Solution to why that nonic is solvable

Previously, we claimed that for any quadruple $a,b,c,d$ of rational integers, not all zero, the nonic polynomial $F(x) = x^9 + a x^8 + b x^7 + c x^6 + d x^5 - (126 - 56 a + 21 b - 6 c + d)x^4$ $- (84 - 28 a + 7 b - c)x^3 - (36 - 8a + c)x^2 - (9 - a)x - 1$

is solvable, meaning that it is possible to determine the roots of $F$ explicitly by radicals. By Galois theory, this is equivalent to the assertion that the Galois group of the Galois closure of $F$ is a solvable group.

To do this, we need the following fact, which was proved by Bhargava and Yang in this paper as Theorem 4. The statement of their theorem is correct, and the proof is mostly correct, but there is a minor issue. The problem is that the stabilizer of $F$ under $\text{GL}_2(\mathbb{C})$, which we will denote by $\text{Aut}_\mathbb{C} F$, need not be realizable as a subgroup of the Galois group of $F$, which we will denote by $\text{Gal} (F)$. However, the argument they gave for the commuting action between elements of $\text{Aut}_\mathbb{C} F$ and $\text{Gal}(F)$ is correct.

We now consider a binary form $F$ of the shape given above. We see that both $\text{Aut}_\mathbb{C} F$ and $\text{Gal}(F)$ act on the roots of $F$ and can therefore be embedded via their action on the roots of $F$ into $S_9$, the symmetric group on nine letters. If we restrict to $\text{GL}_2(\mathbb{Q})$ action and denote by $\text{Aut} F = \text{Aut}_\mathbb{Q} F$, then it follows from Galois theory that $\text{Gal} (F)$ must be a subgroup of the centralizer of any element in $\text{Aut} F$ in $S_9$.

We then check that, miraculously, $\text{Aut} F$ always contains the following element of order 3: $U = \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}.$

We check that the only complex numbers fixed by this element are the roots of $x^2 + x + 1$. Therefore, if $F$ is irreducible, then no root of $F$ can be fixed by $U$. Relabelling the roots if necessary, we can assume that $U$ can be realized in $S_9$ as $U = (123)(456)(789).$

The centralizer $C(U)$ of $U$ is the stabilizer of $U$ under the action by conjugation of $S_9$ on itself. The orbit of $U$ is precisely the set of elements in $S_9$ of the same cycle type, therefore the orbit contains $\displaystyle \binom{9}{3}(2!) \binom{6}{3}(2!) (2!) \frac{1}{3!} = 2240.$

By the orbit-stabilizer theorem, it follows that $C(U)$ contains $9!/2240 = 162 = 2 \times 3^4$ elements. Since $\text{Gal} (F)$ is contained in $C(U)$, it is solvable by Burnside’s theorem, which asserts that any finite group whose order is only divisible by two distinct primes is solvable.