Explicit permutation representation of the Schur double cover of the symmetric group

Question

Main question: How can we describe the double covers $(2\cdot\mathfrak{S}_n)^+$ and $(2\cdot\mathfrak{S}_n)^-$ of the symmetric group $\mathfrak{S}_n$ as permutation groups? (I.e., what sort of set with combinatorial structure do they act faithfully upon?)

I am not necessarily asking for the minimal permutation degree (I think this is an open problem; also see PS at bottom), but for some explicit combinatorial construction, reasonably uniform in $n$, that is hopefully reasonably small (perhaps merely exponential in $n$).

For a long time, I wrongly believed that one could form such a permutation representation of degree $2^{\lfloor n/2\rfloor}$ (or perhaps twice or four times that, or thereabouts) by taking an appropriate basis in the basic spin representation (and their negatives, and perhaps also times the imaginary unit), but I now realize that this does not work (at least, not as I thought it would). So I wonder if something else can be done.

The following question may is closely related to the above (trying to lift some $\mathfrak{S}_n$-set $X$ to a $(2\cdot \mathfrak{S}_n)$-set $\tilde X$ consisting of "signed" elements of $X$):

Alternate question: Given a set $X$ on which $\mathfrak{S}_n$ acts faithfully and transitively, with stabilizer $H$, is there some way to detect (merely by looking at $X$) whether $H$ lifts in $2\cdot \mathfrak{S}_n$ to a direct product $\{\pm 1\}\times H$, or equivalently (see below), whether the wreath product $1 \to \{\pm 1\}^X \to \{\pm 1\}\wr_X\mathfrak{S}_n \to \mathfrak{S}_n\to 1$ contains a non-split extension $1 \to \{\pm 1\} \to 2\cdot \mathfrak{S}_n \to \mathfrak{S}_n\to 1$?

Indeed, if $2\cdot \mathfrak{S}_n$ acts transitively on a set $\tilde X$, with stabilizer $H$, and the action does not factor through $\mathfrak{S}_n$, then $H$ intersects $\{\pm 1\}$ (center of $2\cdot\mathfrak{S}_n$) trivially, so maps isomorphically to a subgroup of $\mathfrak{S}_n$, which we can also call $H$, and the latter lifts to $2\cdot\mathfrak{S}_n$ as a direct product $\{\pm 1\}\times H$ (and we can see $\tilde X = (2\cdot\mathfrak{S}_n)/H$ as a double covering of $X = \mathfrak{S}_n/H$). But then, it follows from Derek Holt's paper "Embeddings of Group Extensions into Wreath Products" (Quarter. J. Math. Oxford 29 (1978) 463–468), theorem 2, that the wreath product contains our $2\cdot \mathfrak{S}_n$. Conversely, when this is the case, $2\cdot \mathfrak{S}_n$ acts faithfully and transitively (through the wreath product) on $\tilde X := \{\pm 1\}\times X$.

For example, if $H$ is generated by an $n$-cycle, and either $n$ is odd or we $n\equiv 2\pmod{4}$ if we are considering $(2\cdot\mathfrak{S}_n)^+$, we get a permutation action of degree $2(n-1)!$ on the set $\mathfrak{S}_n/H$ of cyclic orders with a sign added. But this isn't a great improvement over the regular action ($2n!$) and it's not very explicit, so I'm hoping we can do better.

PS: If my computations with Gap are correct, the minimal permutation representation for (Gap's choice) $(2\cdot\mathfrak{S}_n)^-$ is $16,48,80,240,240,480$ for $n=4,5,6,7,8,9$. This is not in the OEIS, I wonder if it's worth adding.

PPS: The corresponding values for $(2\cdot\mathfrak{S}_n)^+$ are $8,40,80,240,480,480$ (also not in the OEIS), so apparently they're not the same. (The trick I found to represent $(2\cdot\mathfrak{S}_n)^+$ under Gap is to multiply by a fourth root of unity the generators of $(2\cdot\mathfrak{S}_n)^-$ that lift an odd permutation.)

Answer 1

I hope you are correct that minimal degree is an open question, as I am currently working on the related question of minimal degree for $2\cdot A_n$, but as this embeds into $(2\cdot S_n)^\pm$ with index $2$ I can give you a permutation representation which I believe to have degree at most $4$ times the minimal degree (for large enough $n$).

For a particularly good permutation representation assume $n$ is even and take the natural embedding $A_\frac{n}{2}\times A_\frac{n}{2}\hookrightarrow S_n$ or $A_\frac{n}{2}\times A_\frac{n}{2}\hookrightarrow S_{n+1}$ (to deal with odd case) and let $H$ be the subdirect product of $A_\frac{n}{2}\times A_\frac{n}{2}$ consisting of all pairs $(\sigma,\sigma)$ with $\sigma\in A_\frac{n}{2}$. The preimage of $H$ is $S_n$, or $S_{n+1}$ is isomorphic to $C_2\times H$. Hence we may identify $H$ as a subgroup of $2\cdot S_n$ or $2\cdot S_{n+1}$. The action of $G$ on the cosets of $H$ is faithful of degree $4n!/\frac{n}{2}!$ or $4(n+1)!/\frac{n}{2}!$.

It's not a particularly nice combinatorial structure (or at least I can't identify it as one), but it's what I've used to construct $2\cdot A_n$ as a permutation group in MAGMA.

Explanation:

As the socle $Z$ of $(2\cdot S_n)^\pm$ is a simple group, it has a minimal permutation representation which is transitive and is therefore the coset action of $(2\cdot S_n)^\pm$ on one of its subgroups. In fact, one can show that any minimal permutation representation of $(2\cdot S_n)^\pm$ is given by the coset action of $(2\cdot S_n)^\pm$ on one of its largest core-free subgroups, where a subgroup is core-free if it contains no normal subgroup of $(2\cdot S_n)^\pm$. In this case a subgroup $H$ of $(2\cdot S_n)^\pm$ is core-free if and only if it intersects $Z$ trivially. It can therefore be identified with it's image in $S_n$.

The choice of $H$ I've given is a particularly good one, though when $n$ is divisible by $8$ you may take $H\times C_2$ where $C_2$ 'swaps' the two copies of $A_n$.

It's a bit of work, but you can show $H$ is core-free using the finite presentation of $2\cdot S_n$. The main restriction is that whenever an element of order $2$ appears in $H$ it has to be a product of a multiple of $4$ transpositions. Hence two copies of $A_n$. Anything significantly larger than $A_\frac{n}{2}$ will end up having an element of order $2$ which does not satisfy this, so you really can't get much better than this.

Answer 2

According to my calculations, the minimal permutation degrees of $2.A_n$ for $n=5,\ldots, 24$ are as follows. In each case $H$ is a core-free subgroup of $2.A_n$ of largest order. The minimal permutation degree of both versions of $2.S_n$ is at most twice this amount, and I think in each case it is exactly twice it, but I haven't tried to check that. I have only gone up to $n=24$ because that is as far as I have got with exact computations (mainly in Magma). I may be able to do a couple more values of $n$ but I would need to think harder to get much further.

G    H                           |H|             |G:H|
2A5  5                           5               24
2A6  3^2                         9               80
2A7  7.3                         21              240
2A8  L27 or 2^3.7.3              168             240
2A9  L28.3                       1512            240
2A10 L28.3                       1512            2400
2A11 M11                         7920            5040
2A12 M11                         7920            60480
2A13 M11                         7920            786240
2A14 M11x3                       23760           3669120
2A15 M11x3                       23760           55036800
2A16 L27^2.2 or (2^3.7^3)^2.2    56448           370656000
2A17 (L27 or 2^3.7.3) x L28.3    254016          1400256000
2A18 (L28.3)^2                   2286144         2800512000
2A19 (L28.3)^2                   2286144         53209728000
2A20 M11 x L28.3                 11975040        203164416000
2A21 M11 x L28.3                 11975040        4266452736000
2A22 M11^2                       62726400        17919101491200
2A23 M11^2                       62726400        412139334297600
2A24 S12                         479001600       1295295050649600