$S_3 \times 2^2 \cdot PSU_6 (2) \overset ? \subset 2^2 \cdot {}^2 E_6 (2)$

Question

Does $2^2 \cdot {}^2 E_6 (2)$ have a conjugacy class of maximal subgroups isomorphic to $S_3 \times 2^2 \cdot PSU_6 (2)$?

Answer 1

Yes. Here is a sketch why. The Monster $F_1$ contains a four-group $V$ such that $C=C_{F_1}(V)\cong [2^2]\,{^2}\!E_6(2)$ and $N=N_{F_1}(V)$ satisfies $N/C\cong S_3$. $C$ contains a subgroup $T\cong Z_3$ such that $N_{C/V}(TV/V)\cong S_3\times PSU_6(2)$. The issue that you raise is the structure of the preimage $L$ in $C$ of the $PSU_6(2)$ direct factor. Using a $3$-element of $N-C$ that centralizes $T$, one sees that either $L$ splits over $V$ or $L$ is quasisimple, i.e., isomorphic to $[2^2]PSU_6(2)$. It therefore suffices to show that $L$ does not split over $V$.

This can be seen in $C_T:=C_{F_1}(T)$. Since $|C_T|_2\ge |L|_2=2^{17}$, $C_T$ is isomorphic to the $3$-fold cover of $Fi_{24}'$. If $L$ did split over $V$, then $Fi_{24}'$ would have a four-subgroup $V_1$ such that $C_{Fi_{24}'}(V_1)\cong Z_2\times Z_2\times PSU_6(2)$. Write $V_1=\langle v,w\rangle$. Then $C_{Fi_{24}'}(v)\cong 2Fi_{22}2$. But $Fi_{22}2\cong Aut(Fi_{22})$ does not contain $Z_2\times PSU_6(2)$. (It does, of course, contain $2U_6(2)$.)

The relevant information about local subgroups of $F_1$, $Fi_{24}'$, and $Fi_{22}$, as well as the desired maximality, can be found in the Atlas of Finite Groups by Conway, Curtis, Norton, Parker, and Wilson.