A **Frobenius algebra** is a vector space that is both an algebra and a coalgebra in a compatible way. (See [here](https://ncatlab.org/nlab/show/Frobenius+algebra#AsAssociativeAlgebraWithLinearForm) for a precise definition.) I guess that a subalgebra of a Frobenius algebra is not again a Frobenius algebra? What is an instructive example that demonstrates this?



A subalgebra of a Frobenius algebra that is not again a Frobenius algebra?
Any finite dimensional $k$-algebra $A$ is also a subalgebra of its trivial extension $$T(A)=A \oplus \rm{Hom}\_k(A,k),$$ which is Frobenius. In this way you get examples where the Frobenius algebra is not semi-simple.


The (symmetric) $k$-bilinear form on $T(A)$ is given by $$\langle (a,f),(b,g) \rangle = g(a)+f(b),$$ for $(a,f),(b,g) \in T(A)$, see section 3 in:


*Bessenrodt, Christine; Holm, Thorsten; Zimmermann, Alexander*, [**Generalized Reynolds ideals for non-symmetric algebras.**](http://dx.doi.org/10.1016/j.jalgebra.2007.02.028), J. Algebra 312, No. 2, 985-994 (2007). [ZBL1119.16001](https://zbmath.org/?q=an:1119.16001).


