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

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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 $f:G\to H$ 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 $\pi:G\to G/N$ makes the result valid for any $H\neq G/N$? 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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    $\begingroup$ 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. $\endgroup$ – nls Jan 6 '19 at 5:20
  • $\begingroup$ @Hello_World But what If I have $f:G \to H,g: G\to K$ 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 '19 at 5:23
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    $\begingroup$ 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. $\endgroup$ – nls Jan 6 '19 at 5:25

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 $f:G\to H$ 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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