Cannot follow proof that refinement of solvable series is also solvable

Question

My book states that the refinement of a solvable series is still a solvable series. For that, they consider

$$G_n < \cdots < G_0 = G$$

a solvable series and assume $N \triangleleft G_i$ s.t. $G_{i+1}\triangleleft N$ (if $i<n$).

I can understand easily that

$$\frac{N}{G_{i+1}}< \frac{G_i}{G_{i+1}}$$

is abelian, given that $G_i/G_{i+1}$ is abelian, but I cannot understand why $G_i / N$ is also abelian. The book argues as follows:

$G_i/N$ is also abelian because $G_i/G_{i+1}$ being abelian implies $G_i^{(1)} < G_{i+1}$.

I do understand the part $G_i/G_{i+1}$ abelian $\implies G_i^{(1)} < G_{i+1}$ because the commutator of $G_i$ is the smallest normal subgroup with which one has to quotient in order to get an abelian group, but then why does it follow that $G_i/N$ is abelian? Is it because $G_i^{(1)} \triangleleft N \implies G_i/N$ abelian? If so, then we have the equivalence

$$N \triangleleft G, G/N\ \text{abelian} \iff G^{(1)} < N$$

is that right?

Answer 1

Yeah, that's right. We have the Abelianization Lemma:

Lemma. Let $G$ be a group and $N$ a normal subgroup. Then $G/N$ is abelian if and only if $[G,G]\subseteq N$.

Try to prove it, it's not difficult at all.

So if $G/N$ is abelian and $N\subseteq N'$ then $G/N'$ is abelian as well, because it contains the commutator.