Datasets:

ulamai
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UnsolvedMath

	---
	id: AMR-010-0110
	classification: PARTIAL-PROGRESS
	wording_corrected: no
	---
	# AMR-010-0110 — Canary's "power-full subgroups" question for hyperbolic groups

	## Problem (corrected statement if needed)

	Statement verified verbatim against the source PDF
	([Bestvina, Questions in Geometric Group Theory, updated July 2004](https://www.math.utah.edu/~bestvina/eprints/questions-updated.pdf), Q 1.10 — no update note is attached to this question in the list):

	> Q 1.10 (Canary). Let $G$ be word-hyperbolic and $H$ a finitely presented subgroup of $G$.
	> Suppose that for every $g\in G$ there is $n>0$ such that $g^n\in H$. Does it follow that $H$ has finite index in $G$?
	> (Bestvina's note: "Yes if $H$ is quasi-convex, since then $\Lambda(H)=\Lambda(G)$.")

	The dataset wording was accurate; no correction was needed. Write $\sqrt{H}:=\{g\in G : \exists n>0,\ g^n\in H\}$; the hypothesis is $\sqrt{H}=G$.

	## Status / Literature

	- I found no published solution. Multiple search angles (Canary + hyperbolic + power + finite index; Bestvina problem list status; arXiv API; DuckDuckGo/Bing mirrors) turned up nothing resolving the question. Caveat: the search tools were heavily rate-limited during this session, so the sweep was shallower than intended; I am not aware of any resolution from my own knowledge of the literature either. As far as I can tell the problem is open.
	- The obstruction to a counterexample is genuinely famous. If $H\trianglelefteq G$ is a normal counterexample, then $Q=G/H$ is an infinite torsion group (the hypothesis says every element of $G$ has a power in $H$, i.e. every element of $Q$ has finite order), and $Q$ is finitely presented: $G$ is finitely presented and $H$ is finitely generated (being finitely presented), so adding a finite generating set of $H$ as relators to a finite presentation of $G$ gives a finite presentation of $Q$. Hence a normal counterexample would produce an infinite finitely presented torsion group, whose existence is a notorious open problem (see e.g. the MathOverflow discussion
	[“An infinite torsion group $G$ with finite type $K(G,1)$?”](https://mathoverflow.net/questions/239057/), which calls the existence of an infinite finitely presented torsion group “a famous open problem”; also listed in the Kourovka Notebook). The general belief is that such groups exist, but even then one would need one as a quotient of a hyperbolic group with finitely presented kernel, which is much stronger (see “What remains”).
	- Infinite torsion quotients of hyperbolic groups certainly exist (Olshanskii: $G/G^n$ is infinite for large odd $n$ when $G$ is non-elementary hyperbolic), but the kernel $G^n$ is not finitely presented (typically not even finitely generated relative to its normal structure in a controllable way), so this does not touch the question.
	- Positive territory already in the literature (used below): the quasi-convex case (noted in the source itself); limit groups are locally quasi-convex (Wilton, Hall's theorem for limit groups, GAFA 2008); Canary's covering theorem for hyperbolic 3-manifolds (R. Canary, A covering theorem for hyperbolic 3-manifolds and its applications, Topology 1996), which says a finitely generated subgroup of (a finite extension of) a closed hyperbolic 3-manifold group is either geometrically finite (quasi-convex) or a virtual fiber.

	## Work done

	I verified the statement against the source PDF (extracted the text of the July 2004 list and confirmed Q 1.10 carries no “Update”), surveyed the status as above, and proved the following partial results and reductions. All arguments below are my own derivations from standard facts.

	Proposition 1 (full limit set). If $\sqrt{H}=G$ with $G$ non-elementary hyperbolic and $H\le G$ arbitrary, then $\Lambda H=\partial G$.
	Proof. Fixed point pairs of loxodromic elements are dense in $\partial G\times\partial G$; if $g$ is loxodromic and $g^n\in H$, then $g^n$ is loxodromic in $H$ with the same fixed points $g^{\pm\infty}$, so $g^{\pm\infty}\in\Lambda H$. Since $\Lambda H$ is closed and contains a dense subset of $\partial G$, $\Lambda H=\partial G$. $\square$

	Corollary 2 (quasi-convex case — Bestvina's remark). If $H$ is quasi-convex and $\sqrt{H}=G$, then $[G:H]<\infty$.
	Proof. By Proposition 1, $\Lambda H=\partial G$. A quasi-convex subgroup of infinite index in a hyperbolic group has nowhere-dense limit set in $\partial G$ (standard: the orbit $H\!\cdot\!x$ misses a uniform neighborhood of a conical limit point of $G$ lying outside $\Lambda H$; such points exist because $\Lambda H\neq\partial G$ for infinite-index quasi-convex $H$). Hence $[G:H]<\infty$. $\square$

	Proposition 3 (the normal case is exactly a torsion-quotient problem). For $H\trianglelefteq G$ ($H$ f.p.):
	$$\sqrt{H}=G \ \Longleftrightarrow\ G/H\text{ is a torsion group},$$
	and if in addition $[G:H]=\infty$ then $G/H$ is an infinite finitely presented torsion group. Consequently:
	- If no infinite finitely presented torsion group exists (a well-known conjectural answer to a famous open problem), then Canary's question has answer yes for every normal $H$.
	- The argument needs only $H$ finitely generated: any normal counterexample with $H$ f.g. (a fortiori f.p.) yields an infinite f.p. torsion group. So even the f.g. analogue of the normal case is exactly as hard as the famous problem.

	Proposition 4 (Rips obstruction — counterexamples cannot be built cheaply). The Rips construction gives, for any f.p. group $Q$, a short exact sequence $1\to K\to G\to Q\to1$ with $G$ hyperbolic (small-cancellation) and $K$ finitely generated (2-generated). Hence: if an infinite f.p. torsion group $Q$ exists, the f.g. version of Canary's question has a negative answer. But the Rips kernel $K$ is not known (and not expected) to be finitely presented, so this does not refute the question as stated. Upgrading the kernel to f.p. via fiber-product machinery (Baumslag–Bridson–Miller–Short “1-2-3 theorem” style) would require $Q$ of type $F_3$, i.e. an infinite torsion group of type $F_3$ — strictly harder than the famous open problem (and its existence is likewise open; cf. the MathOverflow thread above, which asks exactly about torsion groups with strong finiteness properties).

	Proposition 5 (almost malnormal case). Suppose $G$ is torsion-free hyperbolic, $H\le G$ is almost malnormal, and $\sqrt{H}=G$. Then $H=G$.
	Proof. Suppose $g\notin H$ with $g^n\in H$, $n>1$. Then $g^n = g(g^n)g^{-1}\in H\cap gHg^{-1}$, and $g^n$ has infinite order ($G$ torsion-free), so $H\cap gHg^{-1}$ is infinite with $g\notin H$, contradicting almost malnormality. Hence no such $g$ exists, i.e. $\sqrt{H}=G$ forces $H=G$. $\square$
	(With torsion allowed, the same argument works unless every offending power $g^n$ has finite order.)

	Proposition 6 (closed hyperbolic 3-manifold groups — yes, even for f.g. $H$). Let $G=\pi_1(M)$ with $M$ a closed hyperbolic 3-manifold (or any torsion-free convex-cocompact Kleinian group), and let $H\le G$ be finitely generated with $\sqrt{H}=G$. Then $[G:H]<\infty$.
	Proof. By Canary's covering theorem (the ambient group is topologically tame, being convex cocompact), $H$ is either geometrically finite or a virtual fiber: in the latter case a finite-index subgroup of $H$ is the fiber kernel of a fibration of a finite cover $M'\to S^1$, so $H$ has a quotient surjecting onto $\mathbb{Z}$ (up to finite kernel). Then there is $g\in G$ whose image in that $\mathbb{Z}$-quotient has infinite order, and no power $g^n$ ($n>0$) lies in $H$ — contradicting $\sqrt{H}=G$. So $H$ is geometrically finite, hence quasi-convex in the hyperbolic group $G$ (for closed/convex-cocompact hyperbolic 3-manifold groups, geometric finiteness = quasi-convexity). Now apply Corollary 2. $\square$

	Corollary 7 (locally quasi-convex groups). If $G$ is hyperbolic and every f.g. subgroup is quasi-convex (e.g. free groups, closed surface groups, and more generally limit groups by Wilton's theorem), then the answer is yes for every f.g. $H$, since $H$ is quasi-convex and Corollary 2 applies.

	Why the general case is hard (failed-attempt analysis). Finitely presented subgroups of hyperbolic groups can be extremely distorted: Brady (1999) constructed hyperbolic groups containing f.p. subgroups that are not hyperbolic (not quasi-convex, wildly distorted). So no intrinsic geometry of $H$ is available; the only leverage is the algebraic power condition. The two natural attacks both hit famous walls:
	1. Counterexample route — blocked by Propositions 3–4: one must first produce an infinite f.p. torsion group (open since Novikov–Adian, cf. the MO thread), and then realize it as a quotient of a hyperbolic group with f.p. kernel (apparently harder).
	2. Proof route — the hypothesis gives $\Lambda H=\partial G$ (Proposition 1), and the quasi-convex conclusion would follow from the statement “a f.p. subgroup of a hyperbolic group with full limit set has finite index”; but finitely presented subgroups need not have well-behaved limit-set dynamics (they need not be hyperbolic), and I know of no theorem that promotes “f.p. + full limit set” to finite index. Residual-finiteness arguments fail: proper power-dense subgroups of finite groups exist (e.g. $2\mathbb{Z}/4\subset\mathbb{Z}/4$), so even LERF does not obviously separate a hypothetical $g\notin H$.

	## Result

	The problem appears open; I could not find any published resolution. Rigorous partial progress obtained here:

	- Reduction of the normal case: for $H\trianglelefteq G$ the question is equivalent to “does a hyperbolic group admit an infinite finitely presented torsion quotient?”, and any counterexample (even with $H$ merely finitely generated) would solve the famous open problem on the existence of infinite finitely presented torsion groups (Propositions 3–4).
	- Proved special cases: the answer is yes when $H$ is quasi-convex (Corollary 2, Bestvina's remark made precise via Proposition 1); when $H$ is almost malnormal and $G$ is torsion-free (Proposition 5); when $G$ is a closed hyperbolic 3-manifold group — for every f.g. $H$, via Canary's covering theorem (Proposition 6); and when $G$ is locally quasi-convex, e.g. a limit group (Corollary 7).
	- Structural consequence: any $H$ with $\sqrt H = G$ satisfies $\Lambda H=\partial G$ (Proposition 1), so the question is a strengthening of the (also delicate) question whether f.p. subgroups with full limit set have finite index.

	## What remains

	- The general case: $H$ f.p., non-normal, badly distorted. Nothing seems to be known here beyond the cases above.
	- The normal case is pinned to a notorious problem: decide whether infinite f.p. torsion groups exist, and more specifically whether one can be a quotient of a hyperbolic group with f.p. (or type-$F_3$) kernel. A “no” to the latter settles Canary's normal case affirmatively; a “yes” with f.p. kernel settles Canary's question negatively.
	- A proof route might try to show directly that “f.p. + $\sqrt{H}=G$” forces quasi-convexity of $H$ (which would suffice by Corollary 2), but no current technique (JSJ, combination theorems, cubulation) seems to touch distorted f.p. subgroups without extra hypotheses.
	- Literature follow-up when search tools are not rate-limited: check whether Canary himself, or authors citing Bestvina's list (e.g. via Google Scholar citations of the list), have recorded progress on Q 1.10, and whether the term “power-full/radically dense subgroup” has appeared in print for this property.

	---
	id: AMR-010-0110
	classification: PARTIAL-PROGRESS
	wording_corrected: no
	---
	# AMR-010-0110 — Canary's "power-full subgroups" question for hyperbolic groups

	## Problem (corrected statement if needed)

	Statement verified verbatim against the source PDF
	([Bestvina, Questions in Geometric Group Theory, updated July 2004](https://www.math.utah.edu/~bestvina/eprints/questions-updated.pdf), Q 1.10 — no update note is attached to this question in the list):

	> Q 1.10 (Canary). Let $G$ be word-hyperbolic and $H$ a finitely presented subgroup of $G$.
	> Suppose that for every $g\in G$ there is $n>0$ such that $g^n\in H$. Does it follow that $H$ has finite index in $G$?
	> (Bestvina's note: "Yes if $H$ is quasi-convex, since then $\Lambda(H)=\Lambda(G)$.")

	The dataset wording was accurate; no correction was needed. Write $\sqrt{H}:=\{g\in G : \exists n>0,\ g^n\in H\}$; the hypothesis is $\sqrt{H}=G$.

	## Status / Literature

	- I found no published solution. Multiple search angles (Canary + hyperbolic + power + finite index; Bestvina problem list status; arXiv API; DuckDuckGo/Bing mirrors) turned up nothing resolving the question. Caveat: the search tools were heavily rate-limited during this session, so the sweep was shallower than intended; I am not aware of any resolution from my own knowledge of the literature either. As far as I can tell the problem is open.
	- The obstruction to a counterexample is genuinely famous. If $H\trianglelefteq G$ is a normal counterexample, then $Q=G/H$ is an infinite torsion group (the hypothesis says every element of $G$ has a power in $H$, i.e. every element of $Q$ has finite order), and $Q$ is finitely presented: $G$ is finitely presented and $H$ is finitely generated (being finitely presented), so adding a finite generating set of $H$ as relators to a finite presentation of $G$ gives a finite presentation of $Q$. Hence a normal counterexample would produce an infinite finitely presented torsion group, whose existence is a notorious open problem (see e.g. the MathOverflow discussion
	[“An infinite torsion group $G$ with finite type $K(G,1)$?”](https://mathoverflow.net/questions/239057/), which calls the existence of an infinite finitely presented torsion group “a famous open problem”; also listed in the Kourovka Notebook). The general belief is that such groups exist, but even then one would need one as a quotient of a hyperbolic group with finitely presented kernel, which is much stronger (see “What remains”).
	- Infinite torsion quotients of hyperbolic groups certainly exist (Olshanskii: $G/G^n$ is infinite for large odd $n$ when $G$ is non-elementary hyperbolic), but the kernel $G^n$ is not finitely presented (typically not even finitely generated relative to its normal structure in a controllable way), so this does not touch the question.
	- Positive territory already in the literature (used below): the quasi-convex case (noted in the source itself); limit groups are locally quasi-convex (Wilton, Hall's theorem for limit groups, GAFA 2008); Canary's covering theorem for hyperbolic 3-manifolds (R. Canary, A covering theorem for hyperbolic 3-manifolds and its applications, Topology 1996), which says a finitely generated subgroup of (a finite extension of) a closed hyperbolic 3-manifold group is either geometrically finite (quasi-convex) or a virtual fiber.

	## Work done

	I verified the statement against the source PDF (extracted the text of the July 2004 list and confirmed Q 1.10 carries no “Update”), surveyed the status as above, and proved the following partial results and reductions. All arguments below are my own derivations from standard facts.

	Proposition 1 (full limit set). If $\sqrt{H}=G$ with $G$ non-elementary hyperbolic and $H\le G$ arbitrary, then $\Lambda H=\partial G$.
	Proof. Fixed point pairs of loxodromic elements are dense in $\partial G\times\partial G$; if $g$ is loxodromic and $g^n\in H$, then $g^n$ is loxodromic in $H$ with the same fixed points $g^{\pm\infty}$, so $g^{\pm\infty}\in\Lambda H$. Since $\Lambda H$ is closed and contains a dense subset of $\partial G$, $\Lambda H=\partial G$. $\square$

	Corollary 2 (quasi-convex case — Bestvina's remark). If $H$ is quasi-convex and $\sqrt{H}=G$, then $[G:H]<\infty$.
	Proof. By Proposition 1, $\Lambda H=\partial G$. A quasi-convex subgroup of infinite index in a hyperbolic group has nowhere-dense limit set in $\partial G$ (standard: the orbit $H\!\cdot\!x$ misses a uniform neighborhood of a conical limit point of $G$ lying outside $\Lambda H$; such points exist because $\Lambda H\neq\partial G$ for infinite-index quasi-convex $H$). Hence $[G:H]<\infty$. $\square$

	Proposition 3 (the normal case is exactly a torsion-quotient problem). For $H\trianglelefteq G$ ($H$ f.p.):
	$$\sqrt{H}=G \ \Longleftrightarrow\ G/H\text{ is a torsion group},$$
	and if in addition $[G:H]=\infty$ then $G/H$ is an infinite finitely presented torsion group. Consequently:
	- If no infinite finitely presented torsion group exists (a well-known conjectural answer to a famous open problem), then Canary's question has answer yes for every normal $H$.
	- The argument needs only $H$ finitely generated: any normal counterexample with $H$ f.g. (a fortiori f.p.) yields an infinite f.p. torsion group. So even the f.g. analogue of the normal case is exactly as hard as the famous problem.

	Proposition 4 (Rips obstruction — counterexamples cannot be built cheaply). The Rips construction gives, for any f.p. group $Q$, a short exact sequence $1\to K\to G\to Q\to1$ with $G$ hyperbolic (small-cancellation) and $K$ finitely generated (2-generated). Hence: if an infinite f.p. torsion group $Q$ exists, the f.g. version of Canary's question has a negative answer. But the Rips kernel $K$ is not known (and not expected) to be finitely presented, so this does not refute the question as stated. Upgrading the kernel to f.p. via fiber-product machinery (Baumslag–Bridson–Miller–Short “1-2-3 theorem” style) would require $Q$ of type $F_3$, i.e. an infinite torsion group of type $F_3$ — strictly harder than the famous open problem (and its existence is likewise open; cf. the MathOverflow thread above, which asks exactly about torsion groups with strong finiteness properties).

	Proposition 5 (almost malnormal case). Suppose $G$ is torsion-free hyperbolic, $H\le G$ is almost malnormal, and $\sqrt{H}=G$. Then $H=G$.
	Proof. Suppose $g\notin H$ with $g^n\in H$, $n>1$. Then $g^n = g(g^n)g^{-1}\in H\cap gHg^{-1}$, and $g^n$ has infinite order ($G$ torsion-free), so $H\cap gHg^{-1}$ is infinite with $g\notin H$, contradicting almost malnormality. Hence no such $g$ exists, i.e. $\sqrt{H}=G$ forces $H=G$. $\square$
	(With torsion allowed, the same argument works unless every offending power $g^n$ has finite order.)

	Proposition 6 (closed hyperbolic 3-manifold groups — yes, even for f.g. $H$). Let $G=\pi_1(M)$ with $M$ a closed hyperbolic 3-manifold (or any torsion-free convex-cocompact Kleinian group), and let $H\le G$ be finitely generated with $\sqrt{H}=G$. Then $[G:H]<\infty$.
	Proof. By Canary's covering theorem (the ambient group is topologically tame, being convex cocompact), $H$ is either geometrically finite or a virtual fiber: in the latter case a finite-index subgroup of $H$ is the fiber kernel of a fibration of a finite cover $M'\to S^1$, so $H$ has a quotient surjecting onto $\mathbb{Z}$ (up to finite kernel). Then there is $g\in G$ whose image in that $\mathbb{Z}$-quotient has infinite order, and no power $g^n$ ($n>0$) lies in $H$ — contradicting $\sqrt{H}=G$. So $H$ is geometrically finite, hence quasi-convex in the hyperbolic group $G$ (for closed/convex-cocompact hyperbolic 3-manifold groups, geometric finiteness = quasi-convexity). Now apply Corollary 2. $\square$

	Corollary 7 (locally quasi-convex groups). If $G$ is hyperbolic and every f.g. subgroup is quasi-convex (e.g. free groups, closed surface groups, and more generally limit groups by Wilton's theorem), then the answer is yes for every f.g. $H$, since $H$ is quasi-convex and Corollary 2 applies.

	Why the general case is hard (failed-attempt analysis). Finitely presented subgroups of hyperbolic groups can be extremely distorted: Brady (1999) constructed hyperbolic groups containing f.p. subgroups that are not hyperbolic (not quasi-convex, wildly distorted). So no intrinsic geometry of $H$ is available; the only leverage is the algebraic power condition. The two natural attacks both hit famous walls:
	1. Counterexample route — blocked by Propositions 3–4: one must first produce an infinite f.p. torsion group (open since Novikov–Adian, cf. the MO thread), and then realize it as a quotient of a hyperbolic group with f.p. kernel (apparently harder).
	2. Proof route — the hypothesis gives $\Lambda H=\partial G$ (Proposition 1), and the quasi-convex conclusion would follow from the statement “a f.p. subgroup of a hyperbolic group with full limit set has finite index”; but finitely presented subgroups need not have well-behaved limit-set dynamics (they need not be hyperbolic), and I know of no theorem that promotes “f.p. + full limit set” to finite index. Residual-finiteness arguments fail: proper power-dense subgroups of finite groups exist (e.g. $2\mathbb{Z}/4\subset\mathbb{Z}/4$), so even LERF does not obviously separate a hypothetical $g\notin H$.

	## Result

	The problem appears open; I could not find any published resolution. Rigorous partial progress obtained here:

	- Reduction of the normal case: for $H\trianglelefteq G$ the question is equivalent to “does a hyperbolic group admit an infinite finitely presented torsion quotient?”, and any counterexample (even with $H$ merely finitely generated) would solve the famous open problem on the existence of infinite finitely presented torsion groups (Propositions 3–4).
	- Proved special cases: the answer is yes when $H$ is quasi-convex (Corollary 2, Bestvina's remark made precise via Proposition 1); when $H$ is almost malnormal and $G$ is torsion-free (Proposition 5); when $G$ is a closed hyperbolic 3-manifold group — for every f.g. $H$, via Canary's covering theorem (Proposition 6); and when $G$ is locally quasi-convex, e.g. a limit group (Corollary 7).
	- Structural consequence: any $H$ with $\sqrt H = G$ satisfies $\Lambda H=\partial G$ (Proposition 1), so the question is a strengthening of the (also delicate) question whether f.p. subgroups with full limit set have finite index.

	## What remains

	- The general case: $H$ f.p., non-normal, badly distorted. Nothing seems to be known here beyond the cases above.
	- The normal case is pinned to a notorious problem: decide whether infinite f.p. torsion groups exist, and more specifically whether one can be a quotient of a hyperbolic group with f.p. (or type-$F_3$) kernel. A “no” to the latter settles Canary's normal case affirmatively; a “yes” with f.p. kernel settles Canary's question negatively.
	- A proof route might try to show directly that “f.p. + $\sqrt{H}=G$” forces quasi-convexity of $H$ (which would suffice by Corollary 2), but no current technique (JSJ, combination theorems, cubulation) seems to touch distorted f.p. subgroups without extra hypotheses.
	- Literature follow-up when search tools are not rate-limited: check whether Canary himself, or authors citing Bestvina's list (e.g. via Google Scholar citations of the list), have recorded progress on Q 1.10, and whether the term “power-full/radically dense subgroup” has appeared in print for this property.