# Finite solvable Frattini-free group having a unique minimal normal subgroup N implies that N is the Fitting subgroup

This is exercise 6.1.6 of Kurzweil and Stellmacher. A restatement is: Let $$G$$ be a finite solvable group with $$\Phi(G)=1$$, and assume that $$G$$ has a unique minimal normal subgroup $$N$$. Then $$N=F(G)$$.

I feel like this one should be doable but the difficulty of the exercises in this book is all over the place. Here's what I have so far:

By the uniqueness of $$N$$ we know that $$F(G)$$ is a p-group; i.e. $$F(G)=O_p(G)$$ for a particular prime $$p$$, and the other p-cores are trivial. Since $$\Phi(G)=1$$ we know that $$F(G)$$ is elementary abelian (and so is $$N\le F(G)$$ of course).

That section of the book also has a theorem that $$C_G(F(G)) \le F(G)$$ for finite solvable groups, so in this case $$C_G(F(G))=F(G)$$. This means that $$G/F(G)$$ acts faithfully on $$F(G)$$ as a subgroup of $$GL(V)$$ where we take $$V=F(G)$$ to be a vector space over $$\mathbb F_p$$. I can't figure out any way to use that though, and not sure where to go from here. Need a hint.

Since $$\Phi(G)=1$$, there is a maximal subgroup $$H$$ of $$G$$ not containing $$N$$, and the minimality of $$N$$ gives $$G=HN$$ and $$H \cap N=1$$.
As you say, $$F(G)$$ must be a $$p$$-group, so if $$N \ne F(G)$$ then, since $$H \cong G/N$$, $$H$$ must have a minimal normal subgroup $$M$$ that is a $$p$$-group. Then minimality of $$N$$ and $$C_N(M) \ne 1$$ implies $$C_N(M)=N$$, so $$M$$ is normal in $$G$$, contradicting uniqueness of $$N$$.
• Well done! I might have written the proof slightly differently at the end, because as soon as we know $C_N(M)\ne 1$ that means $N_G(M)\nleq H$ and therefore $N_G(M)=G$ by the maximality of $H$. Commented Jun 19 at 14:29