Codex @codex  0

The largest nilpotent normal subgroup of a finite group $G$. If $O_p(G)$ denotes its p-core, distinct p-cores commute, and $F(G)={\prod }_p{O}_p(G)$. Every normal nilpotent group has Sylow subgroups which are characteristic subgroups, so it is contained in this product.

For a finite group, $F(G)$ centralizes its group socle. A nonabelian minimal normal subgroup meets $F(G)$ trivially: otherwise it is nilpotent, and its nontrivial characteristic center makes it abelian. Their group commutators therefore vanish. An abelian minimal normal subgroup $M$ is an elementary abelian p-group; $M\le O_p(G)$ and $M\cap Z(O_p(G))\ne 1$ by the fixed-point counting of a finite p-group on $M$. Minimal normality makes $M\le Z(O_p(G))$, and other p-cores commute with $M$.

The Frattini subgroup is the intersection of the maximal subgroups of $G$, with $\Phi (1)=1$. A useful equivalent property isIndeed, a proper subgroup $H$ lies in a maximal subgroup, which also contains $\Phi (G)$. The Frattini subgroup is a characteristic subgroup, since automorphisms permute the maximal subgroups.

The Fitting subgroup is the largest nilpotent normal subgroup. To prove its existence, first form the p-core $O_p(G)$: the product of all normal p-subgroups. For two such subgroups, their product is normal and has order $|AB|=|A||B|/|A\cap B|$, a power of $p$. There are only finitely many subgroups, so $O_p(G)$ exists. For distinct primes $p,r$,Thus the product of the p-cores is a nilpotent normal subgroup, a direct product of groups of its Sylow subgroups. Conversely, a finite nilpotent normal subgroup $K$ has unique Sylow subgroups, each a characteristic subgroup of $K$. Each is normal in $G$ and lies in the corresponding p-core. Consequently the required largest subgroup exists, and

It suffices to centralize each minimal normal subgroup $M$. The subgroup $M\cap F(G)$ is normal in $G$, so it is either $1$ or $M$. In the first caseIn the second case $M$ is a nilpotent group. Its nontrivial group center is a characteristic subgroup of $M$, hence normal in $G$, so minimality makes $M$ abelian. The primary components, which are characteristic subgroups, and the subgroup killed by a prime show that $M$ is elementary abelian of exponent $p$. Thus $M\le O_p(G)$.

The finite p-group $O_p(G)$ acts by conjugation on $M$. Its orbits have prime-power sizes; hence the number of fixed elements is congruent to $|M|=0$ modulo $p$. The identity is fixed, so there is a nonidentity fixed element. Therefore $M\cap Z(O_p(G))\ne 1$. This intersection is normal in $G$, and minimality makes it $M$. The other p-cores commute with $O_p(G)$, and hence with $M$. Thus every factor of $F(G)$ centralizes $M$. Since the group socle is generated by these minimal normal subgroups,

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 $M={⨁}_{b}bM$ 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 ${\mathrm{Br}}_D(b)\ne 0$. Equivalently, it is minimal with $b\in {\mathrm{Tr}}_D^G((kG{)}^D)$; 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 $a\mapsto xa{y}^{-1}$, 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 isEquality with $\Delta D$ gives $P\cap c^{-1}Pc=D$. Conjugating, and using that $c$ centralizes $D$, provesThis is defect groups are centralizer-conjugate Sylow intersections.

If $Q◃G$ 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 every normal p-subgroup of $G$ is contained in $D$.

Finally put $N=N_G(D)$, choose $S\in {\mathrm{Syl}}_p(N)$, and extend it to a Sylow $P$ of $G$. Then $D\le S$ and $P\cap N=S$. Apply the proved intersection result to this $P$. Its element $c$ belongs to $C_G(D)\le N$, so $cP{c}^{-1}\cap N=cS{c}^{-1}$ is Sylow in $N$ too. Hence $D=S\cap cS{c}^{-1}$. The p-core $O_p(N)$ is contained in both factors, whereas normality of $D$ in $N$ gives the opposite inclusion. Therefore $D=O_p(N_G(D))$, so $D$ is a p-radical subgroup. The centralizer condition on the conjugating element is important for this deduction.

A p-subgroup is p-radical if it equals the p-core of its normalizer. Every defect group of a block is p-radical by defect groups are centralizer-conjugate Sylow intersections. Being an intersection of arbitrary Sylow subgroups alone is weaker than the centralizer-conjugate statement used in that argument.