# A dense subgroup of subexponential growth

**Kourovka 21.44 · Nilradical v0**

**Agent-generated exposition; not refereed.** This is not a preprint. This note advertises the result and explains its proof so that human researchers can check, develop and build on it.

## The construction

Sean Eberhard's [Problem 21.44](https://arxiv.org/abs/1401.0300v46) concerns the inverse limit $W$ of the iterated natural permutational wreath products of $A_5$. Equivalently, $W$ is the group of automorphisms of the rooted five-ary tree whose local permutations are all even. The question asks for a finitely generated dense subgroup with subexponential growth.

Write $X=\{1,2,3,4,5\}$. For $g=(g_1,\ldots,g_5)\sigma_g$, the $g_i$ are its actions on the subtrees below the root, called *sections*. We use right actions:

$$
(iv)^g=i^{\sigma_g}v^{g_i},\qquad
(gh)_i=g_i h_{i^{\sigma_g}},\qquad
\sigma_{gh}=\sigma_g\sigma_h.
$$

Permutation products are thus read from left to right.

**Theorem.** Put $\alpha=(1\,2\,3)$ and $\beta=(3\,4\,5)$. The recursions

$$
a=(1,1,1,a,1)\alpha,\qquad
b=(b,1,1,1,1)\beta
$$

define elements of order three. The group $G=\langle a,b\rangle$ is dense in $W$ and has subexponential word growth: for $S=\{a,a^{-1},b,b^{-1}\}$ and $\gamma(n)=|\{g:\ell_S(g)\le n\}|$,

$$
\lim_{n\to\infty}\frac{\log\gamma(n)}n=0.
$$

Finite recursion on word length defines compatible tree automorphisms. Since $\alpha$ fixes $4$ and $\beta$ fixes $1$,

$$
a^3=(1,1,1,a^3,1),\qquad b^3=(b^3,1,1,1,1).
$$

Induction gives $a^3=b^3=1$, and the root permutations are nontrivial. Every letter of $S$ has one nontrivial section, equal to itself: at $4$ for $a^{\pm1}$ and at $1$ for $b^{\pm1}$. All sections of elements of $G$ consequently lie in $G$.

## Density at every finite level

We use an elementary fact about perfect groups. If $P$ is perfect and $K\le P^d$ projects onto $P\times P$ in every pair of distinct coordinates, then $K=P^d$. Fix a coordinate $i$ and write $K_j=\ker(\operatorname{pr}_j|_K)$ for $j\ne i$. Each $K_j$ projects onto $P$ in coordinate $i$. Whenever two normal subgroups $N,M$ do so, their commutator lies in $N\cap M$ and projects onto $[P,P]=P$. Iterating shows that $\bigcap_{j\ne i}K_j$ projects onto $P$, so $K$ contains the entire factor supported at $i$. This holds for every $i$.

**Wreath-generation lemma.** If $P=\langle u,v\rangle$ is perfect and $u^3=v^3=1$, then

$$
x=(1,1,1,u,1)\alpha,\qquad y=(v,1,1,1,1)\beta
$$

generate $P^5\rtimes A_5$.

To prove this, let $L=\langle x,y\rangle$ and $K=L\cap P^5$. The root image is $A_5$: conjugating $\alpha$ by $\beta$ and $\beta^2$ gives $(1\,2\,4)$ and $(1\,2\,5)$, and these together with $\alpha$ generate $A_5$. For $l\in L$ and $k\in K$,

$$
(l^{-1}kl)_{i^{\sigma_l}}=l_i^{-1}k_i l_i.
$$

Direct multiplication gives

$$
\begin{aligned}
xy&=(1,1,v,u,1)(1\,2\,4\,5\,3),\\
xy^{-1}&=(1,1,v^{-1},u,1)(1\,2\,5\,4\,3).
\end{aligned}
$$

Their fifth powers lie in $K$, with first coordinates $uv$ and $uv^{-1}$. The quotient $(uv)^{-1}(uv^{-1})=v^{-2}=v$ recovers $v$, then $u$. Root transitivity and the conjugation formula show that every coordinate projection of $K$ is $P$.

We next obtain a full pair projection. Put $Q=\operatorname{pr}_{4,5}(K)$ and

$$
N=\{p\in P:(p,1)\in Q\}.
$$

Surjectivity makes $N$ normal in $P$. Conjugation by $x$ sends $(p,q)\in Q$ to $(u^{-1}pu,q)$, since $\alpha$ fixes $4,5$. Hence $u^{-1}pup^{-1}\in N$ for every $p\in P$. Thus $uN$ is central in $P/N$. This quotient is generated by $uN,vN$, so it is abelian; as a quotient of a perfect group it is trivial. It follows that $Q=P\times P$. Two-transitivity of $A_5$ transports this conclusion to every pair of coordinates, and the preceding perfect-group fact gives $K=P^5$. The root image is also full, proving the lemma.

Now $W_0=1$ and $W_{n+1}=W_n^5\rtimes A_5$. Each $W_n$ is perfect, because both the base and complement of this wreath product lie in its derived subgroup. Applying the lemma inductively shows that the restrictions of the single pair $a,b$ generate every $W_n$. Kernels of the maps $W\to W_n$ form a neighbourhood basis of the identity, so $G$ is dense.

## Sign changes shorten sections

Write $\ell=\ell_S$ and $L(g)=\sum_{i=1}^5\ell(g_i)$. The section formula gives

$$
L(gh)\le L(g)+L(h).
$$

Routing a word into its five sections records each letter exactly once, so the total unreduced section-word length equals the original length.

Since $a^3=b^3=1$, two successive letters from the same generator family can be replaced by at most one. Every geodesic word therefore alternates between $a^{\pm1}$ and $b^{\pm1}$. If its signs are $\varepsilon_1,\ldots,\varepsilon_m$, define

$$
V(w)=\bigl|\{j:1\le j\le m-2,\ \varepsilon_j\ne\varepsilon_{j+2}\}\bigr|.
$$

This counts sign changes between successive occurrences of each generator family.

**Shortening estimate.** An alternating word $w$ of length $m$ representing $g$ satisfies

$$
L(g)+\frac{V(w)}{10}\le m+1.
$$

Consider

$$
b^\epsilon a b^\eta a^{-1}b^\zeta,\qquad
a^\epsilon b a^\eta b^{-1}a^\zeta,
\qquad \epsilon,\eta,\zeta\in\{1,-1\}.
$$

In the first pattern, coordinate $1$ follows $1,1,2,2,1,1$ and records $b^\epsilon b^\zeta$, reducible to at most one letter. The other three letters appear elsewhere, giving total section length at most four. In the second, coordinate $4$ follows $4,4,5,5,4,4$ and records $a^\epsilon a^\zeta$, with the same saving.

Let $D$ count downward sign changes, from $+1$ to $-1$, at distance two. In each parity subsequence the two directions of change alternate, so $V(w)\le2D+2$. Each downward change gives one displayed five-letter pattern unless it is too close to an endpoint; at most two are lost. There are thus at least $M\ge V(w)/2-3$ occurrences. Greedy selection from left to right gives $q\ge M/5$ disjoint occurrences, because each choice discards at most five starting positions. Subadditivity now gives

$$
L(g)\le4q+(m-5q)=m-q
\le m-\frac{V(w)}{10}+\frac35
\le m-\frac{V(w)}{10}+1.
$$

Using disjoint blocks ensures that no saving is counted twice.

## Counting words and forcing zero exponential rate

For $0<\delta<1/2$, put

$$
h(\delta)=-\delta\log\delta-(1-\delta)\log(1-\delta).
$$

An alternating word is determined by its length, first generator family, first two signs and variation positions. The number of words of length at most $n$ with $V(w)\le\delta n$ is therefore at most

$$
8(n+1)\sum_{j\le\delta n}\binom nj
\le8(n+1)e^{nh(\delta)}.
$$

For the last inequality, use
$\delta^j(1-\delta)^{n-j}\ge e^{-nh(\delta)}$ when $j\le\delta n$, and sum the corresponding terms of the binomial probability distribution. Words of length zero or one are covered by arbitrary unused initial bits.

Choose one geodesic for each element in the radius-$n$ ball. Those with few variations are counted above. Every remaining element satisfies
$L(g)\le(1-\delta/10)n+1$ by shortening. Its root permutation and five sections determine it, so

$$
\gamma(n)\le8(n+1)e^{nh(\delta)}
+60\sum_{k_1+\cdots+k_5\le(1-\delta/10)n+1}
\gamma(k_1)\cdots\gamma(k_5).
$$

The sum is over nonnegative integers. This upper bound only needs injectivity of the root-and-sections description; arbitrary tuples need not occur.

Put $\lambda=\limsup_{n\to\infty}\log\gamma(n)/n$. Since $1\le\gamma(n)\le5^n$, it is finite and nonnegative. For each $\varepsilon>0$ there is $C\ge1$ such that
$\gamma(k)\le Ce^{(\lambda+\varepsilon)k}$ for all $k$. The sum has at most $(n+2)^5$ terms. Setting $r=1-\delta/10$, taking logarithms and passing to the limsup gives

$$
\lambda\le\max\{h(\delta),r(\lambda+\varepsilon)\}.
$$

Let $\varepsilon\downarrow0$. If $\lambda>0$, choose $\delta$ small enough that $h(\delta)<\lambda$. Since $r<1$, both entries of $\max\{h(\delta),r\lambda\}$ are strictly below $\lambda$, a contradiction. Thus $\lambda=0$, proving the asserted limit. Changing finite generating set only rescales ball radii by a constant.

## Antecedents and contribution

Brieussel constructed dense two-generated subgroups of intermediate growth for the analogous alternating wreath limits in degrees at least $29$: see [*Amenability and non-uniform growth of some directed automorphism groups of a rooted tree*](https://doi.org/10.1007/s00209-008-0417-3) and [his thesis, §3.6](https://imag.umontpellier.fr/~brieussel/these_brieussel.pdf). His construction uses Wilson's eligible generating pairs of orders two and three from [*Further groups that do not have uniformly exponential growth*](https://doi.org/10.1016/j.jalgebra.2004.01.002). The division into words with few sign changes and words whose sections shorten is Brieussel's method, notably Proposition 3.6.6 of the thesis.

Woryna had obtained dense two-generated amenable subgroups for these wreath limits, including $A_5$, but his groups have exponential growth; see [*On some universal construction of minimal topological generating sets…*, Corollary 2](https://doi.org/10.1007/s10801-015-0584-3).

Nilradical v0's contribution recorded here is the degree-five pair and its complete proof: perfect-group generation establishes density, and the five-letter patterns adapt sign-change shortening to two generators of order three. The conclusion is qualitative subexponential growth; no explicit asymptotic rate or intermediate-growth assertion is made.

## Formalization and resources

The [accepted source](https://github.com/alunik/kourovka-lean/tree/5a6b2c18e326b7b0281f00b64cade629acbfe1f5/Kourovka/Problems/P21_44) contains the complete Lean proof; the [verification record](../nilradical-v0-verification/README.md) documents Lean and Nanoda checks, with [human statement acceptance](../nilradical-v0-acceptance/README.md) recorded separately. Formal foundations credit mathlib and Konstantin Slutsky and contributors' recurrent-sections-lean for general word geometry and logarithmic growth, reused under Apache-2.0 as detailed in the [provenance record](https://github.com/alunik/kourovka-lean/blob/5a6b2c18e326b7b0281f00b64cade629acbfe1f5/Kourovka/Problems/P21_44/Proof/THIRD_PARTY.md). Acceptance concerns the linked proof snapshot; it does not imply human review of this note.