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Is every normal subgroup the kernel of some self-homomorphism?

Is every normal subgroup the kernel of some endomorphism?Necessary and sufficient condition for a normal group to be kernel of a homomorphism from the group to itselfCommutator 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.

4

1

$begingroup$

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

share|cite|improve this question

edited 2 hours ago

Dietrich Burde

80.1k647104

asked 2 hours 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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DietrichBurde that’s fine thanks!
  $endgroup$
  – user56834
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Is every normal subgroup the kernel of some endomorphism?
  $endgroup$
  – Arnaud D.
  12 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also math.stackexchange.com/questions/1826655/…
  $endgroup$
  – Arnaud D.
  8 mins ago
  

  
  
  
  

add a comment | 

4

1

$begingroup$

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

share|cite|improve this question

edited 2 hours ago

Dietrich Burde

80.1k647104

asked 2 hours 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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DietrichBurde that’s fine thanks!
  $endgroup$
  – user56834
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Is every normal subgroup the kernel of some endomorphism?
  $endgroup$
  – Arnaud D.
  12 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also math.stackexchange.com/questions/1826655/…
  $endgroup$
  – Arnaud D.
  8 mins 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

share|cite|improve this question

edited 2 hours ago

Dietrich Burde

80.1k647104

asked 2 hours ago

user56834user56834

3,24821252

$endgroup$

share|cite|improve this question

edited 2 hours ago

Dietrich Burde

80.1k647104

asked 2 hours ago

user56834user56834

3,24821252

share|cite|improve this question

edited 2 hours ago

Dietrich Burde

80.1k647104

asked 2 hours ago

user56834user56834

3,24821252

edited 2 hours ago

Dietrich Burde

80.1k647104

edited 2 hours ago

Dietrich Burde

80.1k647104

asked 2 hours ago

user56834user56834

3,24821252

asked 2 hours ago

user56834user56834

3,24821252

1 Answer
1

active

oldest

votes

6

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

YankoYanko

7,3201729

$endgroup$

add a comment | 

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6

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

YankoYanko

7,3201729

$endgroup$

add a comment | 

6

$begingroup$

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.

share|cite|improve this answer

answered 2 hours 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.

share|cite|improve this answer

answered 2 hours ago

YankoYanko

7,3201729

$endgroup$

share|cite|improve this answer

answered 2 hours ago

YankoYanko

7,3201729

answered 2 hours ago

YankoYanko

7,3201729

answered 2 hours ago

YankoYanko

7,3201729