Is there a finite non-solvable group which is $p$-solvable for every odd $p\in\pi(G)$?

Question

Let $G$ be a finite non-solvable group and let $\pi(G)$ be the set of prime divisors of order of $G$. Can we say that there is $r \in \pi(G)-\{2\}$ such that $G$ is not a $r$-solvable group?

Answer 1

Yes. A finite group $G$ is $p$-solvable if every nonabelian composition factor of $G$ has order coprime to $p$.

If $G$ is not solvable, then it has a nonabelian simple group as a composition factor. This nonabelian simple group must have order divisible by some odd prime $r$ (finite groups of order $2^k$ are solvable). Then $r$ also divides the order of $G$, so $G$ is not $r$-solvable.