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).
 A: 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 $j$.
