TITLE: The minimal group with Fitting length three
QUESTION [6 upvotes]: Let $G$ be a group with Fitting lengt $3$ i.e 
$$e< F_1< F_2 < F_3=G$$ and $F(G)=F_1$ and $\bar {F_2}=F(G/F_1)$.
If every proper subgroup of $G$ and every non-trivial quatient of $G$ has fitting lengt at most $2$ then can we say that $F_i/F_{i-1}$ is abelian ? Or What can we say in that case ?

REPLY [2 votes]: These arguments may prove your conjecture,
$F(G)/\Phi(G)=F(G/\Phi(G))$ hence we can assume that $\Phi(G)=1$. Since, $\Phi(G)\geq \Phi(F_1)$ we have $\Phi(F_1)=1$. Hence $F_1$ is a abelian group.
If $F_3/F_2$ is not abelian we can find characteristic subgroup $\bar C < F_3/F_2$. Then $F_2<C<F_3$ and $C$ is charrecteristic in $G$. $F(C)\geq F_1$ and $F(C)$ is also normal in $G$. Hence, $F(C)=F_1$. As $C$ must have nilpotent with lengt $2$,  $C/F(C)=C/F_1$ is nilpotent. Then, $C/F_1\leq F_2/F_1$ which is a contradiction.