For a finite group $G$, the coefficient of the identity in $xy$ defines a nondegenerate symmetric associative bilinear form on $kG$. Hence $kG\cong(kG)^*$ as bimodules and the <group algebra> is a <symmetric algebra (Frobenius algebra)>.

Group algebra is a symmetric algebra

If $P$ is an indecomposable projective $kG$-module, then its head and socle are simple and naturally isomorphic:
$$
P/J(P)\cong\operatorname{Soc}(P).
$$
This is the identity Nakayama permutation of the <symmetric algebra (Frobenius algebra)> $kG$.

Head and socle of an indecomposable projective group-algebra module

Write $P\cong kGe$ for a <primitive idempotent> $e$. The coefficient-of-identity form makes $kG$ a <symmetric algebra (Frobenius algebra)>. Associativity of this nondegenerate form identifies the orthogonal complement of $J(kG)e$ in $kGe$ with the elements annihilated by $J(kG)$, namely $\operatorname{Soc}(P)$. It therefore induces a nondegenerate $kG$-invariant pairing between
$$
P/J(P)=kGe/J(kG)e
$$
and $\operatorname{Soc}(P)$. 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
$$
\boxed{P/J(P)\cong\operatorname{Soc}(P).}
$$

Solution

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

{disambiguate=Frobenius algebra}

A finite-dimensional algebra $A$ is symmetric when there is a nondegenerate symmetric bilinear form $B:A\times A\to k$ satisfying $B(ab,c)=B(a,bc)$. Equivalently, $A\cong A^*$ as $A$-bimodules.


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 Incoming links: Symmetric algebra (Frobenius algebra)