Gorenstein's proof of the classification of solvable CN-groups

Question

I am reading Gorenstein's Finite Groups. Chapter 14 is about CN-groups, a (finite) group where the centralizer of every non-identity element is nilpotent. Theorem 14.1.5 gives the classification of solvable CN-groups, which states that any such group is either (i) nilpotent, (ii) a Frobenius group whose complement is either cyclic or the product of an odd cyclic group with a generalized quaternion group, or (iii) a 3-step group.

A 3-step group $G$ is a group that satisfies all of: (a) $O_{p,p'}(G)$ is a Frobenius group with kernel $O_p(G)$ and odd cyclic complement; (b) $G = O_{p,p',p}(G)$ and $G$ strictly contains $O_{p,p'}(G)$; (c) $G/O_p(G)$ is a Frobenius group with kernel $O_{p,p'}(G)/O_p(G)$.

I feel like there is a gap in Gorenstein's proof of this classification theorem. In high level, the proof proceeds as follows.

1. Let $G$ be a solvable CN-group and let $F=F(G)$ be its Fitting subgroup. If $F=G$, then $G$ is nilpotent and we are done (case (i)). Otherwise, let $\pi=\pi(F)$ be the set of primes dividing $|F|$. Since $G$ is solvable, then $G$ has a Hall $\pi'$-subgroup $A$ ($\pi'$ denotes the primes dividing $G$ but not $F$). Then we show that $A$ acts regularly on $F$ and hence $FA$ is a Frobenius group if $A \ne 1$.

2. Next, we show that $A$ is nilpotent. Then, by properties of Frobenius complements, $A$ must be one of the alternatives described in case (ii). Now if $G = FA$, then we are done (case (ii)).

3. Otherwise, if $G$ properly contains $FA$, we show that $\pi(F)$ consists of a single prime $p$, i.e., $F$ is a $p$-group. Let $P$ be a Sylow $p$-subgroup of $G$. Then $P$ necessarily contains $F$, and $G = PA$. Note that $A$ is a $p'$-group and $F = O_p(G)$. We may assume $A \ne 1$ else we would be in case (i).

4. Next, we write $\overline{G} = \overline{P}\mbox{ }\overline{A}$ where the bars denote reduction modulo $F = O_p(G)$. Gorenstein shows that no nontrivial element of $\overline{P}$ can centralize a nontrivial element of $\overline{A}$. Then Gorenstein claims that $\overline{P}$ acts regularly on $\overline{A}$ which establishes the necessary 3-step group structure of $G$ for case (iii) where $\overline{G}$ is a Frobenius group with kernel $\overline{A}$ and complement $\overline{P}$. Here we use the restricted structure of $A$ already established above; note that $\overline{A} \cong A$ since $A$ is a $p'$-group and $F$ is a $p$-group.

To me, it looks like there is a gap in this last step. The problem is that I don't see anywhere it is shown that $\overline{P}$ has to act on $\overline{A}$ at all, i.e., why is $\overline{A}$ a $\overline{P}$-invariant subgroup? Equivalently in the original group, why $FA$ is a normal subgroup of $G$? It appears that we are claiming that $FA = O_{p,p'}(G)$ in order to establish the 3-step group structure, but $O_{p,p'}(G)$ is clearly a normal subgroup of $G$ and I don't see how to prove that $FA$ must be.

Can anyone help me understand this step?

Answer 1

I think you are right - there does seem to be a gap there. Let's see if we can fill it!

Let $\bar{B} = F(\bar{G})$. Since $F = O_p(G)$, $\bar{B}$ must be a $p'$-group, so $\bar{B} \le \bar{A}$.

Since $\bar{A}$ is nilpotent and $\bar{B}$ is self-centralizing in $\bar{G}$, it follows that $|\bar{B}|$ is divisible by every prime that divides $|\bar{A}|$.

If $|\bar{B}|$ is even then, since its Sylow $2$-subgroups are cyclic or generalized quaternion, it has a unique element of order $2$, which must therefore be centralized by $\bar{P}$. But the proof in Step 4 shows that cannot happen.

So $|\bar{B}|$ and hence also $|\bar{A}|$ is odd, so all of its Sylow subgroups are cyclic, which means that $\bar{A}$ itself is cyclic. But then $\bar{B}$ is centralized by $\bar{A}$ and so we must have $\bar{B}=\bar{A}$, which is what Gorenstein appears to be assuming without explanation.