Is the following embedding possible?


$\mathrm{Sp}\_{2m}(p)\leqslant S\_{p^m-1}$ where $S\_{p^m-1}$ is a symmetric group and $p$ is prime. I see that when $p=3$ and $m=3$, the order of the former does divide the order of the latter. I was also thinking along the lines of $\mathrm{PSp}$ being simple except for finite cases but haven't gotten an answer. I feel that it is not possible, though.


Thank you.



is the embedding $\mathrm{Sp}_{2m}(p)\leqslant S_{p^m-1}$ possible?
No. The smallest degrees of the faithful permutation representations of the finite simple groups are listed in Table 4.5 of [On the maximum orders of elements of finite almost simple groups and primitive permutation groups](https://arxiv.org/abs/1301.5166) by Guest, Morris, Praeger, and Spiga.


$\DeclareMathOperator\PSp{PSp}\DeclareMathOperator\Sp{Sp}$With a couple of small exceptions, for $\PSp(2m,q)$ with $m \ge 2$, the minimum degree is $(q^{2m}-1)/(q-1)$ for $q \ne 2$ and $2^{m-1}(2^m-1)$ when $q=2$.


So $\Sp(2m,2)$ has no proper subgroup of index less than this, and so it cannot have a faithful permutation representation of smaller degree.


I think the minimum degree for $\Sp(2m,q)$ is probably $q^{2m}-1$ for $q$ odd.


