Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 138 / 1

Let $N={\sum }_{g\in G}g$ be the group norm element. This is a nonzero vector in the group algebra $kG$, and $kN$ is a trivial submodule of its left regular module. If $kG$ were a semisimple algebra, this submodule would have a $kG$-linear projection $\pi :kG\to kN$. Write $\pi (1)=cN$. Since $gN=N$ for every $g$, we would havewhereas a projection onto $kN$ satisfies $\pi (N)=N\ne 0$. Thus $\boxed{kG\text{ is not semisimple}}$.

Put $q=p^n$. A finitely generated module over $A=k[X]/(X^q)$ is finite-dimensional over $k$, and multiplication by $X$ is a nilpotent linear map $T$ with $T^q=0$. Its Jordan normal form exists over $k$ because its minimal polynomial is a power of $X$. Each Jordan block has size $r\le q$ and is the cyclic module $M_r=k[X]/(X^r)$. HenceA submodule of $M_r$ is an ideal of $k[X]/(X^r)$, and its inverse image is an ideal of the principal ideal domain $k[X]$ containing $(X^r)$. The only possibilities are $(X^j)$ with $0\le j\le r$. Thus all submodules form the chainEvery successive quotient is the one-dimensional simple module on which $X$ acts as zero. This is its unique composition series, so $M_r$ is a uniserial module and is an indecomposable representation: two nonzero direct summands would be incomparable submodules. Only $M_1$ is simple.

For $G=⟨g⟩$ of order $q$, the group algebra satisfiessince $(T-1{)}^q=T^q-1$ in characteristic $p$. Consequently, up to isomorphism, $\boxed{M_1,\dots ,M_{p^n}}$ are precisely the finite-dimensional indecomposable modules, and $\boxed{M_1\text{ is the unique simple module}}$. The finite-dimensional convention follows from the finite-generation hypothesis; “exactly” counts isomorphism classes.