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Are there any examples of finite-dimensional complex non-semisimple non-commutative symmetric Frobenius algebras? Or can one show that none exist?

I went through this list of all complex associative algebras up to dimension $4$ and couldn't find any non-commutative non-semisimple ones that can be equipped with a corresponding linear form to make them into a symmetric Frobenius algebra. Are there examples in higher dimensions?

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Given any finite dimensional algebra $A$, consider the linear dual $\hat{A}= \hom(A, k)$ as an $A$-$A$-bimodule. Then $R = A \oplus \hat{A}$ may be equipped with an algebra structure as follows:

$$(a, x) \cdot (b,y) = (ab, x \cdot b + a \cdot y)$$

for $a,b \in A$ and $x,y \in \hat{A}$. The algebra $R$ has a natural symmetric Frobenius algebra structure. So every finite dimensional (possibly non-commutative) algebra embeds into a symmetric Frobenius algebra.

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    $\begingroup$ Is it obvious that there are (non-commutative) examples for which the resulting symmetric Frobenius algebra is non-semisimple? $\endgroup$
    – Andi Bauer
    Commented Mar 13, 2021 at 15:40
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    $\begingroup$ @AndiBauer yes it is, since $\hat{A}$ is then a 2-sided ideal with zero law. So the resulting $R=R_A$ is not semisimple unless $A=0$ (and not commutative as soon as $A$ is not). $\endgroup$
    – YCor
    Commented Mar 13, 2021 at 15:44
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    $\begingroup$ The bilinear form is $\langle (a,f), (b,g) \rangle = f(b) + g(a)$. $\endgroup$
    – Chris Schommer-Pries
    Commented Mar 15, 2021 at 18:22
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    $\begingroup$ @AndiBauer The standard actions of $A,A$ on $Hom(M,k)$ would be $(r\cdot f)(m)=f(mr)$ and $(f\cdot r)(m)=f(rm)$, which is my estimate at what is intended. $\endgroup$
    – rschwieb
    Commented Mar 16, 2021 at 13:07
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    $\begingroup$ @AndiBauer You said "algebras". I assumed you meant unital or would have said otherwise. Of course unital algebras are also not-necessarily unital algebras, so this still gives examples asked for in the OP. If A is unital, then the bilinear form is induced by the counit $\epsilon(a,f) = f(1)$. If $A$ is non-unital, then I guess there is no counit, but the bilinear form still makes sense and is symmetric and non-degenerate. $\endgroup$
    – Chris Schommer-Pries
    Commented Mar 17, 2021 at 13:11
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I just realize that my question is actually rather trivial the way I posed it: There exist non-commutative semisimple examples (with the $2\times 2$ matrix algebra being the smallest example). There also exist commutative non-semisimple examples, the smallest one given by $$e_0\cdot e_0 = e_0,\quad e_0\cdot e_1=e_1,\\ e_1\cdot e_0=e_1,\quad \mu(e_1)=1\;,$$ where $\mu$ is the linear form generating the symmetric non-degenerate pairing and all other entries $0$.

All one needs to do is consider direct sums or tensor products of a non-commutative (but semisimple) and a non-semisimple (but commutative) example, and it will be both non-commutative and non-semisimple. The smallest such example is of dimension $6$.

I guess a more sensible but a bit artificial question to ask would have been whether there are examples that do not come from the tensor product or direct sum of a commutative and a semi-simple example.

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