Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 6

The divisibility in (i) is the standard dimension test for a projective modular representation. If $P$ is a projective $kG$-module and $S$ is a Sylow p-subgroup of order $p^d$, then ${\mathrm{Res}}_S^{G}P$ is projective over $kS$. The group algebra of a p-group in characteristic p is local, so every finitely generated projective $kS$-module is free. ConsequentlyThis will apply to $P=\overline{W}$ in the implication (v)$\Rightarrow$(i).

Assume (ii), and write the ring decomposition asThe natural column module $W\cong {\mathcal{O}}^n$ is projective over $M_n(\mathcal{O})$ by Morita equivalence; extending it by zero across $B$ makes it projective over $\mathcal{O}G$. After extending scalars to $K$, the action factors through $M_n(K)$ on its natural module $K^n$, which is simple. Therefore

Assume (iii). Reduction of the projective lattice $W$ is projective, because a direct-summand decomposition of a free $\mathcal{O}G$-module remains one after tensoring with $k$. Completeness of $\mathcal{O}$ and idempotent lifting show that $\overline{W}$ is indecomposable: otherwise a nontrivial idempotent of ${\mathrm{End}}_{kG}(\overline{W})$ would lift and split $W$, hence split the simple $KG$-module $M$.

Let $\chi$ be the ordinary irreducible character of $M$. Since $W$ is projective, $\chi$ vanishes on p-singular elements and restricts to the projective character $\Phi$ of $\overline{W}$. ThusAn indecomposable projective module that is not simple has, besides its identity, a nonzero noninvertible endomorphism obtained by projecting onto its simple head and embedding the isomorphic simple socle. Hence its endomorphism algebra has dimension at least two. Therefore $\overline{W}$ is simple as well as projective, proving

Under (ii), tensor the direct-product decomposition with the residue field $k$. Since $W$ is free over $\mathcal{O}$,Henceand the reduced action map is the projection onto the first factor. This proves

Assume (v). If the generic fibre $M=K{\otimes }_{\mathcal{O}}W$ had a nonzero proper $KG$-submodule, intersecting it with $W$ and rescaling to obtain a saturated lattice would give a nonzero proper $kG$-submodule of $\overline{W}$. Thus $M$ is simple.

Because $\overline{W}$ is projective, its restriction to a Sylow p-subgroup $S$ is projective. The local algebra $kS$ has only free finitely generated projectives, so $|S|=p^d$ dividesConsequently

Let $D$ act on the group algebra $kG$ by conjugation. Its fixed-point algebra isand the centralizer $C_G(D)$ consists of the elements of $G$ commuting with every element of $D$. The Brauer morphism isThus $\beta$ deletes the coefficients of basis elements outside $C_G(D)$. The nonfixed $D$-orbits have cardinality divisible by $p$, so the usual orbit argument shows that this projection is a unital ring homomorphism. Hence

Put $N=N_G(D)$. For an algebra on which a group acts by conjugation, writefor its transfer ideal of conjugation-fixed elements. In the diagram, $\tau ={\mathrm{Tr}}_D^G$, ${\tau }^{\prime }={\mathrm{Tr}}_D^N$, the upper map is $\beta ={\mathrm{Br}}_D$, and ${\beta }^{\prime }$ is the restriction of ${\mathrm{Br}}_D$ from $(kG{)}^G$ to $(kG{)}_D^G$. Since $N$ normalizes both $D$ and $C_G(D)$, the lower map lands in $(k{C}_G(D){)}_D^N$.

For $a\in (kG{)}^D$, let $D$ act on the left cosets $G/D$. A coset $gD$ is fixed precisely when $g^{-1}Dg\le D$, hence, because the two groups have the same order, precisely when $g\in N$. Every nonfixed orbit has size divisible by $p$. Moreover, after applying ${\mathrm{Br}}_D$, all summands indexed by one $D$-orbit are equal: conjugation by an element of $D$ acts trivially on $k{C}_G(D)$. Those orbits therefore contribute zero in characteristic $p$, while the fixed cosets contribute the trace over $N/D$. ConsequentlyThis is the Brauer morphism and relative trace identity, so the diagram commutes.

The decomposition matrix separates into two connected components. The first contains ${\varphi }_1,{\varphi }_2$ and ${\chi }_1,{\chi }_{3A},{\chi }_{3B},{\chi }_4$; it is the principal block $B_0$. The second contains only ${\varphi }_3$ and ${\chi }_5$; since ${\varphi }_3(1)=5$ contains the full 5-part of $|A_5|=60$, this is a defect-zero representation and its block $B_1$ has defect group $1$.

For $D=1$, the normalizer is $N_G(1)=G$, so the Brauer correspondence is the identity and $B_1$ corresponds to itself.

The defect group of the principal block is a Sylow 5-subgroup $P\cong C_5$. There are six Sylow 5-subgroups in $A_5$, so the orbit-stabilizer theorem givesThe centralizer of a 5-cycle in $A_5$ is $P$, and an involution in the normalizer acts on $P$ by inversion. HenceIn characteristic five the simple $kN$-modules are inflated from $N/P\cong C_2$: their Brauer characters are ${\psi }_{+}=(1,1)$ and ${\psi }_{-}=(1,-1)$ on the identity and involution classes. If $1,\epsilon ,{\rho }_1,{\rho }_2$ are the two one-dimensional and two two-dimensional ordinary characters of $D_{10}$, their reductions areThe resulting decomposition matrix is connected, so these characters form the unique 5-block $b_0$ of $N$, with defect group $P$. By the Brauer first main theorem,This is the complete 5-modular blocks of A5 correspondence.