TITLE: Is the size of the Galois group always $n$ factorial?
QUESTION [8 upvotes]: I am studying field theory, and I just started the chapter on Galois theory.
Since a Galois extension is the splitting field of some polynomial $p(x)$ and this polynomial have exactly $n$ roots in the extension field I think that every automorphism of the extension permutes the roots of $p(x)$ [and for every permutation there exist an automorphism].
So I deduced that that size of the Galois group, that is the number of automorphisms from the extension to itself that fix the field we started with, is $n!$ for some natural $n$.
Am I right ?

REPLY [5 votes]: Benjamin Lim's answer is great. I thought I would add an answer with infinitely many counterexamples. Let $a\geq 1$ be a natural number, and let $K$ be the splitting field over $\mathbf{Q}$ of 
$$x^3-ax^2-(a+3)x-1.$$
The discriminant of this polynomial is $(a^2+3a+9)^2$, a square, and the polynomial is irreducible over $\mathbf{Q}$ since it is a cubic without rational roots (the only possibilities for rational roots are $\pm 1$, and neither works). Therefore  $\operatorname{Gal}(K/\mathbf{Q})\cong \mathbf{Z}/3\mathbf{Z}$, instead of $S_3$, so the size of the Galois group is $3$, and not $3!=6$. These fields are called the "simplest cubic fields", and you can learn more about them in this article by Daniel Shanks.