Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 3 / c / Solution

Decompose the projective module aswhere $T$ ranges over the simple modules. Since $P_T/J(P_T)\cong T$, the multiplicity of $S$ in the head of a module $P/J(P)$ is $m_S$. Part (b) gives $\mathrm{Soc}(P_T)\cong T$, so the multiplicity in $\mathrm{Soc}(P)$ is the same $m_S$.

The invariant submodule and coinvariant module satisfyThus $\dim P^G$ is the multiplicity of the trivial module in $\mathrm{Soc}(P)$, while $\dim P_G=\dim {\mathrm{Hom}}_{kG}(P,k)$ is its multiplicity in the head. Applying the same argument to the projective module $P^{\ast }$ yields

The dual $P_S^{\ast }$ is indecomposable projective. Its head is dual to $\mathrm{Soc}(P_S)\cong S$, hence is $S^{\ast }$. Uniqueness of projective covers proves the dual of a projective cover over a group algebra: