There are only two groups of order six, up to isomorphism: $mathbb Z_6$ and $S_3$. [duplicate]
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This question already has an answer here:
There are 2 groups of order 6 (up to isomorphism) [duplicate]
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Let $G$ be group with order $6$. Prove that either $G$ and $Bbb Z_{6}$ are isomorphic binary structure or $G$ and $S_{3}$ are isomorphic binary structure.
I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function between them. Please help me, if you have any good idea.
abstract-algebra group-theory finite-groups group-isomorphism
edited Dec 4 '18 at 22:50
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This question already has an answer here:
There are 2 groups of order 6 (up to isomorphism) [duplicate]
2 answers
Let $G$ be group with order $6$. Prove that either $G$ and $Bbb Z_{6}$ are isomorphic binary structure or $G$ and $S_{3}$ are isomorphic binary structure.
I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function between them. Please help me, if you have any good idea.
abstract-algebra group-theory finite-groups group-isomorphism
edited Dec 4 '18 at 22:50
amWhy
1
asked Dec 4 '18 at 9:32
mathsstudentmathsstudent
386
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marked as duplicate by Chinnapparaj R, jgon, DRF, A. Pongrácz, amWhy abstract-algebra
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This question already has an answer here:
There are 2 groups of order 6 (up to isomorphism) [duplicate]
2 answers
Let $G$ be group with order $6$. Prove that either $G$ and $Bbb Z_{6}$ are isomorphic binary structure or $G$ and $S_{3}$ are isomorphic binary structure.
I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function between them. Please help me, if you have any good idea.
abstract-algebra group-theory finite-groups group-isomorphism
edited Dec 4 '18 at 22:50
amWhy
1
asked Dec 4 '18 at 9:32
mathsstudentmathsstudent
386
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This question already has an answer here:
There are 2 groups of order 6 (up to isomorphism) [duplicate]
2 answers
Let $G$ be group with order $6$. Prove that either $G$ and $Bbb Z_{6}$ are isomorphic binary structure or $G$ and $S_{3}$ are isomorphic binary structure.
I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function between them. Please help me, if you have any good idea.
This question already has an answer here:
There are 2 groups of order 6 (up to isomorphism) [duplicate]
2 answers
abstract-algebra group-theory finite-groups group-isomorphism
abstract-algebra group-theory finite-groups group-isomorphism
edited Dec 4 '18 at 22:50
amWhy
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asked Dec 4 '18 at 9:32
mathsstudentmathsstudent
386
edited Dec 4 '18 at 22:50
amWhy
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asked Dec 4 '18 at 9:32
mathsstudentmathsstudent
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edited Dec 4 '18 at 22:50
amWhy
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edited Dec 4 '18 at 22:50
amWhy
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edited Dec 4 '18 at 22:50
amWhy
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1
asked Dec 4 '18 at 9:32
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asked Dec 4 '18 at 9:32
mathsstudentmathsstudent
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asked Dec 4 '18 at 9:32
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386
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1 Answer
1
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oldest
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Hint: Either your group has an element of order $6$ or not. If it has, then it is isomorphic to $mathbb{Z}_6$. Otherwise, which are the possible orders of elemets of $G$?
answered Dec 4 '18 at 9:40
José Carlos SantosJosé Carlos Santos
164k22131234
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1
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Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
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– amWhy
Dec 4 '18 at 22:48
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@amWhy This time, I did search for a duplicate.
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– José Carlos Santos
Dec 4 '18 at 22:50
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Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
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– amWhy
Dec 4 '18 at 22:51
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@amWhy I had never thought of that. That's a good idea.
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– José Carlos Santos
Dec 4 '18 at 22:52
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Either your group has an element of order $6$ or not. If it has, then it is isomorphic to $mathbb{Z}_6$. Otherwise, which are the possible orders of elemets of $G$?
answered Dec 4 '18 at 9:40
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
1
$begingroup$
Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
$endgroup$
– amWhy
Dec 4 '18 at 22:48
$begingroup$
@amWhy This time, I did search for a duplicate.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:50
$begingroup$
Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
$endgroup$
– amWhy
Dec 4 '18 at 22:51
$begingroup$
@amWhy I had never thought of that. That's a good idea.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:52
add a comment |
$begingroup$
Hint: Either your group has an element of order $6$ or not. If it has, then it is isomorphic to $mathbb{Z}_6$. Otherwise, which are the possible orders of elemets of $G$?
answered Dec 4 '18 at 9:40
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
1
$begingroup$
Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
$endgroup$
– amWhy
Dec 4 '18 at 22:48
$begingroup$
@amWhy This time, I did search for a duplicate.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:50
$begingroup$
Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
$endgroup$
– amWhy
Dec 4 '18 at 22:51
$begingroup$
@amWhy I had never thought of that. That's a good idea.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:52
add a comment |
$begingroup$
Hint: Either your group has an element of order $6$ or not. If it has, then it is isomorphic to $mathbb{Z}_6$. Otherwise, which are the possible orders of elemets of $G$?
answered Dec 4 '18 at 9:40
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
Hint: Either your group has an element of order $6$ or not. If it has, then it is isomorphic to $mathbb{Z}_6$. Otherwise, which are the possible orders of elemets of $G$?
answered Dec 4 '18 at 9:40
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 4 '18 at 9:40
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 4 '18 at 9:40
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 4 '18 at 9:40
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
1
$begingroup$
Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
$endgroup$
– amWhy
Dec 4 '18 at 22:48
$begingroup$
@amWhy This time, I did search for a duplicate.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:50
$begingroup$
Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
$endgroup$
– amWhy
Dec 4 '18 at 22:51
$begingroup$
@amWhy I had never thought of that. That's a good idea.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:52
add a comment |
1
$begingroup$
Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
$endgroup$
– amWhy
Dec 4 '18 at 22:48
$begingroup$
@amWhy This time, I did search for a duplicate.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:50
$begingroup$
Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
$endgroup$
– amWhy
Dec 4 '18 at 22:51
$begingroup$
@amWhy I had never thought of that. That's a good idea.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:52
1
1
$begingroup$
Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
$endgroup$
– amWhy
Dec 4 '18 at 22:48
$begingroup$
Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
$endgroup$
– amWhy
Dec 4 '18 at 22:48
$begingroup$
@amWhy This time, I did search for a duplicate.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:50
$begingroup$
@amWhy This time, I did search for a duplicate.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:50
$begingroup$
Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
$endgroup$
– amWhy
Dec 4 '18 at 22:51
$begingroup$
Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
$endgroup$
– amWhy
Dec 4 '18 at 22:51
$begingroup$
@amWhy I had never thought of that. That's a good idea.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:52
$begingroup$
@amWhy I had never thought of that. That's a good idea.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 22:52
add a comment |
