simplicity of the Janko group $J_1$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Who discovered that normalizer of an abelian Sylow $p$-subgroup controls $p$-transfer?simplicity of GQuestion about $p$-Sylow subgroups of the quotient groupSimplicity of homeomorphism groupProofs of Simplicity of $A_n$Infinite groups whose non-trivial subgroups are of finite indexGeneralizing the proof of simplicity of the alternating groupsGroup with all Sylow subgroups cyclic (exercise 5C.4 of “Finite group theory”, Isaacs).Simplicity of quotient groupNo simple group of order 4400Groups of order $180$, $540$, $1080$ are not simple.
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simplicity of the Janko group $J_1$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Who discovered that normalizer of an abelian Sylow $p$-subgroup controls $p$-transfer?simplicity of GQuestion about $p$-Sylow subgroups of the quotient groupSimplicity of homeomorphism groupProofs of Simplicity of $A_n$Infinite groups whose non-trivial subgroups are of finite indexGeneralizing the proof of simplicity of the alternating groupsGroup with all Sylow subgroups cyclic (exercise 5C.4 of “Finite group theory”, Isaacs).Simplicity of quotient groupNo simple group of order 4400Groups of order $180$, $540$, $1080$ are not simple.
$begingroup$
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.
- 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.
- 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
asked Mar 27 at 17:26
ritosonnritosonn
435
$endgroup$
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.
- 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.
- 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
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.
- 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.
- 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
asked Mar 27 at 17:26
ritosonnritosonn
435
$endgroup$
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.
- 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.
- 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
group-theory finite-groups simple-groups
asked Mar 27 at 17:26
ritosonnritosonn
435
asked Mar 27 at 17:26
ritosonnritosonn
435
asked Mar 27 at 17:26
ritosonnritosonn
435
asked Mar 27 at 17:26
ritosonnritosonn
435
asked Mar 27 at 17:26
ritosonnritosonn
435
435
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
1 Answer
1
active
oldest
votes
$begingroup$
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.
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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$begingroup$
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.
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.
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.
answered Mar 28 at 0:54
Derek HoltDerek Holt
54.7k53574
$endgroup$
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.
answered Mar 28 at 0:54
Derek HoltDerek Holt
54.7k53574
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
54.7k53574
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 |
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
1
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
$begingroup$
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
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
$begingroup$
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
$begingroup$
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
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
$begingroup$
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
$begingroup$
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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$begingroup$
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