Subgroups of groups of Square-free order 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.)
 A: 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.
A: Yes, $G$ always contains a cyclic subgroup of composite order. Note that all Sylow subgroups of $G$ are cyclic, i.e. $G$ is a Zassenhaus metacyclic group. Such groups have a very precise structure: they are of the form
$$ G = \left\langle a, b \mid a^m = b^n = 1, b^{-1} a b = a^r \right\rangle ,$$
where $m,n,r$ satisfy certain restrictions that are not important now. Such a group has order $G = mn$ (in fact more can be said: the derived subgroup $G'$ has order $m$, and both $G'$ and $G/G'$ are cyclic).
Since at least one of the numbers $m$ or $n$ is composite, it follows that either $\langle a \rangle$ or $\langle b \rangle$ is a cyclic subgroup of $G$ of composite order.
