A small doubt on the connection between homomorphisms and normal subgroups?
$begingroup$
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 Jan 6 at 5:13
Billy RubinaBilly Rubina
10.4k1458134
$endgroup$
add a comment |
$begingroup$
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 Jan 6 at 5:13
Billy RubinaBilly Rubina
10.4k1458134
$endgroup$
3
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:20
$begingroup$
@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 $H,K$?
$endgroup$
– Billy Rubina
Jan 6 at 5:23
2
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:25
add a comment |
$begingroup$
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 Jan 6 at 5:13
Billy RubinaBilly Rubina
10.4k1458134
$endgroup$
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 Jan 6 at 5:13
Billy RubinaBilly Rubina
10.4k1458134
asked Jan 6 at 5:13
Billy RubinaBilly Rubina
10.4k1458134
asked Jan 6 at 5:13
Billy RubinaBilly Rubina
10.4k1458134
asked Jan 6 at 5:13
Billy RubinaBilly Rubina
10.4k1458134
asked Jan 6 at 5:13
Billy RubinaBilly Rubina
10.4k1458134
10.4k1458134
3
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:20
$begingroup$
@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 $H,K$?
$endgroup$
– Billy Rubina
Jan 6 at 5:23
2
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:25
add a comment |
3
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:20
$begingroup$
@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 $H,K$?
$endgroup$
– Billy Rubina
Jan 6 at 5:23
2
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:25
3
3
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:20
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:20
$begingroup$
@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 $H,K$?
$endgroup$
– Billy Rubina
Jan 6 at 5:23
$begingroup$
@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 $H,K$?
$endgroup$
– Billy Rubina
Jan 6 at 5:23
2
2
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:25
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:25
add a comment |
1 Answer
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$begingroup$
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 Jan 6 at 5:25
bofbof
51k457120
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add a comment |
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$begingroup$
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 Jan 6 at 5:25
bofbof
51k457120
$endgroup$
add a comment |
$begingroup$
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 Jan 6 at 5:25
bofbof
51k457120
$endgroup$
add a comment |
$begingroup$
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 Jan 6 at 5:25
bofbof
51k457120
$endgroup$
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 Jan 6 at 5:25
bofbof
51k457120
edited Jan 6 at 5:47
answered Jan 6 at 5:25
bofbof
51k457120
answered Jan 6 at 5:25
bofbof
51k457120
answered Jan 6 at 5:25
bofbof
51k457120
51k457120
add a comment |
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$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:20
$begingroup$
@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 $H,K$?
$endgroup$
– Billy Rubina
Jan 6 at 5:23
2
$begingroup$
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.
$endgroup$
– Hello_World
Jan 6 at 5:25