# Groups with order divisible by $d$ and no element of order $d$

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 $d$?

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Just a comment: every cyclic number is squarefree, so you can reduce to that case. (There is a well-known simple characterization of cyclic numbers, but it may not be directly relevant here. I can supply a reference if anyone wants one.) – Pete L. Clark Apr 17 '11 at 5:48
P.S.: I believe it too, without having a proof off the top of my head! It's a fun question... – Pete L. Clark Apr 17 '11 at 5:49

Let $d$ be your composite, squarefree number. Let $p$ be the largest prime dividing $d$. Then the symmetric group $S_p$ will have order a multiple of $d$ (since its order is $p$-factorial, a multiple of every prime up to $p$), but no element of order $d$. Every element of $S_p$ is a product of disjoint cycles; to have an element of order a multiple of $p$, one of those cycles must be a $p$-cycle; but there's no cycle disjoint from that $p$-cycle.

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Great! Simple and elegant. As Pete mentioned, when $d$ is not squarefree there is clearly a non-cyclic group of order $d$ (say, a direct product of cyclic groups of prime orders with the appropriate multilplicities) so your proof handles all the remaining cases. – Alon Amit Apr 17 '11 at 14:28
Perhaps the simplest example when $d$ is not squarefree is ${\bf Z}_p\times{\bf Z}_{d/p}$, where $p$ is any prime whose square divides $d$. – Gerry Myerson Apr 18 '11 at 0:23