Show that the set of non zero rationals is not a cyclic group under multiplication.

Question

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?

Answer 1

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 $q\notin\langle p\rangle$.

Answer 2

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.