# 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 $$f:G\to H$$ 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.

• 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. – nls Jan 6 '19 at 5:20
• @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 $H,K$? – Billy Rubina Jan 6 '19 at 5:23
• 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. – 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 $$\{e\}$$.