If a prime number $p$ divides the order of a finite group, then there is an element in $G$ of order $p$ — this is Cauchy's Theorem.
For what numbers $n$ does this hold? In other words:
what is the set $C$ of positive integers $n$ such that every group with order divisible by $n$ has an element of order $n$?
If $n\in C$ and $G$ is a group of order $n$ then clearly $G$ is cyclic: it follows that $n$ is what is called a cyclic number, and this tells us that it is square-free and if $p_1\cdots p_r$ is its factorization into primes we have $p_i\not\equiv1\mod p_j$ for all choices of $i$ and $j$ in $\{1,\dots,n\}$.
On the other hand, it is not difficult to show that $C$ is closed under divisors: if $n\in C$ and $m\mid n$, then $m\in C$.
Finally, Cauchy's theorem tells us that $C$ contains all primes. Is there anything else in there?
Best Answer
Your set $C$ contains only the primes (and 1). Proof: as you show, it's a divisor-closed subset of the cyclic numbers. But $C$ does not contain any integer of the form $pq$ where $p < q$ are primes, because the group $S_q$ has order divisible by $pq$ but no elements of order $pq$.