From: Derek Holt 
Date: 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.