# Quartic Equation having Galois Group as $S_4$

Let $$f(x)\in \mathbb{Z}[x]$$ be an irreducible quartic polynomial with $$S_4$$ as Galois Group. Let $$\theta$$ be a root of $$f(x)$$ and set $$K=\mathbb{Q}(\theta)$$. Now, the question is:

Prove that $$K$$ is an extension of $$\mathbb{Q}$$ of degree 4 which has no proper subfields.

Are there any Galois extensions of $$\mathbb{Q}$$ of degree 4 with no proper subfields?

As I have adjoined a root of an irreducible quartic, I can see that $$K$$ is of degree $$4$$ over $$\mathbb{Q}$$.

But, why does there is no proper subfield of $$K$$ containing $$\mathbb{Q}$$?

Suppose $$L$$ is proper subfield of $$K$$, then $$L$$ has to be of degree $$2$$ over $$\mathbb{Q}$$. So, $$L$$ is Galois over $$\mathbb{Q}$$, i.e., $$L$$ is normal. So the corresponding subgroup of Galois group has to be normal.

I tried working in this way but could not able to conclude anything from this.

Any help/suggestion would be appreciated.

Thank You

• You're on the way. The group of the splitting field over $K$ has to be normal of order 6 in the group of the splitting field over $L$, which has to be normal of order 12 in $S_4$. Now, what do you know about subgroups of $S_4$? Commented Aug 10, 2013 at 7:21
• yes, so, corresponding subgroup of order 12 is $A_4$ but, it is normal. So i do not get any contradiction with this :(
– user87543
Commented Aug 10, 2013 at 7:26
• Keep going. You haven't used the "normal of order 6" part. Commented Aug 10, 2013 at 7:28
• There you go. Now, you can write it up and post is as an answer. Commented Aug 10, 2013 at 7:33
• @GerryMyerson Thank You Sir. :)
– user87543
Commented Aug 10, 2013 at 7:35

Convert the question to a problem in permutation groups.

Let $F$ be the splitting field of $$f(x)=(x-\theta_1)(x-\theta_2)(x-\theta_3)(x-\theta_4)$$ with $\theta=\theta_1$.

We were given that the Galois group realizes all the 24 permutations of the roots $\theta_i,i=1,2,3,4.$ Therefore $$\operatorname{Gal}(F/\mathbb{Q}(\theta))=\operatorname{Sym}(\{\theta_2,\theta_3,\theta_4\})$$ contains automorphisms realizing all the six permutations of the other roots.

Galois correspondence then means that the claim is equivalent to:

There are no subgroups $H$ properly between $\operatorname{Sym}(\{\theta_2,\theta_3,\theta_4\})$ and $\operatorname{Sym}(\{\theta_1,\theta_2,\theta_3,\theta_4\})$.

In other words, this is equivalent to proving that the obvious copy of $S_3$ inside $S_4$ is a maximal subgroup. Have you seen that? If not, can you prove it?

• I do not know, but for some reason, i am not able to view maps as elements of permutation group :( I have to work on that i guess
– user87543
Commented Aug 10, 2013 at 7:36
• No,the order of $S_{3}$ is $3! = 6.$ Commented Aug 10, 2013 at 8:04
• @Praphulla: If you know that $A_4$ is the only subgroup of index two in $S_4$, then you can use that simply by observing that $A_4$ does not have $S_3$ as a subgroup, because the latter contains odd permutations. Commented Aug 10, 2013 at 8:21
• More generally, isn't it true for all $n$ that $S_{n-1}$ is a maximal subgroup of $S_n$? Commented Aug 11, 2013 at 0:25
• Nothing wrong with what you posted, and thanks for confirming maximality of $S_{n-1}$. So this means that if $f$ is irreducible of degree $n$ with Galois group $S_n$, and $\alpha$ is a root of $f$ in some extension of the rationals, then the field generated by $\alpha$ has degree $n$ and no subfields but itself and the rationals. That's good to know. Commented Aug 11, 2013 at 9:55

As has been remarked, the non-existence of intermediate fields is equivalent to $S_{3}$ being a maximal subroup of $S_{4}.$ If not, then there is a subgroup $H$ of $S_{4}$ with $[S_{4}:H] = [H:S_{3}] = 2.$ Now $S_{3} \lhd H$ and $S_{3}$ contains all Sylow $3$-subgroup of $H.$ But $S_{3}$ has a unique Sylow $3$-subgroup, which is therefore normal in $H.$ Hence $H$ contains all Sylow $3$-subgroups of $S_{4}$ as $H \lhd S_{4}$ and $[S_{4}:H] = 2.$ Then since $H$ only has one Sylow $3$-subgroup, $S_{4}$ has only one Sylow $3$-subgroup, a contradiction (for example, $\langle (123) \rangle$ and $\langle (124) \rangle$ are different Sylow $3$-subgroups of $S_{4}$).

• More generally, isn't it true for all $n$ that $S_{n-1}$ is a maximal subgroup of $S_n$? Commented Aug 11, 2013 at 0:24
• That is true of course Commented Aug 13, 2013 at 1:44