Subgroups and quotient groups of a solvable group are solvable (D&F 3.4.5)

Prove that subgroups and quotient groups of a solvable group are solvable.

Subgroups: Let \(G\) be a solvable group and \(H\) be any subgroup. By definition, there is a chain of subgroups \(1 = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_s = G\) such that \(G_{i+1}/G_i\) is abelian for all \(0 \le i \le s-1\). Using this notation, let \(H_s = H\) and \(H_i = H_{i+1} \cap G_i\) for \(0 \le i \le s-1\) (note that \(H_0 = G_0 = 1\)). For convenience purposes, we shall also define \(H_{s+1} = G\). Since \(H_{i+1} = H_{i+2} \cap G_{i+1} \le G_{i+1} \le N_G(G_i)\), the Diamond Isomorphism Theorem states that \(H_i = H_{i+1} \cap G_i \trianglelefteq H_{i+1}\). Furthermore, \(G_i \le H_{i+1}G_i \le G_{i+1}\) (the latter inequality following from the fact that \(H_{i+1} = H_{i+2} \cap G_{i+1} \subseteq G_{i+1}\) and \(G_i \subseteq G_{i+1}\)), so \(H_{i+1}/H_i = H_{i+1}/(H_{i+1} \cap G_i) \cong H_{i+1}G_i/G_i\) is a subgroup of \(G_{i+1}/G_i\) (by the Lattice Isomorphism Theorem) and is thus abelian. Hence \(H\) is solvable.

Quotient groups: Let \(N\) be a normal subgroup of \(G\). Using the previously defined notation, we can construct a chain of subgroups \(N = NG_0 \le NG_1 \le \cdots \le NG_s = G\). Since \(G_{i+1}\) normalises \(N\) and \(G_i\), it also normalises \(NG_i\). Thus \(G_{i+1}(NG_i)\) is a subgroup and \(NG_i \trianglelefteq G_{i+1}(NG_i) = N(G_iG_{i+1}) = NG_{i+1}\). By an argument involving commutators (which I admittedly still don't understand) we can also show that \(NG_{i+1}/NG_i\) is abelian. Using the lattice isomorphism theorem, we can construct a chain \(1 = N/N = NG_0/N \trianglelefteq NG_1/N \trianglelefteq \cdots \trianglelefteq NG_s/N = G/N\). Furthermore, \((NG_{i+1}/N)/(NG_i/N) \cong (NG_{i+1}/NG_i)\) is abelian, and we can conclude that \(G/N\) is solvable.