Can we do a module of a non associative algebra? Sorry to ask that since I could just probably workout the answer myself, but I read this phrase on wikipedia

In abstract algebra, a representation of an associative algebra is a
  module for that algebra. article>>

Is there a good reason why they said associative algebra? What is that goes wrong in the definition of a module for a non-associative algebra? If so when we say for example an sl(2)-module we just intend the U(sl(2))-module and it's just a sloppy notation? 
 A: I think the statement is slightly misleading. You can also view "representation" as a homomorphism to a "model example" of the structure in question. For associative algebras, these are the algebras of endomorphisms of a vector space, for Lie algebras the same endomorphisms viewed as a Lie algebra and for groups you take the groups of linear automorphisms of a vector space. In each of these cases, you can equivalently describe this is making the underlying vector space into a module over the given structure. My understanding here would be that the fundamental concept is representation, from which the concept of a module is derived. So to extend either concept to a more general class of non-associative algebras, one would need an appropriate choice of model examples.  
The fact that that both the cases of Lie algebras and the case of groups can be reduced to the case of associative algebras (via the universal enveloping algebra respectively the group ring) in my opinion is rather has the character of a nice technical feature. The concepts of representations and modules for groups and Lie algebras certainly were studied before the connection to associative algebras was formalized. 
A: Let $A$ be a $K$-algebra (arbitrary $K$-bilinear product), $M$ a $K$-module and fix a $K$-module homomorphism $f$ from $A$ to the ring $\mathrm{End}_K(M)$ of $K$-module endomorphisms of $M$ (regardless of the algebra structures). This defines an external law $A\times M\to M$ given by $(a,m)\mapsto a.m:=f(a)(m)$.
Say that $M$ is $A$-faithful if $f$ is injective. Say that this is 


*

*an associative algebra representation if it satifies $a.(b.m)=(ab).m$ for all $a,b\in A$ and $m\in M$;

*a Lie algebra representation if it satisfies $(ab)m=a.(b.m)-b.(a.m)$ for all $a,b\in A$ and $m\in M$.


First, these definition do not coincide. Moreover, although I didn't assume $A$ to be associative in the first case and Lie in the second, these are very mild assumptions. Indeed, for an associative algebra representation, $f$ is a $K$-algebra homomorphism $A\to(\mathrm{End}_K(M),.)$ and hence its image is associative; in particular if $M$ is $A$-faithful then $A$ is associative. Similarly, for a Lie algebra representation, $f$ is a $K$-algebra homomorphism $A\to(\mathrm{End}_K(M),[.,.])$ and hence its image is a Lie algebra; in particular if $M$ is $A$-faithful then $A$ is a Lie algebra.
So this just means that the definition of representation is adapted to the context, and the two previous examples (associative/Lie) are just the most celebrated. 
