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The kernel of a function [closed]

Order of kernel of a homomorphism.Image of subgroup and Kernel of homomorphism form subgroupsEvery normal subgroup is the kernel of some homomorphismHomomorphism from $mathbbZoplus mathbbZoplus mathbbZrightarrow mathbbZoplus mathbbZ$ has non-trivial kernel: elementary argumentNumber of constituents in invariant factor decomposition of kernel of homomorphismkernel of the homomorphism (determinant of matrix)Kernel of injective group homomorphismMake arbitrary normal subgroup be the kernel of some homomorphismHomomorphisms and Normal SubgroupsIf $H$ is normal in $G$ then $H$ is the kernel of a group homomorphism.

-1

$begingroup$

Let $f:G to G'$ be a homomorphism. Let ker $f$ denote the kernel of $f$.Let $x$ $in$ ker $f$ .Does this imply $x$ is the identity of $G$?Give a counterexample if the proposition is not true.

Can anyone help ?I was proving a group theory problem when suddenly looking at my work I thought whether the above is true.I know that the converse is true....

abstract-algebra group-theory

share|cite|improve this question

edited Mar 16 at 9:22

att epl

asked Mar 16 at 9:08

att eplatt epl

27012

$endgroup$

closed as off-topic by Derek Holt, Thomas Shelby, José Carlos Santos, mrtaurho, jgon Mar 16 at 22:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Thomas Shelby, José Carlos Santos, mrtaurho, jgon

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a self thought question and I was not able to make anything out of it
  $endgroup$
  – att epl
  Mar 16 at 9:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Its a doubt....
  $endgroup$
  – att epl
  Mar 16 at 9:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any non-injective homomorphism is a counterexample. Also as an exercise you can try to prove that if $xin$ ker $f$ implies $x=1$ if and only if $f$ is injective.
  $endgroup$
  – tangentbundle
  Mar 16 at 11:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your concern.
  $endgroup$
  – att epl
  Mar 16 at 12:31
  

  
  
  
  

add a comment | 

-1

$begingroup$

Let $f:G to G'$ be a homomorphism. Let ker $f$ denote the kernel of $f$.Let $x$ $in$ ker $f$ .Does this imply $x$ is the identity of $G$?Give a counterexample if the proposition is not true.

Can anyone help ?I was proving a group theory problem when suddenly looking at my work I thought whether the above is true.I know that the converse is true....

abstract-algebra group-theory

share|cite|improve this question

edited Mar 16 at 9:22

att epl

asked Mar 16 at 9:08

att eplatt epl

27012

$endgroup$

closed as off-topic by Derek Holt, Thomas Shelby, José Carlos Santos, mrtaurho, jgon Mar 16 at 22:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Thomas Shelby, José Carlos Santos, mrtaurho, jgon

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a self thought question and I was not able to make anything out of it
  $endgroup$
  – att epl
  Mar 16 at 9:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Its a doubt....
  $endgroup$
  – att epl
  Mar 16 at 9:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any non-injective homomorphism is a counterexample. Also as an exercise you can try to prove that if $xin$ ker $f$ implies $x=1$ if and only if $f$ is injective.
  $endgroup$
  – tangentbundle
  Mar 16 at 11:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your concern.
  $endgroup$
  – att epl
  Mar 16 at 12:31
  

  
  
  
  

add a comment | 

$begingroup$

Let $f:G to G'$ be a homomorphism. Let ker $f$ denote the kernel of $f$.Let $x$ $in$ ker $f$ .Does this imply $x$ is the identity of $G$?Give a counterexample if the proposition is not true.

Can anyone help ?I was proving a group theory problem when suddenly looking at my work I thought whether the above is true.I know that the converse is true....

abstract-algebra group-theory

share|cite|improve this question

edited Mar 16 at 9:22

att epl

asked Mar 16 at 9:08

att eplatt epl

27012

$endgroup$

share|cite|improve this question

edited Mar 16 at 9:22

att epl

asked Mar 16 at 9:08

att eplatt epl

27012

share|cite|improve this question

edited Mar 16 at 9:22

att epl

asked Mar 16 at 9:08

att eplatt epl

27012

asked Mar 16 at 9:08

att eplatt epl

27012

asked Mar 16 at 9:08

att eplatt epl

27012

1 Answer
1

active

oldest

votes

1

$begingroup$

For any group $G$ the unique map $f: G longrightarrow e$ is a group homomorphism, so the proposition is not true.

share|cite|improve this answer

answered Mar 16 at 9:25

ServaesServaes

28.5k34099

$endgroup$

add a comment | 

1

$begingroup$

For any group $G$ the unique map $f: G longrightarrow e$ is a group homomorphism, so the proposition is not true.

share|cite|improve this answer

answered Mar 16 at 9:25

ServaesServaes

28.5k34099

$endgroup$

add a comment | 

1

$begingroup$

For any group $G$ the unique map $f: G longrightarrow e$ is a group homomorphism, so the proposition is not true.

share|cite|improve this answer

answered Mar 16 at 9:25

ServaesServaes

28.5k34099

$endgroup$

add a comment | 

$begingroup$

For any group $G$ the unique map $f: G longrightarrow e$ is a group homomorphism, so the proposition is not true.

share|cite|improve this answer

answered Mar 16 at 9:25

ServaesServaes

28.5k34099

$endgroup$

share|cite|improve this answer

answered Mar 16 at 9:25

ServaesServaes

28.5k34099

answered Mar 16 at 9:25

ServaesServaes

28.5k34099

answered Mar 16 at 9:25

ServaesServaes

28.5k34099