Composition factors of injective indecomposable and projective indecomposable modules Let $A$ be a finite-dimensional algebra over an arbitrary field $K$.
Let $L_1$ and $L_2$ be simple modules such that $L_1 \not \cong L_2$.
Let the $A$-module $Q_1$ be the injective hull of $L_1$, and let $P_2$ be the projective cover of $L_2$. Suppose that $L_2$ appears more than once as a composition factor of $Q_1$. Is it true that $L_1$ appears more than once as a composition factor of $P_2$? I cannot find a counterexample.
 A: In the notation of the question, if $X$ is a finite dimensional $A$-module then by induction on the length of $X$ it's easy to prove that
$$\dim_K\operatorname{Hom}_A(P_2,X)=[X:L_2].\dim_K\operatorname{End}_A(L_2),$$
where $[X:L_2]$ denotes the multiplicity of $L_2$ as a composition factor of $X$, and dually
$$\dim_K\operatorname{Hom}_A(X,Q_1)=[X:L_1].\dim_K\operatorname{End}_A(L_1).$$
So
$$\dim_K\operatorname{Hom}_A(P_2,Q_1)=[Q_1:L_2].\dim_K\operatorname{End}_A(L_2)
=[P_2:L_1].\dim_K\operatorname{End}_A(L_1),$$
and so if $[P_2:L_1]=1$ then $[Q_1:L_2]=1$ if and only if $\dim_K\operatorname{End}_A(L_1)=\dim_K\operatorname{End}_A(L_2)$.
This will always be the case when $K$ is algebraically closed, so that $\dim_K\operatorname{End}_A(L_1)=\dim_K\operatorname{End}_A(L_2)=1$, but over a field that is not algebraically closed it may not be true.
For example, consider right modules over the $\mathbb{R}$-algebra
$$A=\begin{pmatrix}\mathbb{C}&0\\\mathbb{C}&\mathbb{R}\end{pmatrix},$$
where 
$L_1=\begin{pmatrix}\mathbb{C}&0\end{pmatrix}$,
$L_2=\begin{pmatrix}\mathbb{C}&\mathbb{R}\end{pmatrix}/\begin{pmatrix}\mathbb{C}&0\end{pmatrix}$,  and $P_2=\begin{pmatrix}\mathbb{C}&\mathbb{R}\end{pmatrix}$.
Then $[P_2:L_1]=1$, but since $\dim_\mathbb{R}\operatorname{End}_K(L_1)=2$ but  $\dim_\mathbb{R}\operatorname{End}_K(L_2)=1$, the formula above gives $[Q_1:L_2]=2$.
