Reference or idea of a proof (on A-groups)

Question

There is the following result on wiki:

The Fitting subgroup of a solvable $A$-group is equal to the direct product of the centers of the terms of the derived series.

An $A$-group is a (finite) group whose all Sylow subgroup are abelian. The proof of this result can be found on Huppert 1967. Unfortunally I don't have access to the book and I cannot read German.

So, does anyone know another reference or how to prove?

Answer 1

Here is the proof, based on Huppert.

First recall that by transfer, you can prove that if $P$ is an abelian $p$-Sylow, then $G' \cap P \cap Z(N_G(P)) = 1$. Thus if $G$ is a finite $A$-group, then $G' \cap Z(G) = 1$.

Now suppose that $G$ is a finite solvable group.

Let $p_1$, $\ldots$, $p_t$ be the prime divisors of $|G|$. Because $G$ is solvable, it admits a Sylow system, meaning $P_i \in Syl_{p_i}(G)$ such that $P_iP_j = P_jP_i$ for all $j,i$. Denote $D = \cap_{i = 1}^t N_G(P_i)$.

Lemma 1: Let $N \trianglelefteq G$. Then $N = (N \cap G')(N \cap D)$.

Proof: The image of $N$ is central in $G/N \cap G'$, so it normalizes the images of $P_i$. Thus $N \leq (N \cap G')D$, and by Dedekind's lemma $N = (N \cap G')(N \cap D)$. $\blacksquare$

Lemma 2: Suppose that $G$ is an $A$-group and $N \trianglelefteq G$ is abelian. Then $N = (N \cap G') \times (N \cap Z(G))$.

Proof: It suffices to consider the case where $N$ is a $p$-group for some prime $p = p_i$.

By Lemma 1 we have $N = (N \cap G')(N \cap D)$. Also $G' \cap Z(G) = 1$ since $G$ is an $A$-group, so it suffices to check that $N \cap D \leq Z(G)$.

Since $N \leq P_i$ and $P_i$ is abelian, we have $[N \cap D,P_i] = 1$. Now $Q = \prod_{j \neq i} P_j$ is a subgroup of $G$ normalized by $D$, so $[N \cap D, Q] \leq Q \cap N = 1$. By $G = P_iQ$ we conclude $N \cap D \leq Z(G)$.$\blacksquare$

Lemma 3: Suppose that $G$ is an $A$-group and $N \trianglelefteq G$ is abelian. Then $N = (N \cap Z(G)) \times (N \cap Z(G')) \times (N \cap Z(G'')) \times \cdots$

Proof: Apply Lemma 2 and induction on the length of the derived series of $G$. $\blacksquare$

As a corollary of Lemma 3, you get the desired result.

If $G$ is an $A$-group, then $F(G) = Z(G) \times Z(G') \times Z(G'') \times \cdots$

Answer 2

The proofs use the theory of system normalizers in solvable groups.

These results were proved originally in

D. Taunt, On $A$-groups, Cambridge Phil. Soc. 45, 14-42, 1949,

which is in English.