--- id: AMR-010-0207 classification: SOLVED-IN-LITERATURE wording_corrected: yes --- # AMR-010-0207 — Wise's "power alternative" for CAT(0) / automatic groups ## Problem (corrected statement if needed) Source: Bestvina, *Questions in Geometric Group Theory* (updated July 2004), Question 2.7 (attributed to D. Wise), . I fetched this PDF and confirmed the transcription in the worklist is faithful; only the exponents were flattened by formatting. The verbatim wording is: > **Q 2.7 (D. Wise).** Let $G$ act properly discontinuously and cocompactly on a CAT(0) space (or let $G$ be > automatic). Consider two elements $a, b$ of $G$. Does there exist $n > 0$ such that either the subgroup > $\langle a^n, b^n\rangle$ is free or $\langle a^n, b^n\rangle$ is abelian? In the modern literature this property is called **Wise's power alternative** (PA): for every $g,h\in G$ there is $n\ge 1$ such that either $[g^n,h^n]=1$ or $\langle g^n,h^n\rangle\cong F_2$. The two formulations are equivalent as yes/no questions: a 2-generator free group is $1$, $\mathbb{Z}$, or $F_2$, and the first two are abelian, so "free or abelian" $\Leftrightarrow$ "$F_2$ or abelian"; and "commute" $\Rightarrow$ "abelian", so a group failing the modern PA fails Wise's version and vice versa. ## Status / Literature **CAT(0) case: answered NO in the literature (2021).** Ian J. Leary and Ashot Minasyan, *Commensurating HNN extensions: nonpositive curvature and biautomaticity*, **Geom. Topol. 25 (2021), no. 4, 1819–1860** (DOI `10.2140/gt.2021.25.1819`; arXiv:1907.03515). Verified via Crossref (metadata match) and via the MSP journal page abstract. Their Example 9.4 introduces groups $G_{k,m}$ (commensurating HNN extensions of $\mathbb{Z}^2$, with stable letter conjugating a finite-index subgroup by the similitude $\begin{pmatrix}k&-m\\ m&k\end{pmatrix}$), and their Corollary 9.6 shows that for $-2m0$ is $\langle a^n,b^n\rangle$ free or abelian (Leary–Minasyan 2021, Example 9.4 + Corollary 9.6; CAT(0) by Corollary 9.3). This is the accepted resolution of Q 2.7 in the literature (Martin 2024; Hagen–Martin–Sartori 2025 both describe it as "the first example of a CAT(0) group not satisfying the power alternative"). Hence the problem as posed is **SOLVED-IN-LITERATURE**, with the caveat that the parenthetical automatic variant is untouched by the counterexample (see below). ## What remains - **Automatic case of Q 2.7: open.** No automatic (or biautomatic) group is known to fail the power alternative; the known CAT(0) counterexamples are provably non-biautomatic, and their automaticity is unknown. A positive answer for biautomatic groups, or an automatic counterexample, would both be significant. - **Groups acting geometrically on a product of two trees** (Burger–Mozes-type irreducible lattices): PA is open even in the absence of "anti-tori" (Hagen–Martin–Sartori, Example 4.13). - **General Artin groups:** PA is known for RAAGs, even FC-type, two-dimensional hyperbolic-type, and (2,2)-free triangle-free cases; Hagen–Martin–Sartori reduce the general case to free-of-infinity Artin groups modulo two conjectures on parabolic subgroups (parabolic intersection property, normaliser structure property). - **Uniformity:** is there a finitely presented group satisfying PA but with no uniform exponent $N$ (Hagen–Martin–Sartori, Question 1.2)? - Related sibling Q 2.8 (Tits alternative for CAT(0) or (bi)automatic groups) remains open in general; the Leary–Minasyan groups satisfy the ordinary Tits alternative (they are virtually solvable-subgroup-controlled lattices), so the *power* alternative is genuinely sharper.