Let $A$ be a filtered algebra and let $G$ be its associated graded algebra. As discussed in this [question](https://mathoverflow.net/questions/429814/properties-of-a-filtered-algebra-that-can-be-concluded-from-properties-of-its-as), if $G$ is Frobenius, then $A$ is also Frobenius.


If $G$ is a symmetric Frobenius algebra, does this imply that $A$ is also symmetric?



Associated graded algebras and symmetric Frobenius algebras
If we assume that the algebras you are interested in are finite dimensional, then the statement that you refer to about Frobenius algebras can be found in Bongale, "Filtered Frobenius Algebras" and "Filtered Frobenius Algebras II".


Now, a Frobenius algebra has a Nakayama automorphism, which effects a permutation $\nu$ on the isomorphism classes of simple modules. One description of this is that if the head of a projective indecomposable module is a simple module $S$ then the socle is the simple module $\nu(S)$. The Frobenius algebra is said to be *weakly symmetric* if $\nu$ is the identity permutation. Thus symmetric $\Rightarrow$ weakly symmetric $\Rightarrow$ Frobenius. The grading is inherited by projective modules, and the Nakayama permutation is therefore the same before and after taking the associated graded. So if $G$ is weakly symmetric then so is $A$.


However, there are weakly symmetric algebras that are not symmetric. An example can be given as follows. Let $\mathbb{F}\_4=\{0,1,\omega,\bar\omega\}$ be the field of four elements, with bar denoting the non-trivial field automorphism, and let $A$ be the four dimensional algebra over $\mathbb{F}\_2$ consisting of matrices $\left(\begin{smallmatrix}a&0&0&0\\b&\bar a&0&0\\0&0&\bar a&0\\0&0&\bar b&a\end{smallmatrix}\right)$ with entries in $\mathbb{F}\_4$. This algebra has just one simple module, whose dimension is two, so it is weakly symmetric. Then the associated graded $G$ is isomorphic to $\mathbb{F}\_4[t]/(t^2)$ which is symmetric, but $A$ is not symmetric. To see that it is not symmetric, the property of being symmetric is invariant under field extension, and once the field is extended from $\mathbb{F}\_2$ to $\mathbb{F}\_4$ there are now two one-dimensional simples instead of one two-dimensional simple, and the Nakayama permutation after field extension becomes non-trivial. Note that this example also shows that the property of being weakly symmetric is not invariant under field extension.


To summarise, the answer to your question is no for symmetric but yes for weakly symmetric.


