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I know that not every finitely-presented group may be embedded into a one-relator group, for example because of a theorem of Magnus stating that the word problem is solvable in one-relator groups. But does there is a great amount of finitely-generated groups embeddable into a one-relator group?

For instance, is there any (hopefully "large") class of groups known to be embeddable into a one-relator group?

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  • $\begingroup$ I don't know if any hyperbolic group is known not to embed into any 1-relator group (I don't know either for $C'(1/6)$ f.p. groups). At the same time, I don't know if a 1-relator group may contain an infinite Property T subgroup. $\endgroup$
    – YCor
    Commented Mar 15, 2021 at 11:35
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    $\begingroup$ PS the cohomological dimension fact mentioned by Carl in a comment to his answer seems to discard many hyperbolic groups. $\endgroup$
    – YCor
    Commented Mar 15, 2021 at 11:51
  • $\begingroup$ The comment referenced by @YCor. $\endgroup$
    – LSpice
    Commented Mar 15, 2021 at 13:46
  • $\begingroup$ @YCor, see my comments on the answer below. Every infinite finitely generated subgroup of a one-relator group has infinite abelianisation. This rules out many 2-dimensional groups, including all with property T. $\endgroup$
    – HJRW
    Commented Mar 16, 2021 at 0:20
  • $\begingroup$ @HJRW Nice; actually I believe it would be a useful additional answer to mathoverflow.net/questions/218422 $\endgroup$
    – YCor
    Commented Mar 16, 2021 at 0:27

2 Answers 2

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There have been a couple of recent results which demonstrate that the class of subgroups of one-relator groups is very rich. For instance, Calegari–Walker proved that a random 1-relator group contains a surface subgroup, and Calegari and I improved this by demonstrating that a random 1-relator group contains a subgroup isomorphic to the fundamental group of an acylindrical hyperbolic 3-manifold. Both papers are on the arXiv.

A random 1-relator group is hyperbolic (since its relator satisfies the C'(1/6) small-cancellation condition) and 2-dimensional. Results of Bowditch and Kapovich–Kleiner then impose certain restrictions on the (quasiconvex) subgroups that can arise—see the answer to Topology of boundaries of hyperbolic groups for some details. In a sense, the above results show that the quasiconvex subgroups of random 1-relator groups are as rich as possible. This is discussed in the introduction to our paper.

This leaves two strands uncovered. A random 1-relator group will also contain non-quasiconvex finitely generated subgroups. (I think Dunfield–Thurston proved this in the 2-generator case.) And, of course, plenty of 1-relator groups do not behave like a ‘random’ one—for instance, Baumslag's famous example of a non-abelian 1-relator group with every finite quotient cyclic.

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    $\begingroup$ A couple more comments. These results are not embeddability results as asked for in the final paragraph of the question, in the sense that they don't say 'Every group of class X is embeddable into a 1-relator group', but rather 'A random 1-relator group contains a subgroup from class X'. But the proof techniques do generate a very large number of subgroups of this form---a generic 1-relator group contains many 3-manifold groups as subgroups. $\endgroup$
    – HJRW
    Commented Oct 4, 2014 at 18:08
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    $\begingroup$ I changed 'no-abelian' to 'non-abelian', which I hope was correct. $\endgroup$
    – LSpice
    Commented Mar 15, 2021 at 13:48
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    $\begingroup$ It was! :)${}{}$ $\endgroup$
    – HJRW
    Commented Mar 15, 2021 at 18:02
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It is known that a right-angled Artin group $A(\Gamma)$ embeds in some one-relator group if and only if $\Gamma$ is a forest. In fact, any such $A(\Gamma)$ embeds in the one-relator group $\operatorname{Gp}\langle a,t \mid [a, a^t] = 1\rangle$, which contains a copy of $A(P_4)$ (and hence also the group corresponding to any forest by certain embeddability results). This is all proved in Gray, Robert D., Undecidability of the word problem for one-relator inverse monoids via right-angled Artin subgroups of one-relator groups, ZBL07160162.

Note that in the torsion case, the set of subgroups possible is significantly reduced, and explicit analysis can be done to a greater extent. For example, every $2$-generated subgroup of a $2$-generated one-relator group with torsion is either free or is itself a one-relator group with torsion. This is all possible by the B. B. Newman Spelling Theorem, which provides a significant degree of control. In particular, this theorem implies that one-relator groups with torsion are hyperbolic, so the only right-angled Artin groups embedding in some one-relator group with torsion are free. For detailed combinatorial analysis, and some more examples in the torsion case, see much of the work by Stephen Pride, e.g. Pride, Stephen J., The two-generator subgroups of one-relator groups with torsion, Trans. Am. Math. Soc. 234, 483-496 (1977). ZBL0366.20022.

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  • $\begingroup$ The raag statement seems to include as particular case that $\mathbf{Z}^3$ doesn't embed into a 1-relator group. $\endgroup$
    – YCor
    Commented Mar 15, 2021 at 11:29
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    $\begingroup$ @YCor Yes, but this already follows from the fact that the cohomological dimension of any torsion-free one-relator group is $\leq 2$. $\endgroup$ Commented Mar 15, 2021 at 11:40
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    $\begingroup$ (By the way, the point about RAAGs is just the observation that the Euler characteristic of a 2-dimenional RAAG defined by the graph $\Gamma$ is $1-\chi(\Gamma)$.) $\endgroup$
    – HJRW
    Commented Mar 15, 2021 at 18:27
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    $\begingroup$ Sorry for the extended comments. Looking at Gray's paper, I see that the non-embeddability result is in fact proved by appealing to out work, exactly as the above comments suggest! $\endgroup$
    – HJRW
    Commented Mar 15, 2021 at 18:31
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    $\begingroup$ @HJRW Yes, of course you're right. Thanks for writing this out! In fact, an earlier version of Bob's paper on the arXiv did not have that argument, after which Howie wrote to him to provide exactly that argument. $\endgroup$ Commented Mar 15, 2021 at 20:20

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