= Solution
Write $A=RG$. An <idempotent> is an element $e$ with $e^2=e$; it is <central idempotent> when it commutes with every element of the <ring>. In a commutative <ring> centrality is automatic. Two <idempotents> are <orthogonal idempotents> if $ef=fe=0$. A nonzero <primitive central idempotent> cannot be expressed as the sum of two nonzero orthogonal <central idempotents>; this is primitivity in $Z(A)$, not necessarily primitivity in $A$.
A <block of a group algebra> is an indecomposable two-sided direct-summand <ideal> of $A$, equivalently an <ideal> $Ae$ with $e$ a <primitive central idempotent>. Here is the equivalence. If $A=B\oplus C$ as two-sided <ideals>, then $BC=CB=0$. Write $1=e+f$ with $e\in B,f\in C$. Multiplication by this equality shows that $e$ is the identity of $B$, that $f$ is the identity of $C$, and that $e,f$ are orthogonal <central idempotents>. Conversely a <central idempotent> gives $A=Ae\oplus A(1-e)$. Splitting the <ideal> further is therefore equivalent to splitting its <central idempotent> further.
For uniqueness, suppose $1=\sum_i e_i=\sum_j f_j$ are decompositions into <primitive central idempotents>. For each $i$,
$$
e_i=\sum_j e_if_j
$$
is an orthogonal central-idempotent decomposition. Exactly one summand is nonzero, and that summand equals $e_i$. Applying the same reasoning to the corresponding $f_j$ gives $e_i=f_j$. Thus \b[the blocks, and their identities, are unique up to ordering]. Existence in the paper's coefficient <rings> follows from finite-dimensionality over $k$, and over $\mathcal O$ from completeness and <idempotent lifting> of the decomposition modulo its maximal <ideal>.
Over $k$, a <principal indecomposable module> is an indecomposable summand of the regular left <module> $A=kG$, equivalently $P=Ae$ for a <primitive idempotent> $e$ of $A$. Put $J=J(A)$, the <Jacobson radical>. The <head of a module> $P/JP$ is nonzero and semisimple. It is simple: projectivity gives a surjection
$$
\operatorname{End}_A(P)\longrightarrow\operatorname{End}_{A/J}(P/JP),
$$
whose kernel consists of maps into $JP$ and is nilpotent, since a product of $r$ such maps sends $P$ into $J^rP$. If the <head of a module> split, a nontrivial projection on it would lift by <idempotent lifting> to a nontrivial projection on $P$, contradicting indecomposability. Every maximal submodule contains $JP$, since $J$ annihilates simple <modules>. The simplicity of the <head of a module> now proves \b[$J(P)=JP$ is the unique maximal submodule], and $P$ is its <head of a module>'s <projective cover>.
Choose the simple <modules> $S_1,\ldots,S_r$, with corresponding <projective covers> $P_1,\ldots,P_r$. Our convention for the <Cartan matrix of a group algebra> is
$$
\boxed{(C_G)_{ij}=[P_j:S_i],}
$$
where brackets denote the multiplicity in a <composition series>.
For the final two unheaded requests, retain the additional hypothesis $G=NC_G(N)$. Choose a central <subgroup> $Z=\langle z\rangle$ of $N$ of order $p$. Every element of $N$ centralizes $Z$, and so does every element of $C_G(N)$; hence $Z\le Z(G)$. Set $t=z-1$, so $t$ is central and $t^p=0$, and $A/tA\cong k(G/Z)$.
To lift a <central idempotent> $\bar e$ of this quotient, first lift it as an <idempotent> $e$ by <idempotent refinement theorem>. Since $\bar e$ is central, $eA(1-e)\subseteq tA$. Multiplying a representation $e a(1-e)=t b$ on the left and right by $e,1-e$ gives
$$
eA(1-e)\subseteq t eA(1-e)\subseteq\cdots\subseteq t^p eA(1-e)=0.
$$
The other off-diagonal corner vanishes likewise, so $e$ is central. Two central lifts of the same <idempotent> coincide: their mismatched products $e(1-f)$ and $f(1-e)$ are <idempotents> in the nilpotent kernel and therefore zero. Primitivity is preserved because orthogonal <idempotent> decompositions lift. Thus the central $p$-quotient gives a bijection of block <idempotents>.
The quotient $G/Z$ still satisfies the required hypothesis for $N/Z$, because the image of $C_G(N)$ centralizes $N/Z$. Induction on $|N|$ therefore proves \b[$\tau$ bijects the block <idempotents> of $kG$ and $k(G/N)$]. The centrality argument is essential: a general nilpotent <algebra> quotient need not give a bijection of central block <idempotents>.
For the <Cartan matrix of a group algebra>, every projective $A$-module is free over $kZ$: $A$ is free over $kZ$, a projective $A$-module is a summand of a finite free $A$-module, and finite projectives over the <local ring> $kZ\cong k[t]/(t^p)$ are free. If $P$ is a <principal indecomposable module>, $\bar P=P/tP$ is a <principal indecomposable module> over $A/tA$: an <idempotent> splitting of its <endomorphism ring> would lift and split $P$. The $p$ quotients of its filtration
$$
P\supset tP\supset\cdots\supset t^{p-1}P\supset0
$$
are all isomorphic to $\bar P$ as $k(G/Z)$-modules, by multiplication by the corresponding power of $t$. All simple <modules> are annihilated by the <nilpotent ideal> $tA$. Hence $[P:S]=p[\bar P:S]$, with the natural indexing of simple <modules>, and $C_G=pC_{G/Z}$. Iterating over $N$ gives
$$
\boxed{C_G=|N|C_{G/N}.}
$$
This is <blocks and Cartan matrices under central p-quotients>. The extra hypothesis is still in force: for example $N=C_3\trianglelefteq S_3$ in characteristic three does not satisfy it, and its <Group algebra Cartan matrix> is $\left(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\right)$ rather than $3I$.
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