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I think the answers for the first few degrees ($n$) are:

$n=2$, $S_2$

$n=3$, $S_3,A_3$

$n=4$, $S_4,A_4,D_4,\mathbb{Z}_4,K_4$ ($K_4$ is the Klein four group)

$n=5$, $S_5,A_5,D_5,\mathbb{Z}_5,Fr_5$ ($Fr_5$ is a Frobenius group)

What results do we have for higher orders and are there any results for a general $n$?

Edit: Sorry, I have aptly demonstrated that I am still rough on some of my concepts here. I believe I was thinking about the case where $f(x)\in \mathbb{Z}[x]$ and the base field is a finite extension of the rational numbers. I hope I have gotten it right this time and the above answers are right for this situation.

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Surely, the group has to be transitive (that is, as a subgroup of $S_n$, has to act transitively on {1,..,n}). By the way, what is K4 in your list? – Joël Nov 19 '10 at 21:18
Yes, because it acts transitively on the roots. $K_4$ is the Klein group/Klein-four group/Vierergruppe. – Timothy Wagner Nov 19 '10 at 21:25
Gerry---if you can prove that you've solved the inverse Galois problem! – Kevin Buzzard Nov 19 '10 at 22:23
@Timothy: are you asking about a particular base field or over an arbitrary base field? – Qiaochu Yuan Nov 19 '10 at 22:35
Your edit concerns a very artificial situation. If the base field is an extension of the rationals, then why is $f$ required to be defined over $\mathbb{Z}$ and not over the ring of integers of that extension? – Alex B. Nov 20 '10 at 23:05

Let $G$ be a transitive subgroup of $S_n$. Let $L=\mathbb{C}(X_1,\dots,X_n)$. The group $G$ acts on $L$ by permutations of the variables. Let $K=L^{G}$. Then $L/K$ is Galois of Galois group $G$. Let $P$ be the minimal polynomial of $X_1$ over $G$. It is irreducible. It has degree at least $n$ because all the $X_i$ are Galois conjugate of $X_1$ over $K$, by transitivity of $G$. On the other hand, it divides $(X-X_1)\dots(X-X_n)$ which has coefficients in $L^{S_n} \subset K$. Hence $P$ has degree $n$. The splitting field of $P$ over $K$ is $L$ obviously, so its Galois group is $G$.

Conclusion: the subgroups of $S_n$ that are the Galois group of an irreducible polynomial are the transitive ones.

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If $L/K$ is any Galois extension with Galois group $G$, and $G$ has a faithful transitive permutation representation $\pi: G \rightarrow S_n$, then there exists an irreducible polynomial $p(x)$ of degree $n$ whose splitting field is $L$, and whose Galois group $G$ acts on the roots of the polynomial via $\pi$. This is the construction: Let $H$ denote the stabilizer of $1$, and let $L^{H} = K(\alpha)$ (primitive element theorem). Then the minimal polynomial $p(x)$ of $\alpha$ will work.

Thus your problem is essentially equivalent to the inverse Galois problem:

This problem is hard, although many cases are known (for example, $G$ solvable, $S_n$, $A_n$, many other simple groups).

If one allows $K$ to vary, of course, then one can easily find (by Hilbert) an $L/\mathbf{Q}$ with Galois group $S_n$, and hence a $K \subset L$ with $G = \mathrm{Gal}(L/K)$ any subgroup of $S_n$, and hence any group (making $n$ sufficiently large).

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I am not sure I am following you (or Kevin in his comment on the main question): what the relation between the question (where no base field is mentioned) and the inverse Galois problem, which is over Q? – Joël Nov 19 '10 at 23:33
Gerry's comment mentioned the rationals. I was just flagging this. – Kevin Buzzard Nov 19 '10 at 23:56
Hi Beaujolais Nouveau. I plan to visit the IAS during the Spring, at least two weeks, maybe more. But I wonder: who are you? I have googled and mathscineted "Beaujolais Nouveau" but there is no number theorist under that name. But perhaps you want to stay anonymous; then a subtle mathematical evidence leading informed people to your real identity will do. – Joël Nov 20 '10 at 2:10
@Joël: that is not surprising: – Qiaochu Yuan Nov 20 '10 at 22:56

This question is ambiguous and can be understood in various ways, since no mention is made of the base field. If you are asking "for what subgroups $G$ of $S_n$ do there exist fields $L,K$ such that $L/K$ is a Galois with Galois group $G$", then Joël has given a complete answer. If you fix the base field $K$ and ask what groups can arise as Galois groups over $K$, then the question can be anything between trivial and open, depending on $K$. Here are are some examples of the spectrum of hardness:

  • If the base field is finite, then of course finite $G$ can be realised as a Galois group if and only if $G$ is cyclic.
  • If $K$ is a local field with finite residue field, then a necessary condition for $G$ to be realisable as a Galois group over $K$ is that $G$ be solvable. There are other restrictions as well, coming from the filtration on $G$ by higher ramification groups. However, not all finite solvable groups can be realised over a given local field. A fun exercise (if you know about the structure of higher ramification groups) is: classify all primes $p$ and $q$ such that the dihedral group $D_{2q}$ is realisable as a Galois group over the field $\mathbb{Q_p}$. Actually, the absolute Galois groups of local fields are fairly well understood. They are certain pro-solvable groups explicitly given by generators and relations (see Neukirch's book on Cohomology of Number fields). So in principle, given any finite group, you could try checking whether it's a quotient of the absolute Galois group (which might still be very difficult).
  • If $K$ is a number field, then the question you are posing is a famous open problem. Shafarevich proved that all solvable finite groups are Galois groups over $\mathbb{Q}$ and people are chipping away at special cases of insoluble groups, but so far, no uniform picture is apparent.

Note, that a generalisation of the Inverse Galois Problem over a given field, which should help answering the inverse Galois problem over all finite extensions of your given field, is the so-called Embedding problem.

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