Let $G$ be an abelian group which does not have a maximal subgroup. Does it follow that $G$ is a $\mathbb{Q}$-algebra?
It is easy to see that $\mathbb{Q}$-algebras do not admit any maximal subgroups. I am looking for an example which is not of this kind. If such an example does not exist, it would be nice to see a proof that $\mathbb{Q}$-algebras are the only ones with this property.
Assuming that by a $\mathbb{Q}$ algebra you mean a $\mathbb{Q}$ vector space, the abelian groups that admit a $\mathbb{Q}$ - vector space structure are precisely the divisible torsion-free abelian groups, i.e. torsion-free abelian groups A such that $\forall$ $x\in A, n\in \mathbb{N}$, $\exists$ $y\in A$ s.t. $ny=x$. The condition is clearly necessary, and for sufficiency, consider the canonical mapping of $A$ into $\mathbb{Q}\otimes_{\mathbb{Z}}A$. This will be an isomorphism precisely when $A$ satisfies the above mentioned condition.
As for the question regarding maximal subgroups, one can show that an abelian group has no maximal subgroups if and only if it is divisible, in the sense mentioned above. If a group is not divisible, then there will exist a prime $p$ such that $pA$ is a proper subgroup of $A$; we can then use the fact that $A/pA$ is a vector space over $\mathbb{Z}/p\mathbb{Z}$ to pick out a maximal subgroup. On the other hand, maximal subgroups in abelian groups must always be of finite (in fact prime) index. Thus, if we have a maximal subgroup $B$ in $A$, of index $p$, then $pA\subseteq B$. However, by divisibility, $pA=A$, a contradiction.
