# Determine the galois group of a quartic

I'm reading Hungerford's algebra chapter about galois theory. There is the following theorem in p.273 (with some minor changes) about determining the Galois group of a quartic:

Let $K$ be a field and $f\in K[X]$ an (irreducible) separable quartic with Galois group $G$ (Considered as a subgroup of $S_4$). We may assume $f$ is monic and $f=(X-u_1)(X-u_2)(X-u_3)(X-u_4)$ in some extension of $K$. We define $V=\{1,(12)(34),(13)(24),(14)(23)\}$(a normal subgroup of $S_4$) and $\alpha=u_1u_2+u_3u_4,\beta=u_1u_3+u_2u_4,\gamma=u_1u_4+u_2u_3$ (so that $\alpha,\beta,\gamma$ are the roots of the resolvant cubic. We let $m=[K(\alpha,\beta,\gamma):K]$. Then:

1. $m=6\iff G=S_4$
2. $m=3\iff G=A_4$
3. $m=1\iff G=V$
4. $m=2\iff G\approx D_8$ or $G\approx Z_4$ ($D_8$ is the dihedral group); in this case $G\approx D_8$ if $f$ is irreducible over $K(\alpha,\beta,\gamma)$ and $G\approx Z_4$ otherwise.

Implications $\implies$ are left to the reader. I can get 1,2 and 3 but I don't know how to use the fact that $f$ is or is not irreducible in $K(\alpha,\beta,\gamma)$. I also know what are the possible options for $G$ so I need to get $f$ irreducible $\implies |G|=8$ and $f$ reducible $\implies |G|=4$.

• The action of the Galois group on the roots of the resolvent is the action of (subgroups of ) $S_4$ on the 3 partitions of $\{1,2,3,4\}$ in two sets of order 2, i.e. images in the factor group $S_3$ of $S_4$. Nor go through the transitive subgroups of $S_4$ and check what these images are. – ahulpke May 21 '15 at 17:52