Brian Bi
\[ \DeclareMathOperator{\End}{End} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\span}{span} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\ad}{ad} \newcommand\d{\mathrm{d}} \newcommand\pref[1]{(\ref{#1})} \]

Remark: I found the proof of Proposition 4.1.2 difficult to understand, so I am recording some notes here in case I forget how to interpret it and need something to refer to. According to the notes, \(\Hom_{k[G]}(k, k[G])\) and \(\Hom_{k[G]}(k[G], k)\) are one-dimensional vector spaces. The reason for this is that \(k[G]\), being semisimple, has a decomposition into irreducibles which contains exactly one copy of the trivial representation \(k\) (any other one-dimensional irreps of \(k[G]\) would be nonisomorphic to this one). By Schur's lemma, every nonzero module homomorphism from \(k\) to \(k[G]\) must map \(k\) isomorphically to the single copy of \(k\) in \(k[G]\) and every nonzero module homomorphism from \(k[G]\) to \(k\) must act by projecting from \(k[G]\) to the single copy of \(k\) and then mapping that subspace to \(k\). The text also skips over the proof that the \(\epsilon\) and \(\Lambda\) chosen are actually homomorphisms.

Problem 4.1.4 We assume as usual that \(k\) is algebraically closed. We will prove the given statement by induction on \(n\).

If \(n = 0\), then \(G\) is just the identity, so any one-dimensional subspace of \(V\) is a subrepresentation isomorphic to \(k\). Since \(V\) is irreducible, \(V \cong k\).

Now suppose \(n \ge 1\). Since \(V\) is irreducible, it must be finite-dimensional. Since \(G\) is a \(p\)-group, \(Z(G)\) is nontrivial. For each \(z \in Z(G)\), we have by Schur's lemma that \(\rho(z)\) is a scalar operator, \(\lambda \Id\) for some \(\lambda \in k\). The order of \(z\) is \(p^k\) for some \(k\), so \(\rho(z^{p^k}) = \lambda^{p^k} \Id = \Id\), that is, \(\lambda^{p^k} = 1\). Since \(p\)th roots are unique in characteristic \(p\), it follows that \(\lambda = 1\). Thus, \(\rho(z) = \Id\) for all \(z \in Z(G)\). Let \(G' = G/Z(G)\). Since \(Z(G) \subseteq \ker \rho\), we can write \(\rho = \rho' \circ \pi\) for some unique homomorphism \(\rho' : G' \to GL(V)\) where \(\pi : G \to G'\) is the quotient map. Since \(\im \rho = \im \rho'\), the representation \((V, \rho')\) of \(G'\) is also irreducible. Now \(G'\) is a \(p\)-group of smaller order, so by the induction hypothesis, \(\rho'\) is the trivial homomorphism, therefore \(\rho\) is also.

Remark: For a version of the proof that does not require the assumption that \(k\) is algebraically closed, see here.