Finite supersolvable groups with trivial Frattini subgroup

Question

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.

Answer 1

What about a Frobenius group of order 20, i.e. $\langle x,y \mid x^5=y^4=1,y^{-1}xy=x^2 \rangle$?

Answer 2

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 $j < i.$