Endomorphisms of irreducible submodules of simple algebras I am quite new to Representation Theory and I am trying to wrap my head around some basics. I have some difficulties with the following setup:
Suppose I start with a group ring $R=K[G]$, where $K$ is a field of characteristic $0$ (not necessarily algebraically closed) and $G$ is a finite group (not necessarily abelian).


*

*I can decompose $R$ as a direct sum 
$$ R = \bigoplus e_{\rho} R,$$
where each $e_{\rho}$ is a central idempotent and for each $\rho$, $e_{\rho}R$ is a simple algebra.

*Suppose I fix some $\rho$ and take $A=e_{\rho}R$. Then $A$ can be decomposed further as direct sum of simple left-modules
$$ A= \bigoplus\limits_{j=1}^f M_j, $$

*Consider now some $M_j$ as above and let $m \in M_j$. Then, as $M_j \subset A$, multiplication by $m$ seems to induce an $A$-endomorphism of $M_j$ and since $M_j$ is simple, this would imply by Schur's lemma that this map is an isomorphism.


Something seems off with the above reasoning, but I am unable to pinpoint where the error in the reasoning occurs. Any help or suggestions is greatly appreciated.
 A: Multiplication by an element of $A$ is typically not an endomorphism of an $A$-module, if $A$ is not commutative.  Indeed, suppose $M$ is a left $A$-module and $a\in A$.  For $f(x)=ax$ to be an endomorphism $M\to M$, we would need to have $f(bx)=bf(x)$ for all $b\in A$, which means $a(bx)=b(ax)$.  There is no reason for this to be true if $ab\neq ba$.
So, in your setup, left multiplication by $m$ is typically not an endomorphism of $M_j$.
On the other hand, though, right-multiplication by $m$ actually is an endomorphism of $M_j$ (it maps $M_j$ to itself since for any $x\in M_j$, $xm\in M_j$ because $m\in M_j$).  Your argument is then (almost) correct and says that right multiplication by $m$ is an isomorphism on $M_j$ unless it is $0$ (don't forget the latter possibility!).
This might seem surprising but you can verify it quite explicitly.  We can identify $A$ with a matrix ring $M_f(D)$ over some division ring $D$, with $M_j$ being the matrices whose only nonzero column is the $j$th column.  You can then verify that if $x,m\in M_j$ then $xm=xd$ where $d\in D$ is the $(j,j)$ entry of the matrix $m$ (none of the other entries matter since $x$ has only one nonzero column).  Since $D$ is a division ring, $d$ is either $0$ or a unit, and so $x\mapsto xd$ is either $0$ or a bijection on $M_j$.
