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.