Square-free implies solvability-forcing

Statement
Suppose $$n = p_1p_2 \dots p_r$$ where the $$p_i$$ are pairwise distinct prime numbers. In other words, $$n$$ is a square-free number.

Then, $$n$$ is a solvability-forcing number: any fact about::finite group of order $$n$$ is a fact about::solvable group, i.e., a fact about::finite solvable group.

Facts used

 * 1) uses::Every Sylow subgroup is cyclic implies metacyclic
 * 2) uses::Metacyclic implies solvable

Proof
The proof follows from facts (1) and (2), and the observation that in a group of square-free order, every nontrivial Sylow subgroup has prime order, and is hence cyclic.