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One notable family of examples is $G = A_n^m$, where $m$ is chosen as large as possible subject to $G$ being $2$-generated. It is well-known that $m \sim n! / 4$, so the order of $G$ is more than exponential in $n!$ but its exponent is roughly $\exp(\sqrt{n \log n})$. Similar analysis applies to direct powers of any simple group.
@daon You are applying that theorem incorrectly. If $X$ is a random variable with law $\mu$ then $Z = f(X)$ has law $f_*\mu = \mu \circ f^{-1}$. Maybe it is helpful to point out that $z = \exp(-x)$ implies $dz = -\exp(-x) dx = -z \, dx$, so $dx = - dz/z$.
If $n$ is even, $\mathrm{SU}(n)$ contains a copy of $S_{n+1}$ as well as a copy of $\mathrm{SL}_2(5)^{n/2} \rtimes S_{n/2}$ (since $\mathrm{SL}_2(5)$ embeds in $\mathrm{SU}(2)$). The former group has no abelian normal subgroup at all, while the latter group has a normal abelian subgroup of index $60^{n/2} (n/2)!$. I believe the point about $n = 70$ is just that $(n+1)! > 60^{n/2} (n/2)!$ for $n > 70$ but not for $n \le 70$.
Clearly not: that is the point of my previous comment. On the other hand if you restrict to rational entries the situation changes. See mathoverflow.net/questions/15127/…
I am not sure why there is such an emphasis (both in this question and its papertrail) on two presents. The question is simpler and just as "mind-boggling" with one present. In this case the connection with intransitive dice is even closer.
For example if $G$ is $S_3$ acting regularly on $6$ points. The question is quite unclear to me. For $H$ do you mean the product action? Note that a transitive permutation group always has a primitive quotient acting on a system of maximal blocks, but there's no guarantee it's nonabelian.
By induction it is equivalent to ask whether every finite nonsolvable group has an exact factorization (i.e., factorization with unique representation) into some number of proper subgroups. Actually this also holds for solvable groups apart from some $p$-groups, by considering Hall subgroups, so one can replace "nonsolvable" by "not a $p$-group". This seems to be unknown for some of the largest sporadic simple groups.
The notion of limit of finitely generated groups was refined in the work of van den Dries and Wilkie. See sciencedirect.com/science/article/pii/0021869384902230. To apply this in the case of $S_N$ you would need to choose generators.
