Brauers Second Main Theorem — Examples

The conjugacy classes of $S_3$ are the identity, the three transpositions, and the two $3$-cycles. The identity and $3$-cycles have $2$-part $1$, while the $2$-part of a transposition is the transposition itself. All transpositions are conjugate. F1 therefore gives the two displayed sections, and these exhaust the sections indexed by conjugacy classes of $2$-elements.

Direct commutation shows $C_G(t)=⟨t⟩$. In characteristic $2$, $k{C}_G(t)\cong k[X]\mathrm{/}(X^2-1)=k[X]\mathrm{/}((X-1{)}^2),$ which is local: its elements $a+b(X-1)$ are units exactly when $a\mathrm{\ne }0$. It therefore has one primitive central idempotent and one block $c$, the principal block. The group $D=⟨t⟩$ is central in itself; F5 puts it in every defect group of $c$, and since it is already the Sylow $2$-subgroup, $D$ is the defect group of $c$.

Put $a=(123)$, $T=(12)+(13)+(23)$, and $C=a+a^2$ in $kG$. The center of $kG$ has basis $1,T,C$, since central coefficients are constant on conjugacy classes. In characteristic $2$ one computes $T^2=1+C$, $C^2=C$, and $TC=0$. Hence for $z=\alpha 1+\beta T+\gamma C$ the equation $z^2=z$ forces $\beta =0$ and $\alpha ,\gamma \in \{0,1\}$. The only central idempotents are therefore $0,1,e:=1+C$, and $f:=C$; thus $e,f$ are precisely the two block idempotents. The augmentation of $e$ is $1$, so $e$ defines the principal block $B_0$, while $f$ defines $B_1$. F3 gives Sylow defect $D$ for $B_0$. For any nontrivial $2$-subgroup $P$ of $S_3$, one has $C_G(P)=P$ and coefficient projection gives ${\mathrm{Br}}_P(f)=0$, whereas ${\mathrm{Br}}_1(f)=f\mathrm{\ne }0$; F3 therefore gives defect $1$ for $B_1$.

If an element normalizes $D$, it fixes its unique nonidentity element $t$, and hence centralizes $t$. Thus $N_G(D)=C_G(t)=D$. Step 1.2 says that $c$ has defect $D$, so F4 defines $c^G$ and makes it a global block of defect $D$. Step 1.3 says that $B_0$ is the only such global block, because $B_1$ has defect $1$. Hence $c^G=B_0$. F2 now gives the asserted subsection statements at $t$.

For $u=1$, one has $C_G(1)=G$, and induction from $G$ to itself fixes each block. Hence $(1,B_i)$ is a $B_i$-subsection for $i=0,1$. No generalized-decomposition table is being asserted here. AC is used only through F2–F5.

For $u=1$, restriction from $G$ to $C_G(1)=G$ leaves $\chi$ as its sole ordinary constituent with multiplicity one. The scalar by which $1$ acts is $1$. Substitution in F1 gives $d_{\chi ,\phi }^1=d_{\chi ,\phi }\mathrm{.}$

Block induction from a group to itself is the identity in F2, so $c^G=c$. Apply F3: if $d_{\chi ,\phi }=d_{\chi ,\phi }^1$ is nonzero, the local block $c$ of $\phi$ equals the global block of $\chi$. This is exactly the off-block vanishing in F4.

Conversely, F4 verifies this boundary specialization independently of the Second Main Theorem. The calculation does not say that every within-block entry is nonzero. Algebraic closedness and AC are used only to invoke F3; steps 1.1–2.1 themselves are finite substitutions.

Put $a=(123)$, $C=a+a^2$, and let $T$ be the sum of the three transpositions. The center of $kG$ has basis $1,T,C$, and in characteristic $2$ one directly obtains $T^2=1+C$, $C^2=C$, and $TC=0$. Thus its only nonzero primitive central idempotents are $e=1+C$ and $f=C$. The first acts as $1$ on the trivial module, so it is the principal block idempotent for $B_0$; F1 then identifies $f$ with the remaining block $B_1$. In $\mathcal{O}G$, the idempotents $\widehat{e}=(1+a+a^2)\mathrm{/}3$ and $\widehat{f}=1-\widehat{e}$ reduce respectively to $e$ and $f$. By uniqueness in F2 they are the integral block lifts. Let $W=\{(x_1,x_2,x_3)\in K^3:x_1+x_2+x_3=0\}$ with the coordinate-permutation action. The operator $\widehat{e}$ averages over $⟨a⟩$ and projects onto $W^{⟨a⟩}$. An $a$-fixed vector has equal coordinates, and its coordinate sum is $3x_1$, so $W^{⟨a⟩}=0$. If a line in $W$ were $G$-stable, the scalar $\lambda$ by which $a$ acted would satisfy ${\lambda }^3=1$ and, from $tat=a^{-1}$, $\lambda ={\lambda }^{-1}$; hence $\lambda =1$, contrary to $W^{⟨a⟩}=0$. Thus $W$ is irreducible and affords ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}$. Now $\widehat{e}W=0$ and $\widehat{f}W=W$, making ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}$ a row of $B_1$ by F2.

By F1, the only local block over $u=t$ induces to $B_0$, whereas step 1.1 puts the row ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}$ in $B_1$. F3 therefore gives $d_{{\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}},\phi }^t=0$ for every $\phi \in \mathrm{IBr}(C_G(t),c)$.

More explicitly, $k{C}_G(t)\cong k[X]\mathrm{/}((X-1{)}^2)$ has the single simple quotient $k$, so there is exactly one local irreducible Brauer character $\phi$, with $\phi (1)=1$. The only $2$-regular element of $C_G(t)\cong C_2$ is $v=1$. The full generalized-decomposition identity at $tv=t$ therefore is ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}(t)=d_{{\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}},\phi }^t\phi (1)=0.$ Equivalently, F3's restricted sum for the block $B_1$ is empty.

There is also a direct characteristic-zero check. On the basis $v_1=(1,-1,0)$, $v_2=(0,1,-1)$ of $W$, the transposition $t=(12)$ satisfies $t{v}_1=-v_1$ and $t{v}_2=v_1+v_2$, so its matrix has trace $0$. This agrees with step 3.1. The example is a genuine empty-local-support boundary case; it does not purport to compute a larger generalized-decomposition table. Algebraic closedness and AC are used only through F3 and the preceding AC-stated example.