I was reading about group theory in Herstein's book (mentioned below) and I came across a couple of propositions that were not clear to me, in the sense that I couldn't quite figure out why they were true and I would really appreciate if you could help me understand them.
They are the following:
Thus there is a one-to-one correspondence between homomorphic images
of G and normal subgroups of G. […]
The set of groups so constructed yields all homomorphic images of G
(up to isomorphisms).
Herstein, I. N., Topics in algebra, Lexington, TX: Xerox College Publishing. xi, 388 p. (1975). ZBL1230.00004.
The constructed set mentioned in the quote above is the set $ \left\{G/N: N \triangleleft G \right\} $
I can understand why given a normal subgroup $N$ of $G$, one can associate a homomorphic image of G, namely, $G/N$, but is the converse true, I mean, can I associate a normal subgroup $N$ of $G$ given a homomorphic image of a group $G$? How exactly does the given set yield all homomorphic images of $G$?
Thanks in advance!
Best Answer
This is basically the First Ismorphism Theorem. Let's say $H$ is a homomorphic image of $G$, i.e. there exists a morphism $\varphi$ such that $\varphi : G \to H$ is surjective. Then $N:= \text{ker} \ \varphi$ is a normal subgroup of $G$ and $G/N \cong H$. Then the correspondence \begin{align} \{ \text{normal subgroups of } G \} &\to \{ \text{homomorphic image of } G \} \\ N &\mapsto G/N \end{align} is inverse to (up to isomorphism) \begin{align} \{ \text{homomorphic image of } G \} &\to \{ \text{normal subgroups of } G \} \\ H &\mapsto N:= \text{ker} \ \varphi, \text{ where } \varphi : G \to H. \end{align}