# Fractions with numerator and denominator both odd

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 $j$.

• $G$ 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. – Jim Nov 21 '14 at 18:15
• 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. – Reinstate Monica Nov 21 '14 at 18:42
• 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. – Reinstate Monica Nov 21 '14 at 18:51
• @Jim I corrected my answer to reflect the issue you raised. – Reinstate Monica Nov 21 '14 at 19:44
• Looks good. Great trick with the linear independence of log primes, btw. I'd never heard that fact before. – Jim Nov 21 '14 at 21:13