# Richness of the subgroup structure of p-groups

Given a prime $p$ and $n \in \mathbb{N}$, let $f_p(n)$ be the smallest number such that there is a group of order $p^{f_p(n)}$ which all groups of order $p^n$ embed into. What is the asymptotic growth of $f_p(n)$ when $n$ tends to infinity?

The question asks in a certain sense for how dense $p$-groups can be "packed together" as subgroups of a larger group.

Let's give an example for illustration: By the bound by Francois Brunault, all groups of order $2^{20}$ embed into a group of order $2^{2^{20}-1}$, which is a number with 315653 decimal digits. On the other hand, by Nick Gill's bound, they do not embed into a group of order $2^{66}$, which is a 20-digit number. Can these bounds be refined?

Added on Feb 21, 2013: Even if finding precise asymptotics for $f_p(n)$ turns out to be delicate, isn't it at least possible to decide whether $f_p(n)$ grows polynomially or exponentially, or whether its growth rate lies somewhere in between? Or alternatively, are there reasons to believe that this is a difficult problem?

Added on Dec 4, 2013: The question whether it is true that $f_p(n)$ grows faster than polynomially but slower than exponentially when $n$ tends to infinity will appear as Problem 18.51 in:

Kourovka Notebook: Unsolved Problems in Group Theory. Editors V. D. Mazurov, E. I. Khukhro. 18th Edition, Novosibirsk 2014.

• All groups of order $p^n$ embed into the symmetric group on $p^n$ letters, whose p-Sylows have size $p^f$ with $f=v_p((p^n)!)$, which gives $f_p(n) \leq (p^n-1)/(p-1)$. – François Brunault Feb 13 '13 at 16:10
• I think you can rule out linear growth just by looking at abelian p-groups. I think the smallest group that will contain all such abelians is a direct product of cyclics of orders $p^n,p^{n/2},p^{n/3},p^{n/4}$,.... (you should take the floor of exponents of course). But now the size of this group is $p$ raised to $n\sum_{i=1}^n \frac{1}{i}$ and since that sum doesn't converge you can't have linear growth. Is that reasonable? – Nick Gill Feb 14 '13 at 10:39
• Stefan, I reckon a good way to get a better upper bound would be to consider (faithful) linear representations of $p$-groups over a field of char $p$. I guess there should be a result saying that every $p$-group of order $p^n$ admits such a representation of degree $\leq x$ where $x$ is God-knows-what. If such a result exists then the next tricky thing would be to show that such a representation can be realised over a finite field of bounded size... – Nick Gill Feb 15 '13 at 11:58
• The number of groups of order $p^n$ is $p^{\Theta(n^3)}$. A group of order $p^m$ can contain at most $p^{mn}$ groups of order $p^n$, since each is generated by at most $n$ elements. Thus $f_p(n)$ is at least $O(n^2)$. – Will Sawin Feb 17 '13 at 23:27
• The cyclic group of order $p^n$ has a faithful representation of degree $p^{n-1}+1$ in char $p$, but of no smaller degree. The reason is that a matrix of order $p^n$ in char $p$ has the form $1+N$, where $N$ is nilpotent, but $N^{p^{n-1}}\neq 0$. It follows that the God-knows-what-$x$ form Nick Gill's comment is at least $p^{n-1}+1$. So modular representation theory will not be of much help, I'm afraid. – Frieder Ladisch Mar 18 '13 at 18:34

The comment by Frieder Ladisch suggests to me that considering exponents may be relevant. Suppose that we generalize Stefan Kohl's function $f_p(n)$ as follows:

Definition: Fix a prime $p$ and an exponent $e$. Let $F(p,e,n)$ be the smallest integer such that there is a group of order $p^{F(p,e,n)}$ which contains isomorphic copies of every group of order $p^n$ and exponent $p^e$.

Lemma: Then $\max_{1\leq e\leq n} F(p,e,n)\leq f_p(n)\leq \sum_{e=1}^n F(p,e,n)$ for all $n\geq1$ and all primes $p$.

Proof: The upper bound is obtained by considering direct products, and the lower bound is easy.

Clearly $F(p,n,n)=n$ and $F(2,1,n)=n$ as a group order $p^n$ and exponent $p^e$ is cyclic if $n=e$, and is elementary abelian if $p=2$ and $e=1$. A very wild guess is that the asymptotic size of $f_p(n)$ as $n\to\infty$ is governed by $F(p,1,n)$ for $p>2$, and $F(2,2,n)$ for $p=2$.

It is unclear to me how helpful wreath products are. Suppose that $G(p,e,n)$ is a $p$-group that contains isomorphic copies of every group of order $p^n$ and exponent $p^e$. I claim (without proof) that the $p$-group $G(p,e_2,n_2)\;{\rm wr}\;G(p,e_1,n_1)$ contains isomorphic copies of every group of order $p^{n_1+n_2}$ and exponent $p^{e_1+e_2}$. This gives the upper-bound $$F(p,e_1+e_2,n_1+n_2)\leq F(p,e_2,n_2)p^{F(p,\,e_1,\,n_1)}+F(p,e_1,n_1).$$ In terms of the previous discussion, a Sylow $p$-subgroup of $S_{p^e}$ or ${\rm GL}(e+1,p)$ has exponent $e$.

• Am I right that your upper bound on $F(p,2e,2n)$ is larger than $p^{F(p,e,n)}$? -- If so, I think this gives hardly an improvement to what we already know. – Stefan Kohl Jan 2 '14 at 11:16
• The claim by Frieder that "modular representation theory will not be of much help" may not be correct because Nick Gill's dimension $x$ now depends on the exponent $e$. You are right that the wreath product bound is weak. – Glasby Jan 2 '14 at 13:44

Each finite $p$-group $H$ may be embedded in the Frattini subgroup of an appropriate $p$-group $G$ (for example, $G=H \wr C_p$; here $d(G)\le d(H)+1$. Then for this $G$, the Frattini subgroup $\Phi(G)$ contains a subgroup which is isomorphic to $H$.)

Problem. Is it true that one can choose $G$ so that $d(G)=2$?

Yakov

• What is the appropriate $p$-group, then? Also, please put dollar signs around your math (use the edit button). – Sebastian Goette Dec 19 '15 at 17:05
• If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. – Stefan Kohl Dec 19 '15 at 18:56
• I am not sure what appropriate means here. However, every $p$-group can be embedded in the Nottingham group (which is $2$-generated for $p>2$). Thus, can be embedded in a finite quotient of it. I am quite sure that embedding it in the Frattini subgroup is not an issue. This will also give some answer to the original question. – Yiftach Barnea Dec 19 '15 at 22:36
• If the "problem" is meant to be a Socratic question, then it would probably be less confusing if rephrased as a declarative sentence. – Todd Trimble Jun 21 '16 at 0:33