= Solution
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 $\operatorname{Ind}_H^G\operatorname{Res}_H^G M$. 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 \b[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> give
$$
\begin{aligned}
\operatorname{Hom}_{\mathcal OG}(M_1,M_2)
&\cong (\mathcal O[G/H_2])^{H_1}\\
&\cong\bigoplus_{g\in H_1\backslash G/H_2}\mathcal O.
\end{aligned}
$$
The summand for $g$ is generated by the sum of the cosets in the $H_1$-orbit of $gH_2$. 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. Hence
$$
\boxed{\operatorname{Hom}_{\mathcal OG}(M_1,M_2)\twoheadrightarrow
\operatorname{Hom}_{kG}(\bar M_1,\bar M_2).}
$$
The 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> $f\in\operatorname{End}_{kG}(\bar X)$. The endomorphism reduction is surjective by the preceding calculation. Its kernel is $\mathfrak p\operatorname{End}_{\mathcal OG}(X)$, and the endomorphism <algebra> is complete over $\mathcal O$. By <idempotent refinement theorem>, $f$ lifts to an <idempotent> $\hat f$. The lattice $\hat fX$ 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 $L,L'$ 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 $L\cong L'$. Thus \b[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.
Back to article page
