Date: 02/27/2003 at 20:07:48 From: Jenn Subject: Cyclic Groups Prove that the group of nonzero rational numbers under multiplication is not cyclic.
Date: 02/28/2003 at 03:33:08 From: Doctor Jacques Subject: Re: Cyclic Groups Hi Jenn, If the group in question is cyclic, it is generated by a single element g, i.e. any element of the group is some power of g or its inverse. As g is rational, we can write it as a fraction: g = p/q where p and q are integers, and we may assume that they are relatively prime, i.e. that the fraction is in lowest terms. All powers of g and its inverse are of the form g^k = (p^k)/(q^k) (1/g)^k = (q^k)/(p^k) Now, p and q are integers, so they have at most a finite set of prime factors, and it is possible to find a prime number r that divides neither p or q. Can you show that r does not belong to the group generated by g? Please feel free to write back if you want to discuss this further. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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