How to show $(G/N)' = NG'/N$ How to show $(G/N)' = NG'/N$ where $G' = \text{commutator of the group G}$, and $N$ is a normal subgroup of $G$.
 A: Suppose we have a commutator of $G/N$. This is:
$(Na)(Nb)(Na)^{-1}(Nb)^{-1} = (Na)(Nb)(Na^{-1})(Nb^{-1}) = Naba^{-1}b^{-1}$
Since this is equal to: $N(e)(aba^{1-}b^{-1})$ with $e \in N$ and $aba^{-1}b^{-1} \in G'$, every commutator of $G/N$ is an element of $NG'/N$, so by closure, $(G/N)'$ is contained in $NG'/N$.
On the other hand, given $N(ng) = Ng \in NG'/N$, we have from $g \in G'$ that:
$g = (a_1b_1a_1^{-1}b_1^{-1})(a_2b_2a_2^{-1}b_2^{-1})\cdots(a_kb_ka_k^{-1}b_k^{-1})$ so that:
$Ng = N[(a_1b_1a_1^{-1}b_1^{-1})(a_2b_2a_2^{-1}b_2^{-1})\cdots(a_kb_ka_k^{-1}b_k^{-1})]$
$ = (Na_1b_1a_1^{-1}b_1^{-1})(Na_2b_2a_2^{-1}b_2^{-1})\cdots(Na_kb_ka_k^{-1}b_k^{-1})$
$ = (Na_1Nb_1(Na_1)^{-1}(Nb_1)^{-1})(Na_2Nb_2(Na_2)^{-1}(Nb_2)^{-1})\cdots(Na_kNb_k(Na_k)^{-1}(Nb_k)^{-1}) \in (G/N)'$
A: A slightly different approach (in particular for the first direction). Take
$$\overline x,\overline y\in\left(G/N\right)/\left(NG'/N\right)\cong G/NG'$$
Thus, we get that
$$\overline x\overline y=\overline y\overline x\Longleftrightarrow \overline {x^{-1}}\,\overline{y^{-1}}\overline x\overline y\in NG'\;\;\;(**)$$
Putting $\,\overline x=aN\;\;,\;\;\overline y=bN\,\,,\,\,a,b\in G\,$ , we have that   $\,(**)\,$ is true iff
$$a^{-1}b^{-1}abN:=[a,b]N\in NG'=G'N$$
But since $\,G'\,$ is generated by elements of the form $\,[a,b]\,\,,\,\,a,b\in G\,$ , then $\,(**)\,$ is true indeed, and this means $\,\left(G/N\right)/\left(NG'/N\right)\cong G/NG'\,$ is abelian, and this happens iff $\,\left(G/N\right)'\leq NG'/N\,$ ...
For the other direction: let $\,xN\in G'N/N\,\,\,,\,\,x=cn\,\,,\,\,c\in G'\,,\,n\in N\Longrightarrow$
$$xN=cnN=cN=(c_1\cdot...\cdot c_m)N\;\;,\;\;c_i=[a_i,b_i]\,\,,\,\,a_i,b_i\in G\Longrightarrow$$
$$xN=\prod_{i=1}^mc_iN\in\left(G/N'\right)\,\,,\,\text{since}\,\,c_iN=[a_i,b_i]N\in\left(G/N\right)'\,\,,\,\forall i=1,...,m$$
A: If $f: G \rightarrow H$ is a homomorphism, then $f(G') = f(G)'$. This follows from the fact that $f([a,b]) = [f(a), f(b)]$. Thus when $f$ is the natural homomorphism $f: G \rightarrow G/N$, we get $$(G/N)' = f(G)' = f(G') = NG'/N.$$
More generally, you can show that $f(G^{(n)}) = f(G)^{(n)}$, where $G^{(1)} = G' = [G,G]$ and $G^{(k+1)} = [G^{(k)}, G^{(k)}]$. Using this you can prove many basic facts about solvable groups.
