I wonder if it is known that there are $\textit{infinitely}$ many number fields $F$ (up to isomorphism) with fixed degree $[F:\mathbb{Q}]=n$ and fixed a transitive group $G$ of $S_n$ such that $G=\textrm{Gal}(F^c/\mathbb{Q})$ (if we assume inverse Galois holds for $G$).

Obviously, there are finitely many number fields $F$ if we make a restriction for $F$ such that $|d_{F}|<M$ for some $M>0$. This is from Hermite's theorem.

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    $\begingroup$ My guess is that this is an open question, but you may get a better response on Mathoverflow. $\endgroup$ – Mathmo123 Oct 25 '18 at 6:51
  • $\begingroup$ @Mathmo123 Thank you for your suggestion. $\endgroup$ – Thomas Oct 26 '18 at 1:06

Basically, no.

One comparison is with (a special case of) Dirichlet's Theorem. For example, given an integer $n$, suppose you know there exists a prime $p \equiv 1 \bmod n$, then you can ask whether there exists infinitely many primes $p \equiv 1 \bmod n$? There is not really any mechanism to go from one to the other.

Of course, there is a variant. If you knew that there existed at least one prime $p \equiv 1 \bmod n$ for all $n$ then you can deduce that there are infinitely many primes. Proof: Given primes $p_1, \ldots, p_k$ which are $1 \bmod n$, we are guaranteed one prime $p \equiv 1 \mod p_1 p_2 \ldots p_k n$. Certainly $p \ne p_i$ for any $i$, so we have a new prime $p \equiv 1 \bmod n$.

The same thing is true in your case. Clearly if you knew there was at least one extension with Galois group $G$ for any $G$, you can find infinitely many with Galois group $G$. (Consider fields with Galois group $G \times G \times \ldots G$ for larger and larger products of $G$.)

Note the analogy with Dirichet's theorem is stronger than one might expect: Finding infinitely many fields with Galois group $\mathbf{Z}/n \mathbf{Z}$ is exactly the same as finding infinitely many primes $p \equiv 1 \bmod n$.

Now there are a few cheap tricks in some cases. Suppose we knew the inverse Galois problem for a class of groups (say abelian groups, which is true). And suppose that $G$ has the property that there is a surjection

$$\phi: G \times A \rightarrow G$$

for some abelian group $A$ for which the image of $A$ is non-trivial. Then given one Galois extension $F$ with Galois group $G$ one can find infinitely many --- by assumption there are infinitely many fields $E$ with Galois group $A$, and then using $\phi$ we can find a subfield of the compositum $E.F$ which has Galois group $G$ and is not isomorphic to $F$ (using basic Galois theory). For an example, let $G$ be any group with a non-trivial center. Then there is a map

$$\phi: G \times Z \rightarrow G$$

sending $(g,z)$ to $gz$. So this works if $Z(G) \ne 1$. This is not unrelated to the following example. Let $E$ be an elliptic curve over $\mathbf{Q}$, and suppose that the Galois extension of $\mathbf{Q}(E[p])$ has Galois group $G \subset \mathrm{GL}_2(\mathbf{F}_p)$. If $p > 2$, the center of $G$ has an element of order $2$. Hence there is a map

$$\phi: \mathrm{GL}_2(\mathbf{F}_p) \times \mathbf{Z}/2 \mathbf{Z} \rightarrow \mathrm{GL}_2(\mathbf{F}_p)$$

as above, and one gets a different such extension for each quadratic field. In fact, the corresponding extension also comes from another elliptic curve, namely the twist of $E$ by the quadratic character of the quadratic extension giving rise to the degree $2$ extension.

OTOH, if $G$ is simple, then the only way to get non-trivial maps from $G \times A \rightarrow G$ is for $A$ to surject onto $G$. (p.s. this last construction is known as "crossing" of extensions, and first turned up in Cebotarev's proof of his theorem)

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    $\begingroup$ This is a really great answer! $\endgroup$ – Mathmo123 Oct 25 '18 at 22:01
  • $\begingroup$ @user608470 Thank you for this careful explanation first! I wonder how to know when you get any $G$-extension number field $F$ in $G\times G\times\cdots$-extension, you will $\textit{not}$ get isomorphic number fields $F$ in your construction. So, you can definitely get infinitely many non-isomorphic fields $F$. Sorry maybe this is dumb. $\endgroup$ – Thomas Oct 26 '18 at 1:24
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    $\begingroup$ @Thomas It follows from the Galois correspondence. Two subfields will be isomorphic if and only if the correspondence subgroups are conjugate. $\endgroup$ – Mathmo123 Oct 26 '18 at 7:35
  • $\begingroup$ @Mathmo123 Got you! $\endgroup$ – Thomas Oct 27 '18 at 8:37

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