Fitting subgroup is abelian and has complement if the mother group is Frattini free I am currently reading on some theorems about relations between finite group and its largest cardinality of independent generating sequence. One assumed-well-known result is that if given a finite Frattini free group $G$ (Frattini free means the intersection of all maximal subgroups is trivial), then the Fitting subgroup $F(G)$ is abelian and complemented. Also, $F(G)$ is a direct product of minimal normal subgroups of G. 
Definition: if $G$ is finite, then the Fitting subgroup is the subgroup generated by all nilpotent normal subgroups of G. 
Currently, I do not know why the result stated above is true. Any comment or guide is greatly appreciated. 
 A: I think the following works.
Since $F(G)$ is normal in $G$, we have $\Phi(F(G))\leq\Phi(G)=1$, so $F(G)$ is also Frattini-free.  Since $F(G)$ is nilpotent, $F(G)/\Phi(F(G))\cong F(G)$ is abelian (indeed, a direct product of elementary abelian groups).
As to the complement, choose a subgroup $H$ of $G$ minimal with respect to $G = HF(G)$.  Then $H\cap F(G)$ is normal in $G$, since $F(G)$ is normal and abelian. Suppose that $H\cap F(G)$ is not contained in $\Phi(H)$.  Then there is a maximal (proper) subgroup $M$ of $H$ for which $H = M(H\cap F(G))$.  Then $G = HF(G) = MF(G)$, contradicting minimality of $H$. Thus, $H\cap F(G)\leq\Phi(H)\leq\Phi(G)\cap F(G) = 1$.
ADDED
The last step uses the following fact.  If $N$ is a normal subgroup of a finite group $G$ and $H$ is a subgroup of $G$ such that $N\leq\Phi(H)$, then $N\leq\Phi(G)$.
To see this, suppose that $N$ is not contained in the Frattini subgroup $\Phi(G)$ of $G$.  Then we can write $G = NM$, for some maximal subgroup $M$ of $G$ which does not contain $N$.  Then $H = G\cap H = NM\cap H = N(M\cap H)$, because $N\leq H\leq NM$.  Since $N\leq\Phi(H)$, we get $H = M\cap H$, so that $N\leq H\leq M$, a contradiction.
