A small doubt on the connection between homomorphisms and normal subgroups?












2














I'm reading Robert Ash's Basic Abstract Algebra. Here:




enter image description here




I've tried to find the proof that every normal subgroup is the kernel of a homomorphism before looking in the book. I got stuck thinking about an $f:Gto H$ and failing, eventually I gave up.



When I looked at the proof, it was easy to follow but he picks $pi: G to G/N$ and I guess I'd never have guessed that. I am confused with the following: I was expecting the result to be valid for any homomorphism but he picks that exact one. My doubt may be a little bit silly, but why this proof for $pi:Gto G/N$ makes the result valid for any $Hneq G/N$? I guess It's clear that that proposition depends only on $G$ and hence, we can pick any H but I am unsure why this is legitimate.










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  • 3




    The result is valid because you started with an arbitrary normal subgroup and then constructed a homomorphism whose kernel was that normal subgroup. Note that you do not need the homomorphism to be arbitrary as you think it is.
    – Hello_World
    yesterday










  • @Hello_World But what If I have $f:G to H,g: Gto K$ and the kernels of both are different? Could they be different or $G$ determines the kernels for all $H,K$?
    – Billy Rubina
    yesterday








  • 2




    All we know is that the kernel of $f$ and $g$ is a normal subgroup of $G.$ That's it. They can be different or same.
    – Hello_World
    yesterday
















2














I'm reading Robert Ash's Basic Abstract Algebra. Here:




enter image description here




I've tried to find the proof that every normal subgroup is the kernel of a homomorphism before looking in the book. I got stuck thinking about an $f:Gto H$ and failing, eventually I gave up.



When I looked at the proof, it was easy to follow but he picks $pi: G to G/N$ and I guess I'd never have guessed that. I am confused with the following: I was expecting the result to be valid for any homomorphism but he picks that exact one. My doubt may be a little bit silly, but why this proof for $pi:Gto G/N$ makes the result valid for any $Hneq G/N$? I guess It's clear that that proposition depends only on $G$ and hence, we can pick any H but I am unsure why this is legitimate.










share|cite|improve this question


















  • 3




    The result is valid because you started with an arbitrary normal subgroup and then constructed a homomorphism whose kernel was that normal subgroup. Note that you do not need the homomorphism to be arbitrary as you think it is.
    – Hello_World
    yesterday










  • @Hello_World But what If I have $f:G to H,g: Gto K$ and the kernels of both are different? Could they be different or $G$ determines the kernels for all $H,K$?
    – Billy Rubina
    yesterday








  • 2




    All we know is that the kernel of $f$ and $g$ is a normal subgroup of $G.$ That's it. They can be different or same.
    – Hello_World
    yesterday














2












2








2


1





I'm reading Robert Ash's Basic Abstract Algebra. Here:




enter image description here




I've tried to find the proof that every normal subgroup is the kernel of a homomorphism before looking in the book. I got stuck thinking about an $f:Gto H$ and failing, eventually I gave up.



When I looked at the proof, it was easy to follow but he picks $pi: G to G/N$ and I guess I'd never have guessed that. I am confused with the following: I was expecting the result to be valid for any homomorphism but he picks that exact one. My doubt may be a little bit silly, but why this proof for $pi:Gto G/N$ makes the result valid for any $Hneq G/N$? I guess It's clear that that proposition depends only on $G$ and hence, we can pick any H but I am unsure why this is legitimate.










share|cite|improve this question













I'm reading Robert Ash's Basic Abstract Algebra. Here:




enter image description here




I've tried to find the proof that every normal subgroup is the kernel of a homomorphism before looking in the book. I got stuck thinking about an $f:Gto H$ and failing, eventually I gave up.



When I looked at the proof, it was easy to follow but he picks $pi: G to G/N$ and I guess I'd never have guessed that. I am confused with the following: I was expecting the result to be valid for any homomorphism but he picks that exact one. My doubt may be a little bit silly, but why this proof for $pi:Gto G/N$ makes the result valid for any $Hneq G/N$? I guess It's clear that that proposition depends only on $G$ and hence, we can pick any H but I am unsure why this is legitimate.







abstract-algebra group-theory






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asked yesterday









Billy Rubina

10.3k1458134




10.3k1458134








  • 3




    The result is valid because you started with an arbitrary normal subgroup and then constructed a homomorphism whose kernel was that normal subgroup. Note that you do not need the homomorphism to be arbitrary as you think it is.
    – Hello_World
    yesterday










  • @Hello_World But what If I have $f:G to H,g: Gto K$ and the kernels of both are different? Could they be different or $G$ determines the kernels for all $H,K$?
    – Billy Rubina
    yesterday








  • 2




    All we know is that the kernel of $f$ and $g$ is a normal subgroup of $G.$ That's it. They can be different or same.
    – Hello_World
    yesterday














  • 3




    The result is valid because you started with an arbitrary normal subgroup and then constructed a homomorphism whose kernel was that normal subgroup. Note that you do not need the homomorphism to be arbitrary as you think it is.
    – Hello_World
    yesterday










  • @Hello_World But what If I have $f:G to H,g: Gto K$ and the kernels of both are different? Could they be different or $G$ determines the kernels for all $H,K$?
    – Billy Rubina
    yesterday








  • 2




    All we know is that the kernel of $f$ and $g$ is a normal subgroup of $G.$ That's it. They can be different or same.
    – Hello_World
    yesterday








3




3




The result is valid because you started with an arbitrary normal subgroup and then constructed a homomorphism whose kernel was that normal subgroup. Note that you do not need the homomorphism to be arbitrary as you think it is.
– Hello_World
yesterday




The result is valid because you started with an arbitrary normal subgroup and then constructed a homomorphism whose kernel was that normal subgroup. Note that you do not need the homomorphism to be arbitrary as you think it is.
– Hello_World
yesterday












@Hello_World But what If I have $f:G to H,g: Gto K$ and the kernels of both are different? Could they be different or $G$ determines the kernels for all $H,K$?
– Billy Rubina
yesterday






@Hello_World But what If I have $f:G to H,g: Gto K$ and the kernels of both are different? Could they be different or $G$ determines the kernels for all $H,K$?
– Billy Rubina
yesterday






2




2




All we know is that the kernel of $f$ and $g$ is a normal subgroup of $G.$ That's it. They can be different or same.
– Hello_World
yesterday




All we know is that the kernel of $f$ and $g$ is a normal subgroup of $G.$ That's it. They can be different or same.
– Hello_World
yesterday










1 Answer
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Read the passage carefully. Does the author claim (a) that if $N$ is a normal subgroup of $G$ then for every group $H$ there is a homomorphism $f:Gto H$ with kernel $N$? Or is he merely claiming (b) that if $N$ is a normal subgroup of $G$ then $N$ is the kernel of some homomorphism of $G$ into some group $H$?



In case (a) your objection is justified and the claim is wrong: ${e}$ is a normal subgroup of $G$, and if $H$ is smaller than $G$ then there is no homomorphism from $G$ to $H$ whose kernel is ${e}$.



In case (b) your objection goes away.






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    7














    Read the passage carefully. Does the author claim (a) that if $N$ is a normal subgroup of $G$ then for every group $H$ there is a homomorphism $f:Gto H$ with kernel $N$? Or is he merely claiming (b) that if $N$ is a normal subgroup of $G$ then $N$ is the kernel of some homomorphism of $G$ into some group $H$?



    In case (a) your objection is justified and the claim is wrong: ${e}$ is a normal subgroup of $G$, and if $H$ is smaller than $G$ then there is no homomorphism from $G$ to $H$ whose kernel is ${e}$.



    In case (b) your objection goes away.






    share|cite|improve this answer




























      7














      Read the passage carefully. Does the author claim (a) that if $N$ is a normal subgroup of $G$ then for every group $H$ there is a homomorphism $f:Gto H$ with kernel $N$? Or is he merely claiming (b) that if $N$ is a normal subgroup of $G$ then $N$ is the kernel of some homomorphism of $G$ into some group $H$?



      In case (a) your objection is justified and the claim is wrong: ${e}$ is a normal subgroup of $G$, and if $H$ is smaller than $G$ then there is no homomorphism from $G$ to $H$ whose kernel is ${e}$.



      In case (b) your objection goes away.






      share|cite|improve this answer


























        7












        7








        7






        Read the passage carefully. Does the author claim (a) that if $N$ is a normal subgroup of $G$ then for every group $H$ there is a homomorphism $f:Gto H$ with kernel $N$? Or is he merely claiming (b) that if $N$ is a normal subgroup of $G$ then $N$ is the kernel of some homomorphism of $G$ into some group $H$?



        In case (a) your objection is justified and the claim is wrong: ${e}$ is a normal subgroup of $G$, and if $H$ is smaller than $G$ then there is no homomorphism from $G$ to $H$ whose kernel is ${e}$.



        In case (b) your objection goes away.






        share|cite|improve this answer














        Read the passage carefully. Does the author claim (a) that if $N$ is a normal subgroup of $G$ then for every group $H$ there is a homomorphism $f:Gto H$ with kernel $N$? Or is he merely claiming (b) that if $N$ is a normal subgroup of $G$ then $N$ is the kernel of some homomorphism of $G$ into some group $H$?



        In case (a) your objection is justified and the claim is wrong: ${e}$ is a normal subgroup of $G$, and if $H$ is smaller than $G$ then there is no homomorphism from $G$ to $H$ whose kernel is ${e}$.



        In case (b) your objection goes away.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        bof

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