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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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    P.S.: I believe it too, without havi$n$g $a$ proof off the top of my head! It's a fu$n$ question...2011-04-17

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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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    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 $d$.2011-04-18