The images of commutator subgroup under homomorphisms, and images of solvable groups

Question

Let $f:G \rightarrow H$ be a group homomorphism. Let $G'$ be the commutator subgroup generated by all commutators. Prove that

(a) $(f(G))'=f(G')$

(b) if $H$ is soluble and $f$ is injective then $G$ is soluble; and

(c) if $G$ is soluble and $f$ is surjective then $H$ is soluble.

This question is in 17th section of Humphreys' A Course in Group Theory book. There he gives definition of a soluble group and mentions about the basic properties of soluble groups. And by reading the section, I didn't get any intuition about how to approach to this question.

Answer 1

(a) $(f(G))'=\{f(x)f(y)f^{-1}(x)f^{-1}(y)|x,y\in G\}=\{f(xyx^{-1}y^{-1})|x,y \in G\}=f(G')$ , since $f$ is a homomorphism.

(b) We (inductively) define $G^{(n)}=[G^{(n-1)},G^{(n-1)}]$ where $G^{1}=G'$. Now by part a) note that $f(G^{n})=H^{n}$ (again, check by induction). If $H$ is solvable $H^n=\{1_{H}\}$ for some $n$ and thus $G^{n}=f^{-1}(H^n)=f^{-1}(1_H)=1_G$ by injectivity. This shows $G$ is solvable.

(c) If $G$ is solvable and $f$ surjective then $f(G)=H$. Hence $H'=f(G')$ by part a). Since $G$ is solvable $G^{n}=\{1\}$ for some $n$ and thus $H^{n}=f(G^{n})=f(1_{G})=1_{H}$ since $f$ is a homomorphism and this shows $H$ is solvable.

Answer 2

Let $xyx^{-1}y^{-1}=[x,y]$, $f([x,y])=[f(x),f(y)])$ implies a)

b) if $H$ is slvable and $f$ injective, $G\simeq f(G)\subset H$ is solvable since $G'\subset H'$ is nilpotent since $H'$ is nilpotent.

c) Suppose that $G$ is surjective, $f(G)'=H'=f(G')$ is nilpotent since there exists $n$ such that $G^{n}=1$, we have $f(G^{n})=f(G)^{n}=H^n=1$ where $G^2=[G',G']$ and $G^{n}=[G^{n-1},G^{n-1}]$.