Solvability of Groups and the Group Order [closed]
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The theorem of Feit-Thompson on group solvability states that if $G$ is a finite group of odd order, then $G$ is solvable. And yet we know the case of abelian groups of prime order (isomorphic to cyclic groups of prime order) which are in fact simple and thus not solvable.
Can somebody explain this discrepancy to me ?
abstract-algebra group-theory
asked Dec 28 '18 at 13:54
user249018user249018
398127
$endgroup$
closed as off-topic by Dietrich Burde, Shaun, Cesareo, amWhy, onurcanbektas Dec 28 '18 at 20:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Shaun, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
The theorem of Feit-Thompson on group solvability states that if $G$ is a finite group of odd order, then $G$ is solvable. And yet we know the case of abelian groups of prime order (isomorphic to cyclic groups of prime order) which are in fact simple and thus not solvable.
Can somebody explain this discrepancy to me ?
abstract-algebra group-theory
asked Dec 28 '18 at 13:54
user249018user249018
398127
$endgroup$
closed as off-topic by Dietrich Burde, Shaun, Cesareo, amWhy, onurcanbektas Dec 28 '18 at 20:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Shaun, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
4
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All abelian groups are solvable.
$endgroup$
– lulu
Dec 28 '18 at 13:56
1
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Simple nonabelian implies not solvable.
$endgroup$
– Randall
Dec 28 '18 at 13:57
1
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(Simple and solvable) is equivalent to (Abelian of prime order).
$endgroup$
– Nicky Hekster
Dec 28 '18 at 13:57
1
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If G is simple, it does not contain, by definition, a proper normal subgroup. Let $Z/pZ$ be the cyclic group of prime order $p.$ All subroups are of order $p,$(exept ${e}$) and thus not proper. I dont understand the shock I provoked. I very probably misunderstood something, but I dont think it deserved the voting. And I didn't get any explanation I was looking for.
$endgroup$
– user249018
Dec 28 '18 at 14:15
add a comment |
$begingroup$
The theorem of Feit-Thompson on group solvability states that if $G$ is a finite group of odd order, then $G$ is solvable. And yet we know the case of abelian groups of prime order (isomorphic to cyclic groups of prime order) which are in fact simple and thus not solvable.
Can somebody explain this discrepancy to me ?
abstract-algebra group-theory
asked Dec 28 '18 at 13:54
user249018user249018
398127
$endgroup$
The theorem of Feit-Thompson on group solvability states that if $G$ is a finite group of odd order, then $G$ is solvable. And yet we know the case of abelian groups of prime order (isomorphic to cyclic groups of prime order) which are in fact simple and thus not solvable.
Can somebody explain this discrepancy to me ?
abstract-algebra group-theory
abstract-algebra group-theory
asked Dec 28 '18 at 13:54
user249018user249018
398127
asked Dec 28 '18 at 13:54
user249018user249018
398127
asked Dec 28 '18 at 13:54
user249018user249018
398127
asked Dec 28 '18 at 13:54
user249018user249018
398127
asked Dec 28 '18 at 13:54
user249018user249018
398127
398127
closed as off-topic by Dietrich Burde, Shaun, Cesareo, amWhy, onurcanbektas Dec 28 '18 at 20:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Shaun, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Dietrich Burde, Shaun, Cesareo, amWhy, onurcanbektas Dec 28 '18 at 20:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Shaun, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
All abelian groups are solvable.
$endgroup$
– lulu
Dec 28 '18 at 13:56
1
$begingroup$
Simple nonabelian implies not solvable.
$endgroup$
– Randall
Dec 28 '18 at 13:57
1
$begingroup$
(Simple and solvable) is equivalent to (Abelian of prime order).
$endgroup$
– Nicky Hekster
Dec 28 '18 at 13:57
1
$begingroup$
If G is simple, it does not contain, by definition, a proper normal subgroup. Let $Z/pZ$ be the cyclic group of prime order $p.$ All subroups are of order $p,$(exept ${e}$) and thus not proper. I dont understand the shock I provoked. I very probably misunderstood something, but I dont think it deserved the voting. And I didn't get any explanation I was looking for.
$endgroup$
– user249018
Dec 28 '18 at 14:15
add a comment |
4
$begingroup$
All abelian groups are solvable.
$endgroup$
– lulu
Dec 28 '18 at 13:56
1
$begingroup$
Simple nonabelian implies not solvable.
$endgroup$
– Randall
Dec 28 '18 at 13:57
1
$begingroup$
(Simple and solvable) is equivalent to (Abelian of prime order).
$endgroup$
– Nicky Hekster
Dec 28 '18 at 13:57
1
$begingroup$
If G is simple, it does not contain, by definition, a proper normal subgroup. Let $Z/pZ$ be the cyclic group of prime order $p.$ All subroups are of order $p,$(exept ${e}$) and thus not proper. I dont understand the shock I provoked. I very probably misunderstood something, but I dont think it deserved the voting. And I didn't get any explanation I was looking for.
$endgroup$
– user249018
Dec 28 '18 at 14:15
4
4
$begingroup$
All abelian groups are solvable.
$endgroup$
– lulu
Dec 28 '18 at 13:56
$begingroup$
All abelian groups are solvable.
$endgroup$
– lulu
Dec 28 '18 at 13:56
1
1
$begingroup$
Simple nonabelian implies not solvable.
$endgroup$
– Randall
Dec 28 '18 at 13:57
$begingroup$
Simple nonabelian implies not solvable.
$endgroup$
– Randall
Dec 28 '18 at 13:57
1
1
$begingroup$
(Simple and solvable) is equivalent to (Abelian of prime order).
$endgroup$
– Nicky Hekster
Dec 28 '18 at 13:57
$begingroup$
(Simple and solvable) is equivalent to (Abelian of prime order).
$endgroup$
– Nicky Hekster
Dec 28 '18 at 13:57
1
1
$begingroup$
If G is simple, it does not contain, by definition, a proper normal subgroup. Let $Z/pZ$ be the cyclic group of prime order $p.$ All subroups are of order $p,$(exept ${e}$) and thus not proper. I dont understand the shock I provoked. I very probably misunderstood something, but I dont think it deserved the voting. And I didn't get any explanation I was looking for.
$endgroup$
– user249018
Dec 28 '18 at 14:15
$begingroup$
If G is simple, it does not contain, by definition, a proper normal subgroup. Let $Z/pZ$ be the cyclic group of prime order $p.$ All subroups are of order $p,$(exept ${e}$) and thus not proper. I dont understand the shock I provoked. I very probably misunderstood something, but I dont think it deserved the voting. And I didn't get any explanation I was looking for.
$endgroup$
– user249018
Dec 28 '18 at 14:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A group $G$ is solvable if there is a decomposition series in which sucesive quotients are abelian. So, for instance, if you take an abelian group $A$ (work for a cyclic group of order $p$ for example), then you have ${e}leq A$ as a such composition series, so that $A$ is solvable.
A group that is simple not need to be non-solvable, as we have seen above. And a group that is non-solvable, need not to be non-simple (see $S_5$ the symmetric group).
answered Dec 28 '18 at 14:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A group $G$ is solvable if there is a decomposition series in which sucesive quotients are abelian. So, for instance, if you take an abelian group $A$ (work for a cyclic group of order $p$ for example), then you have ${e}leq A$ as a such composition series, so that $A$ is solvable.
A group that is simple not need to be non-solvable, as we have seen above. And a group that is non-solvable, need not to be non-simple (see $S_5$ the symmetric group).
answered Dec 28 '18 at 14:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
A group $G$ is solvable if there is a decomposition series in which sucesive quotients are abelian. So, for instance, if you take an abelian group $A$ (work for a cyclic group of order $p$ for example), then you have ${e}leq A$ as a such composition series, so that $A$ is solvable.
A group that is simple not need to be non-solvable, as we have seen above. And a group that is non-solvable, need not to be non-simple (see $S_5$ the symmetric group).
answered Dec 28 '18 at 14:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
A group $G$ is solvable if there is a decomposition series in which sucesive quotients are abelian. So, for instance, if you take an abelian group $A$ (work for a cyclic group of order $p$ for example), then you have ${e}leq A$ as a such composition series, so that $A$ is solvable.
A group that is simple not need to be non-solvable, as we have seen above. And a group that is non-solvable, need not to be non-simple (see $S_5$ the symmetric group).
answered Dec 28 '18 at 14:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
A group $G$ is solvable if there is a decomposition series in which sucesive quotients are abelian. So, for instance, if you take an abelian group $A$ (work for a cyclic group of order $p$ for example), then you have ${e}leq A$ as a such composition series, so that $A$ is solvable.
A group that is simple not need to be non-solvable, as we have seen above. And a group that is non-solvable, need not to be non-simple (see $S_5$ the symmetric group).
answered Dec 28 '18 at 14:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
edited Dec 28 '18 at 22:33
answered Dec 28 '18 at 14:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Dec 28 '18 at 14:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Dec 28 '18 at 14:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
802110
add a comment |
add a comment |
4
$begingroup$
All abelian groups are solvable.
$endgroup$
– lulu
Dec 28 '18 at 13:56
1
$begingroup$
Simple nonabelian implies not solvable.
$endgroup$
– Randall
Dec 28 '18 at 13:57
1
$begingroup$
(Simple and solvable) is equivalent to (Abelian of prime order).
$endgroup$
– Nicky Hekster
Dec 28 '18 at 13:57
1
$begingroup$
If G is simple, it does not contain, by definition, a proper normal subgroup. Let $Z/pZ$ be the cyclic group of prime order $p.$ All subroups are of order $p,$(exept ${e}$) and thus not proper. I dont understand the shock I provoked. I very probably misunderstood something, but I dont think it deserved the voting. And I didn't get any explanation I was looking for.
$endgroup$
– user249018
Dec 28 '18 at 14:15