Derived Subgroup and Factor Groups

Question

Let $N \unlhd G$, does it holds that $(G/N)' = G'N /N$, or more generally for any subgroup $H \le G$ we have $(HN/N)' = H'N/N$? Does anyone has a proof of this fact?

PS: Side-Question: Is it wrong to write $(G/N)' = G'/N$?

Answer 1

Write $x=hn$ and $y=km$ with $h,k\in H,\ n,m\in N$, and use equations like $nk=k(k^{-1}nk)\in HN$ to pull each $h,k,h^{-1},k^{-1}$ in front of all $N$-elements. Thus you will see that $[x,y]\in H'N$.
(So, it means $(HN)'\subseteq H'N$.)

Conversely, if $g\in H'$ then $g$ is a product of commutators $[h_i,k_i]$, then any $sN\in H'N/N$ can be written as $s=gn$ for some $g\in H'$ and $n\in N$, but then $sN=gnN=gN=[h_1N,k_1N][h_2N,k_2N]\dots\ \in (HN/N)'$.

Answer 2

Yes this is correct, that is $(G/N)'=G'N/N$, but it is wrong to write $(G/N)'=G'/N$, since $N$ does not have to be a subgroup of $G'$.

Proofs can be given by writing out everything mod $N$. Can give it later if you want, but first try it yourself. And $(HN/N)'=H'N/N$ is correct too.

Answer 3

Let $f$ be the canonical homomorphism $G \rightarrow G/N$. Every element of $(G/N)'$ is of the form $[f(x_1), f(y_1)]\cdots [f(x_n), f(y_n)] = [x_1, y_1]\cdots [x_n, y_n]N$, where $[a, b]$ denotes the commutator of $a,b$. Hence $(G/N)' = G'N /N$. A similar argument can be applied for $(HN/N)' = H'N /N$.