A small doubt on the connection between homomorphisms and normal subgroups?
I'm reading Robert Ash's Basic Abstract Algebra. 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
asked yesterday
Billy RubinaBilly Rubina
10.3k1458134
add a comment |
I'm reading Robert Ash's Basic Abstract Algebra. 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
asked yesterday
Billy RubinaBilly Rubina
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
add a comment |
I'm reading Robert Ash's Basic Abstract Algebra. 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
asked yesterday
Billy RubinaBilly Rubina
10.3k1458134
I'm reading Robert Ash's Basic Abstract Algebra. 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
abstract-algebra group-theory
asked yesterday
Billy RubinaBilly Rubina
10.3k1458134
asked yesterday
Billy RubinaBilly Rubina
10.3k1458134
asked yesterday
Billy RubinaBilly Rubina
10.3k1458134
asked yesterday
Billy RubinaBilly Rubina
10.3k1458134
asked yesterday
Billy RubinaBilly 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
add a comment |
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
add a comment |
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.
answered yesterday
bofbof
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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.
answered yesterday
bofbof
50.5k457119
add a comment |
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.
answered yesterday
bofbof
50.5k457119
add a comment |
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.
answered yesterday
bofbof
50.5k457119
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.
answered yesterday
bofbof
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edited yesterday
answered yesterday
bofbof
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answered yesterday
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answered yesterday
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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