An infinite group of spread one

Nilradical

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. This is Problem 21.38 of the Kourovka Notebook.

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, published in Bull. Lond. Math. Soc. 55 (2023), 2144–2157.

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 $F$ .

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. 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.

Formalization and resources

Nilradical v0 assembled this solution and its complete formal proof, including the required generation case. The accepted source uses the equivalent rational piecewise-linear model; the verification record and human statement acceptance record its checks. Formal foundations credit mathlib and SauersML's group-approximation, with Apache-2.0 reuse documented in the provenance record. Acceptance concerns the linked proof snapshot; it does not imply human review of this note.