Existence of a “$pq$-free” subgroup of a solvable group.

Question

Let $G$ be a finite solvable group and denote by $\pi(G)$ the set of prime divisors of $|G|$.

Suppose there is a $p\in \pi(G)$ for which there is an element of order $pq$ in $G$ for every $q\in \pi(G)\setminus\{p\}$. If we assume that $P\in \operatorname{Syl}_p(G)$ is not a direct factor of $G$, is it true that there exists a subgroup $H\leqslant G$ such that

1. $\pi(H)=\pi(G)$, and

2. there exists a $q\in \pi(H)\setminus\{p\}$ such that there is no element of order $pq$ in $H$?

Note that we may assume without loss of generality that every Sylow subgroup of $G$ is elementary abelian (though I'm not sure whether this will help).

Answer 1

I think that ${\rm SL}(2,3)$ is a counterexample with $p =3.$ The only proper subgroup of order divisible by $6$ is cyclic of order $6,$ and contains an element of order two in its centre.