# A group of spread one

**Kourovka 21.38 · 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 result

The *spread* $s(H)$ of a group $H$ is the largest $k$ such that every $k$ nonidentity elements $h_1,\ldots,h_k$ have a common *companion*: an element $y$ with $\langle h_i,y\rangle=H$ for every $i$. Repetitions are permitted, and the spread is infinite if this holds for every $k$. Donoven and Harper asked whether an infinite group can have spread exactly one in [*Infinite $3/2$-generated groups*, Question 2](https://doi.org/10.1112/blms.12356). This is [Problem 21.38 of the Kourovka Notebook](https://arxiv.org/abs/1401.0300v46).

Let $F$ be Thompson's group of increasing piecewise-linear homeomorphisms of $[0,1]$, with finitely many dyadic breakpoints and slopes integral powers of two. Its elements are linear near the endpoints, and

$$
\pi:F\longrightarrow\mathbb Z^2,\qquad
\pi(f)=\bigl(\log_2 f'(0^+),\log_2 f'(1^-)\bigr)
$$

is a homomorphism.

**Theorem.** The group

$$
G=\{f\in F:f'(0^+)=f'(1^-)\}
 =\pi^{-1}\bigl(\mathbb Z(1,1)\bigr)
$$

is infinite and has ordinary spread exactly one.

The two bounds use the endpoint condition differently. Golan-Polak's generation theorem supplies a companion for every nonidentity element. Equal endpoint slopes also force an interior fixed point, which prevents two suitably supported elements from having a common companion.

## Every nonidentity element has a companion

We use the following form of Gili Golan-Polak's [*Thompson's group $F$ is almost $3/2$-generated*, Theorem 2](https://arxiv.org/abs/2210.03564v1), published in [Bull. Lond. Math. Soc. 55 (2023), 2144–2157](https://doi.org/10.1112/blms.12841).

**Generation theorem.** Let $(a,b),(c,d)\in\mathbb Z^2$, with $(a,c)\ne(0,0)$ and $(b,d)\ne(0,0)$. If $1\ne f\in F$ and $\pi(f)=(a,b)$, there is $g\in F$ such that

$$
\pi(g)=(c,d),\qquad
\langle f,g\rangle
=\pi^{-1}\bigl(\mathbb Z(a,b)+\mathbb Z(c,d)\bigr).
$$

In particular, $(a,b)$ may be zero, and the two vectors may be linearly dependent. Given $1\ne f\in G$, write $\pi(f)=(n,n)$ and choose $(c,d)=(1,1)$. Both hypotheses hold, including when $n=0$, and the theorem gives

$$
\langle f,g\rangle
=\pi^{-1}\bigl(\mathbb Z(n,n)+\mathbb Z(1,1)\bigr)=G.
$$

Thus $s(G)\ge1$. This applies the published generation theorem, including its prescribed-endpoint construction in Proposition 16(1). The local generation methods have earlier foundations in Golan's [*The generation problem in Thompson group $F$*](https://arxiv.org/abs/1608.02572v2).

## A fixed pair with no common companion

First construct small supported elements. The map

$$
u(t)=
\begin{cases}
2t,&0\le t\le\tfrac14,\\
t+\tfrac14,&\tfrac14\le t\le\tfrac12,\\
\tfrac12t+\tfrac12,&\tfrac12\le t\le1
\end{cases}
$$

belongs to $F$ and satisfies $u(t)>t$ for $0<t<1$. For dyadic $0<r<s<1$, conjugate $u$ onto $[r,s]$ by $\varphi(t)=r+(s-r)t$ and extend by the identity. The resulting map $h_{r,s}$ has dyadic breakpoints and slopes among $1,2,1/2$, so

$$
h_{r,s}\in G,\qquad
\operatorname{supp}(h_{r,s})=(r,s),\qquad
h_{r,s}(t)>t\quad(r<t<s).
$$

Here support means the set of moved points. Its positive powers move an interior point through a strictly increasing sequence; hence $G$ is infinite. Also, every $z\in(0,1)$ lies inside some such dyadic interval, so an element of $G$ moves $z$. Every interior point stabilizer $G_z$ is therefore proper.

On the other hand, each $y\in G$ fixes an interior point. Put $\lambda=y'(0^+)=y'(1^-)$. Near the endpoints,

$$
y(t)-t=(\lambda-1)t\quad\text{near }0,\qquad
y(t)-t=(\lambda-1)(t-1)\quad\text{near }1.
$$

If $\lambda\ne1$, these have opposite signs at interior points, so continuity supplies a fixed point between them. If $\lambda=1$, an entire endpoint neighbourhood is fixed.

Now fix

$$
a=h_{1/8,\,1/4},\qquad b=h_{3/4,\,7/8}.
$$

Their supports are disjoint. For any proposed common companion $y$, take an interior point $z$ fixed by $y$. At least one of $a,b$ also fixes $z$, and the subgroup generated by that element and $y$ lies in $G_z<G$. The same pair $a,b$ therefore defeats every companion. This proves $s(G)\le1$ and completes the theorem.

This is the singleton case of the disjoint-support argument in the **proof** of Bleak, Donoven, Harper and Hyde's [*Generating simple vigorous groups*, Proposition 4.5, p. 22](https://arxiv.org/pdf/2607.21754v2). Their stated proposition excludes strong infinite uniform spread; the fixed-point hypothesis here lets one fixed pair obstruct all companions and gives the ordinary-spread bound. Related fixed-point obstructions appear in Golan Polak's [*The “spread” of Thompson's group $F$*, Lemma 2 and Remark 11](https://arxiv.org/abs/2402.19444v1).

## Formalization and resources

Nilradical v0 assembled this solution and its complete formal proof, including the required generation case. The [accepted source](https://github.com/alunik/kourovka-lean/tree/5a6b2c18e326b7b0281f00b64cade629acbfe1f5/Kourovka/Problems/P21_38) uses the equivalent rational piecewise-linear model; the [verification record](../nilradical-v0-verification/README.md) and [human statement acceptance](../nilradical-v0-acceptance/README.md) record its checks. Formal foundations credit mathlib and SauersML's group-approximation, with Apache-2.0 reuse documented in the [provenance record](https://github.com/alunik/kourovka-lean/blob/5a6b2c18e326b7b0281f00b64cade629acbfe1f5/Kourovka/Problems/P21_38/Proof/THIRD_PARTY.md). Acceptance concerns the linked proof snapshot; it does not imply human review of this note.