Let $$G :=\left\{\frac {a}{b}\in\mathbb{Q}\; ;\; a,b\in\mathbb{Z}, a \text{ odd}, b \text{ odd}\right\}$$

Clearly, $G$ is a subgroup of the multiplicative group $\mathbb{Q}^*$. I was wondering if $G$ is isomorphic to a "known" group (or a direct/semi-direct product of "known" groups).


I will first show that the groups $(\mathbb{Q}^*)^+$ and $G^+$ of positive elements of $\mathbb{Q}^*$ and $G$ are isomorphic. Converting $(\mathbb{Q}^*)^+$ to log scale (taking logs of its elements and working additively instead of multiplicatively), an arbitrary element of the form $$\frac{\prod_i p_i^{m_i}}{\prod_i q_i^{n_i}}$$ becomes $$\sum_i m_i \log(p_i) - \sum_i n_i \log(q_i)$$ where $m_i, n_i \in \mathbb{Z}$. Call this group $\log((\mathbb{Q}^*)^+)$. Clearly $(\mathbb{Q}^*)^+ \cong \log((\mathbb{Q}^*)^+)$ with isomorphism $\log$. We can do a similar transformation for $G^+$, obtaining $G^+ \cong \log(G^+)$.

As logarithms of primes are linearly independent (see below), $\log((\mathbb{Q}^*)^+) \cong \mathbb{Z}^\mathbb{N}$ via the isomomorphism that maps the coefficient of $\log(p_i)$ to the $i$th component, where $p_i$ is the $i$th prime. Similarly, $\log(G^+) \cong \mathbb{Z}^\mathbb{N}$ via the isomorphism that maps the coefficient of $\log(p_{i+1})$ to the $i$th component.

Putting all this together, $$(\mathbb{Q}^*)^+ \cong \log((\mathbb{Q}^*)^+) \cong \mathbb{Z}^\mathbb{N} \cong \log(G^+) \cong G^+.$$

Let $\phi$ be such an isomorphism from $(\mathbb{Q}^*)^+$ to $G^+$. Then $\phi$ induces an isomorphism $\Phi$ from $\mathbb{Q}^*$ to $G$ where $\Phi(\pm x) = \pm \phi(x)$ as follows. First of all, $\Phi$ is a bijection. Second of all, $\Phi((\pm_x x) (\pm_y y)) = \Phi((\pm_x) (\pm_y) x y) = (\pm_x) (\pm_y) \phi(x y) = (\pm_x) (\pm_y) \phi(x) \phi(y) = \Phi(\pm_x x) \Phi(\pm_y y)$. Thus $\Phi$ is an isomorphism so $\mathbb{Q}^* \cong G$.

There's a very simple proof that $\{\log(p_i) | i\in\mathbb{N}\}$ is linearly independent. If $$0 = \sum_j a_j \log(p_{i_j}) = \log\left(\prod_j (p_{i_j})^{a_j}\right)$$ then $$\prod_j (p_{i_j})^{a_j} = 1$$ which implies that $a_j = 0$ for all $j$.

  • $\begingroup$ $G$ contains negative numbers, so you probably need a semi-direct with $\mathbb Z/2$ or something like that in there. Those might balance such that your final conclusion still holds, but what you've written doesn't look quite right yet. $\endgroup$ – Jim Nov 21 '14 at 18:15
  • $\begingroup$ You are right. The issue is that $\mathbb{Q}^*$ contains negative numbers. The sign must be carried over in some way to the subsequent groups. I will think about this and edit in a bit. $\endgroup$ – Reinstate Monica Nov 21 '14 at 18:42
  • $\begingroup$ My argument should work for $(\mathbb{Q}^*)^+$ and $G^+$, the positive elements of the respective groups. I believe that that implies that $\mathbb{Q}^*$ and $G$ are isomorphic. $\endgroup$ – Reinstate Monica Nov 21 '14 at 18:51
  • 1
    $\begingroup$ @Jim I corrected my answer to reflect the issue you raised. $\endgroup$ – Reinstate Monica Nov 21 '14 at 19:44
  • $\begingroup$ Looks good. Great trick with the linear independence of log primes, btw. I'd never heard that fact before. $\endgroup$ – Jim Nov 21 '14 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.