The general $4$-deg and some $8$-deg (such as the [Schein octic](https://mathoverflow.net/q/145145/12905)) 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?



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


