Consider $G := \{ a/b : \gcd(a, b) = 1 \text{ and } |b| \leq 2 \} \subset \mathbb{Q}$. This is not a subgroup of $\mathbb{Q}^\times$ since $\frac 1 2 \cdot \frac 3 2= \frac 3 4$. However, addition in $(\mathbb{Q}, +)$ works as $(a, b) + (c, d) = (ad + bc, bd)$ (writing $(a, b)$ instead of $a/b$), and
$$
\begin{align*}
(a, 1) + (c, 1) &= (a + c, 1) \\
(a, 1) + (c, 2) &= (2a + c, 2) \\
(a, 2) + (c, 2) &= (a + c, 2)
\end{align*}
$$
shows that $G$ is closed under addition; indeed, $G$ is a subgroup of $(\mathbb{Q}, +)$. I wrote the operation as tuples because I suspected that $G \cong \mathbb{Z} \times \mathbb{F}_2$, but that's incorrect.
Why is it unsurprising that $G$ is a group?
I use "unsurprising" to mean that I am looking for a construction that gives rise to $G$ (e.g. a quotient group or semidirect product), or a large class of groups that $G$ belongs to, or simply a familiar group isomorphic to $G$. For example, if I asked this question about why $\{ a/b : \text{$b$ is odd} \}$ is a group under addition, then an excellent answer would be "Because $\{ a/b : b \in \mathbb{Z} – p \mathbb{Z} \}$ is a ring for any prime $p$ since it is the localization of $\mathbb{Z}$ at the multiplicative set $\mathbb{Z} – p \mathbb{Z}$."
Best Answer
As J. W. Tanner notes, it's just $\frac 1 2 \mathbb{Z}$. I fooled myself into thinking something deep and mysterious was happening here...
More generally, if $n$ is a nonzero integer, then $$\frac 1 n \mathbb{Z} = \{ a/n : a \in \mathbb{Z} \} = \{ a/b : \gcd(a, b) = 1 \text{ and } b \mid n \} = \langle 1/n \rangle$$ is a subgroup of $(\mathbb{Q}, +)$ that is isomorphic to $\mathbb{Z}$.