Let $G=AB$ be a semidirect product of groups $A$ and $B$ with $B\lhd G$ and $A\cap B=\{e\}$. Let $N=[A,B]$ be the subgroup generated by commutators $[a,b]=aba^{-1}b^{-1}$, $a\in A, b\in B$.
1.) Show that $N$ is normal subgroup of $G$ contained in $B$.
Proof: Let $n\in N$ and then for any $g\in G$ we have $gng^{-1}=gng^{-1}(n^{-1}n)=(gng^{-1}n^{-1})n.$ Thus $gng^{-1}\in N$ for all $g\in G$. Further, for any $n\in N$ there exist $a\in A$ and $b\in B$ such that $n=aba^{-1}b^{-1}=(aba^{-1})b^{-1}\in B.$ Thus $N$ is a normal subgroup of $G$ contained in $B$.
EDIT: We want to show that for an arbitrary element $n\in N$ and an arbitrary element $g\in G$ that $gng^{-1}\in N$. Every $n\in N$ has the form $n=[a_1,b_1][a_2,b_2]\cdots[a_k,b_k]$ so that $$ \begin{align*} gng^{-1}&=g([a_1,b_1][a_2,b_2]\cdots[a_k,b_k])g^{-1}\\ &=g[a_1,b_1]g^{-1}g[a_2,b_2]g^{-1}g\cdots g^{-1}g[a_k,b_k]g^{-1}\\ &=(g[a_1,b_1]g^{-1})(g[a_2,b_2]g^{-1})\cdots (g[a_k,b_k]g^{-1}). \end{align*} $$ Thus it suffices to show that $g[a,b]g^{-1}\in N$. Then $$ \begin{align*} g[a,b]g^{-1}&=g(aba^{-1}b^{-1})g^{-1}\\ &=ga(g^{-1}g)b(g^{-1}g)a^{-1}(g^{-1}g)b^{-1}g^{-1}\\ &=(gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})\\ &=(gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}\\ &=[gag^{-1},gbg^{-1}] \end{align*} $$ Clearly $gng^{-1}$ is a commutator, but is it a commutator in $N$? Since $B\lhd G$, then $gbg^{-1}\in B$, but what can I say about $gag^{-1}$? I need to be able to say that $gag^{-1}\in A$ to complete this part of the proof, right?
2.) Prove that $G/N\cong A\times B/N$.
Now I am stuck. All I seem to know right now is that $G/N$ is going to be abelian, so it makes sense to find a direct product structure for $G/N$.