It is well-known that the symmetric group S4 has two Schur covering groups, S4-tilde and S4-hat. There are explicit presentations for both groups, and we know that S4-hat is isomorphic to GL(2,3). Question: Is there an alternative, explicit description/realization of S4-tilde?


The lowest degree d for which S4-tilde can be realized as a transitive permutation group is d=16. Is there a concrete (geometric?) construction of this permutation action? (other than via coset action...)


According to wiki, S4-tilde can be embedded inside GL(2,9). What is a good way to "see" this (geometric?) action? (transitive, I presume?)


Last but not least: Is there a "geometric" way to think about/realize the two Schur covers of S\_n for general n? And what the six-fold covers of A\_n for n=6,7?


Thanks!



Schur covering group for S4
What you call $\tilde{S}\_{4}$ is also known as the binary octahedral group. Probably the easiest way to construct it is to take the natural matrix representation of $G = {\rm GL}(2,3)$ and leave the elements of ${\rm SL}(2,3)$ as they are, but replace the extra generator $\left(\begin{array}{clcr} 1 & 0\\0& -1 \end{array} \right)$ by the scalar multiple $\left(\begin{array}{clcr} i & 0\\0& -i \end{array} \right),$ where $i$ is a primitive $4$-th root of unity in ${\rm GF}(9).$ This does not change $G/Z(G)$, but the new group $\tilde{G} \subseteq {\rm SL}(2,9)$ has a generalized quaternion Sylow $2$-subgroup with only one element of order $2$.


Later edit: The general picture for double covers of $S\_{n}$ (as constructed by Schur himself) is similar. In one, transpositions become involutions in the double cover, and in the other, they become elements of order $4$.


