I'm curious about the quotient group $\frac{\mathbb{Q}^* }{(\mathbb{Q}^*)^2}$. What do its elements look like? Is it isomorphic to some infinite group?
** For clarification on notation, $\mathbb{Q}^* = \mathbb{Q} \setminus \{0 \} $, and $(\mathbb{Q}^*)^2 = \{ q^2 | q \in \mathbb{Q^*} \}$.
By unique factorization, $(\mathbb Q^*,\times)\cong \mathbb Z^{\oplus\mathbb N}\times\mathbb Z_2$. Thus, $\mathbb Q^*/(\mathbb Q^*)^2\cong \mathbb Z_2^{\oplus\mathbb N}\times\mathbb Z_2\cong\mathbb Z_2^{\oplus\mathbb N}$.
P.S.:
The isomorphism $(\mathbb Q^*,\times)\cong \mathbb Z^{\oplus\mathbb N}\times\mathbb Z_2$ is given by $\mathbb Z^{\oplus\mathbb N}\times\mathbb Z_2\to\mathbb Q^*:((n_i)_{i\in\mathbb N},m)\mapsto (-1)^m\prod_{i\in\mathbb N}p_i^{n_i}$, where $p_i$ is the $i$th prime number, which is well-defined since all but finitely many $n_i$s are $0$.
Each element $x\in \mathbb Q^*/(\mathbb Q^*)^2$ can thus be represented as $\pm q_1q_2\cdots q_k$ for some distinct prime numbers $q_1,\dots,q_k$.