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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

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  • Write $u=$"$UG$". Exercise: $u(p)=pu(p-1)$ for $p$ prime; $u(4)=24$, $u(6)=120=6!/6$. $u(8)$, $u(9)$ and $u(10)$ should be not too hard to determine. – YCor Dec 25 '19 at 23:06
  • At least for small cases, this question basically boils down to this question:https://mathoverflow.net/questions/121719/richness-of-the-subgroup-structure-of-p-groups – verret Dec 26 '19 at 00:02
  • All the groups of order $8$ embed in a group of order $32$ and no smaller. It easily follows that $u(8)=3235*7$. Also this sequence is not in the OEIS. – verret Dec 26 '19 at 03:43
  • I think it's then not hard to show that $u(9)=u(10)=2^53^35*7$. Clearly it can be no smaller. To construct an example, start with $C_7\times C_5\times C_9\times C_3$, then make your group of order $32$ act on this, where an index $2$ subgroup centralises everything, and something outside the index $2$ subgroup acts by inversion. – verret Dec 26 '19 at 03:47
  • I didn't actually check carefully, but I wouldn't be surprised if $u(12)=u(11)$. It fact, it seems plausible that $u(n)$ is just the product of $u(p^m)$ where $p$ ranges over the primes less or equal to $n$ and $p^m$ is the largest power of $p$ less or equal to $n$, in which case the question reduces to the prime power case (which seems hard anyway, according to the discussion on MO). – verret Dec 26 '19 at 03:51
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    See this paper where $UG(n)$ is determined for $1\leq n\leq 15$. Perhaps of particular interest to @verret is that $u(11)\neq u(12)$. (It turns out that $u(12) = 2u(11)$ but $u(13)=u(14)=u(15) = 13u(12)$.) – James Dec 27 '19 at 13:05
  • @James You could probably make that an answer, as I doubt we're going to get much better than that. – verret Dec 27 '19 at 19:08
  • @verret Okay. I've added some summary notes which I hope will improve the quality of the "answer". – James Dec 29 '19 at 06:44

1 Answers1

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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 $n$.)

James
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