If there exists an epimorphism from a group $G_1$ to $G_2$ that is not one-to-one, then can $G_1$ and $G_2$...
This question already has an answer here:
Does $Gcong G/H$ imply that $H$ is trivial?
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I am trying to show that two groups are isomorphic only if a certain condition holds. I can show that a specific epimorphism between the two groups is one-to-one only if this condition holds, but it isn't obvious that for infinite groups that this should mean that there can't exist an isomorphism between the two groups.
If there is a non-one-to-one epimorphism from a group $G_1$ to a group $G_2$, does this mean that all epimorphisms from $G_1$ to $G_2$ are not one-to-one?
group-theory group-isomorphism group-homomorphism infinite-groups
asked Nov 22 '18 at 1:32
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marked as duplicate by Paul Plummer, user10354138, Lord Shark the Unknown, Derek Holt group-theory
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Nov 22 '18 at 7:55
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This question already has an answer here:
Does $Gcong G/H$ imply that $H$ is trivial?
10 answers
I am trying to show that two groups are isomorphic only if a certain condition holds. I can show that a specific epimorphism between the two groups is one-to-one only if this condition holds, but it isn't obvious that for infinite groups that this should mean that there can't exist an isomorphism between the two groups.
If there is a non-one-to-one epimorphism from a group $G_1$ to a group $G_2$, does this mean that all epimorphisms from $G_1$ to $G_2$ are not one-to-one?
group-theory group-isomorphism group-homomorphism infinite-groups
asked Nov 22 '18 at 1:32
Information Aether
946
marked as duplicate by Paul Plummer, user10354138, Lord Shark the Unknown, Derek Holt group-theory
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Nov 22 '18 at 7:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
No. The free abelian group $mathbb Z^infty$ has a quotient (e.g. quotienting out the first component) isomorphic to itself.
– Cave Johnson
Nov 22 '18 at 1:35
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This question already has an answer here:
Does $Gcong G/H$ imply that $H$ is trivial?
10 answers
I am trying to show that two groups are isomorphic only if a certain condition holds. I can show that a specific epimorphism between the two groups is one-to-one only if this condition holds, but it isn't obvious that for infinite groups that this should mean that there can't exist an isomorphism between the two groups.
If there is a non-one-to-one epimorphism from a group $G_1$ to a group $G_2$, does this mean that all epimorphisms from $G_1$ to $G_2$ are not one-to-one?
group-theory group-isomorphism group-homomorphism infinite-groups
asked Nov 22 '18 at 1:32
Information Aether
946
This question already has an answer here:
Does $Gcong G/H$ imply that $H$ is trivial?
10 answers
I am trying to show that two groups are isomorphic only if a certain condition holds. I can show that a specific epimorphism between the two groups is one-to-one only if this condition holds, but it isn't obvious that for infinite groups that this should mean that there can't exist an isomorphism between the two groups.
If there is a non-one-to-one epimorphism from a group $G_1$ to a group $G_2$, does this mean that all epimorphisms from $G_1$ to $G_2$ are not one-to-one?
This question already has an answer here:
Does $Gcong G/H$ imply that $H$ is trivial?
10 answers
group-theory group-isomorphism group-homomorphism infinite-groups
group-theory group-isomorphism group-homomorphism infinite-groups
asked Nov 22 '18 at 1:32
Information Aether
946
asked Nov 22 '18 at 1:32
Information Aether
946
asked Nov 22 '18 at 1:32
Information Aether
946
asked Nov 22 '18 at 1:32
Information Aether
946
asked Nov 22 '18 at 1:32
Information Aether
946
946
marked as duplicate by Paul Plummer, user10354138, Lord Shark the Unknown, Derek Holt group-theory
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Nov 22 '18 at 7:55
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marked as duplicate by Paul Plummer, user10354138, Lord Shark the Unknown, Derek Holt group-theory
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Nov 22 '18 at 7:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
No. The free abelian group $mathbb Z^infty$ has a quotient (e.g. quotienting out the first component) isomorphic to itself.
– Cave Johnson
Nov 22 '18 at 1:35
add a comment |
1
No. The free abelian group $mathbb Z^infty$ has a quotient (e.g. quotienting out the first component) isomorphic to itself.
– Cave Johnson
Nov 22 '18 at 1:35
1
1
No. The free abelian group $mathbb Z^infty$ has a quotient (e.g. quotienting out the first component) isomorphic to itself.
– Cave Johnson
Nov 22 '18 at 1:35
No. The free abelian group $mathbb Z^infty$ has a quotient (e.g. quotienting out the first component) isomorphic to itself.
– Cave Johnson
Nov 22 '18 at 1:35
add a comment |
1 Answer
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No, that would be too much to ask. For instance, consider the map $$g:(Bbb Z[X],+)to (Bbb Z[X],+)\ g(p)=frac{p-p(0)}{X}$$
Then, $ker g=Bbb Z$, the map $pmapsto Xp$ is a section of $g$ and $id:Bbb Z[X]to Bbb Z[X]$ is an isomorphism.
answered Nov 22 '18 at 1:35
Saucy O'Path
5,8721626
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1 Answer
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1 Answer
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No, that would be too much to ask. For instance, consider the map $$g:(Bbb Z[X],+)to (Bbb Z[X],+)\ g(p)=frac{p-p(0)}{X}$$
Then, $ker g=Bbb Z$, the map $pmapsto Xp$ is a section of $g$ and $id:Bbb Z[X]to Bbb Z[X]$ is an isomorphism.
answered Nov 22 '18 at 1:35
Saucy O'Path
5,8721626
add a comment |
No, that would be too much to ask. For instance, consider the map $$g:(Bbb Z[X],+)to (Bbb Z[X],+)\ g(p)=frac{p-p(0)}{X}$$
Then, $ker g=Bbb Z$, the map $pmapsto Xp$ is a section of $g$ and $id:Bbb Z[X]to Bbb Z[X]$ is an isomorphism.
answered Nov 22 '18 at 1:35
Saucy O'Path
5,8721626
add a comment |
No, that would be too much to ask. For instance, consider the map $$g:(Bbb Z[X],+)to (Bbb Z[X],+)\ g(p)=frac{p-p(0)}{X}$$
Then, $ker g=Bbb Z$, the map $pmapsto Xp$ is a section of $g$ and $id:Bbb Z[X]to Bbb Z[X]$ is an isomorphism.
answered Nov 22 '18 at 1:35
Saucy O'Path
5,8721626
No, that would be too much to ask. For instance, consider the map $$g:(Bbb Z[X],+)to (Bbb Z[X],+)\ g(p)=frac{p-p(0)}{X}$$
Then, $ker g=Bbb Z$, the map $pmapsto Xp$ is a section of $g$ and $id:Bbb Z[X]to Bbb Z[X]$ is an isomorphism.
answered Nov 22 '18 at 1:35
Saucy O'Path
5,8721626
edited Nov 22 '18 at 1:42
answered Nov 22 '18 at 1:35
Saucy O'Path
5,8721626
answered Nov 22 '18 at 1:35
Saucy O'Path
5,8721626
answered Nov 22 '18 at 1:35
Saucy O'Path
5,8721626
5,8721626
add a comment |
add a comment |
1
No. The free abelian group $mathbb Z^infty$ has a quotient (e.g. quotienting out the first component) isomorphic to itself.
– Cave Johnson
Nov 22 '18 at 1:35