G(P)|$ is divisible by $7$, then the involutions in $P$ are all conjugate under an element of order $7$. That is what we are trying to prove.

Now $|rm GL(3,2)| = 8 times 3 times 7$, and $|N_G(P)/C_G(P)|$ is odd, so the only other options are $|N_G(P)/C_G(P)| = 1$ or $3$. In both cases, the action is reducible with at least one element of order $2$ centralizex by $N_G(P)$, and then $N_G(P)$ has a quotient group of order $2$. Hence so does $G$, contradicting the assumption.

share|cite|improve this answer

edited Mar 28 at 4:03

answered Mar 28 at 0:54

Derek HoltDerek Holt

54.7k53574

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. I doubt this answer uses the simplicity of $J_1$, is this a circular reasoning?
  $endgroup$
  – ritosonn
  Mar 28 at 3:53
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes you are right. But we are assuming that $G$ has no subgroup of index $2$, which is enough for this argument to work.
  $endgroup$
  – Derek Holt
  Mar 28 at 4:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have another question. In the latter of the proof, you use the result "the Sylow $2$-group of $J_1$ is $(mathbbZ_2)^3$". Is it necessary of the proof (or is the result easy to show) ?
  $endgroup$
  – ritosonn
  Mar 29 at 17:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This follows from the given assumptions that the Sylow $2$-subgroups are abelian, and that the centralizer of an element $t$ of order $2$ has the structure $C_G(t) = C_2 times A_5$.
  $endgroup$
  – Derek Holt
  Mar 29 at 20:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had misunderstood, but we could calculate the concrete structure of Sylow 2-subgroup... . As you say we can take the subgroup of $C_G(t)$ isomorphic to the $(mathbbZ_2)^3$, and $|C_G(t)|=120=2^3times 15$, so it's the maximal 2-subgroup of $C_G(t)$. In addition to that, if we take the Sylow 2-group $P$ which includes $langle trangle cong mathbbZ_2$, then $P$ must be the subgroup of $C_G(t)$ because $P$ is abelian from the assumption (a). So the Sylow 2-group of $J_1$ is $(mathbbZ_2)^3$.
  $endgroup$
  – ritosonn
  Mar 29 at 21:39
  
  

  
  
  
  

add a comment | 

6

$begingroup$

The group in question is simple and has elementary abelian Sylow 2-subgroups of order 8.

A crucial fact that you need to know is that, if finite group $G$ has an abelian Sylow $p$-subgroup $P$, then $p$-transfer is controlled by $N_G(P)$ - see here for example.

This means that the largest $2$-quotient of $G$ (which is trivial by assumption) is isomorphic to the largest $2$-quotient of $N_G(P)$.

Now $N_G(P)/C_G(P) le rm Aut(P) cong rm GL(3,2)$. If $|N_G(P)/C_G(P)|$ is divisible by $7$, then the involutions in $P$ are all conjugate under an element of order $7$. That is what we are trying to prove.

Now $|rm GL(3,2)| = 8 times 3 times 7$, and $|N_G(P)/C_G(P)|$ is odd, so the only other options are $|N_G(P)/C_G(P)| = 1$ or $3$. In both cases, the action is reducible with at least one element of order $2$ centralizex by $N_G(P)$, and then $N_G(P)$ has a quotient group of order $2$. Hence so does $G$, contradicting the assumption.

share|cite|improve this answer

edited Mar 28 at 4:03

answered Mar 28 at 0:54

Derek HoltDerek Holt

54.7k53574

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. I doubt this answer uses the simplicity of $J_1$, is this a circular reasoning?
  $endgroup$
  – ritosonn
  Mar 28 at 3:53
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Yes you are right. But we are assuming that $G$ has no subgroup of index $2$, which is enough for this argument to work.
  $endgroup$
  – Derek Holt
  Mar 28 at 4:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have another question. In the latter of the proof, you use the result "the Sylow $2$-group of $J_1$ is $(mathbbZ_2)^3$". Is it necessary of the proof (or is the result easy to show) ?
  $endgroup$
  – ritosonn
  Mar 29 at 17:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This follows from the given assumptions that the Sylow $2$-subgroups are abelian, and that the centralizer of an element $t$ of order $2$ has the structure $C_G(t) = C_2 times A_5$.
  $endgroup$
  – Derek Holt
  Mar 29 at 20:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had misunderstood, but we could calculate the concrete structure of Sylow 2-subgroup... . As you say we can take the subgroup of $C_G(t)$ isomorphic to the $(mathbbZ_2)^3$, and $|C_G(t)|=120=2^3times 15$, so it's the maximal 2-subgroup of $C_G(t)$. In addition to that, if we take the Sylow 2-group $P$ which includes $langle trangle cong mathbbZ_2$, then $P$ must be the subgroup of $C_G(t)$ because $P$ is abelian from the assumption (a). So the Sylow 2-group of $J_1$ is $(mathbbZ_2)^3$.
  $endgroup$
  – ritosonn
  Mar 29 at 21:39
  
  

  
  
  
  

add a comment | 

$begingroup$

The group in question is simple and has elementary abelian Sylow 2-subgroups of order 8.

A crucial fact that you need to know is that, if finite group $G$ has an abelian Sylow $p$-subgroup $P$, then $p$-transfer is controlled by $N_G(P)$ - see here for example.

This means that the largest $2$-quotient of $G$ (which is trivial by assumption) is isomorphic to the largest $2$-quotient of $N_G(P)$.

Now $N_G(P)/C_G(P) le rm Aut(P) cong rm GL(3,2)$. If $|N_G(P)/C_G(P)|$ is divisible by $7$, then the involutions in $P$ are all conjugate under an element of order $7$. That is what we are trying to prove.

Now $|rm GL(3,2)| = 8 times 3 times 7$, and $|N_G(P)/C_G(P)|$ is odd, so the only other options are $|N_G(P)/C_G(P)| = 1$ or $3$. In both cases, the action is reducible with at least one element of order $2$ centralizex by $N_G(P)$, and then $N_G(P)$ has a quotient group of order $2$. Hence so does $G$, contradicting the assumption.

share|cite|improve this answer

edited Mar 28 at 4:03

answered Mar 28 at 0:54

Derek HoltDerek Holt

54.7k53574

$endgroup$

share|cite|improve this answer

edited Mar 28 at 4:03

answered Mar 28 at 0:54

Derek HoltDerek Holt

54.7k53574

answered Mar 28 at 0:54

Derek HoltDerek Holt

54.7k53574

answered Mar 28 at 0:54

Derek HoltDerek Holt

54.7k53574