Galois group of irreducible Quartic polynomial over $\mathbb{Q}$ Actual Question is  : 

What are all possible galois groups of an irreducible Quartic polynomial over $\mathbb{Q}$

As polynomial is irreducible, Galois group is transitive subgroup of $S_4$.
I have no idea what (actually why) are all transitive subgroups of $S_4$.
Firstly, I would distinguish what are all possible subgroups (and then see for transitive subgroups) of $S_4$.
As $|S_4|=24$ only possible orders of subgroups of $S_4$ are $\{1,2,3,4,6,8,12,24\}$
Subgroup of order $24$


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*$S_4$ is subgroup of order $24$ in $S_4$ (Trivial)


Subgroup of order $12$


*

*I have proved sometime back that $A_4$ is the only subgroup of $S_4$ having order $12$ but i could not recall the proof now (I would be thankful if some one can give some hint). 


Subgroups of order $8$


*

*As $|S_4|=2^3.3$, there does exist a sylow $2$ subgroup i.e., subgroup of order $8$ in $S_4$. Only groups (upto isomorphism) of order $8$ are : $\mathbb{Z}_8,\mathbb{Z}_4\times \mathbb{Z}_2, \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2, D_8,Q_8$.


*

*$\mathbb{Z}_8$ can not be a subgroup of  $S_4$ as $S_4$ does not have an element of order $8$.

*With lot of labor work i could see that no element of order $4$ commutes with an element of order $2$ thus $\mathbb{Z}_4\times \mathbb{Z}_2$ can not be subgroup of $S_4$.(Help needed)

*As of now i can not conclude why  (I know it is but not why) $ \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ is a subgroup of $S_4$. (Help needed)

*I know $D_8$ can be seen as subgroup of $S_4$.

*I know $Q_8$ can not be a subgroup of $S_4$.
Subgroups of order $6$


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*Only groups of order $6$ are $\mathbb{Z}_6$ and $S_3$.


*

*$\mathbb{Z}_6$ can not be subgroup of $S_4$ as $S_4$ do not have an element of order $6$.

*$S_3$ is subgroup of $S_4$. This is not transitive because  : Suppose $S_3$ fixes $4$ then there is no way i can get an element which maps say $3$ to $4$. Thus $S_3$ is not transitive subgroup of $S_4$.. So, this $S_3$ is out of this game.



Subgroups of order $4$


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*Only subgroups of order $4$ are $\mathbb{Z}_4$ and $\mathbb{Z}_2\times \mathbb{Z}_2$.


*

*$\mathbb{Z}_4$ is a subgroup of $S_4$ as $\{Id, (1234),(13)(24),(1432)\}$ which can be easily seen to be transitive.. 


In $\mathbb{Z}_4$,  i can send  $1$ to $2$ with $(1234)$ ; $1$ to $3$ with $(13)(24)$ and $1$ to $4$ with $(1432)$ similarly i can other elements to every othe element.. So, $\mathbb{Z}_4$ is seen to be transitive subgroup of $S_4$.

*$\mathbb{Z}_2\times \mathbb{Z}_2$ is a subgroup of $S_4$ seen as $\{Id, (12)(34),(13)(24),(14)(23)\}$ . In this also i can send each element to all other elements and thus $\mathbb{Z}_2\times \mathbb{Z}_2$ is a transitive subgroup of $S_4$.
Subgroups of order $3$


*

*No subgroup of order $3$ can be transitive for the same reason (actually for a more simple reason) as why $S_3$ is not transitive.


Subgroups of order $2$


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*No subgroup of order $2$ can be transitive for the same reason (actually for a more simple reason)as why $S_3$ is not transitive.


Subgroups of order $1$


*

*Trivial subgroup is not transitive.


So, Only subgroups of importance (transitive subgroups) i am worried about are.... 


*

*$S_4$ which is trivially transitive 

*$A_4$ is transitive.

*$D_8$ is  transitive. 

*$\mathbb{Z}_4$ is  transitive. 

*$\mathbb{Z}_2\times \mathbb{Z}_2$ is  transitive. 


I would be thankful if some one can help me to make this a bit more clear and simple..
 A: A transitive subgroup has to have order divisible by the number of elements $n$ permuted, in this case $n=4$. Not all such subgroups need be transitive. 
The first part is because of the orbit-stabiliser theorem - we know that the orbit has size $n$ and the order of the group is then $n \times |S|$ where $S$ is the stabiliser of an element, hence is divisible by $n$.
People don't always know or use the full versions of the Sylow Theorems, which say for a group $G$ of order $p^nq$ with $p$ prime, $(p,q)=1$ and $n\ge 1$, that there is a subgroup of order $p^n$, that all the subgroups of order $p^n$ in $G$ are conjugate to each other, that the number of such subgroups is congruent to $1$ modulo $p$ and is a factor of $q$ (and is equal to the index of the normaliser of the subgroup in $G$ - which is just the orbit-stabiliser result again).
For the subgroup of order $8$ you can use that all Sylow $2$-Subgroups are conjugate to each other. Together these subgroups contain every element whose order is a power of $2$, and the union of them consists of complete conjugacy classes. Since the conjugacy classes in a symmetric group are determined by cycle type, and all the subgroups of order $8$ are conjugate to each other, each group of order $8$ must contain at least one representative of each conjugacy class of elements of order $2$ or $4$ (there are no elements of order $8$) - these are types $(1234), (12)(34), (12)$
So you need a $4$-cycle and note that its square is a product of two transpositions. And the subgroup also contains a transposition. The product of a transposition with a $4$-cycle is an even permutation, which is either a $3$-cycle or a product of two transpositions (the identity is impossible). A group of order $8$ cannot contain a $3$-cycle, so you need to generate a product of two transpositions.
This double transposition will be different from the square of the 4-cycle, so it is easy to see that all three such elements will be contained in the subgroup (since each one is the product of the other two). Thus the group contains a the identity, a $4$-cycle and its inverse, the three elements of type $(12)(34)$ - and therefore two transpositions, which can be computed by multiplying the $4$-cycle by the double transposition elements.
You can then determine the type of this group. Because it contains a $4$-cycle any such group is guaranteed to be transitive. Let $v$ be the four cycle and $d$ a double transposition with $v^2\neq d$, then $vd$ is a transposition and $vdvd=id$ whence $v^4=d^2=id$ and $dvd=v^{-1}$ and you get the dihedral group.
Note that every subgroup of order $4$ is a subgroup of a Sylow subgroup of order $8$ - which is another way of identifying possibles (less good in this case). Also note that $\{id , (12), (34), (12)(34)\}$ is a subgroup of order $4$ which is not transitive.
For the subgroup of order $12$, note that a subgroup of index $2$ is always normal, and therefore consists of whole conjugacy classes:
Id - $1$ element
$(12)$ - $6$ transpositions
$(12)(34)$ - $3$ products of two transpositions
$(123)$ - $8$ $3$-cycles
$(1234)$ - $6$ $4$-cycles 
$A_4$ is the only possibility, because a conjugacy class of size $6$ won't fit.
A: All i wanted to show is that :


*

*$S_4$can not have a subgroup of order $8$ of the form $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ 

*Any subgroup of order $12$ in $S_4$ is $A_4$...
Half of the credit goes to Gerry Myerson and the other half to Mark Bennet...
Now, I want to first prove that  any element in that $S_4$can not have a subgroup of order $8$ of the form $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$..
any element of $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ has to be either a transposition or product of two transposition...
I can not have a transposition in that subgroup because it commutes with only 3 non identity elements where as it has to commute with 7 non identity elements for it to be in that group...
I can not form a subgroup of order $8$ with only three elements :D (I am talking about remaining elements $(12)(34),(13)(24),(14)(23)$.. so, I am done :D :D :D 
Credits for this half is for Gerry Myerson..
Now i want to prove that Any subgroup of order $12$ in $S_4$ is $A_4$
Any subgroup of order $12$ in $S_4$ is normal..
Any normal group must contain whole conjugacy class... 
Representatives (and cardinality) of conjugacy classes are 
Id - $1$ element
$(12)$ - $6$ transpositions
$(12)(34)$ - $3$ products of two transpositions
$(123)$ - $8$ $3$-cycles
$(1234)$ - $6$ $4$-cycles 
Now, this subgroup $H$ of $S_4$ can not have conjugacy class having  $6$ elements(for that to be in $H$ we need another conjugacy class of $5$ elements which is not possible)
Thus, $H$ can not no element of the form $(12)$ or $(1234)$  
There is a possibility of $H$ to consist of a conjugacy class of cardinality $8$ as there is another conjugacy class of cardinality $3$ adding upto $11$ elements and including with identity we get the whole group.
Thus $H=\{Id, Cl\{(123)\}, Cl\{(12)(34)\}\}=A_4$
Credits for this half goes to Mark Bennet
