It occurred to me that I somehow believe the following statement without actually knowing how to prove it: for every composite natural number $d$ there is a group whose order is divisible by $d$ yet it contains no element of order $d$. This is a kind of converse to Cauchy's Theorem that every group of order divisible by a prime $p$ contains an element of order $p$.
This is obviously true when $d$ is such that there is more than a single isomorphism type of groups of order $d$; just take any such non-cyclic group and you're done. But there are composite numbers $d$ for which every group of order $d$ is cyclic (I think these numbers, along with the primes themselves, are called "Cyclic numbers").
For example, every group of order $15$ is cyclic. However, the group $A_5$ has order divisible by $15$ but contains no element of that order.
How would one proceed to construct such a group for any given composite $d$?