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.)



Subgroups of groups of Square-free order
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.


