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### Cyclic Groups

```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/
```
Associated Topics:
College Modern Algebra

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