The following comes from Hungerford's Algebra.
[Prove that if] $f: G \to H$ is a homomorphism, $H$ is abelian and $N$ is a subgroup of $G$ containing $\ker f$, then $N$ is normal in $G.$
A solution found here (at I.5.16) begins by saying "since $H$ is abelian, $f(H)$ is an abelian subgroup." Obviously, this makes sense, but the domain of $f$ is $G.$ Why does $f(H)$ make sense here? This is really the part of the proof I don't understand.
Is there also another way to proceed with the proof?
That entire solution is wrong, in my opinion. For instance following sentence in the proof don’t make any sense: Thus for every $g\in G$, $n\in N$, $(gH)nH(g^{-1}H)=(gH)(g^{-1}H)nH=nH$. Product between elements of $G$ and $H$ are not defined. There are two approaches to solve this exercise. One of the solution is bit harder relative to another. First (easy) approach and second approach.