Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 4 / 3

Put $q=p^n$ and choose a generator $h$ of $H$. In characteristic $p$, $(h-1{)}^q=h^q-1=0$, soA module is a vector space with a nilpotent operator $t$ whose nilpotence index is at most $q$. The nilpotent Jordan normal form, valid over every field, splits it into Jordan blocks. One block of size $i$ isIt is indecomposable: its endomorphism ring is $k[t]/(t^i)$, a local ring with no nontrivial idempotents. Distinct $i$ have distinct dimensions. Conversely a direct sum of at least two Jordan blocks is decomposable. Thus there are exactly $q$ indecomposable isomorphism classes.

A homomorphism $V_i\to V_j$ is determined by the image of $1$, which can be any element of $V_j$ annihilated by $t^i$. If $i\ge j$ this is all of $V_j$, of dimension $j$; if $i<j$ it is the span of $t^{j-i},\dots ,t^{j-1}$, of dimension $i$. HenceThis realizes the indecomposable modules for a cyclic p-group.

For generators $g,h$ of $P$, write $x=g-1$, $y=h-1$. ThenFor each $\lambda \in k$, define a two-dimensional module $M_{\lambda }$ on basis $v,w$ by $xv=w$, $xw=0$, $yv=\lambda w$, $yw=0$. Every product of two of these operators is zero, so the defining relations hold even when $p=2$.

The action of $x$ is one nonzero nilpotent Jordan block. A decomposition of $M_{\lambda }$ would decompose its restriction to $k[x]/(x^p)$, which is already indecomposable. Thus $M_{\lambda }$ is indecomposable. An isomorphism $F:M_{\lambda }\to M_{\mu }$ commutes with $x$ and $y$, and consequentlySince $F$ is invertible and $x\ne 0$, $\lambda =\mu$. The infinite field supplies infinitely many parameters, proving $kP$ has infinite representation type. This is the parameter family for an elementary abelian p-group.

Let $B=kGe$ have defect group $D$. Use two vertex-and-source facts, with their relevant content made explicit. Every indecomposable module in $B$ has a module vertex conjugate into $D$. Also $B$, as a $k[G\times G]$-module with $(g,h)a=ga{h}^{-1}$, is an indecomposable permutation-module summand with module vertex $\Delta D=\{(d,d):d\in D\}$ and trivial module source. The second fact is the bimodule characterization of a defect group of a block; the first follows by applying its relative-$\Delta D$ induction splitting to $B{\otimes }_{kG}M=M$.

If $D$ is cyclic, every module source for every module vertex $Q\le D$ occurs among the finitely many indecomposable modules for a cyclic p-group. Each module in the block is a summand of an induction of one of these finitely many module sources. There are finitely many subgroups $Q$ and each induced representation has finitely many indecomposable summands by Krull–Schmidt theorem. Thus cyclic defect implies finite representation type of the block.

For the converse, the restriction $B{\downarrow }_{D\times D}$ has the regular $kD$-bimodule as a direct summand. To see this from module vertices and module sources, its Brauer quotient of a module at $\Delta D$ is nonzero. Its summands are permutation modules for the $p$-group $D\times D$ and have module vertices of order at most $|D|$. A summand detected at $\Delta D$ therefore has module vertex $\Delta D$ and trivial module source, hence isHere a transitive permutation module for a $p$-group is indecomposable: it has one-dimensional head of a module over the local group algebra and cannot split into two nonzero modules. This also explains the Brauer-quotient detection used in the argument.

For any $kD$-module $U$, tensor that split bimodule inclusion with $U$. It giveswhere $\mid$ denotes being a direct summand. If $B$ has only finitely many indecomposable modules $M_1,\dots ,M_s$, each $B{\otimes }_{kD}U$ is a sum of these. Therefore every indecomposable $U$ must occur among the summands of the finite list $M_i{\downarrow }_D$. Thus $kD$ has finite representation type.

A noncyclic $p$-group has a quotient $C_p\times C_p$: its Frattini quotient has rank at least two, since a one-generator Frattini quotient forces the group itself to be cyclic. Inflation along this quotient embeds all the indecomposable modules of that elementary abelian group, preserving isomorphisms and indecomposability. Part (b) rules out finite type when $k$ is infinite.

The conclusion remains valid if the splitting field $k$ is finite. For completeness, replace the scalar parameter in part (b) by an $m\times m$ companion matrix $T_f$ of an irreducible polynomial $f$. On $k^m\oplus k^m$ useAn endomorphism commuting with these has form $\left(\begin{array}{cc}A& 0\\ C& A\end{array}\right)$ with $A{T}_f=T_{f}A$. Its endomorphism ring has a square-zero ideal of the $C$-matrices and quotient $k[T_f]\cong k[t]/(f)$, a field, so it is local and the module is indecomposable. Finite fields have irreducible polynomials of arbitrarily large degrees, giving infinitely many dimensions. This also excludes noncyclic $D$ in that case. Altogether,