Fitting subgroup of a finite solvable group with trivial center and trivial Frattini subgroup.

Question

$\textbf{The question is as follows:}$

Let $G$ be a finite solvable group with trivial center and trivial Frattini subgroup. Show that its Fitting subgroup $F(G)$ is contained in $G'$.

$\textbf{This question has a hint:}$

[Hint: show that there exists $A \triangleleft G$ such that $A \cap G' = 1$ and $F(G) \subseteq AG'$.]

$\textbf{Definition:}$

The Fitting subgroup of a group is defined as the subgroup generated by all the nilpotent normal subgroups of the group.

Frattini subgroup of a group $G$, denoted $Frat G$ is defined to be the intersection of all maximal subgroups of $G$. When $G$ has no maximal subgroup, $Frat G$ is set to be $G$ itself.

$\textbf{What I can do is as follows:}$

If the Frattini subgroup is trivial, then the Fitting subgroup is a direct product of Abelian, minimal normal subgroups of $G$, and it is complemented by some subgroup $M$. On each Abelian, minimal normal subgroup $A$ of $G$, $M$ acts irreducibly so either $[M,A]=1$ (and $M$ centralizes $A$ so that $A \le Z(G))$ or $[M,A]=A$. In particular, if $Z(G)=1$ then $[M,A]=A$ and $[M,Fit(G)]=Fit(G)$ is contained in $[G,G]$.

As you see, here we do not use the solvability of the group $G$, which is also part of the assumption.

Can you please let me know which part of this proof is wrong?

Can you please give me a detailed proof?

Thanks!

Answer 1

Your proof is OK, and yes the solvability is a bit hidden. But you are using it with the fact that is $\Phi(G)=1$, then $Fit(G)$ is the direct product of minimal (abelian) subgroups. This is true for solvable $G$ and hinges on the fact that in general, if $N$ is a solvable minimal normal subgroup of a finite group $G$, then $N$ is an elementary abelian $p$-group for some prime $p$.

Note: the hint is void, if $A \unlhd G$ and $A \cap G'=1$, then $A \subseteq Z(G)$, but $Z(G)=1$, so $A=1$ ... which leads to the original question.

Alternative proof I am using the general fact that if $N \unlhd G$ and $N \cap G'=1$, then $N \subseteq Z(G)$ (proof: $[N,G]$ lies in the normal $N$ and is also a subset of $G'$). Now, as you remarked, since $\Phi(G)=1$, the Fitting subgroup is a direct product of (abelian), minimal normal subgroups of $G$. Let $M$ be one of those minimal normal subgroups. Then $M \cap G' \gt 1$, for $M \cap G'=1$ would lead to $M=1$ by the fact above and that $Z(G)=1$. By the minimality of $M$, we get $M \cap G'=M$, that is, $M \subseteq G'$. And since this holds for any such $M$, this implies $Fit(G) \subseteq G'$.