Let $G$ be a group. A proper subgroup $H$ is called irreducible if $H$ can't be written as an intersection of two subgroups which contain it properly.
I'd like to know if $(\mathbb Q,+)$ (and more general, any divisible abelian group) has irreducible subgroups.
On
Yes, there do exist such irreducible subgroups - $0$ is such a subgroup. There are also nontrivial examples - note that this is the same as asking if there exist quotients of $\mathbb{Q}$ such that every two nonzero subgroups (in the quotient) intersect nontrivially.
Let $H$ be a subgroup of $\mathbb{Q}$ with $\mathbb{Q}/H \cong E(\mathbb{Z}/p)$ for some nonzero prime $p$ (such an $H$ exists, e.g. since $\mathbb{Q}/\mathbb{Z} \cong \bigoplus_p E(\mathbb{Z}/p)$ and taking an appropriate preimage). Since $\mathbb{Z}/p \subseteq E(\mathbb{Z}/p)$ is an essential extension, every nonzero subgroup of $E(\mathbb{Z}/p)$ meets $\mathbb{Z}/p$ nontrivially. Since $\mathbb{Z}/p$ is a simple $\mathbb{Z}$-module, this implies $\mathbb{Z}/p$ is contained in every nonzero subgroup of $E(\mathbb{Z}/p)$, i.e. is the unique minimal subgroup. Thus no two nonzero subgroups of $\mathbb{Q}/H$ intersect trivially, i.e. $H$ is an irreducible subgroup of $\mathbb{Q}$.
In fact, the above reasoning and the structure theorem of injectives over $\mathbb{Z}$ yield the following classification of sorts (which should also work over PIDs in general):
Proposition: Let $A$ be a divisible abelian group, $H \le A$ a subgroup. Then
1) $H$ is irreducible iff $A/H \cong E(\mathbb{Z}/p)$ for some prime $p$ (including $p = 0$)
2) $A$ has infinitely many irreducible subgroups
How about the subgroup consisting of all fractions with odd denominator?
Proof: Let $H$ be that subgroup of $\mathbb{Q}$, and let $K$ be any subgroup of $\mathbb{Q}$ containing $H$ properly. Then there exists an element $k$ of $K\setminus H$ which has to be of the form $k = \frac{u}{2^e v}$ where $e>0$ and $u$ and $v$ are odd and coprime.
Since $u$ and $2^ev$ are coprime, there exist integers $a$ and $b$ with $a\cdot u+b\cdot 2^ev=1$.
Now since $\mathbb{Z}\le H < K$, $K$ contains $avk+bv=(a\cdot u+b\cdot 2^e v)/2^e=1/2^e$. Therefore $K$ also contains $1/2$, i.e. the group $\langle H,1/2\rangle$ is contained in $K$, so $H$ is irreducible.