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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 $L \supset K$?

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    $\begingroup$ Do you mean $L\subset K?$ Don’t see how a bigger field gives a smaller group. $\endgroup$ Jul 20, 2021 at 4:12
  • $\begingroup$ @ThomasAndrews I thought things were inclusion-reversing here. $\endgroup$ Jul 20, 2021 at 4:21
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    $\begingroup$ 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 $L = \mathbb Q$. $\endgroup$
    – Mathmo123
    Jul 20, 2021 at 12:20
  • $\begingroup$ Good question! Let's chat about this. $\endgroup$ Aug 24, 2021 at 0:04

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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.

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    $\begingroup$ 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 $2$. $\endgroup$ Jul 20, 2021 at 4:15

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