# Finite solvable groups with abelian Fitting subgroup

Let $G$ be any finite solvable group with Fitting subgroup $F(G)$. Which conditions on $F(G)$ makes $G$ to be supersolvable? (It is well-known that any finite solvable group with cyclic Fitting subgroup is supersolvable). In particular, if $F(G)$ is abelian which conditions on $F(G)$ (and $G$) are needed for $G$ being supersolvable?

The relation between supersolvability of a finite group and its Fitting subgroup is that for a finite group $G$ the following are equivalent:

1. $G$ is supersolvable.

2. $G' \leq {\rm Fit}(G)$ and ${\rm Fit}(G)$ is the product of cyclic and weak $S$-quasinormal subgroups of $G$ of prime power orders.

Here a subgroup $H$ of $G$ is called weak $S$- quasinormal in $G$ if, for every $p \in \pi(G)$, there is at least one Sylow $p$-subgroup of $G$ that permutes with $H$. -- See Theorem 2 in

Piroska Csörgö, On Supersolvability of Finite Groups, Glasgow Math. J. 43(2001), 327-333.

• Dear Stefan, many thanks for the answer and reference. – majid arezoomand Apr 5 '16 at 15:10

Just a remark really. If $G$ is any finite solvable group, it is well-known that $F(G/\Phi(G)) = F(G)/\Phi(G)$. Since $\Phi(F(G)) \leq \Phi(G)$, we see that $H = G/\Phi(G)$ has $F(H)$ Abelian of squarefree exponent. Hence the requirement that $F(G)$ be Abelian in the question of itself puts almost no restriction on the structure of $G$.

• Dear Geoff, Thank you for your answer. – majid arezoomand Apr 6 '16 at 3:11
• Stefan Kohl gave a much more comprehensive answer than I did, so I do not think you should have changed the acceptance of his answer. – Geoff Robinson Apr 6 '16 at 3:11
• Yes, but you focused on my question.However, I changed the acceptance of answer again. Many thanks again. – majid arezoomand Apr 6 '16 at 3:22
• Obviously, $G$ is supersolvable iff all indices of a chief series of $G$ containing $F(G)$ and lying below $F(G)$ are primes (i.e., $F(G)$ is supersolvable immersed in $G$, in terminology of R. Baer), – yakov Jun 28 '16 at 22:25