# Bounds for the number of edges in an Alperin diagram

If $$A$$ is an algebra over a field $$k$$ and $$M$$ is a finite-dimensional $$A$$-module, then Alperin showed in a paper [Diagrams for modules, JPAA, 1980] how to associate a diagram to $$M$$ with the vertices being the composition factors, putting all simple submodules (the socle) on the bottom, all simple quotients (the head) on the top and connecting vertices in the middle if there is a subquotient in which the two vertices form an indecomposable module; that is, there is a non-trivial element of $$Ext^1$$ creating a non-split extension between them. I would guess that there are various issues with the well-definedness of this process for general $$A$$ and $$M$$, but one can make sense of it in various situations.

My question is: Suppose the number of composition factors in $$M$$ is a fixed $$n$$, so that the number of vertices of the Alperin diagram of $$M$$ is $$n$$. Are there any conditions on the algebra $$A$$ (homological or otherwise) which impose bounds on the number of \emph{edges} there can be in the Alperin diagram of $$M$$? For example, do we know anything if $$A=kG$$ is the group algebra of a finite group $$G$$?

For a more specific question, is it true that for each $$n$$ there is a finite group $$G$$ and a $$kG$$-module $$M$$ such that the Alperin diagram is a complete graph?

• Alperin's definition of module diagrams appears to rule out the possibility of triangles ($K_3$'s) as subgraphs: if $y_1,\ldots,y_n$ are nodes, $n>2$, and there is an edge from $y_i$ to $y_{i+1}$, $1 \leq i <n$, then there is no edge from $y_1$ to $y_n$.'' So complete graphs on $n>2$ vertices cannot occur. – Alex Dugas Apr 10 '19 at 23:19