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I don't believe this statement, so I think I'm making a mistake. I've asked ~10 people, and nobody found a mistake in the proof. Here it is.

Let $L/K$ be a Galois extension with Galois group $S_n$. (There are various ways to get such Galois extensions, the most common one uses indeterminates. See Constructing a Galois extension field with Galois group $S_n$.)

Since $L/K$ is a finite and separable extension, there's a primitive element $\alpha$ such that $L = K(\alpha)$.

Then, the minimal polynomial of $\alpha$ has degree $n!$.

The Galois group acts transitively on the roots of the polynomial, and therefore $S_n$ acts transitively on a set of size $n!$.

If this is true, every element of $S_n$ should act differently on every element of the set $\{1,2,...,n!\}$. Even if such an action exists, it's highly counterintuitive.

Edit: @coiso pointed out that $S_n$ acting on itself is transitive and is such an action, which makes sense. I just never put 2 and 2 together and noticed that this means that $S_n$ is a transitive subgroup of $S_{n!}$.

Boran Erol
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    How are you defining "transitive subgroup"? – Danny Duberstein Jun 04 '24 at 22:05
  • Your definition would require a surjection from a group of size n! to one of (n!)! ... The galois theory example says nothing about acting on S_{n!} – Danny Duberstein Jun 04 '24 at 22:30
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    Where is this action on $S_{n!}$ coming from? Are you assuming a group acting transitively on $k$ objects acts transitively on $S_k$? – Steve D Jun 04 '24 at 22:32
  • @MikeDaas Since $L/K$ has group $S_n$, the roots are in bijection with $n!$ not $n$. – coiso Jun 04 '24 at 22:58
  • OP, the group $S_n$ is acting on itself transitively (indeed, regularly). It is not acting on $S_{n!}$. Indeed, this action means we identify $S_n$ with a transitive subgroup of $S_{n!}$, but again the action is on $S_n$ not $S_{n!}$. – coiso Jun 04 '24 at 22:59
  • @coiso That makes sense. It's just super counter-intuitive to me that $S_n$ is a transitive subgroup of $S_{n!}$. Can this action be made explicit? – Boran Erol Jun 04 '24 at 23:06
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    Yes. As Cayley's theorem states, any group $G$ acts on itself (hence is identified with a transitive subgroup of $S_G$) by left multiplication. That is, there is a map $G\to S_G$ given by $g\mapsto L_g$, where $L_g\in S_G$ is defined by $L_g(x)=gx$. (One can also get a left action via right multiplication using $R_{g^{-1}}$, or speak of right actions instead.) In this case, $G=S_n$. – coiso Jun 04 '24 at 23:09
  • @coiso Last question, does this mean that $S_{n-1}$ is a transitive subgroup of $S_{n}$? Can this also be proved? Or, would $S_n$ be a transitive subgroup of $S_m$ forall $m \leq n!$ ? – Boran Erol Jun 04 '24 at 23:17
  • Let's be clearer. Let $G=S_n$ act on itself. This yields a copy $H$ of $G$ within $S_G$. This is a transitive subgroup, i.e. $H$ acts on $S_n$ transitively. The subgroup $S_{n-1}$ of $S_n$ corresponds to some subgroup $K<H$ within $S_G$. This $K$ never acts transitively: indeed, the orbit of $e\in S_n$ is the proper subset $S_{n-1}$. For a subset $X\subseteq G$ we can let $S_X$ be the subgroup of $S_G$ fixing points of $G\setminus X$: not only is $H$ not a transitive subgroup of $S_X$ for any proper subset $X$, it isn't even a subgroup of any such $S_X$. Do you see why? – coiso Jun 04 '24 at 23:26
  • I think you misunderstood my question. Your response shows that $S_{n-1}$ is not a transitive subgroup of $S_{n!}$. I asked whether $S_{n-1}$ is a transitive subgroup of $S_{n}$. I see why the first one would be false. – Boran Erol Jun 05 '24 at 00:14
  • For the question "is $S_{n-1}$ a transitive subgroup of $S_n$" to make sense you will need to specify which group action on which set you are considering – Mike Daas Jun 05 '24 at 10:15

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