Is a group isomorphic to the internal product of its Sylow p-subgroups

Question

Let $G$ be a group such that its order is a product of distinct primes $p_1, \dots, p_n$ and let $P_i$ denote each Sylow $p_i$-subgroup. Is $P_1 \dots P_n$ (the internal or Frobenius product) equal to $G$, that is, $G = P_1 \dots P_n$?

Answer 1

For $G$ finite, one of the many theorems of P. Hall is that your condition holds whenever $G$ is solvable (regardless of order). A note of Rowley and Holt discusses the general problem (and includes a reference for said result of Hall), and provides a few non-solvable examples. They also show that such a product does not exist for the finite simple group $G=U_3(3)$. So in full generality the answer is "no". However, if $|G|$ is square-free, as you assume, then $G$ is solvable (see also these notes for more details), so the answer will be "yes" in this case.