Szamuely's "Galois groups and fundamental groups" exercise I.7 I am working through the exercises in Tamás Szamuely's book "Galois group and fundamental groups". Exercise 7 from the first chapter is the following.
Let $k$ be a field and $\bar{k}$ be a fixed algebraic closure of $k$. Given a finite etale $k$-algebra $A$ with a finite group $G$ acting via $k$-algebra automorphisms on it, say it is "Galois" if $A^G = k$, and $\mathrm{dim}_k(A) = |G|$. Show that such an algebra $A$ is Galois if and only if $A \otimes_k \bar{k} \simeq \bar{k}[G]$ as a $G$-module.
I'm currently stuck at showing that if $A$ is Galois, then $A \otimes_k \bar{k} \simeq \bar{k}[G]$ as a $G$-module. I get that they both have dimension $|G|$ as $\bar{k}$-algebra, that the $G$-invariants are $\bar{k}$ in both case, and that this boils down to finding a basis of the form $(x, g_1.x,\ldots,g_{n-1}.x)$ of $A \otimes_k \bar{k}$. I know that in the case that $A$ is a finite separable extension of $k$, then being Galois and being a normal extension is the same, and $G$ is then the Galois group of $A$. In this case the normal basis theorem gives a slightly stronger result, i.e $A \simeq k[G]$ as a $G$-module. I tried using this to get the general case where $A$ is a product of finite separable extensions of $k$, but this is getting nowhere. I also tried to see if I could adapt the proof of the normal basis theorem to get this, but I couldn't as well.
Any hint would be appreciated.
 A: Write $\bar A := A\otimes_k\bar k$. Since $A$ is étale we have $\bar A = \prod_{i\in I}\bar k$, where
$$
\lvert I\rvert = \dim_{\bar k}\bar A = \lvert G\rvert.
$$
Since $G$ acts on $\bar A$ by $\bar k$-automorphisms, this action must come from an action on $I$: Indeed, the canonical basis $\{e_i\}_{i\in I}$ is the unique maximal system of orthogonal idempotents. But for each $g\in G$ also $\{g(e_i)\}_{i\in I}$ is a system of orthogonal idempotents; it is maximal for cardinality reasons. Hence $G$ permutes $\{e_i\}_{i\in I}$. But this precisely means that the action on $\bar A$ is induced by an action on $I$.
If $\emptyset\neq J\subseteq I$ is a $G$-stable subset, it follows that $\sum_{j\in J}e_j \in \bar A^G$. But $\bar A^G$ is one-dimensional and spanned by $1 = \sum_{i\in I}e_i$. Therefore, $\sum_{j\in J}e_j = \sum_{i\in I}e_i$, which means $J=I$. This proves that $G$ acts transitively on $I$, so that $I\cong G$ as $G$-sets.
Therefore, $\bar A = \prod_{g\in G}\bar k = \bar k[G]$ as $G$-modules, where the last equality is basically the definition of $\bar k[G]$.
