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.
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 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.
$\DeclareMathOperator\Sp{Sp}$Derek's answer is definitive. I think, at least for $p> 3$ odd, we can also see that the answer is no by considering the irreducible characters of $\Sp(2m,p)$. The smallest non-trivial irreducible character degrees are $\frac{p^{m}-1}{2}$ and $\frac{p^{m}+1}{2}$, coming from the so-called Weil representations. One of these (the one whose degree is odd) contains the central involution in its kernel, and the other represents the central involution by $-I$. Neither of these characters is rational-valued, and all other non-trivial irreducible characters (not algebraically conjugate to either of these) have degree at least $p^{m}-1$. On the other hand, the symmetric group $S_{p^{m}-1}$ has a faithful irreducible rational valued character of degree $p^{m}-2$, while $\Sp(2m,p)$ has no faithful rational-valued character (irreducible or not) of degree $p^{m}-2$.
