Brauers First Main Theorem — Examples

Write $s=(12)$ and $r=(1324)$. Then $r^4=s^2=1$, $srs=r^{-1}$ and $s\mathrm{\notin }⟨r⟩$, giving exactly the eight elements $r^j,s{r}^j$. A normalizer preserves the common fixed set $\{5,6,7\}$, so it is $N_{S_4}(D)\times S_3$. The first factor has order $8$ or $24$ by divisibility. It is not $S_4$: conjugating $(12)$ to $(13)$ leaves $D$, whose only transpositions are $(12)$ and $(34)$. Therefore it is $D$.

In the last $S_3$ write $a^3=t^2=1$, $tat=a^{-1}$. The element $e_0=1+a+a^2$ is a central idempotent in characteristic $2$. Its ideal has basis $e_0,t{e}_0$ and is $k{C}_2$, since $(t{e}_0{)}^2=e_0$. The complementary idempotent is $e_1=a+a^2$. To identify its four-dimensional ideal, represent $a$ by $A=\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$ and $t$ by $T=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Direct multiplication gives $A^3=T^2=I$, $TAT=A^2$ and $I+A+A^2=0$. The image contains $E_{22}=A+T$, $E_{11}=I+E_{22}$, $E_{21}=AT+I$, and $E_{12}=T+E_{21}$. Thus it is all $M_2(k)$; dimension gives $k{S}_3{e}_1\cong M_2(k)$.

Hence $kN\cong k[D\times C_2]\times M_2(kD)$. F4 makes the first factor local and gives no nontrivial idempotents in $kD$. A central matrix commuting with every matrix unit is a scalar matrix over $Z(kD)$; thus the second factor has no nontrivial central idempotents either. These are exactly the two blocks. Augmentation is $1$ on $e_0$ and $0$ on $e_1$, so F3 places the trivial module in the first, the principal block. Its defect is $D\times ⟨t⟩$ by F2.

Regard $e_1=a+a^2$ as an element of $kN$. Both support elements centralize $D$, so ${\mathrm{Br}}_D(e_1)=e_1\mathrm{\ne }0$. If a $2$-subgroup $R$ of $N$ properly contains $D$, its projection to $S_3$ is a subgroup of order $2$. Its involution centralizes neither $a$ nor $a^2$. Thus neither support element centralizes $R$ and ${\mathrm{Br}}_R(e_1)=0$. F5 and F6 prove that $D$ is a defect group of the second block.

F1 now gives exactly one global block having $D$ as a defect group, paired with this nonprincipal local block. It is not the global principal block: the $2$-part of $7!=5040$ is $16$, whereas $\mathrm{\mid }D\mathrm{\mid }=8$, and F2 gives Sylow defect for the principal block. The larger local principal defect has order $16$ as well. All idempotents and matrix units were displayed; there is no zero block, empty correspondence, endpoint parameter or choice operation in this calculation. The correspondence is bijective in both directions by F1.

Counting a nonzero first column and then the $p$ second columns with determinant one gives $\mathrm{\mid }G\mathrm{\mid }=p(p^2-1)$, so $D$ is Sylow of order $p$. Its common fixed line is $kY$. A normalizer must preserve that line, hence is lower triangular. Conversely the lower triangular matrices normalize $D$ by direct conjugation. Thus $N=D⋊T$, where $T=\{\mathrm{diag}(a,a^{-1}):a\in {\mathbf{F}}_p^{\times }\}$. Such a diagonal conjugates $c$ to $a^{-2}c$, so $C_N(D)=D\times ⟨-I⟩$.

Put $n=i-1<p$. On the basis $X^j{Y}^{n-j}$, the operator $u-1$, for $u$ with $c=1$, lowers the highest $X$-degree by one with leading coefficient $j\mathrm{\ne }0$ when $j>0$. Its successive powers on $X^n$ therefore give a triangular basis. It is one Jordan block of size $i$, with kernel $k{Y}^n$. Every nonzero invariant subspace contains a nonzero kernel vector, by applying a maximal nonvanishing power of this nilpotent operator. The upper unipotent acts in the reverse way, and its powers of $h-1$ on $Y^n$ span the whole space. Hence every nonzero $G$-submodule is all of $V_i$: these $p$ modules are simple and have distinct dimensions. This also covers $n=0$.

The $p$-regular matrices in $G$ are exactly the semisimple ones. Indeed finite order prime to $p$ gives a square-free annihilating polynomial; conversely a semisimple matrix has eigenvalues in ${\mathbf{F}}_{p^2}^{\times }$ and hence order prime to $p$. For every trace $t\mathrm{\ne }\pm 2$, the polynomial $Z^2-tZ+1$ has distinct roots, and its companion matrix gives one ${\mathrm{G}\mathrm{L}}_2({\mathbf{F}}_p)$ class. It gives one $G$ class too: the determinant map from its centralizer is onto. In the split case this follows from diagonal matrices. In the nonsplit case the centralizer is ${\mathbf{F}}_{p^2}^{\times }$ acting by multiplication, with determinant $x^2-d{y}^2$ for a nonsquare $d$. For each $c\mathrm{\ne }0$, the sets of squares and $c+d$ times squares, meaning $\{x^2\}$ and $\{c+d{y}^2\}$, each have $(p+1)\mathrm{/}2$ elements and intersect; thus $x^2-d{y}^2=c$ has a solution. Multiplying a conjugator by a centralizer element adjusts its determinant to one. Trace $\pm 2$ gives only the semisimple matrices $\pm I$. There are exactly $p$ regular classes. F4 proves that the list in step 1.2 exhausts all simple modules.

The subspaces spanned by $Y^n,X{Y}^{n-1},\dots ,X^j{Y}^{n-j}$ are $N$-stable and give a full composition flag. Since $u-1$ is a single Jordan block, its invariant subspaces are precisely these: viewing the module as $k[z]\mathrm{/}(z^i)$, submodules are ideals $(z^j)$. Thus restriction to $N$ is indecomposable and has a unique head, the line represented by $X^n$. Define $U_j$ on $N$ by trivial $D$ action and $\mathrm{diag}(a,a^{-1})$ acting by $a^j$. The head is $U_n$, and the factors from head down are $U_n,U_{n-2},\dots ,U_{-n}$. The element $-I$ acts throughout by $(-1{)}^n$.

By F7 every central block idempotent of $kN$ lies in $k{C}_N(D)=kD\otimes k⟨-I⟩$. This commutative algebra is $kD\times kD$, whose only primitive idempotents are $e_{+}=(1+(-I))\mathrm{/}2$ and $e_{-}=(1-(-I))\mathrm{/}2$, by F5. These are central in $kN$, so they are exactly its two block idempotents. Each has nonzero Brauer projection at $D$ (all its support centralizes $D$); since $D$ is Sylow, F8 gives defect $D$ for both.

For either $L=G$ or $N$, every finite $kL$-module is relatively $D$-projective: the relative trace of $[L:D{]}^{-1}\mathrm{i}\mathrm{d}$ is the identity, so F6 applies. Also $kD\cong k[z]\mathrm{/}(z^p)$, with $z=u-1$. Every finite projective module over this algebra is free. To see this, lift a basis of $P\mathrm{/}zP$ to obtain a surjection $A^r\to P$, since its cokernel $C=zC$ vanishes by $z^p=0$. Split this surjection; its kernel $K$ has $K\mathrm{/}zK=0$ by dimensions and thus is zero by the same nilpotence argument. Consequently the single Jordan module of dimension $i<p$ is not projective on $D$, and cannot be projective on $L$, since restriction preserves finite free modules and their summands. F10 and F11 give it vertex $D$, the only alternative inside $D$ being $1$, which would imply projectivity by F6's finite counit. This applies to both $V_i$ and its restriction. For $i=p$ the restriction to $D$ is regular free; the split counit from relative $D$-projectivity makes $V_p$ a summand of a free induced module, hence projective.

F1 gives exactly two global blocks with defect $D$; every positive defect is conjugate to $D$ since its order divides the $p$-part $p$ of $\mathrm{\mid }G\mathrm{\mid }$. For $i<p$, step 3.1 and F2 identify the block of $V_i$ with induction of the local block containing its restriction. Step 2.2 identifies this as $e_{+}$ when $i$ is odd and $e_{-}$ when $i$ is even. Both sets occur, since $1,2<p$ or $p=3$ with these same two indices. Under A1, F12 identifies the indecomposable vertex-$D$ restriction itself as the Green correspondent. AC is used only in this invocation; the trace, sign and block calculations did not use it.

By F9 the projective simple $V_p$ is alone in a defect-zero block. The exhaustive list in step 2.1 and the partition in F3 leave no further blocks: every nonzero finite block has a simple quotient by a proper left ideal of largest dimension. Thus the two positive-defect blocks and this Steinberg block are all the blocks. The endpoint $i=1$ was included in the constant-polynomial calculation, and $i=p$ is precisely the projective exception. The hypothesis $p>2$ ensures the two sign idempotents exist and are distinct; the assertion does not extend this calculation to $p=2$.

The averaging element $e=n^{-1}{\sum }_{j=0}^{n-1}{r}^j$ exists since $n$ is odd. Counting the $n$ occurrences of each $r^j$ in the square proves $e^2=e$. Conjugation by $r,s$ permutes its terms, so it is central in $kG$. Its ideal in $kH$ is the one-dimensional algebra $ke$, since $r^{j}e=e$, and augmentation sends $e$ to $1$. Hence $b_0(H)=ke$. F3 gives defect $1$, since $H$ has odd order.

The ideal $kGe$ has basis $e,se$, supported on the disjoint cosets $H,sH$. Its multiplication satisfies $(se{)}^2=e$, so it is $k{C}_2$. By F4 it has no nontrivial idempotents; consequently $e$ is primitive central in $kG$. As it acts by identity on the trivial module, $kGe=b_0(G)$. On restriction to $H\times H$ both $e$ and $se$ are fixed: $H$ is normal and $he=e$ for all $h\in H$. Thus the restriction is two copies of $ke$ as a bimodule.

For any other block idempotent $f$, $ef=0$, so left multiplication by $e$ is zero on the whole bimodule $kGf$ and its restriction. On $ke$ that operator is identity. Therefore $ke$ cannot be isomorphic to a summand of such a restriction. F1 proves uniqueness and $b_0(H{)}^G=b_0(G)$. Yet the centralizer of the trivial subgroup is $G$, while $H$ has index two. This contradicts the proposed necessity, without contradicting F2's sufficient implication. For $n=1$ the same computation is $H=1$, $G=C_2$; for the smallest nonabelian case $n=3$, $D_6\cong S_3\cong {\mathrm{S}\mathrm{L}}_2({\mathbf{F}}_2)$, the latter via its faithful action on the three nonzero vectors and its order six. No semisimplicity or choice assumption is needed.

If $N_G(D)=G$, restriction from $G\times G$ to itself changes no module. A block bimodule $B$ occurs in itself. It cannot occur in another block $B^{\mathrm{\prime }}$: the central idempotent of $B$ acts as identity on $B$ and as zero on $B^{\mathrm{\prime }}$, hence on all its summands. Thus F3 gives $B^G=B$, and F1's map and inverse are the identity on the set of blocks with defect group $D$. This proves both directions of the correspondence directly.

For $D=1$, every group element normalizes $D$, so step 1.1 applies. The numerical defect is ${\log }_p\mathrm{\mid }D\mathrm{\mid }=0$, and F2 says exactly these blocks are full matrix algebras with one projective simple module. If there are no such blocks, the identity has empty domain and codomain: its injectivity condition and the assertion that every codomain element has a preimage are both vacuous. The same reasoning handles an empty defect-$D$ set whenever $D$ is normal. The zero algebra is not inserted as a block. All these identity and idempotent calculations are choice-free.