Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 4 / 2

Use the usual Krull–Schmidt theorem setting, in which indecomposable summands have local endomorphism rings; the paper's rings $\mathcal{O},k$ satisfy it for the subgroups involved. The relative projective module condition for $H$ means that $M$ is a direct summand of a module induced from $H$, equivalently, $M$ is a summand of ${\mathrm{Ind}}_H^G{\mathrm{Res}}_H^{G}M$. A vertex of an indecomposable module is an inclusion-minimal such subgroup. A source of an indecomposable module at its module vertex $D$ is an indecomposable $RD$-summand $S$ of $M{\downarrow }_D$ such that $M\mid S{\uparrow }^G$.

Existence of a module source follows by decomposing $M{\downarrow }_D$ and using Krull–Schmidt theorem on the induced decomposition: the indecomposable summand $M$ must occur in the induction of one indecomposable summand $S$. Such $S$ has module vertex $D$ as a $D$-module, since relative projectivity of $S$ for a proper subgroup would make $M$ relatively projective for that subgroup.

In the paper's modular coefficient rings, an indecomposable module is a trivial source module when its module source is the rank-one trivial module $R$ of its module vertex. It is then a summand of $R[G/D]$, a permutation module. Conversely suppose $M$ is an indecomposable summand of a permutation module. Restriction to a module vertex $D$ is a sum of modules $R[D/E]$, one for each $D$-orbit of the permutation basis. For a $p$-group $D$, each transitive permutation module is indecomposable: over $k$ it has simple head of a module over the local ring $kD$, and over $\mathcal{O}$ a nontrivial splitting would give a nontrivial splitting after reduction. A module source $S$ must be one of these summands by Krull–Schmidt theorem. Its module vertex is $D$, whereas $R[D/E]$ is relatively $E$-projective. Thus $E=D$, and $S\cong R$. Therefore trivial module source is equivalent to being an indecomposable permutation-module summand.

The rank-one formulation uses the paper's coefficient rings. If the earlier general coefficient-ring wording were applied to a disconnected ring, it would need qualification: for $R=k\times k$ and $G=1$, $(k,0)$ is an indecomposable summand of the permutation module $R$, but is not the rank-one trivial module $R$. One can instead allow indecomposable summands of the trivial coefficient module as module sources in that generality.

For the homomorphism request, take $M_i=\mathcal{O}[G/H_i]$. Frobenius reciprocity and Mackey decomposition giveThe summand for $g$ is generated by the sum of the cosets in the $H_1$-orbit of $g{H}_2$. The identical orbit-sum construction over $k$ gives one copy of $k$ per double coset. These are free bases in both characteristics, since invariance means coefficients are constant on orbits, with no division by orbit size. HenceThe same argument applies to direct sums of coset permutation modules.

If $M$ is a trivial module source $kG$-module, choose a permutation lattice $X$ with $M\mid \bar{X}$. Its projection is an idempotent $f\in {\mathrm{End}}_{kG}(\bar{X})$. The endomorphism reduction is surjective by the preceding calculation. Its kernel is $\mathfrak{p}{\mathrm{End}}_{\mathcal{O}G}(X)$, and the endomorphism algebra is complete over $\mathcal{O}$. By idempotent refinement theorem, $f$ lifts to an idempotent $\hat{f}$. The lattice $\hat{f}X$ is a permutation-module summand, hence a trivial module source lattice, and reduces to $M$.

To prove uniqueness within the class of trivial module source lattices, let $L,L^{\prime }$ be two such lifts. They are summands of permutation lattices. Lift a modular isomorphism and its inverse between them by the same homomorphism-surjectivity argument, inserting the summand projections on either side. The lifted composites are identity modulo $\mathfrak{p}$ and therefore invertible over $\mathcal{O}$, by Nakayama lemma or the determinant criterion. The lifted maps are consequently inverse up to an invertible composite and give $L\cong L^{\prime }$. Thus each trivial module source module has a unique trivial module source lift up to isomorphism. This is permutation homomorphisms lift through modular reduction; uniqueness is not claimed among arbitrary lattice lifts.

Let $D,E$ be module vertices of $M$, and choose a module $T$ such that $M\mid {\mathrm{Ind}}_E^{G}T$. Relative $D$-projectivity then givesApply Mackey decomposition to the inner restriction and use induction transitivity. The resulting module is a direct sum of modules induced to $G$ from subgroups $D\cap {}^{g}E$, for representatives $g\in D\backslash G/E$. By Krull–Schmidt theorem, the indecomposable module $M$ is a summand of one of these induced representations. It is therefore relatively projective for one intersection contained in $D$. Minimality of $D$ forces $D=D\cap {}^{g}E$, so $D\le {}^{g}E$.

Interchanging $D,E$ gives $E\le {}^{h}D$ for some $h$. The two inclusions force equal orders, and henceThus all vertices of an indecomposable module are conjugate in $G$.

Let $S,T$ be module sources at $D$. Since $S\mid M{\downarrow }_D$ and $M\mid T{\uparrow }^G$, Mackey decomposition givesThe module source $S$ has module vertex $D$. By Krull–Schmidt theorem it is a summand of one term, and minimality of its module vertex forces $D\cap {}^{g}D=D$. As both subgroups have order $|D|$, $g\in N_G(D)$. The corresponding term is simply ${}^{g}T$, which is indecomposable. ThusThis proves conjugacy of sources of an indecomposable module at the specified module vertex.

Choose a Sylow $p$-subgroup $P$ of a module vertex $D$. Its index $[D:P]$ divides the $p^{\prime }$-part of $|G|$, so it is a unit in $R$ under the hypothesis. For every $RD$-module $U$, the induction counit from $P$ splits by averaging: equivalently, the relative trace satisfiesThe D. Higman criterion therefore makes every $RD$-module relatively $P$-projective. Since $M$ is relatively $D$-projective, transitivity of induction makes it relatively $P$-projective. Minimality of the module vertex now forces $D=P$. Thus every module vertex is a $p$-subgroup.