Is a group isomorphic to the internal product of its Sylow p-subgroups? 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, $G = P_1 \dots P_n$?
 A: There is a more direct proof than quoting the fairly deep Theorem of P. Hall, but you do need to know a little transfer theory. The argument that follows is well-known and may be found in many group theory texts. We proceed by induction, there being nothing to prove when $n = 1$. Suppose then that $n > 1$ and that the result is true for smaller values of $n$.
If $|G| = p_{1}p_{2} \ldots p_{n}$ where $p_{1} < p_{2} < p_{3} < \ldots < p_{n}$ are primes, and if we let $P_{i}$ be a Sylow $p_{i}$-subgroup of $G$ for each $i$, then we note that the order of $N_{G}(P_{1})/C_{G}(P_{1})$ divides $p_{1}-1.$ But since $p_{1}$ is the smallest prime divisor of $|G|$, we see that $N_{G}(P_{1}) = C_{G}(P_{1})$.
By Burnside's transfer theorem, $G$ has a normal $p_{1}$-complement, which means that $G$ has a normal subgroup $H_{1}$ of order $p_{2}p_{3} \ldots p_{n}.$ Then $H_{1}$ contains all elements of $G$ of order coprime to $p_{1}$, and we have $G = H_{1}P_{1} = P_{1}H_{1}$, since $H_{1} \lhd G$.
By induction, we have $H_{1} = P_{2} P_{3} \ldots P_{n}$, so that $G = P_{1}H_{1}
= P_{1}P_{2} \ldots P_{n}.$
A: 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.
