Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 4

An $RG$-module $M$ is relative projective module for $H$ when it has the lifting property for every $H$-split epimorphism: whenever the solid arrows form a commutative diagramwith $\pi :X\twoheadrightarrow Y$ an $RG$-map possessing an $RH$-linear section, there is an $RG$-map $\tilde{f}:M\to X$ such that $\pi \tilde{f}=f$. Equivalently, every $H$-split epimorphism onto $M$ has a $G$-linear section, or

For $\alpha \in {\mathrm{End}}_{RH}(M)$ define the relative traceThe D. Higman criterion is

SetEvery conjugate of $\alpha$ equals $\alpha$, soThe D. Higman criterion therefore gives

If $M$ is projective over $RG$, then its restriction is projective over $RH$ because $RG$ is a free right $RH$-module and a free $RG$-module restricts to a free $RH$-module.

Conversely, suppose ${\mathrm{Res}}_H^{G}M$ is projective. Thenis projective over $RG$. Part (i) says that $M$ is a direct summand of this induced module, so $M$ is projective. Hence projectivity detected on a subgroup of invertible index gives

Let $U\le {\mathrm{SL}}_2(p)$ be the Sylow p-subgroup of upper unitriangular matrices. It is cyclic of order $p$, generated byRealize $S^r(V_2)$ as the homogeneous polynomials of degree $r$ in $X,Y$, with $u$ acting by $X\mapsto X$ and $Y\mapsto X+Y$. For $0\le r<p$, the only vectors fixed by $u$ are the multiples of $X^r$: successive comparison of the coefficients of $Y^r,Y^{r-1},\dots$ proves this. Thus the nilpotent operator $u-1$ has one-dimensional kernel. Its Jordan normal form therefore has a single block, soThis also follows from the indecomposable modules of a cyclic p-group in characteristic p.

For $r=p-1$, the restriction has dimension $p$ and is the regular $kU$-module, hence is projective. Since $[G:U]$ is prime to $p$, part (b)(ii) makes $S^{p-1}(V_2)$ a simple projective $kG$-module. It is therefore a defect-zero representation and lifts to an ordinary irreducible representation of the same dimension. Consequently

Let $P$ be a Sylow p-subgroup. Since $[G:P]$ is invertible in $k$, every $kG$-module is relatively $P$-projective.

Suppose first that $kP$ has only finitely many indecomposable modules $U_1,\dots ,U_t$. For every indecomposable $kG$-module $M$, decompose ${\mathrm{Res}}_P^{G}M$ into the $U_i$. Relative projectivity makes $M$ a summand of the corresponding finite direct sum of the ${\mathrm{Ind}}_P^G{U}_i$. The Krull–Schmidt theorem leaves only finitely many possible indecomposable summands, so $kG$ has finite representation type.

Conversely, suppose $kG$ has finitely many indecomposables. For an indecomposable $kP$-module $U$, the identity double coset in the Mackey restriction formula shows that $U$ is a direct summand ofDecomposing the induced module into the finitely many $kG$-indecomposables and restricting them shows, again by Krull–Schmidt, that only finitely many $U$ can occur. Thus

If $P$ is cyclic, the indecomposable modules of a cyclic p-group in characteristic p form a finite list. If $P$ is noncyclic, its Frattini quotient has rank at least two and therefore has a quotient $C_p\times C_p$. Inflation preserves indecomposability and nonisomorphism, while $k(C_p\times C_p)$ has infinitely many indecomposable modules. The Higman criterion for finite representation type of a group algebra now gives

Existence of a minimal subgroup follows because $G$ has only finitely many subgroups and every module is relatively $G$-projective. Suppose an indecomposable module $M$ is relatively projective for both $H$ and $K$. Then $M$ is a summand of ${\mathrm{Ind}}_H^{G}U$ and of ${\mathrm{Ind}}_K^{G}V$ for suitable modules $U,V$. Applying the Mackey restriction formula and the Krull–Schmidt theorem shows that $M$ is relatively projective for some subgroupIf $H$ and $K$ are minimal, this forces $H\subseteq {}^{g}K$. Reversing their roles gives the reverse containment after conjugacy; since the groups are finite, $H$ and $K$ are conjugate. Thus vertices form a unique conjugacy class.

Let $Q$ be a vertex and let $P$ be a Sylow p-subgroup of $Q$. Since $[Q:P]$ is invertible in $k$, every $kQ$-module is relatively $P$-projective. Transitivity of relative projectivity makes $M$ relatively $P$-projective, so minimality forces $Q=P$. Hence every vertex is a p-group.

For the trivial module $k$, every $kH$-endomorphism is scalar, and its relative trace to $G$ is multiplication by $[G:H]$. The D. Higman criterion says that $k$ is relatively $H$-projective exactly when $p\nmid [G:H]$. The minimal such subgroups are precisely the Sylow p-subgroups. Therefore the vertex of an indecomposable module gives