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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?

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    $\begingroup$ Can you please say what you mean by symmetric k-algebra? $\endgroup$ – ArB Jun 21 at 3:33
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    $\begingroup$ 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. $\endgroup$ – Master Gang Jun 21 at 6:02
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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.

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    $\begingroup$ 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. $\endgroup$ – Jeremy Rickard Jun 21 at 10:43
  • $\begingroup$ @Jemery Rickard Very thanks for your nice counterexample. To be honest, I am considering the general case here. As there are some references for the case when char(k) does not divide |G|. I am not really interested in that situation. $\endgroup$ – Master Gang Jun 21 at 13:12

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