# 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!

• Sorry, what's $k$, what's $N$?
– YCor
Oct 13 '20 at 13:54
• hi! $k$ is a field and $N$ is a group. Oct 13 '20 at 15:53
• You probably also implicitly assumed that $N$ is finite.
– YCor
Oct 13 '20 at 16:23

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 $$M \rightarrow soc(M)$$.
• thank you for your input! question: what is $top(Pi)$? pardon please Oct 13 '20 at 12:04
• @steven It is the top composition factor, namely $P_i /rad(P_i)$.
• hang on sir. Do we need "the two coincide" to conclude that there is a onto map from $M$ to the socle? what is the connection here? thx! Oct 13 '20 at 12:16