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$\DeclareMathOperator\ncl{ncl}$When I attended a geometric group theory summer school, a question asked by the speaker reminded me of an old question but shaped in a different manner:

Given two nontrivial groups $A,B$ and $w \in A \ast B$. For any $k\geq 2$, is the quotient $(A\ast B)/\ncl(w^k)$ nontrivial, where $\ncl(w^k)$ is the normal closure of $w^k$?

This is true if $B = \mathbb{Z}$ and the projection of $w$ to $B =\mathbb{Z}$ is $\pm 1$. For $k\geq 4$, this was handled by a theorem of Howie.

It is not true in general when $k =1$. For example, let $A = \mathbb{Z}_2$ and $B = \mathbb{Z}_3$ and $w = ab^{-1}$. When $w$ becomes a relation, $a,b$ must be conjugate, but $a,b$ originally have orders $2$ and $3$, resp. It follows that $a = b$ are trivial elements and the group is trivial. I'm particularly interested in the following question

When $k=1$, are there counterexamples if $A,B$ are torsion-free?

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    $\begingroup$ There are no counter-examples for $A$ torsion-free and $B\cong\mathbb{Z}$. This corresponds to the Kervaire-Laudenbach Conjecture for torsion-free groups, which was proven by Klyachko; see Fenn and Rourke, Klyachko's methods and the solution of equations over torsion-free groups, Enseign. Math. (2) 42 (1996), no. 1-2, 49–74. $\endgroup$
    – ADL
    Commented Aug 22, 2022 at 8:33
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    $\begingroup$ @ADL: since this answers the question, you should post it as an answer! $\endgroup$
    – HJRW
    Commented Aug 23, 2022 at 2:26
  • $\begingroup$ @HJRW It isn't a complete solution, so wasn't sure! I've posted it now. $\endgroup$
    – ADL
    Commented Aug 23, 2022 at 9:19
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    $\begingroup$ I do not know. The question is a weaker form of Levin’s conjecture: if $A$ and $B$ are torsion-free groups and $w$ is not conjugate to elements of $A$, then $A$ embeds into the quotient. $\endgroup$ Commented Aug 24, 2022 at 20:29
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    $\begingroup$ Corollary V.9.4 of Lyndon-Schupp implies that for a C’(1/6) free product of groups, each factor embeds and hence the group is non-trivial. link.springer.com/book/10.1007/978-3-642-61896-3 $\endgroup$
    – Ian Agol
    Commented Aug 24, 2022 at 20:40

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Here is a partial answer: The Kervaire-Laudenbach Conjecture states that, for any group $A$, $(A\ast\mathbb{Z})/\operatorname{ncl}(w)$ is non-trivial. This was proven by Klyachko for torsion free groups $A$; see Fenn and Rourke, Klyachko's methods and the solution of equations over torsion-free groups, Enseign. Math. (2) 42 (1996), no. 1-2, 49–74 (doi). Hence:

There are no counter-examples for $A$ torsion-free and $B\cong Z$.

This can be easily generalised: A group is indicable if it surjects onto $\mathbb{Z}$. If $B$ is indicable, so $\phi:B\twoheadrightarrow\mathbb{Z}$, then the map $\phi$ induces a map $\overline{\phi}:(A\ast B)/\operatorname{ncl}(w)\twoheadrightarrow(A\ast\mathbb{Z})/\operatorname{ncl}(u)$ for some word $u\in A\ast\mathbb{Z}$. By Klyachko's theorem, we know the image is non-trivial so we immediately have:

There are no counter-examples for $A$ torsion-free and $B$ indicable.


Klyachko's proof is really pretty, based on a "funny property" of spheres which "is so simple and funny" that, he claims, it could be included in a collection of puzzles or suggested as a problem for a school mathematical tournament. He phrased it in terms of car crashes, and you can find a MathOverflow discussion on it here. The paper of Fenn and Rourke I cited above is the best place to read about it though.


Added later: I asked Jim Howie if he knew what the status of this problem is, and he said he thought it open and very difficult (so there are no known counter-examples).

He also pointed to two papers of his - one with Brodskii from 1993, and the other with Edjvet from 2021. These prove that when $A,B$ are torsion-free, then they embed into $(A*B)/\operatorname{ncl}(w)$ if $w$ has free product length $\le 6$ (first paper) and $\le 8$ (second paper). These are innocuous-sounding results, but the one extra step took ~28 years, and both took quite a bit of work to prove.

Embedding results like these are typically referred to as Freiheitssatz and seem to be the main approach to problems of this form. For example the main result of Klyachko's paper is a Freiheitssatz for when $w$ has a specific form; the other cases follow from an easy observation. A Freiheitssatz is a priori stronger than a non-triviality result would be, however Jim points out that it is not clear how much stronger.

The papers are:

S. D. Brodskii and J. Howie, One-relator products of torsion-free groups, Glasgow Math. J. 35 (1993) 99-104 (doi).

M. Edjvet and J. Howie, On singular equations over torsion-free groups Internat. J. Algebra Comput. 31 (2021), 551-580 (doi, arXiv.

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  • $\begingroup$ Thanks. It does cover many interesting cases so for sure an upvote. But since it’s a partial answer, I’d like to wait to see if someone can come up with a complete solution if there is one. $\endgroup$
    – Shijie Gu
    Commented Aug 23, 2022 at 14:01
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    $\begingroup$ Note that if A is hyper linear, then Kervaire-Laudenbach also holds. doi.org/10.1515/jgth-2019-0167 $\endgroup$
    – Ian Agol
    Commented Aug 24, 2022 at 16:59
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    $\begingroup$ Thanks for reporting on Howie’s comments. He is the expert on these things! $\endgroup$
    – HJRW
    Commented Aug 29, 2022 at 0:24

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