Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Do the positive rationals under multiplication contain a subgroup of finite index?

Similar questions usually rely the fact a subgroup is divisible, however, this is not the case in this question. I have a feeling that the answer should be "no", but have been unable to show this.

share|improve this question
1  
Please try to make the body of your posts self-contained; don't rely on the title for content. Think of the subject line as the spine of a book; it should be informative, help put it in place, but readers should be be asked to go look at the spine for content. –  Arturo Magidin Apr 12 '11 at 13:12

3 Answers 3

Yes. This is perhaps clearer if you notice that $\mathbb{Q}^+$ is isomorphic to the group $\mathbb{Z}^{\infty}$ of eventually zero integer sequences under addition, the isomorphism being given by prime factorization. You can mod out all but one factor of $\mathbb{Z}^{\infty}$ to get $\mathbb{Z}$ as a quotient, then mod out the squares to get $\mathbb{Z}/2\mathbb{Z}$. Thus $\mathbb{Z}^{\infty}$ has a finite quotient, hence a finite index subgroup.

Translating this back to $\mathbb{Q}^+$, this subgroup is the set of rationals which, in lowest terms, contain some fixed prime $p$ to an even power.

share|improve this answer

The group $(\mathbb{Q}^{>0},\cdot )$ is a free abelian group with a countably infinite basis. Hence there exists a surjective homomorphism $h: \mathbb{Q}^{>0}\rightarrow G$ to every countably generated abelian group $G$. In particular $G$ can be any finite abelian group, so that subgroups of any finite index exist.

share|improve this answer

Remember that if $p$ is a prime and $z$ is a nonzero integer, we define the $p$-order of $z$ to be $n$, $\mathrm{ord}_p(z) = n$, if and only if $p^n|z$ but $p^{n+1}$ does not divide $z$. For a rational $\frac{r}{s}$, we let $\mathrm{ord}_p(\frac{r}{s}) = \mathrm{ord}_p(r) - \mathrm{ord}_p(s)$. Define the ord

Fix a prime $p$, and let $n$ be any positive integer. Define $$H(p,n) = \{q\in \mathbb{Q}_{\gt0} \mid \mathrm{ord}_p(q)\in n\mathbb{Z}\}.$$

$H$ is a multiplicative subgroup of $\mathbb{Q}$, since the $p$-order satisfies $$\mathrm{ord}_p(rs) = \mathrm{ord}_p(r) + \mathrm{ord}_p(s)\quad\text{and}\quad \mathrm{ord}_p\left(\frac{1}{r}\right) = -\mathrm{ord}_p(r).$$

Finally, note that $H$ is of finite index in : $\mathrm{ord}_p$ is an abelian group homomorphism from $\mathbb{Q}_{\gt0}$ onto $\mathbb{Z}$; $H$ is the pre-image of $n\mathbb{Z}$, hence the index of $H$ is equal to $n$.

This is just a way to make explicit the comment of Hagen Knaf: the basis of the positive rationals under multiplication can be taken to be the primes.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.