# Top and bottom composition factors of $M$ are isomorphic

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

For 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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