Let $G$ be a finite group and $KG$ its group algebra over some field $K$ with $\mathrm{char}\ K$ dividing the order of $G$. It's well-known that the Green correspondence is compatible with the Brauer correspondence. Suppose we are dealing with Green correpondence between indecomposable modules of a block $B$ and its Brauer correpondent $b$. My question is the following:


Is there always a simple $B$-module whose Green correspondent is a simple $b$-module?


Here a module always means a left module. And the question is trivial for the principal block. For blocks with Klein four defect group, Craven and his coauthor have proved it in a published paper. I guess the question is true for blocks with abelian defect groups since all examples I have found in the literature satisfy this property. Or there might be counterexamples due to my lack of knowledge. I will be very grateful if anyone could provide me with one.



Is there always a simple module whose Green correspondent is a simple module under some conditions?
The answer is "no" in general. I presume you mean that $B$ is a block of $KG$, and $b$ is its local Brauer correspondent.


Consider the case $G = {\rm SL}(2,3)$ with $p = 3.$ Then $G$ has three $3$-blocks.
One is the principal $3$-block, one is $3$-block of defect zero. The third block is a non-principal $3$-block $B$ of full defect.


Because $G$ is a $3$-nilpotent group, each $3$-block of $G$ contains just one simple module.


In the case of the block $B$ the unique simple $B$-module is the $2$-dimensional ``natural" module, which we label $V$. Let $D$ be a Sylow $3$-subgroup of $G$, which is a defect group for $B$, and a vertex for $V$.
Let $H = N\_{G}(D)$. Then Green correspondence tells us that
${\rm Res}^{G}\_{H}(V) = U \oplus P$ where $U$ is indecomposable with vertex $D$ and $P$ is projective (or zero). By dimension, $P = 0.$


Hence $U$ is indecomposable, but $U$ is not simple ( since $O\_{3}(H) = D$ acts non-trivially on $U$). Thus no simple $B$-module has a simple Green correspondent.


