I am reading a proof for exercise 5.13 in Atiyah and Macdonald, given below.
Exercise 12 is the following
I understand everything except one thing, why is $\prod_{\sigma \in G} \sigma^{-1}(x) \in A^{G}$? I understand that the product lies in $\mathfrak{p}$, because one of all the $\sigma$ is the identity, but why is the product a $G$-invariant?
If $\tau \in G$, $\tau(\prod_{\sigma \in G} \sigma^{-1}(x)) = \prod_{\sigma \in G} (\tau\sigma^{-1})(x)= \prod_{\sigma \in G} \sigma^{-1}(x)$ because multiplication by $\tau$ just permutes the elements of $G$.