I thought of this question the other day and have not been able to get any traction on references or results along its lines, so I finally caved and decided to ask it here. I am no expert on Galois Theory or the Inverse Galois Problem (IGP), but have more interest in the representation theory side of this.

Motivating Question: For each finite group $G$ and each transitive permutation representation $\pi_G$, is there a polynomial $\mathfrak{p}$ whose Galois group over $\mathbb{Q}$ is precisely $G$ and for which the action of $G$ on the roots of $\mathfrak{p}$ is given by $\pi_G$?

This could be thought of as the Strong IGP over $\mathbb{Q}$ since the usual IGP could be true and yet this stronger form could be false.

For example, consider the symmetric group $S_3$. It has 3 nontrivial transitive permutation representations, of degrees 2,3, and 6.

$\bullet$ Its degree 2 transitive permutation representation is realized by the action of $S_3$ (really its quotient $\mathbb{Z}/2\mathbb{Z}$ in this case) on the roots of any irreducible quadratic.

$\bullet$ Its degree 3 transitive permutation representation is realized by the action of $S_3$ on the roots of any irreducible cubic with two complex roots.

$\bullet$ Its degree 6 transitive permutation representation is realized by the action of $S_3$ on the polynomial $X^6+3$.

Since the ordinary inverse Galois Problem is still not resolved over $\mathbb{Q}$, the strong form is not either. With that in mind, I have some questions:

Question 1:Is there a standard name used in the literature for the strong version of the IGP described above? Is the strong form studied at all, or are folks who care about the IGP content to just find $some$ polynomial realizing a given $G$?

Assuming that the answer to Q1 is affirmative:

Question 2: Are there any groups (whether or not they are currently known to be Galois groups over $\mathbb{Q}$) which have some transitive permutation representation known $not$ to be realizable over $\mathbb{Q}$?

Since most groups have multiple transitive permutation representations, an example of what Q2 seeks could be known without the status of IGP being known.

Question 3: Same as Question 2, but for an arbitrary ground field. For example, are counterexamples of the type sought in Q2 known over $\mathbb{C}(t)$ (where the IGP is already known to be true)? Conversely, is there some ground field (perhaps $\mathbb{C}(t)$ itself) for which an affirmative answer to the Strong IGP is known?

Any relevant references would also be appreciated.


1 Answer 1


Given a group $G$, the regular IGP for $G$ and the strong IGP for $G$ are equivalent. More precisely, if $L/K$ is a Galois extension with ${\rm Gal}(L/K) \cong G$, and $\pi_{G} : G \to S_{n}$ is a faithful, transitive, permutation representation, let $M$ be the fixed field of a stabilizer of a point. Choose $\beta \in M$ so that $M = K(\beta)$ and let $f(x)$ be the minimal polynomial of $\beta$ over $K$. Then, the action of $G$ on the roots of $f(x)$ is given precisely by $\pi_{G}(G)$. [This is almost the same as exercise 18.6 in Marty Isaacs "Algebra: A Graduate Course".]

As a consequence, there is no special name given for this strong form of the IGP. People do sometimes give this some attention (for example, Kluners and Malle's database of number fields includes many non-minimal degree representations - for example, degree $10$ fields of small discriminant whose Galois group is isomorphic to $A_{5}$).

  • $\begingroup$ You are starting with some field $L$ such that $Gal(L/K)\cong G$... so you get some faithful $\pi_G$ corresponding to this field. But how does this realize all the other faithful permutation representations of $G$ as the action of $G$ on various subfields of $L$? I feel like if the strong form is true, it should follow from some sort of construction analogous to taking subrepresentations of tensor products of representations (which is one standard way to obtain all faithfuls from one starting faithful in the representation setting). $\endgroup$
    – ARupinski
    May 29, 2014 at 0:39
  • $\begingroup$ In your case it sounds like you are taking an arbitrary $\pi_G$ and letting it act on $L$ and taking the fixed subfield of this action as your $M$; but how are we letting such an $\pi_G$ act on $L$ since a priori the degree of $L/K$ may not match the degree of $\pi_G$? So what aspect of your argument am I misinterpreting here? $\endgroup$
    – ARupinski
    May 29, 2014 at 0:43
  • $\begingroup$ Think of $G$ as an abstract group that acts on $L$. If $G$ has a faithful and transitive action on $n$ letters, a point stabilizer $H$ has index $n$ and trivial core. By the fundamental theorem of Galois theory, there is a subfield $K \subseteq M \subseteq L$ so that $Gal(L/M) = H$. Let $M = K(\beta_{1})$, and let $\beta_{2}$, $\ldots$, $\beta_{n}$ be the conjugates of $\beta_{1}$. We can see that the set of $g \in G$ with $g(\beta_{1}) = \beta_{i}$ is a left coset of $H$ in $G$ - so we get a pairing between left cosets of $H$ and conjugates of $\beta_{1}$. $\endgroup$ May 29, 2014 at 3:20
  • $\begingroup$ Because the faithful, transitive action $\pi_{G}$ is equivalent to the action of $G$ on the left cosets of $H$, we get the action we want if we pick the right subfield. $\endgroup$ May 29, 2014 at 3:22
  • $\begingroup$ Here's an example. Let $f(x) = x^6 + x^4 - 2x^2 - 1$. If $L$ is the splitting field of $f$, then $L/\mathbb{Q}$ has degree $12$ and $Gal(L/\mathbb{Q}) \cong A_{4}$. Because the Galois group has an index $4$ subgroup $H$, there is a field $\mathbb{Q} \subseteq M \subseteq L$ with $|M : \mathbb{Q}| = 4$ so if $M = \mathbb{Q}(\beta)$, the action on $\beta$ is the usual $A_{4}$ action. In fact, we can take $\beta$ to be a root of $x^{4} + 2x^{2} + 8x + 9$. $\endgroup$ May 29, 2014 at 3:30

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