From: mareg@csv.warwick.ac.uk (Dr D F Holt) Newsgroups: sci.math Subject: Re: group theory question Date: 15 Oct 1997 17:56:44 +0100 In article , please@no.email (L. Zer) writes: >I made a mistake in my earlier posting. Here's >the correct question: > > >The following *was* a homework problem, but no one, >including the instructor could do it: > > Given positive integers r,s,t show that there > is a finite group G with elements x and y > so that the elements x, y and xy have > orders being r,s,t respectively. > I think, in fact, that you need to assume r,s,t are at least 2 to make this true. I am surprised this is given as an exercise - I have been looking for a nice solution to exactly this problem for some time, and have not found one, although for any given r,s,t I never have any difficulty finding permutations that satisfy the hypotheses. However, recently, I was sent a solution by a man named B. Sury. which seems to work. The elements are found in the group PSL(2,K), for a suitable finite field K. Specifically, you choose K so that it contains primitive 2rst - th roots of unity (so the characteristic is relatively prime to 2rst). You then choose x and y to be the images of suitable elements in SL(2,K) of the form X = a 0 and Y = b c respectively, 1 a^-1 0 b^-1 where a and b are primitive (2r)-th and (2s)-th roots of unity. So these matrices have orders 2r and 2s. The point is that the order of a matrix in SL(2,K) is determined by its trace, provided the trace is not +2 or -2, which correspond to elements of order divisible by the characteristic of K. The trace of XY is equal to ab + a^-1b^-1 + c and so c can be chosen so that XY has any desired trace, and we can choose c to make this trace equal to the that of an element of order 2t. Then XY has order 2t, and the images of x, y and xy of X, Y XY in PSL(2,K) have the desired orders r,s and t. Derek Holt.