# 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 at 0:33

PROBLEM 1. Given an abelian $p$-group $G$, present an overgroup of $G$ of order $p|G|$ of maximal possible class. Consider in detail the following cases: (i) $G$ is homocyclic (i.e., direct product of cyclic subgroups of equal order), (ii) all invariants of $G$ are pairwise distinct.

PROBLEM 2. Classify the nonabelian groups $G$ of exponent $p$ that are not covered by nonabelian $p$-groups of order $p$. (For example, if $G$ is of order $p^4$, it is not covered by nonabelian subgroups of order $p^3$.)

-
Is this an answer to the question? – Gerry Myerson Jun 20 at 22:59
Again, answers in the form of Socratic questions (if that what this "problem" is) are less than optimal. Please give straightforward answers until you are more familiar with how the site generally works. – Todd Trimble Jun 21 at 0:34
The abelian $p$-group of type $(p^{k_1},p^{k_2},\dots,p^{k_n})$, where $k_i= [n/i]$ for all $i\le n$, is the unique abelian $p$-group of minimal order containing all abelian $p$-groups of orders $\le p^n$. Thus, its oerder is equal to $p^N$, where $N=\sum_{i=1}^n[n/i]$ (Moshe Roitman, Univ. of Haifa). It is interesting to find the minimal order of a $p$-group containing all abelian subgroups of orders $\le p^n$. – yakov Jul 5 at 11:04