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The principal block $B_0(G)$ is the unique block of a group algebra containing the trivial $kG$-module. Its central character of a block is augmentation $\epsilon :Z(kG)\to k$, and its block idempotent $e_0$ is the unique primitive central idempotent with $\epsilon (e_0)=1$. Its defect groups of a block are Sylow subgroups: if $e_0={\mathrm{Tr}}_D^G(a)$, augmentation gives $1=[G:D]\epsilon (a)$, so $p\nmid [G:D]$.

The Brauer third main theorem states that, if $b$ is a block of $kH$ with defect group $D$ and $D{C}_G(D)\le H\le G$, thenWe give a proof, including the normal-subgroup ingredient needed for the difficult implication.

First, block idempotents centralize a normal p-subgroup. If $P◃L$ is a p-subgroup, the ideal $J(kP)kL$ is nilpotent: conjugation preserves $J(kP)$, so its powers are $J(kP{)}^{m}kL$, eventually zero. For a central idempotent $e$ of $kL$, the difference $e-{\mathrm{Br}}_P(e)$ is a sum of nonfixed $P$-conjugation orbit sums. Such an orbit maps to a single element of $L/P$ multiplied by its size, which is zero in $k$. Thus the difference belongs to $J(kP)kL$. Since $P$ is normal, its centralizer is normal too, and the Brauer image is $L$-invariant; consequently it is itself a central idempotent of $kL$. Commuting central idempotents congruent modulo a nilpotent ideal coincide: their off-diagonal products are nilpotent idempotents, hence zero. ThereforeIt follows that central block idempotents of $kL$ are exactly the sums over $L$-orbits of primitive central idempotents of $k{C}_L(P)$. The principal block idempotent of $k{C}_L(P)$ is invariant under every automorphism preserving augmentation, so its singleton orbit is the principal block idempotent of $kL$.

In particular, if $P{C}_G(P)\le K\le L\le N_G(P)$, the groups $K,L$ have the same centralizer $C_G(P)$. Their principal block idempotents are both the principal block idempotent of $k{C}_G(P)$. By the definition using the central character of a block of block induction, a block of $K$ inducing to the principal block of $L$ must be principal: its idempotent, a sum of centralizer block idempotents, has nonzero overlap with this single principal idempotent, and is therefore precisely that idempotent. The forward implication also holds, by the same observation.

For an arbitrary subgroup $H$ in the theorem, use the Brauer first main theorem inside $H$ to replace $b$ by its correspondent $\beta$ in $K=N_H(D)$. This preserves principalness. Indeed a Brauer image preserves augmentation on central elements, since nonfixed $D$-orbits have sizes divisible by $p$; hence the principal block of $H$, when it has defect $D$, corresponds to the principal block of $K$. Conversely, if $\beta$ is principal, its induced central character of a block isso ${\beta }^H$ is principal. Transitivity gives ${\beta }^G=b^G$. Thus we may work with $D{C}_G(D)\le K\le N_G(D)$ and a block $\beta$ of defect $D$.

The forward implication is already proved by this augmentation calculation, now with $H$ replaced by $G$. For the converse, fix $G$ and suppose there is a counterexample with $\beta$ nonprincipal but ${\beta }^G$ principal. Choose one whose defect group $D$ has maximal order. Put $N=N_G(D)$ and $c={\beta }^N$. To verify the transitivity needed here, write $B={\beta }^G$. Its defining character gives $e_{\beta }{\mathrm{Br}}_D(e_B)=e_{\beta }$. The image ${\mathrm{Br}}_D(e_B)$ is $N$-invariant and central in $kN$, so the definition of $c$ gives $e_c{\mathrm{Br}}_D(e_B)=e_c$. If $E\ge D$ is a defect group of $c$, applying ${\mathrm{Br}}_E$ gives ${\mathrm{Br}}_E(e_c){\mathrm{Br}}_E(e_B)={\mathrm{Br}}_E(e_c)$, since $C_G(E)\subseteq C_G(D)$. The character defining $c^G$ therefore selects $e_B$. Thus $c^G={\beta }^G$. The normal-subgroup observation shows that $c$ is nonprincipal, since otherwise $\beta$ would be principal. Every defect group of $c$ contains the normal $p$-subgroup $D$ up to conjugacy: ${\mathrm{Br}}_D(e_c)=e_c\ne 0$. Choose a defect group $E\ge D$ of $c$. Then $C_G(E)\le C_G(D)\le N$, so $(G,N,c)$ is another admissible counterexample, now with defect group $E$. Maximality forces $E=D$.

The first main theorem applied to $N=N_G(D)$ therefore implies that the principal block $c^G$ has defect $D$. Thus $D$ is a Sylow subgroup of $G$, and also of $N$. The principal block of $N$ then has defect $D$ and, by the forward implication, induces to the principal block of $G$. Injectivity in the first main theorem gives $c=B_0(N)$, contradicting the established nonprincipalness of $c$. This proves the third main theorem.

A useful equivalent consequence, with its justification, isLet $a$ be a block idempotent of $k{C}_G(Q)$ occurring in the left side, and let $E$ be a defect group of $a$ containing the central $p$-subgroup $Q$. Then $C_G(E)\le C_G(Q)$, so $a^G$ is defined. Applying ${\mathrm{Br}}_E$ to $a{\mathrm{Br}}_Q(e_0)=a$ shows that the central character of a block defining $a^G$ selects $e_0$: deleting coefficients outside $C_G(E)$ can be done after deleting those outside $C_G(Q)$. Hence $a^G=B_0(G)$, and the theorem makes $a$ principal. The image is nonzero because its augmentation is $1$, so it is exactly the principal centralizer idempotent.

Now put $C=C_G(x)$ and $P=O_p(C)$, the p-core of a finite group. The hypothesis $O_{p^{\prime }}(C)=1$, using the p-prime core of a finite group, and the definition of a p-constrained group say $C_C(P)\le P$. The normal-subgroup observation puts every block idempotent of $kC$ inside $k{C}_C(P)\subseteq kP$. A group algebra of a p-group in characteristic p is local, so its only idempotents are $0,1$. Thus $kC$ has a single block, its principal block.

Write $Q=⟨x⟩$. By the trace and Brauer characterization of defect in Question 1,For the forward direction one can conjugate $Q$ into $D$; the nonzero image at $D$ then ensures the image at $Q$ is nonzero. The reverse direction follows because every subgroup with nonzero image is conjugate into a defect group. Since $kC$ has only one block, a nonzero image must equal $1=e_0(C)$. The preceding local-block argument then gives $B=B_0(G)$ by the third main theorem. Conversely, if $B$ is principal, its defect group is a Sylow subgroup and every $p$-element is conjugate into it. ThereforeThis includes $x=1$: in that case the hypotheses force $kG$ itself to have only its principal block.

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 $i$,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 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 surjectionwhose kernel consists of maps into $JP$ and is nilpotent, since a product of $r$ such maps sends $P$ into $J^{r}P$. 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 $J(P)=JP$ is the unique maximal submodule, and $P$ is its head of a module's projective cover.

Choose the simple modules $S_1,\dots ,S_r$, with corresponding projective covers $P_1,\dots ,P_r$. Our convention for the Cartan matrix of a group algebra iswhere brackets denote the multiplicity in a composition series.

For the final two unheaded requests, retain the additional hypothesis $G=N{C}_G(N)$. Choose a central subgroup $Z=⟨z⟩$ 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 $ea(1-e)=tb$ on the left and right by $e,1-e$ givesThe 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 $\tau$ 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 filtrationare 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 $C_G=p{C}_{G/Z}$. Iterating over $N$ givesThis is blocks and Cartan matrices under central p-quotients. The extra hypothesis is still in force: for example $N=C_3⊴S_3$ in characteristic three does not satisfy it, and its Group algebra Cartan matrix is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ rather than $3I$.

A block idempotent is a primitive central idempotent $e\in R$. The module $M$ lies in the corresponding block of a group algebra whenequivalently when every other block idempotent annihilates $M$.

If $V$ lies in $e$, then the centrality of $e$ makes both its submodule $U$ and quotient $W$ lie in $e$. Conversely, suppose $eU=U$ and $eW=W$. Then $(1-e)V$ maps to zero in $W$, so $(1-e)V\subseteq U$. But $(1-e)U=0$, and applying the idempotent $1-e$ once more givesThus $eV=V$, proving

The principal block of a group algebra is the unique block of a group algebra containing the trivial representation. Its primitive central idempotent is the unique one with augmentation $1$. Its defect groups of a block are the Sylow subgroups for the residue characteristic.

For a $p$-subgroup $D$ of a finite group $G$, set $I_D={\mathrm{Tr}}_D^G((kG{)}^D)$, a transfer ideal of conjugation-fixed elements in $Z(kG)$. A block of a group algebra idempotent $b$ lies in $I_D$ if and only if a defect group of a block of $b$ is conjugate to a subgroup of $D$. In particular, defect groups can equivalently be defined as minimal subgroups $D$ with $b\in I_D$.

Here is a proof using the Brauer morphism. A subgroup of smallest order with $b\in I_D$ exists, since a Sylow subgroup $P$ gives $b={\mathrm{Tr}}_P^G(b/[G:P])$. Write $b={\mathrm{Tr}}_D^G(a)$ and replace $a$ by $ba$. If ${\mathrm{Br}}_D(b)=0$, then $ba$ belongs to the kernel of the Brauer morphism, which is the sum of proper-subgroup relative traces. Hence $b\in {\sum }_{Q<D}{I}_Q$. The algebra $bZ(kG)$ is a commutative Artinian ring and a local ring, so multiplying this equality by $b$ forces some ideal $b{I}_Q$ to contain its identity $b$. This contradicts minimality. On the other hand, ${\mathrm{Br}}_E(I_D)=0$ unless $E$ is conjugate into $D$: the $E$-action on $G/D$ has no fixed coset otherwise, and nonfixed orbit sizes vanish in characteristic $p$. Thus these minimal subgroups are precisely the maximal subgroups with nonzero Brauer image. They are all conjugate. Trace transitivity proves the stated criterion for larger $D$.