Does there exist a group $G$ that satisfies the following conditions:
- Any proper subgroup of $G$ is contained in a maximal subgroup.
- There is some $N\unlhd G$ such that $\frac{G}{N}$ is divisible.
?
How about if $G$ is abelian?
My definition of divisibility is the same as what is used in this post: https://math.stackexchange.com/a/419101/138171
There does not exist an abelian group with the prescribed properties.
Assume contrariwise that $G$ is such a group and that $N\lhd G$ gives rise to a divisible quotient group. The first assumption implies that $N$ is contained in a maximal subgroup $M$. If $G/M$ is infinite, then it has non-trivial subgroups, so correpondence principle tells us that there are subgroups properly between $M$ and $G$. So $G/M$ is finite, say $n=[G:M]$.
This implies that for all $x\in G$ we have $x^n=M$. So if $yN\in G/N$ is chosen in such a way that $y\notin M$, then $(zN)^n\neq yN$ for all $z\in G$, because $(zN)^n\in M/N$.
If we look for a non-abelian example, the same argument does not work, because $M$ is not necessarily normal. The argument only disallows those pairs $(G,N)$ such that $N$ is contained in a normal subgroup of a finite index.