This question arose in observations made while trying to answer Intersecting cubic equations and roots.. Consider a (depressed) cubic polynomial $P(x) = x^3 + ax + b$, either over $\mathbb{Q}$ with $a, b \in \mathbb{Q}$, or over $\mathbb{Q}(a, b)$ (or over $\overline{\mathbb{Q}}(a, b)$ or $\mathbb{C}(a, b)$ if that's easier). Then the set $\{ c \mid P(P(c)) = c, P(c) \ne c \}$ is given by the roots of the polynomial $$Q(x) = \frac{P(P(x)) – x}{P(x) – x} = x^6 + (2 a + 1) x^4 + 2 b x^3 + (a^2 + a + 1) x^2 + (2 a + 1) b x + a + b^2 + 1$$
@dxiv's answer to that question shows that $Q$ becomes a cubic in $z^2$ under the Tschirnhaus transformation $z = P(x) – x$, in particular its Galois group is always solvable. For $a, b \in \mathbb{Z}$ with $|a|, |b| \le 100$ the following observations hold:
- If $Q$ is irreducible then so is $R(x) = x^3 + (a+2)x + b$ (a previous version of the question incorrectly also claimed the converse). @dxiv's comment notes that the Tschirnhaus transformation $z = P(x) + x$ converts $Q(x)$ to $R(-z)^2$, which seems relvant.
- Let $G = \operatorname{Gal}(Q)$ and $H = \operatorname{Gal}(R)$ over $\mathbb{Q}$ when $Q$ is irreducible. Then $G$ is one of: $C_2 \times S_4$, $C_2 \times A_4$, $S_4$ and $D_6 = C_2 \times S_3$, with orders $48$, $24$, $24$ and $12$ respectively.
- If $G = C_2 \times S_4$ then $H = S_3$.
Apparently $C_2 \times S_4$ is the unique transitive subgroup of $S_6$ of order $48$, though I am not sure if it's unique up to conjugacy. From these observations, and my informal impression of Hilbert's irreducibility type arguments, I expect that the Galois group of $Q$ over $\mathbb{Q}(a, b)$ is $C_2 \times S_4$, and the smaller groups occur as further transitive subgroups.
My broad informal question is, what is going on? To ask a single precise question: is the Galois group of $Q$ over $\mathbb{Q}(a, b)$ isomorphic to $C_2 \times S_4$?
Related/follow-up questions that I might ask separately if not answered here:
- Can the irreducibility implication be inferred from the Tschirnhaus transformations? Is there an a priori reason to expect such a polynomial of degree smaller than $Q$ to exist?
- (How) does any of this generalize to $P$ of higher degrees and higher order periods (ie $P^n(c) = c$)? In particular, what construction yields $C_2 \times S_4$ (or whatever the correct Galois group is) from the parameters $(3, 2)$?
Best Answer
Let $L$ be the splitting field of $Q(x)$ over $\mathbf{Q}(a,b)$. If $\alpha \in L$ is a root of $Q(x)$, then $P(\alpha)$ is also a root of $Q(x)$. Thus the roots of $Q(x)$ are divided up into three pairs:
$$\alpha, P(\alpha), \beta, P(\beta), \gamma, P(\gamma).$$
These are distinct because $Q(x)$ has non-zero discriminant and so is separable. Any automorphism sending $\alpha$ to $\sigma \alpha$ sends $P(\alpha)$ to $P(\sigma \alpha)$, and thus must preserve these three pairs. This puts a restriction on the Galois group $G = \mathrm{Gal}(L/\mathbf{Q}(a,b)) \subset S_6$, namely that it lies inside $\Gamma \subset S_6$, the maximal subgroup of $S_6$ preserving this decomposition. We shall see that this group is $S_4 \times S_2$.
So far this all generalizes to roots of $P^{(n)}(x) = x$ with $P(x)$ general of degree $d$; the roots of the corresponding polynomial $Q(x)$ will come in packets of $n$ roots which are iterates under $\alpha \rightarrow P(\alpha)$, and since $\sigma P(\alpha) = P(\sigma \alpha)$, the Galois group will commute with this automorphism, and thus give a restriction on the Galois group as lying inside the centralizer of this permutation.
Here are some more details in the case $(d,n) = (d,2)$. We have $P^{(2)}(x) = x$ and $Q(x)$ has degree $d^2 - d$, the maximal group $\Gamma$ will preserve the $m:=(d^2-d)/2$ pairs. The largest such group naturally surjects to $S_m$ which describes how it acts on the $m$ pairs. The kernel consists of elements which just permute each pair which is the group $(\mathbf{Z}/2 \mathbf{Z})^m$. This is a group of order $2^m \cdot m!$ and is usually denoted
$$ S_2 \wr S_m =(S_2)^m \ltimes S_m.$$
The symbol on the LHS is the wreath product and this group is called $S_2$ wreath $S_m$; it is defined/identified with the semidirect product preserving $m$ pairs of elements described above. If $\alpha$ is a root of $Q(x)$, the quotient $S_m$ corresponds to considering the Galois group of $\theta = \alpha + P(\alpha)$. You already noticed this when $m = 3$. If you take a random example with $d = 4$, say $P(x)=x^4+x+1$, you will find the Galois group of $Q(x)$ has order $2^6 \cdot 6! = 46080$, whereas if $d = 5$, say $P(x)=x^5+x+1$, it will have order $2^{10} 10! = 3715891200$. In neither case will it be solvable because the groups surject onto $S_6$ and $S_{10}$ respectively.
So far we have seen that $G$ is contained within this group $\Gamma$, but it is also equal to this group, since the group only gets smaller under specialization and one can compute at least one example with $a,b \in \mathbf{Q}$ for which this extension has Galois group $S_4 \times S_2$.
When $d=3$ and so $m=3$ something additional and special happens, namely that there is an isomorphism
$$S_2 \wr S_3 \simeq S_2 \times S_4.$$
This is an exceptional isomorphism which has no analog for higher $d$. You can see that $\Gamma \simeq S_4 \times S_2$ in a number of ways:
Label the pairs of roots which sum to zero as $(1,6)$, $(2,5)$, and $(3,4)$, the condition of being in $\Gamma$ is equivalent to commuting with $\sigma = (1,6)(2,5)(3,4)$. This is because
$$g \sigma g^{-1} = (g(1),g(6))(g(2),g(5))(g(3),g(4)).$$
So it suffices to show that the centralizer of $\sigma$ is $S_4 \times S_2$.
The order of the centralizer is the stabilizer of $G$ acting on $\sigma$ by conjugation. Since $\sigma$ has $15$ conjugates, the order of the stabilizer is $6!/15 = 48$ by the Orbit-Stabilizer theorem. Alternatively the previous description of $\Gamma$ shows that its order is $2^3 \cdot 3! = 48$.
Claim: The centralizer of $\sigma = (16)(25)(34) \in S_6$ is isomorphic to $S_4 \times S_2$.
Proof One: There is an outer automorphism of $S_6$ sending $\sigma$ to a $2$-cycle. The centralizer of $(1,2)$ certainly contains $S_2 \times S_4$ with the first $S_2$ generated by $(1,2)$ and the second $S_4$ acting on $3,4,5,6$. This has order $48$ so we are done.
As a remark, any $S_4 \times S_2$ inside $S_6$ has to normalize an element of order $2$. There are three conjugacy classes of order $2$ elements; $(**)$, $(**)(**)$, and $(**)(**)(**)$. The normalizer of the second conjugacy class has order $6!/45 = 16$, but the first two give the two conjugacy classes of $S_4 \times S_2 \subset S_6$, one that acts transitively and the other that doesn't, but which are interchanged by the exotic automorphism of $S_6$.
Proof Two: Let $C \subset \mathrm{SO}(3)$ denote the group of rotations that preserve the cube (the cube group), with faces labelled in the usual way. It's well known that $C \simeq S_4$ by the action on the $4$ diagonals of the cube. Now $C$ also acts faithfully on the faces, which realizes $S_4 = C$ as a subgroup of $S_6$. Moreover the image commutes with $\sigma$, because $C$ preserves the pairs of opposite faces of the cube. Finally, the reflection $-I \notin C$ both commutes with $C$ and maps to $\sigma$. The group $S_4 \times S_2 \simeq C \times \langle -I \rangle \subset \mathrm{O}(3)$ acts faithfully on the six faces of the cube and also commutes with $\sigma$, and so its image in $S_6$ is isomorphic to $S_4 \times S_2$ and is the centralizer of $\sigma$.