What is the minimal possible order of an $n$-universal group? Suppose $G$ is a finite group. We call $G$ $n$-universal iff any group $H$, such that $|H| \leq n$ is isomorphic to some subgroup of $G$. Here are some examples of universal groups:

$S_n$ is $n$-universal.

Proof of this fact (usually known as Cayley theorem) can be found in any group-theory textbook.

Suppose $p$ is a prime and $n \in \mathbb{N}$, that satisfies the conditions:

*

*$p^n+1$ is composite


*If $p = 2$, then $n \geq 4$
Then $S_{p^n}$ is $p^n + 1$ universal

That follows from Cayley theorem and the fact that all incompressible groups are $p$-groups.

If $G$ is $n$-universal, then it is $(n-1)$-universal.

Trivially follows from the definition
Let's define $UG(n)$ as the minimal possible order of an $n$-universal group.

Does there exist some explicit formula (or at least asymptotic) for $UG(n)$?

I only managed to prove the three following facts:

$UG(n)$ is monotonously non-decreasing

Follows from the third example

$UG(n) \leq n!$

Follows from Cayley theorem

$UG(n) \geq e^{\psi(n)}$, where $\psi$ stands for the Second Chebyshev function

Follows from Lagrange theorem
 A: This 2017 paper by Heffernan, MacHale and McCann determines $UG(n)$ for $1\leq n\leq 15$.  Their numeric results appear in their Table 5 at the end of the paper, just before the references.
In addition to those numeric values, they determine the groups of order $UG(n)$ showing that, in particular, the group is not necessarily unique.  For $6\leq n\leq 15$, it is smaller than the symmetric group $S_n$.
Along the same lines that @verret had discussed in the comments, they prove an interesting lower bound.  If $p_1, p_2, \ldots, p_m$ are the primes not exceeding $n$ and if, for each $i$, $p_i^{k_i}$ is the largest power of $p_i$ less than or equal to $n$, then a group $G$ that embeds all groups of order at most $n$ is divisible by $p_1^{2k_1 - 1}p_2^{2k_2 - 1}\cdots p_m^{2k_m - 1}$.  (Their Lemma 2.)
Also of interest is their Lemma 6 which asserts that this lower bound is not achieved for $p^k$ with $p$ an odd prime and $k\geq 3$.  (The order of a group must be divisible by at least $p^6$ to embed all groups of order  $p^3$.)
(They also consider the question of the minimal order of a group that embeds all groups of order $n$ (again, for $1\leq n\leq 15$), but not necessarily all groups of order at most $n$.)
