Question on monoid algebras Let $G$ be a finite monoid.

Question 1: In case the monoid algebra $A=kG$ is weakly symmetric (meaning soc(P)=top(P) for each indecomposable projective modules), is $kG$ even symmetric (meaning $A \cong D(A)$)?
Question 2: In case $G=M_n(F_q)$ the monoid of matrices over a finite field $F_q$ with $q$ elements under matrix multiplication, are the dimensions of the simple $kG$-modules known? (for certain fields $k$ such as $\mathbb{C}$, finite fields or even in general)

 A: The answer to Question 2 is that if the characteristic of $k$ is not that of $F_q$ then $kM_n(F_q)$ is isomorphic to the direct 
product of $k$ with the algebras $M_{n_j}(kGL(j,q))$ where $n_j$ is the number of $j$-dimensional subspaces of $F_q^n$ as $j$ ranges from $1$ to $n$ and so we know the dimensions of the simples as long as we know them for the general linear groups, so we know them over $\mathbb C$ for example. This was proved for $k=\mathbb C$ by Okninski and Putcha and in general by Kovacs. A proof appears in my book. 
If the characteristic of $k$ is the same as that of  $F_q$ it is known that every irreducible representation of $M_n(F_q)$ restricts to an irreducible representation of $GL(n,q)$ by a result of Kuhn.  Also every irreducible representation of $GL(n,q)$ extends to $M_n(F_q)$ but not uniquely.  So the dimensions are the same as for the general linear group but we don’t know a description of all simples in this case.
The answer to your first question I don’t immediately know.  What if you take a quiver and divide out the paths of some length greater than or equal to $2$.  Do we get weakly symmetric without symmetric?  This is a semigroup algebra.    I think if the the monoid is von Neumann regular, these are equivalent.
