Question- If $G$ is a finite group, $N$ a normal nilpotent subgroup of $G$ such that $G/[N,N]$ is nilpotent. Prove that $G$ is nilpotent.
How i did it in my exam today- (I know my solution had to be faulty because it didn't make any use of $N$ being normal and nilpotent)
Denote $[N,N]=N'$ i.e. commutator of $N$
I tried to prove that any maximal subgroup of $G$ is normal. To see that, first note that $G/N'$ is nilpotent and let $K/N'$ is some maximal subgroup of $G/N'$ then $K$ is maximal in $G$ w.r.t containment of $N'$ and we know every subgroup is contained in some maximal subgroup of $G$ implies $K\subseteq P$ where $P$ is some maximal subgroup of $G$ but $N'\subseteq P$ also then, and $K$ is maximal w.r.t containment of $N'$ implies $K=P$. so every maximal subgroup of $G/N'$ gives a maximal subgroup of $G$.
Next using canonical homomorphism $\eta: G\to G/N'$ showed that any maximal subgroup of $G$ goes to a maximal subgroup of $G/N'$ and thus we now have a $1-1$ correspondence between maximal subgroups of $G$ and $G/N'$ and as $G/N'$ is nilpotent each of its maximal subgroup is normal and thus each maximal subgroup of $G$ is also normal as there is also a 1-1 correspondence between normal subgroups of $G$ and $G/N'$.
Now tell me what did I do wrong or what did I miss. I didn't have time check my solution again in exam today. so now i was thinking on it. what can i expect with this solution if it was a 10 marks question?
There is no 1-1 correspondence between the maximal subgroups of $G$ and $G/N'$. Now see Prove: If $N\lhd G$ and $N$ and $\frac{G}{N'}$ are nilpotent, then $G$ is nilpotent.