Let $k$ be a field and $N$ a finite group. Let $M$ be a projective indecomposable $kN$-module. Since the algebra $kN$ is symmetric, it follows that the top and bottom composition factors of $M$ are isomorphic. In particular, there is a nonzero endomorphism of

$M$ sending $M$ onto the socle $operatorname{soc}(M)$.

I cannot see the connection here. How does being symmetric implies composition factors? Any help would be appreciated!

MathOverflow Asked by user666 on January 2, 2021

1 AnswersFor every Frobenius algebra $A$ there is a bijection $pi$ such that $top(P_i) cong soc(P_{pi (i)})$ when $P_i$ denote the indecomposable projective $A$-modules. Being symmetric implies that $A$ is weakly symmetric (meaning that $pi$ is the identity). Thus top and socle of every $P_i$ coincide which is what you asked for when I understand your question correct. For proofs and more on this see the book "Frobenius algebras I" in chapter IV. by Skowronski and Yamagata. When M is an indecomposable projective $A$-module, let $S:=soc(M)$ be the socle of $M$. Then we have a surjective map $M rightarrow S$ (since top and socle of $M$ coincide) that induces an isomorphism $top(M)=M/rad(M) rightarrow S$. Thus we have a surjective map $M rightarrow soc(M)$.

Correct answer by Mare on January 2, 2021

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