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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.

Martin Brandenburg
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S_j
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1 Answers1

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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.

James
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  • for the conclusion, why is it that $Frat(H)$ subgroup of $Frat(G)$$ \cap$ $F(G)$ – S_j Jun 30 '17 at 03:24
  • Is there any explanation for the last sentence ? I don't think the deduction is trivial. – S_j Jul 02 '17 at 20:15
  • Sorry for the delay; I was away for the holiday weekend. In general, if $N$ is a normal subgroup of a finite group $G$, and if $H$ is a subgroup of $G$ such that $N\leq\Phi(H)$, then $N\leq\Phi(G)$. Since $H\cap F(G)$ is normal and contained in $\Phi(H)$, we get that $H\cap F(G)\leq\Phi(G)$. – James Jul 04 '17 at 16:37
  • thank you but I still dont see why that is necessarily true, maybe it is true for nilpotent groups. Somehow I couldn't arrive at a general proof. – S_j Jul 05 '17 at 18:55
  • Yes, it's not obvious. I've edited the answer to include a proof of the additional fact that I used. – James Jul 05 '17 at 19:32
  • Thank you very much, I figured out the same proof while waiting. This is good. – S_j Jul 05 '17 at 21:22