# Why doesn't "$S_n$ appears as a Galois group over $\Bbb{Q}$" wrap up the Inverse Galois Problem?

Since every finite group $$G$$ is embedded in $$S_n$$ for $$n = |G|$$ and Hilbert showed that $$S_n$$ appears as a Galois group of $$K/\Bbb{Q}$$ for some Galois extension $$K$$, then how does that not wrap up the Inverse Galois Problem?

Why couldn't we somehow take any subgroup of $$S_n$$ and show that it must be the Galois group over $$\Bbb{Q}$$ of some $$L \supset K$$?

• Do you mean $L\subset K?$ Don’t see how a bigger field gives a smaller group. Jul 20, 2021 at 4:12
• @ThomasAndrews I thought things were inclusion-reversing here. Jul 20, 2021 at 4:21
• You've just proven that every group occurs as a Galois group of some extension $K/L$: if $H\subset S_n$ and $\mathrm{Gal}(K/\mathbb Q) = S_n$, then $H = \mathrm{Gal}(K/L)$ for some subfield $L$. The difficult bit is showing that we can take $L = \mathbb Q$. Jul 20, 2021 at 12:20
Subgroups $$H$$ of $$Gal(E/F)$$ correspond to intermediate fields $$K$$ where $$H = Gal(E/K)$$ Therefore, if we apply it to $$F = \mathbb{Q}$$, then all we can conclude is that $$H$$ is the Galois group of some field extension $$E/K$$, neither of which are required to be $$\mathbb{Q}$$.
To solve the inverse Galois problem, it is sufficient to show that every finite group is a quotient of $$S_n$$, not a subgroup.
• It's probably useful to remark that most groups aren't quotients of $S_n$: for large $n$ ($n>6$?), the only nontrivial quotient of $S_n$ is the group of order $2$. Jul 20, 2021 at 4:15