# 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 $24$

• $S_4$ 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 $6$

• 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 $4$

• 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 $2$

• 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..

• For what it's worth, you have correctly identified all the possible groups of an irreducible quartic. – Gerry Myerson May 17 '14 at 13:18
• @GerryMyerson : Thank you for seeing through the text... I am not sure why would $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$ would not occur as subgroup of $S_4$ and why would only subgroup of $S_4$ of order $12$ is $A_4$... though it is discussed in below answer i would be happy to hear some comment of you on this... – user87543 May 17 '14 at 15:01
• @GerryMyerson : please have a look at this and help me... – user87543 May 20 '14 at 10:45
• There is no argument for order 12 better than the one Mark Bennet gives --- I urge you to study it and understand it, or ask a question about it. – Gerry Myerson May 20 '14 at 11:32
• @GerryMyerson : Ok... I will ask him... :) – user87543 May 20 '14 at 11:47

## 2 Answers

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 transitive subgroup has to have order divisible by the number of elements permuted, in this case 4." sound so nice than what i have said though both represents same idea.... I do not understand second paragraph... can you please edit just that part.. – user87543 May 17 '14 at 16:41
• @PraphullaKoushik I have filled out the detail, which I hope will help you follow. – Mark Bennet May 17 '14 at 18:15
• Thank you for your interest... I am actually getting confused... all i wanted to see is that any subgroup of $S_4$ can not be $\mathbb{Z}_4\times \mathbb{Z}_2$ or $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$... :O – user87543 May 19 '14 at 11:00
• your explanation of why $A_4$ is the only subgroup of order $12$ in $S_4$ is very clear and i am very happy with that... Thank you so much :) – user87543 May 19 '14 at 11:05
• I understand "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" but not the very next step.. "and the union of them consists of complete conjugacy classes" why does it contain? Let $H=\cup H_{2i}$ union of all sylow subgroups... $gHg^{-1}=\cup gH_{2i}g^{-1}$ I know any two sylow subgroups are conjugate but i am not able to relate it here.... please help me – user87543 May 20 '14 at 13:33

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