I was asked to prove the following statement: Let $A$ be a finite dimensional semisimple algebra, $S$ a simple $A$-module and $D=\text{End}_A(S)$. Then $S$ may be regarded as a module over $D$ and the multiplicity of $S$ as a summand of $_AA$ equals $\dim_DS$.
I understand that $S$ is a $D$-module. However, I don't see why $\dim_DS$ is equal to the multiplicity of $S$. Pick any nonzero $x\in S$. Since $S$ is a simple $A$-module, $x$ is a generator of $S$. $\forall y\in S$, $y=ax$ for some $a\in A$. $\forall a\in A, \psi_a:S\rightarrow S$ where $x\mapsto ax$ is an $A$-homomorphism. So $y=\psi_a(x)$ for some $\psi_a\in D$. This implies that $x$ is also a generator of $S$ as a $D$-module. $\dim_DS=1$. The multiplicity of $S$ as a summand of $_AA$ is not necessarily $1$.