# Is the fixed subring a symmetric algebra?

Let A be a finite dimensional symmetric k-algebra over some field k. The set of units of A is denoted by U(A). Suppose G is a cyclic group of prime order which acts via inner algebra automorphism on A, say, there is a homomorphism $$\phi : G \rightarrow U(A)$$ such that $$a\cdot g:= a ^{\phi(g)}$$ for all $$a\in A, g\in G$$ and this action preserves scalar multiplication by k. We donote the fixed subring under the action of G by $$A^G$$. My question is the following:

Is $$A^G$$ always a symmetric k-algebra in this condition?

• Can you please say what you mean by symmetric k-algebra? – ArB Jun 21 at 3:33
• Here I mean an algebra with a symmetric, associative, and nondegenerate k-bilinear form or an algebra which is isomorphic to its k-dual as bimodule. – Master Gang Jun 21 at 6:02

Let $$k$$ be a field of characteristic $$2$$, and let $$A$$ be the path algebra over $$k$$ of the quiver with two vertices, $$v_1$$ and $$v_2$$, and arrows $$a:v_1\to v_2$$ and $$b:v_2\to v_1$$, modulo the relations $$aba=0$$ and $$bab=0$$.
Then $$A$$ is a symmetric algebra (with symmetrizing form given by $$\varphi(ab)=\varphi(ba)=1$$ and $$\varphi(p)=0$$ for all paths $$p$$ of length $$0$$ or $$1$$).
Let $$\phi(G)$$ be the subgroup of $$U(A)$$ generated by $$1+a$$, which has order $$2$$.
Then $$A^G$$ is the span of $$\{1,a,ab,ba\}$$, which is isomorphic to $$k[x,y,z]/(x,y,z)^2$$ and is not symmetric.
• Possibly the answer is different if you add the requirement that $\operatorname{char}(k)$ does not divide $|G|$, although I've not really thought about it. Lots of statements of this kind require that condition. – Jeremy Rickard Jun 21 at 10:43