Let $G$ be a group and $G'$ denote the subgroup generated by all commutators $[x,y]=xyx^{-1}y^{-1}$ of $G$. If $G=G'$, then $G$ is called a perfect group.
My question:
If $G/N$ is a perfect group for a normal subgroup $N\lhd G$ and $f:G \longrightarrow H$ is a group homomorphism, where $H$ is soluble, then $f(G)=f(N)$.
proof: Assume that $G/N$ is an abelian group. Then, $(G/N)'=\{e\}$. Since $G/N$ is perfect, $N=G$. So I obtain $f(G)=f(N)$, clearly.
If $G/N$ is non abelian, then how can I prove my assertion above.
Since a subgroup of a soluble group is soluble, you may as well assume that $f$ is surjective, that is, $f(G) = H$.
Since $H$ is soluble, there is $n$ such that $H^{(n)} = 1$, where $H^{(1)} = H$, and $H^{(i+1)} = [H^{(i)}, H^{(i)}]$.
Since $G/N$ is perfect, you have $G/N = (G/N)' = G'N/N$, and repeating $G / N = G^{(n)} N / N$, that is $G = G^{(n)} N$.
Thus $$H = f(G) = f(G^{(n)} N) = f(G^{(n)}) f(N) = f(G)^{(n)} f(N) = H^{(n)} f(N) = 1 \cdot f(N) = f(N).$$