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Show that the set of non zero rationals is not a cyclic group under multiplication.I know the set of rationals is not a cyclic group under addition. In an exercise of my book it is given that the set of non zero rationals is not a cyclic group under multiplication. I do not know how to prove it. Can someone give me a hint?

Rudy the Reindeer
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abc
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2 Answers2

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Hint: $-1 \in \mathbb{Q} \setminus \{ 0 \}$. What can you say about the generator, if it exists?

Edit: An easier way is to note that $-1$ has order $2$ but infinite cyclic groups don't have finite order elements.

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HINT: Let $p$ be a non-zero rational; the multiplicative subgroup of $\Bbb Q$ generated by $p$ is $\langle p\rangle=\{p^n:n\in\Bbb Z\}$. Clearly $p=0$ does not generate $\Bbb Q$.

  • If $p>0$, there’s a very simple reason that $\langle p\rangle\ne\Bbb Q$; what is it?

  • If $p<0$, you have to work a little harder. It helps first to show that $\langle p\rangle=\langle p^{-1}\rangle$; then you can assume that $|p|\ge 1$. You can eliminate $p=-1$ immediately, so you can assume that $|p|>1$. Thus, if $m,n\in\Bbb Z$ with $m<n$, then $p^{2m}<p^{2n}$. Let $q$ be any rational number between $1$ and $p^2$, and show that $q\notin\langle p\rangle$.

Brian M. Scott
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  • sorry i did mistake in writing .Actually iwas writing that set of rationals under addition is not cyclic.i know it if someone help when it is a group under multipication. – abc Sep 04 '13 at 05:12
  • @abc: It happens: I’ve made my share of typos. I’ve removed the part of my answer that addressed that. – Brian M. Scott Sep 04 '13 at 05:15
  • sorry i cannot understand your hint about Q/{0} under multipication. will you please explain your hint a little bit more. – abc Sep 04 '13 at 05:26
  • @abc: You’ll have to be a bit more specific about what you don’t understand. – Brian M. Scott Sep 04 '13 at 05:27
  • @BrianM.Scott, good night... I try to solve this problem and understand your hint. If p>0 only generate positive integers not rationals numbers. I dont understand when p<0 specifically when I eliminate p=-1 – Salvattore Apr 13 '16 at 03:04
  • @Salvattore: You can eliminate $-1$ by seeing that it generates the subgroup ${-1,1}$, which is certainly not all of the non-zero rationals. – Brian M. Scott Apr 13 '16 at 06:27
  • @BrianM.Scott Can I use the same arguments to solve that $<Q,+>$ is not cyclic? – Salvattore Apr 13 '16 at 15:47
  • @Salvattore: Perhaps the easiest way to do that is to show that the subgroup generated by $q$ does not contain any rational $p$ such that $|p|<|q|$. – Brian M. Scott Apr 13 '16 at 19:59
  • @BrianM.Scott thank you!!! for your hint!!! – Salvattore Apr 13 '16 at 22:34
  • @Salvattore: You’re welcome! – Brian M. Scott Apr 13 '16 at 22:37