I struggle with the following theorem: given an abelian group $(G,+)$ such that $\mathbb{Q} < G$, show that if $H$ is a maximal subgroup of $G$ such that $H\cap \mathbb{Q} = \{ 0 \}$, then $G = \mathbb{Q}\oplus H$.
I think that we only need the fact that these are $\mathbb{Z}$-modules and that $\mathbb{Q}$ is injective $\mathbb{Z}$-module (to use Baer's criterion). However, I get stuck on the details. Can you prove this theorem?
You already have a direct sum since the intersection is zero. Now if you were to take $g \in G \setminus \mathbb{Q} \oplus H$ you could consider $K = \mathbb{Z} g \oplus H$ which contradicts the maximality of $H$:
Take $mg +h \in \mathbb{Q}$ with $m$ non zero. You have $a=(mg+h)/m \in \mathbb{Q}$, so $a- g = \in \mathbb{Q} \oplus \mathbb{Z}g \leq G$ and $ma = h \in H$.
Consider $H' = H + \mathbb{Z} a$, by maximality, $H' = H$, so $a \in H$.
We thus have $m(g + a) = mg + h \in \mathbb{Q}$, so $g+a \in \mathbb{Q}$ thus $g \in \mathbb{Q} \oplus H$ which is absurd. Thus $G = \mathbb{Q} \oplus H$.
Is it more clear now?