Question about the quotientgroup and index of normal subgroups. $G$ group , $N \lhd G$ en $K \lhd G/N$ then there is a group $N \leq K' \lhd G$ such that $K'/N = K$, i.e. $K'$ is the group with all representants of $K$ in $G/N$. Is it then true that $[G:K'] = [G/N : K]$.
Let $A' := \{gK' \mid g\in G\}$ and $A := \{(gN)K \mid gN \in G/N \}$. Then define
$$
 f: A' \rightarrow A: gK' \mapsto (gN)K
$$
Assume $f(gK') = f(hK')$ then $(gN)K = (hN)K$ thus $(hN)^{-1}(gN) = (h^{-1}gN) \in K$ and thus $h^{-1}g \in K'$ so that $hK' = gK'$ such that $f$ is injective.
Further is $f$ surjective because let $(gN)K \in A$ then $f(gK') =(gN)K$. This proves that $|A| = |A'|$ and thus $[G:K']=[G/N:K]$.
Is this a correct proof ?

Your statement is equivalent to $\frac{|G|}{|K'|}=\frac{|G|/|N|}{|G|/|K'|}=\frac{|K'|}{|N|}$, which is not true in general.
For example let $G = \mathbb{Z}_{12}$, $N=\langle 6 \rangle$, so $G/N\cong \mathbb{Z}_6$.  The subgroup $K$ isomorphic to $\mathbb{Z}_2$ of $G/N$ corresponds to $K'=\langle 2 \rangle$ in $G$, but $$2=[G:K] \not= [G/N:K]=3.$$
I think the theorem you're looking for is $[G:K']=[G:K'][K':N]$.  You should be able to prove this with the same sort of method you were trying, by setting up a surjection $G/N\rightarrow G/K'$.
