fully faithful inclusion of Q-vector spaces into Ab

Let \(\varphi: V \rightarrow W\) be a group-homomorphism between \(\mathbb{Q}\)-vector spaces.
So let \(\frac{p}{q} \in \mathbb{Q} \setminus \{0\}\) be a scalar and \(v \in V\).
Then we want to show that

\begin{align*} \varphi( \frac{p}{q} v) = \frac{p}{q} \varphi(v) \end{align*}

We know that \(\frac{p}{q} = \sum_{i=1}^p \frac{1}{q}\) and hence

\begin{align*} \varphi( \frac{p}{q} v) =& \varphi( \sum_{i=1}^p \frac{1}{q} v) \\ =& \sum_{i=1}^p \varphi(\frac{1}{q} v) \\ =& p \cdot \varphi( \frac{1}{q} v) \end{align*}

Now \(V,W\) are in particular uniquely divisible abelian group, so \(\frac{1}{q} v\) is characterized by the property that \(\sum_{i=1}^q \frac{1}{q} v = v\).
Analogously \(\frac{1}{q} \varphi(v)\) is characterized by \(\sum_{i=1}^q \frac{1}{q} \varphi(v) = \varphi(v)\).

Now this shows that \(\varphi(\frac{1}{q} v) = \frac{1}{q} \varphi(v)\) as both elements satisfy \(\sum_{i=1}^q x = v\) for \(x \in \{\frac{1}{q} \varphi(v), \varphi(\frac{1}{q} v)\}\), which determines them uniquely.

special case