= Solution
Take finite-dimensional modules over $k$ and finite $R$-free lattices over the complete valuation ring, the usual integral <modular representation theory> convention. If a completion has several local factors, use the factor supporting the <modular block>. A <block of a group algebra> is an indecomposable two-sided ideal $B=RGb$, where $b$ is a <primitive central idempotent>. The identity decomposes as the sum of orthogonal block idempotents, and $M=\bigoplus_b bM$ for every <module>. Hence an <indecomposable module> belongs to $B$ exactly when $bM=M$.
A <defect group of a block> in characteristic $p$ is a maximal <p-subgroup> $D$ with $\operatorname{Br}_D(b)\ne0$. Equivalently, it is minimal with $b\in\operatorname{Tr}_D^G((kG)^D)$; the <trace criterion for defect groups> and conjugacy are proved below. For a complete <discrete valuation ring>, use the corresponding reduced <modular block>; <modular reduction of block idempotents> preserves its <defect groups>. The equivalent <bimodule> description says that the <module vertices> of $B$, with $(x,y)$ acting by $a\mapsto xay^{-1}$, are conjugates of $\Delta D=\{(d,d):d\in D\}$. This basic characterization follows from the relative-projectivity trace criterion for the <bimodule> action.
Fix a <Sylow subgroup> $P$ containing $D$. The regular <group algebra> is the <permutation module> $R[G]$ for $G\times G$, and multiplication by $b$ is its block projection. Restriction to a subgroup containing a chosen <module vertex> retains a summand with that <module vertex>: a <module source> at that <module vertex> is a summand of the <module restriction>, and the <Mackey restriction formula> and minimality prevent all the restricted summands from having smaller <module vertices>. Thus $B\downarrow_{P\times P}$ has an indecomposable summand with <module vertex> $\Delta D$.
For a <finite p-group> $Q$, the transitive <permutation module> $k[Q/L]$ is indecomposable: the <group algebra of a p-group in characteristic p is local>, and this module has a one-dimensional top, incompatible with a sum of two nonzero modules. Its <module vertex> is $L$, because it is induced from the trivial $L$-module and its <Brauer quotient of a module> at $L$ has the nonzero fixed <coset> $L$. The fixed-basis rule for this quotient follows by removing nonsingleton orbit sums, which are proper-subgroup traces. It excludes a smaller <module vertex>. Reduction gives the same statements for permutation lattices. Therefore <Krull-Schmidt decomposition> identifies the chosen block summand with an entire transitive $P\times P$ orbit in $G$.
Choose a point $c$ in that orbit whose stabilizer is exactly $\Delta D$. Being fixed by every $(d,d)$ gives $c\in C_G(D)$. Its full stabilizer is
$$
\{(cyc^{-1},y):y\in P\cap c^{-1}Pc\}.
$$
Equality with $\Delta D$ gives $P\cap c^{-1}Pc=D$. Conjugating, and using that $c$ centralizes $D$, proves
$$
\boxed{D=P\cap cPc^{-1},\qquad c\in C_G(D).}
$$
This is <defect groups are centralizer-conjugate Sylow intersections>.
If $Q\triangleleft G$ is a <p-subgroup>, then $QP$ is a <p-subgroup> for every Sylow $P$, so $QP=P$. It lies in both Sylow factors above. Thus \b[every normal <p-subgroup> of $G$ is contained in $D$].
Finally put $N=N_G(D)$, choose $S\in\operatorname{Syl}_p(N)$, and extend it to a Sylow $P$ of $G$. Then $D\leq S$ and $P\cap N=S$. Apply the proved intersection result to this $P$. Its element $c$ belongs to $C_G(D)\leq N$, so $cPc^{-1}\cap N=cSc^{-1}$ is Sylow in $N$ too. Hence $D=S\cap cSc^{-1}$. The <p-core> $O_p(N)$ is contained in both factors, whereas normality of $D$ in $N$ gives the opposite inclusion. Therefore \b[$D=O_p(N_G(D))$], so $D$ is a <p-radical subgroup>. The <centralizer> condition on the conjugating element is important for this deduction.
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