*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.