From: Derek HoltDate: Fri, 25 Oct 1996 17:42:23 +0100 Subject: assigning orders to permutations I have more or less convinced myself that the following is true, but would be interested to hear whether anyone knows or can come up with a slick proof. Let l,m,n be positive integers. Then there exists a finite set Omega (which may be as large as you like), and permutations g and h of Omega, such that g, h and gh have orders l, m and n, respectively. My attempt at a proof is first to note that we may assume that l <= m <= n, and that all three are prime powers. I then treat the 8 cases separately, according to the parity of l, m and n. The easiest is when they are all odd, when it seems to be possible to combine cycles of g and h to get a single cycle of length n for gh. When some of l, m and n are even, we have to be more careful, because of parity restraints on the permutations. Anyway, I would prefer to see a more uniform or more informative argument. Of course, by Cayley's theorem, one could look for suitable elements g, h in any finite group. -------- Addendum: From: Derek Holt Date: Fri, 25 Oct 1996 18:48:04 +0100 Thanks to Keith Dennis for pointing out that I need to assume that l, m and n are at least 2. Derek Holt.