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.


1 Answer 1


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 $N$.

  • $\begingroup$ 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 $H$. $\endgroup$ Commented Jun 19 at 14:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .