So let $G$ be a finite group and let $\iota:G\rightarrow S_n$ be an embedding of $G$ in a symmetric group of degree $n$ for some fixed integer $n$. Let $K$ be a fixed field of characteristic $0$. The group $S_n$ permutes the variables $\{x_1,\ldots,x_n\}$ and therefore acts on the field $$ L:=K(x_1,\ldots,x_n). $$ One may look at the invariant subfield $L^{G}\subseteq L$. From Galois theory one has that $L/L^G$ is a Galois extension with Galois group $G$. In particular, the transcendence degree of $L^G$ over $K$ is equal to $n$. In general, the field $L^G$ is not purely transcendental so the following question makes sense:

Q: Does the isomorphism class of $L^G$ depend on the embedding $\iota$ ?

Intuitively I would say no, but this is really just a guess!

`$L^G$`

over $K$ is $n$, not $|G|$. $\endgroup$ – Andreas Blass Jun 29 '11 at 0:41