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Let $A$ be a finite dimensional Frobenius algebra and $e$ and idempotent of $A$. It is well known that the algebra $eAe$ does not have to be a Frobenius algebra. But if $A$ is additionally symmetric, then $eAe$ is also a symmetric Frobenius algebra for any idempotent $e$.

The Frobenius algebra $A$ is called weakly symmetric if for every indecomposable projective module $P$: $top(P)=soc(P)$.

Question: If $A$ is just weakly symmetric, is $eAe$ also always weakly-symmetric for any idempotent $e$?

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Yes.

Using left modules, the indecomposable projective $eAe$-modules are $eP$ for $P$ an indecomposable projective $A$-module such that $e\operatorname{top}(P)=\operatorname{top}(P)$, and in this case $\operatorname{top}(eP)=\operatorname{top}(P)$.

For any finite-dimensional $A$-module $M$ such that $\operatorname{soc}(M)=e\operatorname{soc}(M)$, we have $\operatorname{soc}(eM)=\operatorname{soc}(M)$. Thus for the relevant indecomposable projectives we also have $\operatorname{soc}(eP)=\operatorname{soc}(P)$, and the result follows.

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    $\begingroup$ Thanks, this should show more generally that a selfinjective algebra $A$ with Nakayama permutation $\nu$ has the property that $eAe$ is also selfinjective when $e$ is a sum of primitive orthogonal idempotents that is fixed by $\nu$. $\endgroup$
    – Mare
    Commented Feb 5, 2021 at 23:11

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