Does every $\mathbb{Q}$-module contain a copy of $\mathbb{Q}$?

Question

Every abelian group is a $\mathbb{Z}$-module. I was curious what it would take for one to also be a $\mathbb{Q}$-module. This would mean in particular that an element of the group would have to multiplied by fractions $1/n$, and associativity of scalar multiplication requires that for any $g \in G$, $$ n \left(\frac{1}{n} \left( g \right) \right) = \left( n \cdot \frac{1}{n} \right) g = 1_\mathbb{Q} g = g $$ Therefore, for every integer $n > 0$ and $g \in G$, there must be an element $h$ such that $nh = g$. Equivalently, the map $f_n : G \to G$ sending $g \mapsto ng$ must be a surjection. It follows immediately that $G$ cannot be a periodic group; otherwise $f_{e}$ (where $e$ is the exponent of $G$) would send every $g$ to $1_G$. In particular, it contains an isomorphic copy of $\mathbb{Z}$ via the embedding $n \mapsto n \cdot 1_G$. My question is, does it also contain an isomorphic copy of $\mathbb{Q}$? It seems intuitively clear that it must, since it contains the integers and also must contain a multiplicative inverse of every integer -- i.e. an element $n^{-1} = \frac{1}{n} \cdot 1_G$ such that $n \cdot n^{-1} = n^{-1} + \cdots + n^{-1} = 1$. However, I can imagine that maybe for some groups $G$ there are other constructions that add in such inverses without ending up with an embedding of $\mathbb{Q}$.