On reducing degree-$12$ equations with Mathieu group $M_{11}$ to its degree-$11$ resolvent?

Question

The general $4$-deg and some $8$-deg (such as the Schein octic) when a linear transformation is done so their $x^{n-1}$ term vanishes can have a neat solution as,

$$x = \sqrt{z_1}+\sqrt{z_2}+\sqrt{z_3}$$

$$x = \sqrt{z_1}+\sqrt{z_2}+\dots+\sqrt{z_7}$$

where the $z_i$ are the roots of their cubic and septic resolvents, respectively. Of course, the sign of the square root is chosen appropriately. At first, I assumed this is only doable when the resolvent $2^m-1$ is a Mersenne prime.

However, inspecting $12$-deg, it seems it is also doable if its Galois group is the smallest sporadic group, the Mathieu group $M_{11}$, of order 7920. (Or, in Magma notation, $12T272$.) For example, the $12$-deg,

$$x^{12} - 21x^{10} - 20x^9 + 210x^8 + 12x^7 - 670x^6 + 108x^5 + 1305x^4 - 940x^3 - 189x^2 - 120x + 4 = 0$$

its $11$-deg resolvent (a $11T6$),

$$z^{11} - 126 z^{10} + 4815 z^9 - 36180 z^8 - 680625 z^7 - 1853982 z^6 + 3094497 z^5 + 280910160 z^4 + 1168901280 z^3 - 16329427200 z^2 + 30682457856 z -119664^2 = 0$$

and the two are related by,

$$x_k = \frac{\sqrt{z_1}+\sqrt{z_2}+\dots+\sqrt{z_{11}}}6$$

analogous to the ones for $4$-deg and $8$-deg.

Question: So, given a $12$-deg whose Galois group is the Mathieu group $M_{11}(12)$. If needed, do a linear transformation such that the $x^{11}$ term vanishes, hence the sum of the roots equals zero. Then is it true there is an ordering of its roots $x_k$ such that,
$$z=(x_1+x_2+x_3+x_4+x_5+x_6)^2 = (x_7+x_8+x_9+x_{10}+x_{11}+x_{12})^2$$ and $z$ is a root of an $11$-deg equation?

Answer 1

The answer is yes: $M_{11}$ in its action of degree $12$ has a subgroup of index $11$ (some people call it $M_{10}$) such that there is a $6$-element subset $A$ of the $12$ points which $M_{11}$ acts on such that $A^g=A$ or $A\cap A^g=\emptyset$ for all $g\in M_{10}$. From that the claim follows immediately.