Does the invariant ring determine the group?

Question

Let $G$ be a finite group $n = |G|$. Let $\sigma : G \rightarrow GL(n,\mathbb{C})$ be the regular representation. Hence every element of $G$ can be seen as a permutation matrix. Let $I_G := \mathbb{Q}[x_1,...,x_n]^G$ be the invariant ring. Let $H \le S_n$ be a subgroup with invariant ring $I_H$. Does $I_H \le I_G$ imply that $G \le H$?

Answer 1

Yes. Once you show that taking invariants commutes with taking fraction fields, this follows from observing that $\mathbb{Q}(x_1, \dots x_n)$ is a Galois extension of $\mathbb{Q}(x_1, \dots x_n)^{S_n}$ with Galois group $S_n$, then applying the Galois correspondence.