$G/K$ is solvable $\iff G^{n}⊆K$

Question

There is a detail of this proof that isn't clear to me.

$G/K$ is solvable $\iff \exists n \in \mathbb{N}:(G/K)^{n}=\{1\} \iff \exists n \in \mathbb{N}:G^{n}⊆K$

I know that the first relation is true however the second one is not clear to me. Maybe I should try to proof it by induction?

Answer 1

Hint: in general, $(G/K)'=G'K/K$ and use induction.

As you asked: somewhat more elaboration. Let me give you some steps. Let $K \unlhd G$.

In general, if $x,y \in G$, then in $G/K$ we have $[xK,yK]=[x,y]K$. With this you can prove that if $$H,J \leq G, \text { then } [HK/K,JK/K]=[H,J]K/K.$$ In particular (take $H=J=G$) it follows that $(G/K)'=G'K/K$. With this and taking $H=J=G'$, it follows that $(G/K)''=G''K/K$ and by induction $(G/K)^{(n)}=G^{(n)}K/K$.
Finally, if $(G/K)^{(n)}=G^{(n)}K/K=\bar{1}$ (this is the identity in $G/K$), this means $G^{(n)}K=K$ being equivalent to $G^{(n)} \subseteq K$.