Is every normal subgroup the kernel of some self-homomorphism?Commutator Subgroup is Normal Subgroup of...
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Is every normal subgroup the kernel of some self-homomorphism?
Commutator Subgroup is Normal Subgroup of Kernel of HomomorphismIs every normal subgroup the kernel of some endomorphism?What is an example of a proper normal subgroup of the kernel of a homomorphism?Every normal subgroup is the kernel of some homomorphismImage of normal subgroup under surjective homomorphism as kernelNormal subgroup implies quotient group?Understanding normal subgroup using invariance under conjugationQuestion about normal subgroup that contains Kernel of a homomorphismRestriction of homomorphism to finite index subgroup gives finite index subgroup of kernelIf $H$ is normal in $G$ then $H$ is the kernel of a group homomorphism.
$begingroup$
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 1 hour ago
Dietrich Burde
80.1k647104
asked 1 hour ago
user56834user56834
3,24821252
$endgroup$
add a comment |
$begingroup$
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 1 hour ago
Dietrich Burde
80.1k647104
asked 1 hour ago
user56834user56834
3,24821252
$endgroup$
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
1 hour ago
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
1 hour ago
add a comment |
$begingroup$
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 1 hour ago
Dietrich Burde
80.1k647104
asked 1 hour ago
user56834user56834
3,24821252
$endgroup$
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
group-theory normal-subgroups group-homomorphism
edited 1 hour ago
Dietrich Burde
80.1k647104
asked 1 hour ago
user56834user56834
3,24821252
edited 1 hour ago
Dietrich Burde
80.1k647104
asked 1 hour ago
user56834user56834
3,24821252
edited 1 hour ago
Dietrich Burde
80.1k647104
edited 1 hour ago
Dietrich Burde
80.1k647104
edited 1 hour ago
Dietrich Burde
80.1k647104
80.1k647104
asked 1 hour ago
user56834user56834
3,24821252
asked 1 hour ago
user56834user56834
3,24821252
asked 1 hour ago
user56834user56834
3,24821252
3,24821252
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
1 hour ago
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
1 hour ago
add a comment |
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
1 hour ago
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
1 hour ago
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
1 hour ago
$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
1 hour ago
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
1 hour ago
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
1 hour ago
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
That's false.
If $G=mathbb{Z}$ then any homomorphism $f:mathbb{Z}rightarrowmathbb{Z}$ takes the form $f(a)=ma$ for some $minmathbb{Z}$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $ninmathbb{N}$ we have that $nmathbb{Z}$ is a normal subgroup of $mathbb{Z}$. In particular $2mathbb{Z}$ is not a kernel of any homomorphim from $mathbb{Z}$ 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 1 hour ago
YankoYanko
7,3201729
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
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$begingroup$
That's false.
If $G=mathbb{Z}$ then any homomorphism $f:mathbb{Z}rightarrowmathbb{Z}$ takes the form $f(a)=ma$ for some $minmathbb{Z}$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $ninmathbb{N}$ we have that $nmathbb{Z}$ is a normal subgroup of $mathbb{Z}$. In particular $2mathbb{Z}$ is not a kernel of any homomorphim from $mathbb{Z}$ 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 1 hour ago
YankoYanko
7,3201729
$endgroup$
add a comment |
$begingroup$
That's false.
If $G=mathbb{Z}$ then any homomorphism $f:mathbb{Z}rightarrowmathbb{Z}$ takes the form $f(a)=ma$ for some $minmathbb{Z}$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $ninmathbb{N}$ we have that $nmathbb{Z}$ is a normal subgroup of $mathbb{Z}$. In particular $2mathbb{Z}$ is not a kernel of any homomorphim from $mathbb{Z}$ 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 1 hour ago
YankoYanko
7,3201729
$endgroup$
add a comment |
$begingroup$
That's false.
If $G=mathbb{Z}$ then any homomorphism $f:mathbb{Z}rightarrowmathbb{Z}$ takes the form $f(a)=ma$ for some $minmathbb{Z}$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $ninmathbb{N}$ we have that $nmathbb{Z}$ is a normal subgroup of $mathbb{Z}$. In particular $2mathbb{Z}$ is not a kernel of any homomorphim from $mathbb{Z}$ 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 1 hour ago
YankoYanko
7,3201729
$endgroup$
That's false.
If $G=mathbb{Z}$ then any homomorphism $f:mathbb{Z}rightarrowmathbb{Z}$ takes the form $f(a)=ma$ for some $minmathbb{Z}$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $ninmathbb{N}$ we have that $nmathbb{Z}$ is a normal subgroup of $mathbb{Z}$. In particular $2mathbb{Z}$ is not a kernel of any homomorphim from $mathbb{Z}$ 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 1 hour ago
YankoYanko
7,3201729
answered 1 hour ago
YankoYanko
7,3201729
answered 1 hour ago
YankoYanko
7,3201729
answered 1 hour ago
YankoYanko
7,3201729
7,3201729
add a comment |
add a comment |
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$begingroup$
I edited the title to clarify the question. Let me know if you agree with it.
$endgroup$
– Dietrich Burde
1 hour ago
$begingroup$
@DietrichBurde that’s fine thanks!
$endgroup$
– user56834
1 hour ago