TITLE: Subgroups of groups of Square-free order
QUESTION [7 upvotes]: If $G$ is a group of square-free order with at-least three prime factors, $|G|=p_1p_2....p_r$, $(2< p_i < p_{i+1})$, does $G$ contain a cyclic subgroup of composite order?
(As groups of square-free order are solvable, $G$ will necessarily have a proper subgroup of composite order.)

REPLY [2 votes]: Adding a bit to Tom's very complet answer you can in fact find a cyclic subgroup of oder $pq$ where $p,q$ are among the three bigest primes in the factorization of $|G|$. First notice that $G$ contains a subgroup $H$ of order $p_{r-2}p_{r-1}p_r$. This follows inductively from:
Fact: If $|G|=pm$ where $(p,m)=1$ and $p$ is the smallest prime dividing $|G|$ then $G$ contains a subgroup, in fact normal, of order $m$.
Now you've got a group $H$ of order $p_{r-2}p_{r-1}p_r$, so you use Tom's argument with $H$ and obtain that $m$ or $n$ is $pq$. I believe that for groups of order $pqr$ one could actually show by hand what you want without invoking the known presentation of Z-groups.