Finite supersolvable groups with trivial Frattini subgroup

Let $G$ be a finite supersolvable group with trivial Frattini subgroup. Is it true that all Sylow subgroups of $G$ are elementary abelian?

EDIT: Many Thanks to Derek for his answer. Let me say some words on the motivation. I am asked of: Is it true that in a finite supersolvable group with trivial Frattini subgroup, every subgroup is supplemented? A subgroup $H$ of a group $G$ is called supplemented if there exists a proper subgroup $K$ of $G$ such that $G=KH$. If $K\cap H=1$, $H$ is called complemented. I learnt that in a finite group if every subgroup is supplemented then every subgroup is complemented [This is Corollary 3.7 of L.-C. Kappe and J. KIRTLAND, SUPPLEMENTATION IN GROUPS, Glasgow Math. J. 42 (2000) 37-50] and it follows from what is mentioned in the paragraph after the statement of Corollary 3.7 in the latter paper that in a supersolavble group, every subgroup is supplemented if all Sylow subgroups of the group are elementary abelian. This is a result due to P. Hall, Complemented groups, J. London Math. Soc. 12 (1937), 201-204.

• Regarding the edit, it is true that in a solvable group with trivial frattini subgroup, the fitting subgroup is elementary abelian. And of course in the supersolvable case, the commutator is nilpotent. Perhaps that is enough for the supplemented results. I think maybe even the group could split over the fitting subgroup, using some Hall-Higman type results, but I haven't tried to prove it. – Steve D Jan 27 '13 at 23:59
• @Steve. By a result of Kappe and Kirtland, Groups with trivial Frattini subgroup, Archiv der Math. (Basel) 2003, it is proved that a finite group has a trivial Frattini subgroup if and only if every nontrivial normal subgroup has a proper supplement. – Alireza Abdollahi Jan 28 '13 at 10:33

What about a Frobenius group of order 20, i.e. $\langle x,y \mid x^5=y^4=1,y^{-1}xy=x^2 \rangle$?
In a finite solvable group $G,$ one has $F(G)/\Phi(G) = F(G/\Phi(G)),$ while also $\Phi(F(G)) \leq \Phi(G).$ It follows when $\Phi(G) = 1,$ that $F(G)$ is completely reducible as a module for $G/F(G),$ that is to say, it is a direct product of minimal normal subgroups of $G,$ on which $F(G)$ clearly acts trivially by conjugation. If, in addition, $G$ is supersolvable, all of these minimal normal subgroups must be cyclic of prime order, from which it follows that $G/F(G)$ is Abelian. It also follows that when $p$ is the largest prime divisor of $|G|,$ the Sylow $p$-subgroup of $G$ is normal, hence elementary Abelian. It is possible to bound the rank of the other Sylow subgroups of $G/F(G),$ and the exponents, but the answer is not pretty: If the prime divisors of $|G|$ are $p_{1} >p_{2} > \ldots > p_{k}$ and $|F(G)| = p_{1}^{r_{1}}\ldots p_{k}^{r_{k}},$ then the rank of the Sylow $p_{i}$ subgroup of $G/F(G)$ is at most $\sum_{j=1}^{i-1} r_{j},$ and its exponent is at most the largest power of $p_{i}$ dividing any $p_{j}-1$ with $j < i.$