Galois group of a quartic polynomial
Consider a general (monic) quartic polynomial over $\mathbb{Q}$
$$f(x)={x}^{4}+a{x}^{3}+b{x}^{2}+cx+d$$ 
and denote the Galois group^{} of $f(x)$ by $G$.
The Galois group $G$ is isomorphic^{} to a subgroup^{} of ${S}_{4}$ (see the article on the Galois group of a cubic polynomial for a discussion of this question).
If the quartic splits into a linear factor and an irreducible cubic, then its Galois group is simply the Galois group of the cubic portion and thus is isomorphic to a subgroup of ${S}_{3}$ (embedded in ${S}_{4}$)  again, see the article on the Galois group of a cubic polynomial.
If it factors as two irreducible quadratics, then the splitting field^{} of $f(x)$ is the compositum of $\mathbb{Q}(\sqrt{{D}_{1}})$ and $\mathbb{Q}(\sqrt{{D}_{2}})$, where ${D}_{1}$ and ${D}_{2}$ are the discriminants of the two quadratics. This is either a biquadratic extension and thus has Galois group isomorphic to ${V}_{4}$, or else ${D}_{1}{D}_{2}$ is a square, and $\mathbb{Q}(\sqrt{{D}_{1}},\sqrt{{D}_{2}})=\mathbb{Q}(\sqrt{{D}_{1}})$ and the Galois group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.
This leaves us with the most interesting case, where $f(x)$ is irreducible. In this case, the Galois group acts transitively on the roots of $f(x)$, so it must be isomorphic to a transitive^{} (http://planetmath.org/GroupAction) subgroup of ${S}_{4}$. The transitive subgroups of ${S}_{4}$ are
${S}_{4}$  
${A}_{4}$  
${D}_{8}$  $\cong \{e,(1234),(13)(24),(1432),(12)(34),(14)(23),(13),(24)\}\text{and its conjugates}$  
${V}_{4}$  $\cong \{e,(12)(34),(13)(24),(14)(23)\}$  
$\mathbb{Z}/4\mathbb{Z}$  $\cong \{e,(1234),(13)(24),(1432)\}\text{and its conjugates}$ 
We will see that each of these transitive subgroups actually appears as the Galois group of some class of irreducible quartics.
The resolvent cubic^{} of $f(x)$ is
$$C(x)={x}^{3}2b{x}^{2}+({b}^{2}+ac4d)x+({c}^{2}+{a}^{2}dabc)$$ 
and has roots
$${r}_{1}=({\alpha}_{1}+{\alpha}_{2})({\alpha}_{3}+{\alpha}_{4})$$  
$${r}_{2}=({\alpha}_{1}+{\alpha}_{3})({\alpha}_{2}+{\alpha}_{4})$$  
$${r}_{3}=({\alpha}_{1}+{\alpha}_{4})({\alpha}_{2}+{\alpha}_{3})$$ 
But then a short computation shows that the discriminant $D$ of $C(x)$ is the same as the discriminant of $f(x)$. Also, since ${r}_{1},{r}_{2},{r}_{3}\in \mathbb{Q}({\alpha}_{1},{\alpha}_{2},{\alpha}_{3},{\alpha}_{4})$, it follows that the splitting field of $C(x)$ is a subfield^{} of the splitting field of $f(x)$ and thus that the Galois group of $C(x)$ is a quotient^{} of the Galois group of $f(x)$. There are four cases:

•
If $C(x)$ is irreducible, and $D$ is not a rational square, then $G$ does not fix $D$ and thus is not contained in ${A}_{4}$. But in this case, where $D$ is not a square, the Galois group of $C(x)$ is ${S}_{3}$, which has order $6$. The only subgroup of ${S}_{4}$ not contained in ${A}_{4}$ with order a multiple of $6$ (and thus capable of having a subgroup of index $6$) is ${S}_{4}$ itself, so in this case $G\cong {S}_{4}$.

•
If $C(x)$ is irreducible but $D$ is a rational square, then $G$ fixes $D$, so $G\le {A}_{4}$. In addition^{}, the Galois group of $C(x)$ is ${A}_{3}$, so $3$ divides the order of a transitive subgroup of ${A}_{4}$, which means that $G\cong {A}_{4}$ itself.

•
If $C(x)$ is reducible, suppose first that it splits completely in $\mathbb{Q}$. Then each of ${r}_{1},{r}_{2},{r}_{3}\in \mathbb{Q}$ and thus each element of $G$ fixes each ${r}_{i}$. Thus $G\cong {V}_{4}$.

•
Finally, if $C(x)$ splits into a linear factor and an irreducible quadratic, then one of the ${r}_{i}$, say ${r}_{2}$, is in $\mathbb{Q}$. Then $G$ fixes ${r}_{2}=({\alpha}_{1}+{\alpha}_{3})({\alpha}_{2}+{\alpha}_{4})$ but not ${r}_{1}$ or ${r}_{3}$. The only possibilities from among the transitive groups are then that $G\cong {D}_{8}$ or $G\cong \mathbb{Z}/4\mathbb{Z}$. In this case, the discriminant of the quadratic is not a rational square, but it is a rational square times $D$.
Now, $G\cap {A}_{4}$ fixes $\mathbb{Q}(\sqrt{D})$, since $G$ fixes $\sqrt{D}$ up to sign and ${A}_{4}$ restricts our attention to even permutations^{}. But $G:G\cap {A}_{4}=2$, so the fixed field of $G\cap {A}_{4}$ has dimension $2$ over $\mathbb{Q}$ and thus is exactly $\mathbb{Q}(\sqrt{D})$. If $G\cong {D}_{8}$, then $G\cap {A}_{4}\cong {V}_{4}$, while if $G\cong \mathbb{Z}/4\mathbb{Z}$, then $G\cap {A}_{4}\cong \mathbb{Z}/2\mathbb{Z}$; in the first case only, $G\cap {A}_{4}$ acts transitively on the roots of $f(x)$. Thus $G\cap {A}_{4}\cong {V}_{4}$ if and only if $f(x)$ is irreducible over $\mathbb{Q}(\sqrt{D})$.
So, in summary, for $f(x)$ irreducible, we have the following:
Condition  Galois group 

$C(x)$ irreducible, $D$ not a rational square  ${S}_{4}$ 
$C(x)$ irreducible, $D$ a rational square  ${A}_{4}$ 
$C(x)$ splits completely  ${V}_{4}$ 
$C(x)$ factors as linear times irreducible quadratic, $f(x)$ irreducible over $\mathbb{Q}(\sqrt{D})$  ${D}_{8}$ 
$C(x)$ factors as linear times irreducible quadratic, $f(x)$ reducible over $\mathbb{Q}(\sqrt{D})$  $\mathbb{Z}/4\mathbb{Z}$ 
References
 1 D.S. Dummit, R.M. Foote, Abstract Algebra, Wiley and Sons, 2004.
Title  Galois group of a quartic polynomial 

Canonical name  GaloisGroupOfAQuarticPolynomial 
Date of creation  20130322 17:41:35 
Last modified on  20130322 17:41:35 
Owner  rm50 (10146) 
Last modified by  rm50 (10146) 
Numerical id  7 
Author  rm50 (10146) 
Entry type  Topic 
Classification  msc 12D10 