Problem with IsSymmetricAlgebra?

[Nakayamaalgebra]

The command IsSymmetricAlgebra says that a group algebra is not symmetric:

G:=DihedralGroup(8);Size(G); K:=GF(2);R:=GroupRing(K,G); A:=AlgebraAsQuiverAlgebra(R)[1]; gap> IsSymmetricAlgebra(A); false

gap> C:=AlgebraAsModuleOverEnvelopingAlgebra(A);D:=DualOfAlgebraAsModuleOverEnvelopingAlgebra(A);IsomorphicModules(C,D); <[ 8 ]> <[ 8 ]> true

[sunnyquiver]

I have gone back to the old and hopefully correct implementation of the function. It might be slow for big examples.

[Nakayamaalgebra]

Thanks. The old function is really slow. It might be faster to check whether the order of the Nakayama automorphism is equal to 1 (which is equivalent to the algebra being symmetric).

[sunnyquiver]

What is the order of the Nakayama autormorphism? Is it defined as the order of a Nakayama automorphism viewed as an element in the group of outer automorphisms of the algebra? In the above example QPA computes a Nakayama automorphism of the algebra which is not the identity (it has order 2). However as an element in the group of outer automorhisms of the algebra, one can show that it is equivalent to the identity. I see no easy way to compute the group of outer auto's of the algebra. So I see no alternative to deciding when an algebra is symmetric than the current implementation.

[Nakayamaalgebra]

Thank you for the answer. There might be no easy solution then. A good test whether a selfinjective algebra A is symmetric is to check whether the center of A (zeroth Hochschild cohomology) has the same vector space dimension as A/[A,A] (zeroth Hochschild homology). In fact, I do not know whether this is equivalent to a selfinjective algebra A being symmetric.