Example of a subalgebra of infinite-generated algebra which cannot be extended to a maximal subalgebra First of all, sorry for my English, especially mathematical one. The problem is that I know how to call such things in Ukrainian but unfortunately did not manage to find proper translations to English. So, I will give some brief definitions of terms in order to not confuse you with the names.
So, an algebra (maybe, partial) is a set of elements together with a set of operations defined on these elements $ (A, \Omega) $.
Let $ B \subset A $. Then a closure $[B]_f$ of $A$ by an $n$-ary operation $f \in \Omega$ is a set defined by two rules:


*

*$ B \subseteq [B]_f$

*$ \forall (a_1, a_2, \ldots, a_n) \subset [B]_f $  if  $f(a_1, a_2, \ldots, a_n)$ is defined  then $f(a_1, a_2, \ldots, a_n) \in [B]_f$


A closure $[B]$ of $A$ is $ B^0 \cup B^1 \cup B^2 \ldots $ where $B^0 = B$, $B^{i+1} = \bigcup_{f \in \Omega}[B^i]_f$.
The definition of subalgebra and therefore algebra extension is obvious, I think.
$B$ is a system of generators (I believe, it is close to the basis) for algebra $(A,\Omega)$ if $[B] = A$.
And, finally, subalgebra $(B,\Omega)$ is called maximal subalgebra of $(A,\Omega)$ if there is no subalgebra $(C,\Omega)$ such that $B \subset C \subset A, B \neq A, C \neq A, C \neq B$.
It can be proved that for a algebra with existing finite system of generators each its subalgebra can be extended to some maximal subalgebra. 
But for algebras with infinite systems of generators, this statement is not always true. And I need a counter-example. Therefore, I need an example of infinite-based algebra and a subalgebra which is impossible to extend to some maximum subalgebra. 

The answer is an abelian group of rational numbers $Q$. It has no maximal subgroups at all. Proof can be found here.
Thanks, Qiaochu Yuan, for the direction.
