Group Isomorphism regarding Sylow Subgroups
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Suppose I have given two groups say, $G_1,G_2$ such that they have same order.I'm assuming they are non commutative.Then their Syllow subgroups has same order clearly.If I'm given that the number of Syllow subgroups of these are also same then "are $G_1,G_2$ isomorphic"? I have always find this statement as true considering lower order groups but can't proved it. Is it true or there are some counterexamples too! Thanks for reading.
abstract-algebra group-isomorphism sylow-theory
edited Dec 5 at 19:17
Ethan Bolker
40.4k545107
asked Dec 5 at 19:06
Subhajit Saha
255113
add a comment |
up vote
3
down vote
favorite
Suppose I have given two groups say, $G_1,G_2$ such that they have same order.I'm assuming they are non commutative.Then their Syllow subgroups has same order clearly.If I'm given that the number of Syllow subgroups of these are also same then "are $G_1,G_2$ isomorphic"? I have always find this statement as true considering lower order groups but can't proved it. Is it true or there are some counterexamples too! Thanks for reading.
abstract-algebra group-isomorphism sylow-theory
edited Dec 5 at 19:17
Ethan Bolker
40.4k545107
asked Dec 5 at 19:06
Subhajit Saha
255113
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose I have given two groups say, $G_1,G_2$ such that they have same order.I'm assuming they are non commutative.Then their Syllow subgroups has same order clearly.If I'm given that the number of Syllow subgroups of these are also same then "are $G_1,G_2$ isomorphic"? I have always find this statement as true considering lower order groups but can't proved it. Is it true or there are some counterexamples too! Thanks for reading.
abstract-algebra group-isomorphism sylow-theory
edited Dec 5 at 19:17
Ethan Bolker
40.4k545107
asked Dec 5 at 19:06
Subhajit Saha
255113
Suppose I have given two groups say, $G_1,G_2$ such that they have same order.I'm assuming they are non commutative.Then their Syllow subgroups has same order clearly.If I'm given that the number of Syllow subgroups of these are also same then "are $G_1,G_2$ isomorphic"? I have always find this statement as true considering lower order groups but can't proved it. Is it true or there are some counterexamples too! Thanks for reading.
abstract-algebra group-isomorphism sylow-theory
abstract-algebra group-isomorphism sylow-theory
edited Dec 5 at 19:17
Ethan Bolker
40.4k545107
asked Dec 5 at 19:06
Subhajit Saha
255113
edited Dec 5 at 19:17
Ethan Bolker
40.4k545107
asked Dec 5 at 19:06
Subhajit Saha
255113
edited Dec 5 at 19:17
Ethan Bolker
40.4k545107
edited Dec 5 at 19:17
Ethan Bolker
40.4k545107
edited Dec 5 at 19:17
Ethan Bolker
40.4k545107
40.4k545107
asked Dec 5 at 19:06
Subhajit Saha
255113
asked Dec 5 at 19:06
Subhajit Saha
255113
asked Dec 5 at 19:06
Subhajit Saha
255113
255113
add a comment |
add a comment |
2 Answers
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4
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accepted
This is easily seen to fail for abelian groups, since all abelian groups of a given order have the same number of Sylow subgroups. For a nonabelian example, consider two distinct nonabelian groups of order $p^n$ for some prime $p$ and integer $n$.
answered Dec 5 at 19:18
Matt Samuel
36.4k63464
Ok! But what about those cases if we have distinct syllow subgroups of different order
– Subhajit Saha
Dec 5 at 19:55
@Sub I don't have an example, but I'm $100%$ certain the numbers of Sylow subgroups doesn't classify the group, abelian or not. It would make the group isomorphism problem easy, which it isn't.
– Matt Samuel
Dec 5 at 20:14
add a comment |
up vote
2
down vote
Groups with Identical Subgroup Lattices in All Powers shows there are many, many examples even when the Sylow subgroups are required to be cyclic.
answered Dec 5 at 19:16
Eric Towers
31.5k22265
@MattSamuel D'oh. You are correct.
– Eric Towers
Dec 5 at 19:20
Sir,I said both of them as non Abelian , further you have done a mistake saying $S_3$ has unique $2-$ Syllow subgroups.
– Subhajit Saha
Dec 5 at 19:25
@SubhajitSaha : All Sylow subgroups cyclic implies metacyclic, which does not imply abelian.
– Eric Towers
Dec 5 at 21:43
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
This is easily seen to fail for abelian groups, since all abelian groups of a given order have the same number of Sylow subgroups. For a nonabelian example, consider two distinct nonabelian groups of order $p^n$ for some prime $p$ and integer $n$.
answered Dec 5 at 19:18
Matt Samuel
36.4k63464
Ok! But what about those cases if we have distinct syllow subgroups of different order
– Subhajit Saha
Dec 5 at 19:55
@Sub I don't have an example, but I'm $100%$ certain the numbers of Sylow subgroups doesn't classify the group, abelian or not. It would make the group isomorphism problem easy, which it isn't.
– Matt Samuel
Dec 5 at 20:14
add a comment |
up vote
4
down vote
accepted
This is easily seen to fail for abelian groups, since all abelian groups of a given order have the same number of Sylow subgroups. For a nonabelian example, consider two distinct nonabelian groups of order $p^n$ for some prime $p$ and integer $n$.
answered Dec 5 at 19:18
Matt Samuel
36.4k63464
Ok! But what about those cases if we have distinct syllow subgroups of different order
– Subhajit Saha
Dec 5 at 19:55
@Sub I don't have an example, but I'm $100%$ certain the numbers of Sylow subgroups doesn't classify the group, abelian or not. It would make the group isomorphism problem easy, which it isn't.
– Matt Samuel
Dec 5 at 20:14
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
This is easily seen to fail for abelian groups, since all abelian groups of a given order have the same number of Sylow subgroups. For a nonabelian example, consider two distinct nonabelian groups of order $p^n$ for some prime $p$ and integer $n$.
answered Dec 5 at 19:18
Matt Samuel
36.4k63464
This is easily seen to fail for abelian groups, since all abelian groups of a given order have the same number of Sylow subgroups. For a nonabelian example, consider two distinct nonabelian groups of order $p^n$ for some prime $p$ and integer $n$.
answered Dec 5 at 19:18
Matt Samuel
36.4k63464
answered Dec 5 at 19:18
Matt Samuel
36.4k63464
answered Dec 5 at 19:18
Matt Samuel
36.4k63464
answered Dec 5 at 19:18
Matt Samuel
36.4k63464
36.4k63464
Ok! But what about those cases if we have distinct syllow subgroups of different order
– Subhajit Saha
Dec 5 at 19:55
@Sub I don't have an example, but I'm $100%$ certain the numbers of Sylow subgroups doesn't classify the group, abelian or not. It would make the group isomorphism problem easy, which it isn't.
– Matt Samuel
Dec 5 at 20:14
add a comment |
Ok! But what about those cases if we have distinct syllow subgroups of different order
– Subhajit Saha
Dec 5 at 19:55
@Sub I don't have an example, but I'm $100%$ certain the numbers of Sylow subgroups doesn't classify the group, abelian or not. It would make the group isomorphism problem easy, which it isn't.
– Matt Samuel
Dec 5 at 20:14
Ok! But what about those cases if we have distinct syllow subgroups of different order
– Subhajit Saha
Dec 5 at 19:55
Ok! But what about those cases if we have distinct syllow subgroups of different order
– Subhajit Saha
Dec 5 at 19:55
@Sub I don't have an example, but I'm $100%$ certain the numbers of Sylow subgroups doesn't classify the group, abelian or not. It would make the group isomorphism problem easy, which it isn't.
– Matt Samuel
Dec 5 at 20:14
@Sub I don't have an example, but I'm $100%$ certain the numbers of Sylow subgroups doesn't classify the group, abelian or not. It would make the group isomorphism problem easy, which it isn't.
– Matt Samuel
Dec 5 at 20:14
add a comment |
up vote
2
down vote
Groups with Identical Subgroup Lattices in All Powers shows there are many, many examples even when the Sylow subgroups are required to be cyclic.
answered Dec 5 at 19:16
Eric Towers
31.5k22265
@MattSamuel D'oh. You are correct.
– Eric Towers
Dec 5 at 19:20
Sir,I said both of them as non Abelian , further you have done a mistake saying $S_3$ has unique $2-$ Syllow subgroups.
– Subhajit Saha
Dec 5 at 19:25
@SubhajitSaha : All Sylow subgroups cyclic implies metacyclic, which does not imply abelian.
– Eric Towers
Dec 5 at 21:43
add a comment |
up vote
2
down vote
Groups with Identical Subgroup Lattices in All Powers shows there are many, many examples even when the Sylow subgroups are required to be cyclic.
answered Dec 5 at 19:16
Eric Towers
31.5k22265
@MattSamuel D'oh. You are correct.
– Eric Towers
Dec 5 at 19:20
Sir,I said both of them as non Abelian , further you have done a mistake saying $S_3$ has unique $2-$ Syllow subgroups.
– Subhajit Saha
Dec 5 at 19:25
@SubhajitSaha : All Sylow subgroups cyclic implies metacyclic, which does not imply abelian.
– Eric Towers
Dec 5 at 21:43
add a comment |
up vote
2
down vote
up vote
2
down vote
Groups with Identical Subgroup Lattices in All Powers shows there are many, many examples even when the Sylow subgroups are required to be cyclic.
answered Dec 5 at 19:16
Eric Towers
31.5k22265
Groups with Identical Subgroup Lattices in All Powers shows there are many, many examples even when the Sylow subgroups are required to be cyclic.
answered Dec 5 at 19:16
Eric Towers
31.5k22265
edited Dec 5 at 19:20
answered Dec 5 at 19:16
Eric Towers
31.5k22265
answered Dec 5 at 19:16
Eric Towers
31.5k22265
answered Dec 5 at 19:16
Eric Towers
31.5k22265
31.5k22265
@MattSamuel D'oh. You are correct.
– Eric Towers
Dec 5 at 19:20
Sir,I said both of them as non Abelian , further you have done a mistake saying $S_3$ has unique $2-$ Syllow subgroups.
– Subhajit Saha
Dec 5 at 19:25
@SubhajitSaha : All Sylow subgroups cyclic implies metacyclic, which does not imply abelian.
– Eric Towers
Dec 5 at 21:43
add a comment |
@MattSamuel D'oh. You are correct.
– Eric Towers
Dec 5 at 19:20
Sir,I said both of them as non Abelian , further you have done a mistake saying $S_3$ has unique $2-$ Syllow subgroups.
– Subhajit Saha
Dec 5 at 19:25
@SubhajitSaha : All Sylow subgroups cyclic implies metacyclic, which does not imply abelian.
– Eric Towers
Dec 5 at 21:43
@MattSamuel D'oh. You are correct.
– Eric Towers
Dec 5 at 19:20
@MattSamuel D'oh. You are correct.
– Eric Towers
Dec 5 at 19:20
Sir,I said both of them as non Abelian , further you have done a mistake saying $S_3$ has unique $2-$ Syllow subgroups.
– Subhajit Saha
Dec 5 at 19:25
Sir,I said both of them as non Abelian , further you have done a mistake saying $S_3$ has unique $2-$ Syllow subgroups.
– Subhajit Saha
Dec 5 at 19:25
@SubhajitSaha : All Sylow subgroups cyclic implies metacyclic, which does not imply abelian.
– Eric Towers
Dec 5 at 21:43
@SubhajitSaha : All Sylow subgroups cyclic implies metacyclic, which does not imply abelian.
– Eric Towers
Dec 5 at 21:43
add a comment |
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