The book states that:"Let G and K be groups, let H be a subgroup of G and let $\phi : G \rightarrow K$ be a surjective homomorphism.
(2)$\phi(G^{(i)}) = K^{(i)}.$ In particular, homomorhpic images and quotient groups of solvable groups are solvable (of solvable length $\leq$ to that of the domain group)
(3) If N is normal in G and both N and G/N are solvable then so is G."
The proof of (3) of the book is as follows: Finally, if G/N and N are solvable, of lengths n and m respectively then by (2) applied to the natural projection $\phi : G \rightarrow G/N$ we obtain $$\phi(G^{(n)}) = (G/N)^{(n)} = 1N$$
i.e., $G^{(n)}\leq N.$
And then the book completes the proof.
-But, I did not understand in the statement of the proof in (2) this statement "(of solvable length $\leq$ to that of the domain group)" I need an example that shows this. Could anyone explain this for me please?
-Also, I did not understand in the proof of (3), How $\phi(G^{(n)}) = (G/N)^{(n)} = 1N$ means $G^{(n)}\leq N.$ Could anyone explain this for me please?
thanks.
Solvable means that $G^{(n)}=\{1\}$ for some $n$. The solvable length is the minimal $n$ that satisfies that, let me represent it with a capital $N$. So, if $\phi(G^{(n)})=\phi(G)^{(n)}$ for all $n$, then in particular $\phi(G)^{(N)}=\phi(G^{(N)})=\phi(\{1\})=\{1\}$, so $\phi(G)$ is solvable with solvable length $\leq N$. Note that in the statement, since $\phi$ is surjective, $K=\phi(G)$.
For the second part, where $\phi$ is the projection, the fact that the map takes every element of $G^{(n)}$ to the identity in the image means that every element of $G^{(n)}$ is in the kernel of $\phi$, ie. $N$.