Green exceptional intersection families

Fix a finite group $G$, a field $k$ of characteristic $p>0$, a $p$-subgroup $P$, and $N_G(P)\le H\le G$. All modules considered on this page are finite dimensional left modules; indecomposable means nonzero and not a direct sum of two nonzero submodules. Put ${}^{g}Q=gQ{g}^{-1}$ and define the finite subgroup families

$\mathcal{X}=\{P\cap {}^{g}P:g\in G\setminus H\},\mathcal{Y}=\{H\cap {}^{g}P:g\in G\setminus H\}\mathrm{.}$

For a subgroup $Q$ and a family $\mathcal{F}$, write $Q{\le }_K\mathcal{F}$ if some $x\in K$ and $T\in \mathcal{F}$ satisfy ${}^{x}Q\le T$. Define

$\mathcal{Z}=\{Q\le P:Q\mathrm{̸}{\le }_G\mathcal{X}\}\mathrm{.}$

For a family of subgroups of an ambient group $K$, a $kK$-module is relatively $\mathcal{F}$-projective when each of its indecomposable direct summands is relatively $T$-projective for some $T\in \mathcal{F}$, in the sense of A module is relatively H-projective when it is a direct summand of one induced from H. This quantifies over summands without choosing a decomposition. The zero module qualifies for every family, including the empty family.

Having a vertex in $\mathcal{Z}$ means that at least one vertex, as defined in A vertex is a minimal p-subgroup for relative projectivity, and a source is an indecomposable inducing summand there, is literally a member of $\mathcal{Z}$. For $H=G$ the families $\mathcal{X},\mathcal{Y}$ are empty and $\mathcal{Z}$ consists of all subgroups of $P$. If $P=1$, then $N_G(P)=G$, so $H=G$. Members of $\mathcal{Y}$ are subgroups of $H$; this definition makes no smaller-order assertion about them.

The definition uses only the inducing-summand clause of relative projectivity, not its arbitrary-dimensional projectivity comparison. No choice principle or existence theorem for vertices is asserted here.