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

Give the dual $M^{\ast }={\mathrm{Hom}}_k(M,k)$ the contragredient action $(g\lambda )(m)=\lambda (g^{-1}m)$. If ${\delta }_h$ is the functional dual to the basis element $h\in G$, thenThus the map $h\mapsto {\delta }_h$ extends to a $kG$-isomorphismThis is also the left-module form of the fact that a group algebra is a symmetric algebra.

If $P$ is finitely generated and projective, it is a direct summand of $(kG{)}^r$. Dualizing makes $P^{\ast }$ a direct summand of $((kG{)}^{\ast }{)}^r\cong (kG{)}^r$, so $P^{\ast }$ is projective. The converse follows by dualizing again and using $P\cong P^{\ast \ast }$.

For any finite-dimensional algebra $A$, the module $A^{\ast }$ is injective becauseis exact. Since $kG\cong (kG{)}^{\ast }$, free $kG$-modules are injective, and so are their projective direct summands. Conversely, duality sends injectives to projectives, so every finite-dimensional injective is projective. Hence projective modules over a finite group algebra are injective.

Finally, let $P$ be indecomposable projective. It is also an indecomposable injective. Its nonzero socle contains a simple module $S$, and the injective hull $E(S)$ is a direct summand of $P$. Indecomposability forces $P=E(S)$. Since $S$ is essential in its injective hull, every simple submodule of $P$ equals $S$. Therefore

Write $P\cong kGe$ for a primitive idempotent $e$. The coefficient-of-identity form makes $kG$ a symmetric algebra. Associativity of this nondegenerate form identifies the orthogonal complement of $J(kG)e$ in $kGe$ with the elements annihilated by $J(kG)$, namely $\mathrm{Soc}(P)$. It therefore induces a nondegenerate $kG$-invariant pairing betweenand $\mathrm{Soc}(P)$. Both are simple by projectivity and part (a), and the symmetric form has identity Nakayama permutation. Consequently the head and socle of an indecomposable projective group-algebra module satisfy

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:

Put $N_G={\sum }_{g\in G}g$. Its image on every module lies in the invariant submodule, since $h{N}_G=N_G$. In the regular module,and this line is the socle of the projective cover $P_k$ of the trivial module.

Suppose $N_{G}M\ne 0$. Choose $m\in M$ with $N_{G}m\ne 0$ and consider the homomorphismIts restriction $f{|}_{P_k}$ is nonzero on $\mathrm{Soc}(P_k)=k{N}_G$. Because $P_k$ is the injective hull of its simple socle, that socle is essential: every nonzero submodule meets it. Hence $\ker (f{|}_{P_k})=0$. The resulting embedding $P_k\hookrightarrow M$ splits because $P_k$ is injective. Since $M$ is indecomposable, $M\cong P_k$.

Conversely, on $P_k$ the image of $N_G$ is its one-dimensional socle. Thus the group norm element detects the trivial projective cover: