Is every normal subgroup the kernel of some self-homomorphism? [duplicate]
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This question already has an answer here:
Is every normal subgroup the kernel of some endomorphism?
3 answers
Let $G$ be a group. If there is a homomorphism $f:Gto G$ (special case of the codomain being arbitrary group), then the kernel $f^-1(id)$ is a normal subgroup of $G$.
But now the other way around: Start out with the existence of a normal subgroup $H$ of $G$. Is there necessarily a homomorphism $f:Gto G$ such that the kernel of $f$ is $H$?
group-theory normal-subgroups group-homomorphism
edited Mar 1 at 12:08
Dietrich Burde
81.5k648106
asked Mar 1 at 11:55
user56834user56834
3,35921253
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marked as duplicate by Arnaud D., John Coleman, Adrian Keister, Community♦ Mar 1 at 17:00
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.
add a comment |
$begingroup$
This question already has an answer here:
Is every normal subgroup the kernel of some endomorphism?
3 answers
Let $G$ be a group. If there is a homomorphism $f:Gto G$ (special case of the codomain being arbitrary group), then the kernel $f^-1(id)$ is a normal subgroup of $G$.
But now the other way around: Start out with the existence of a normal subgroup $H$ of $G$. Is there necessarily a homomorphism $f:Gto G$ such that the kernel of $f$ is $H$?
group-theory normal-subgroups group-homomorphism
edited Mar 1 at 12:08
Dietrich Burde
81.5k648106
asked Mar 1 at 11:55
user56834user56834
3,35921253
$endgroup$
marked as duplicate by Arnaud D., John Coleman, Adrian Keister, Community♦ Mar 1 at 17:00
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.
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
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– Dietrich Burde
Mar 1 at 12:09
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@DietrichBurde that’s fine thanks!
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– user56834
Mar 1 at 12:25
$begingroup$
See also math.stackexchange.com/questions/1826655/…
$endgroup$
– Arnaud D.
Mar 1 at 14:09
1
$begingroup$
A more interesting question would be: for which groups is this true? Trivially true for simple groups and also true for finite cyclic groups. Are there any others?
$endgroup$
– John Coleman
Mar 1 at 14:55
add a comment |
$begingroup$
This question already has an answer here:
Is every normal subgroup the kernel of some endomorphism?
3 answers
Let $G$ be a group. If there is a homomorphism $f:Gto G$ (special case of the codomain being arbitrary group), then the kernel $f^-1(id)$ is a normal subgroup of $G$.
But now the other way around: Start out with the existence of a normal subgroup $H$ of $G$. Is there necessarily a homomorphism $f:Gto G$ such that the kernel of $f$ is $H$?
group-theory normal-subgroups group-homomorphism
edited Mar 1 at 12:08
Dietrich Burde
81.5k648106
asked Mar 1 at 11:55
user56834user56834
3,35921253
$endgroup$
This question already has an answer here:
Is every normal subgroup the kernel of some endomorphism?
3 answers
Let $G$ be a group. If there is a homomorphism $f:Gto G$ (special case of the codomain being arbitrary group), then the kernel $f^-1(id)$ is a normal subgroup of $G$.
But now the other way around: Start out with the existence of a normal subgroup $H$ of $G$. Is there necessarily a homomorphism $f:Gto G$ such that the kernel of $f$ is $H$?
This question already has an answer here:
Is every normal subgroup the kernel of some endomorphism?
3 answers
group-theory normal-subgroups group-homomorphism
group-theory normal-subgroups group-homomorphism
edited Mar 1 at 12:08
Dietrich Burde
81.5k648106
asked Mar 1 at 11:55
user56834user56834
3,35921253
edited Mar 1 at 12:08
Dietrich Burde
81.5k648106
asked Mar 1 at 11:55
user56834user56834
3,35921253
edited Mar 1 at 12:08
Dietrich Burde
81.5k648106
edited Mar 1 at 12:08
Dietrich Burde
81.5k648106
edited Mar 1 at 12:08
Dietrich Burde
81.5k648106
81.5k648106
asked Mar 1 at 11:55
user56834user56834
3,35921253
asked Mar 1 at 11:55
user56834user56834
3,35921253
asked Mar 1 at 11:55
user56834user56834
3,35921253
3,35921253
marked as duplicate by Arnaud D., John Coleman, Adrian Keister, Community♦ Mar 1 at 17:00
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.
marked as duplicate by Arnaud D., John Coleman, Adrian Keister, Community♦ Mar 1 at 17:00
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.
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
Mar 1 at 12:09
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
Mar 1 at 12:25
$begingroup$
See also math.stackexchange.com/questions/1826655/…
$endgroup$
– Arnaud D.
Mar 1 at 14:09
1
$begingroup$
A more interesting question would be: for which groups is this true? Trivially true for simple groups and also true for finite cyclic groups. Are there any others?
$endgroup$
– John Coleman
Mar 1 at 14:55
add a comment |
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
Mar 1 at 12:09
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
Mar 1 at 12:25
$begingroup$
See also math.stackexchange.com/questions/1826655/…
$endgroup$
– Arnaud D.
Mar 1 at 14:09
1
$begingroup$
A more interesting question would be: for which groups is this true? Trivially true for simple groups and also true for finite cyclic groups. Are there any others?
$endgroup$
– John Coleman
Mar 1 at 14:55
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
Mar 1 at 12:09
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
Mar 1 at 12:09
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
Mar 1 at 12:25
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
Mar 1 at 12:25
$begingroup$
See also math.stackexchange.com/questions/1826655/…
$endgroup$
– Arnaud D.
Mar 1 at 14:09
$begingroup$
See also math.stackexchange.com/questions/1826655/…
$endgroup$
– Arnaud D.
Mar 1 at 14:09
1
1
$begingroup$
A more interesting question would be: for which groups is this true? Trivially true for simple groups and also true for finite cyclic groups. Are there any others?
$endgroup$
– John Coleman
Mar 1 at 14:55
$begingroup$
A more interesting question would be: for which groups is this true? Trivially true for simple groups and also true for finite cyclic groups. Are there any others?
$endgroup$
– John Coleman
Mar 1 at 14:55
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
That's false.
If $G=mathbbZ$ then any homomorphism $f:mathbbZrightarrowmathbbZ$ takes the form $f(a)=ma$ for some $minmathbbZ$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $ninmathbbN$ we have that $nmathbbZ$ is a normal subgroup of $mathbbZ$. In particular $2mathbbZ$ is not a kernel of any homomorphim from $mathbbZ$ to itself.
However, it is possible to "correct" this statement. If you only request an homomorphism from $G$ to some other group. Given any normal subgroup $H$, the quotient homomorphism $f:Grightarrow G/H$ which sends $gmapsto g+H$ has $H$ as it's kernel. In other words, every normal subgroup is a kernel of some homomorphism, not necessarily from $G$ to itself.
answered Mar 1 at 11:59
YankoYanko
8,1282830
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That's false.
If $G=mathbbZ$ then any homomorphism $f:mathbbZrightarrowmathbbZ$ takes the form $f(a)=ma$ for some $minmathbbZ$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $ninmathbbN$ we have that $nmathbbZ$ is a normal subgroup of $mathbbZ$. In particular $2mathbbZ$ is not a kernel of any homomorphim from $mathbbZ$ to itself.
However, it is possible to "correct" this statement. If you only request an homomorphism from $G$ to some other group. Given any normal subgroup $H$, the quotient homomorphism $f:Grightarrow G/H$ which sends $gmapsto g+H$ has $H$ as it's kernel. In other words, every normal subgroup is a kernel of some homomorphism, not necessarily from $G$ to itself.
answered Mar 1 at 11:59
YankoYanko
8,1282830
$endgroup$
add a comment |
$begingroup$
That's false.
If $G=mathbbZ$ then any homomorphism $f:mathbbZrightarrowmathbbZ$ takes the form $f(a)=ma$ for some $minmathbbZ$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $ninmathbbN$ we have that $nmathbbZ$ is a normal subgroup of $mathbbZ$. In particular $2mathbbZ$ is not a kernel of any homomorphim from $mathbbZ$ to itself.
However, it is possible to "correct" this statement. If you only request an homomorphism from $G$ to some other group. Given any normal subgroup $H$, the quotient homomorphism $f:Grightarrow G/H$ which sends $gmapsto g+H$ has $H$ as it's kernel. In other words, every normal subgroup is a kernel of some homomorphism, not necessarily from $G$ to itself.
answered Mar 1 at 11:59
YankoYanko
8,1282830
$endgroup$
add a comment |
$begingroup$
That's false.
If $G=mathbbZ$ then any homomorphism $f:mathbbZrightarrowmathbbZ$ takes the form $f(a)=ma$ for some $minmathbbZ$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $ninmathbbN$ we have that $nmathbbZ$ is a normal subgroup of $mathbbZ$. In particular $2mathbbZ$ is not a kernel of any homomorphim from $mathbbZ$ to itself.
However, it is possible to "correct" this statement. If you only request an homomorphism from $G$ to some other group. Given any normal subgroup $H$, the quotient homomorphism $f:Grightarrow G/H$ which sends $gmapsto g+H$ has $H$ as it's kernel. In other words, every normal subgroup is a kernel of some homomorphism, not necessarily from $G$ to itself.
answered Mar 1 at 11:59
YankoYanko
8,1282830
$endgroup$
That's false.
If $G=mathbbZ$ then any homomorphism $f:mathbbZrightarrowmathbbZ$ takes the form $f(a)=ma$ for some $minmathbbZ$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $ninmathbbN$ we have that $nmathbbZ$ is a normal subgroup of $mathbbZ$. In particular $2mathbbZ$ is not a kernel of any homomorphim from $mathbbZ$ to itself.
However, it is possible to "correct" this statement. If you only request an homomorphism from $G$ to some other group. Given any normal subgroup $H$, the quotient homomorphism $f:Grightarrow G/H$ which sends $gmapsto g+H$ has $H$ as it's kernel. In other words, every normal subgroup is a kernel of some homomorphism, not necessarily from $G$ to itself.
answered Mar 1 at 11:59
YankoYanko
8,1282830
answered Mar 1 at 11:59
YankoYanko
8,1282830
answered Mar 1 at 11:59
YankoYanko
8,1282830
answered Mar 1 at 11:59
YankoYanko
8,1282830
8,1282830
add a comment |
add a comment |
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
Mar 1 at 12:09
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
Mar 1 at 12:25
$begingroup$
See also math.stackexchange.com/questions/1826655/…
$endgroup$
– Arnaud D.
Mar 1 at 14:09
1
$begingroup$
A more interesting question would be: for which groups is this true? Trivially true for simple groups and also true for finite cyclic groups. Are there any others?
$endgroup$
– John Coleman
Mar 1 at 14:55