Decompose the projective module as
$$
P\cong\bigoplus_T P_T^{,m_T},
$$
where $T$ 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 $\operatorname{Soc}(P_T)\cong T$, so the multiplicity in $\operatorname{Soc}(P)$ is the same $m_S$.

The <invariant submodule> and <coinvariant module> satisfy
$$
P^G\cong\operatorname{Hom}_{kG}(k,P),
\qquad
(P_G)^*\cong(P^*)^G.
$$
Thus $\dim P^G$ is the multiplicity of the trivial module in $\operatorname{Soc}(P)$, while $\dim P_G=\dim\operatorname{Hom}_{kG}(P,k)$ is its multiplicity in the head. Applying the same argument to the projective module $P^*$ yields
$$
\boxed{\dim P^G=\dim P_G=\dim(P^*)^G=\dim(P^*)_G.}
$$

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

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-138/3/c/solution