= Solution
{c}
An <integral form of a group representation> $V$ is a $G$-stable finite free $\mathcal O$-submodule $W\subset V$ with $K\otimes_{\mathcal O}W\cong V$. Such a form exists: take a basis lattice $L$ and replace it by $\sum_{g\in G}gL$, which is finite and torsion-free and therefore free over the <discrete valuation ring> $\mathcal O$.
Choose ordinary <simple modules> $V_i$ with forms $W_i$, and modular <simple modules> $S_j$ with <projective covers> $P_j$. With $\pi$ a <uniformizer>, put $\overline W_i=W_i/\pi W_i$. The <decomposition matrix (modular representation theory)> and <Cartan matrix of a group algebra> have entries
$$
d_{ij}=[\overline W_i:S_j],\qquad c_{\ell j}=[P_j:S_\ell].
$$
The first numbers do not depend on the integral form: the <Brauer character> of its reduction is the ordinary character restricted to <p-regular elements>, and part 2(a) determines all composition multiplicities from this restriction.
For the assertion in (i), set $I=\operatorname{Hom}_{\mathcal OG}(W,W')$. It is a <submodule> of the <finite free module> $\operatorname{Hom}_{\mathcal O}(W,W')$, hence is finite free over $\mathcal O$. There is a natural injective map
$$
K\otimes_{\mathcal O}I\longrightarrow\operatorname{Hom}_{KG}(M,M').
$$
For any homomorphism $f$ in the target, clearing the finitely many denominators of its values on an $\mathcal O$-basis of $W$ gives $\pi^Nf(W)\subset W'$. Thus $\pi^Nf\in I$, and the map is surjective. Therefore $\boxed{I\text{ is an }\mathcal O\text{-form of }\operatorname{Hom}_{KG}(M,M')}$, regarding this Hom space as a vector space. The original PDF supplies part (i), which is missing from the supplied TeX.
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