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.