Talk:Second Isomorphism Theorem/Groups

Explaining the change. Previously, we had:

$\map \ker \phi = \set {h \in H: \map \phi {h} = e_{H N / N}}$

$\map \ker \phi = \set {h \in H: hN = N}$

However, we could only assume that if we were sure h = e_H where e_H is the identity in H. What we know is that h multiplied by some n gives e_{H_N/N}. If you see in the new version, we do not assume h = e_H. We simply reach the conclusion that there is some n in N that is equals to h. n and h don't need to be identities.

So all we're short of is a link to the page proving it.

Reverted changes, so as for the link to be added to the appropriate c parameter. --prime mover (talk) 10:52, 22 December 2023 (UTC)

Yes. that would be perfect. We proved one way (==>). For completeness, we would need to add in that link, the way back (<==) since the goal is to prove that $hN = e_{H N / N }$. That means, we have proved given $h$, $\exists h$. We need to prove $n$, $\exists h$.

All this stuff should be there in existing results. Just a matter of looking for them, and that person will not be me -- at least not today, I have other things to do, like deliver parcels. --prime mover (talk) 11:54, 22 December 2023 (UTC)

I will try to have a look later today or tomorrow. I started looking at the Proofwiki because I found errors or inconsistencies in textbooks. I like the detailed way of Proofwiki.

So I found this page which contains a note why nN = N. https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/First-Semester_Abstract_Algebra%3A_A_Structural_Approach_(Sklar)/09%3A_The_Isomorphism_Theorem/9.02%3A_The_Second_and_Third_Isomorphism_Theorems

Quote from that page:

$\map \phi {h}=hN$ (notice that when $h \in H$, $h \in HN$ since $h=h e_G$).

The reason is that $e_G \in H$ since $H$ is subgroup of $G$ and thus, the identity of $G$ is also identity of $H$. I think that is a good enough explanation.