Let $F$ be a field and $f\in F[x]$ a polynomial that is irreducible, separable and of degree $4$.

How do I prove that there are not more than $5$ different possibilities for the Galois group Gal($f$) (not counting isomorphisms)?

I found multiple answers/sites that determine exactly what the Galois groups can be. But those were a little too advanced for me, since I just got introduced to Galois theory. I think that's also the reason why the question says 'not more than 5' instead of being more specific.
So maybe somebody can help?

Edit: @DietrichBurde has helped a lot here below by giving a source that says that Gal($f$) must be a transitive subgroup of $S_4$, of which there are only five. However, it's not clear for me why it must be a transitive subgroup of $S_4$.


A very detailed and elementary reference are the notes of Keith Conrad on Galois groups of cubics and quartics. In Section $3$ it is proved that we have exactly $5$ possibilities, for $Gal(f)$, namely the groups $S_4,A_4,D_4,C_4$ and $C_2\times C_2$, which are just the transitive subgroups of $S_4$. In Theorem $1.1$ it is proven that $f$ is irreducible if and only if $Gal(f)$ is a transitive subgroup of $S_4$. However, it is easy to see that there are at most $5$ transitive subgroups of $S_4$, see Section $3$.

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    $\begingroup$ The last sentence of this answer is probably what the OP needs since that's what implies the "not more than 5". $\endgroup$ – Ethan Bolker Apr 1 '16 at 13:29
  • $\begingroup$ I'm not exactly sure what discs and squares are in your source. Also where is it proven that Gal(f) must be a transitive subgroup of $S_4$? $\endgroup$ – user314965 Apr 1 '16 at 14:44
  • $\begingroup$ @DietrichBurde I think OP might be referring to the fact that Theorem $1.1$ is not, in fact, proved in the notes. $\endgroup$ – Ravi Apr 1 '16 at 22:28

It is not difficult to prove that if $f$ is irreducible over $F$, then $G=Gal(f)$ must act transitively on the roots of $f$. Briefly, suppose $\alpha$ is a root of $f$ in a splitting field $K$, and $\underline{\alpha}=G\cdot\alpha$ is the orbit of $\alpha$. Then $G$ fixes any symmetric polynomial in $\underline{\alpha}$ which shows that $e_\lambda(\underline{\alpha})\in F$ for every elementary symmetric function (since this is precisely the fixed point space for the action of $G$).

But, then $g(x)=\prod_{\beta\in G\cdot\alpha}(x-\beta)\in F[x]$ because when we multiply out the product, the coefficients are elementary symmetric functions in the elements of $G\cdot\alpha$.

Now, $g(x)$ is non-constant (because $g(\alpha)=0$) and $g(x)|f(x)$ (because all the roots of $g(x)$ in $K$ are roots of $f(x)$ in $K$). Hence, $g(x)=a f(x)$ for some nonzero $a\in F$. It now follows that $G\cdot\alpha$ must contain every root of $f$.


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