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I understand that:

$\Bbb Z^0 = \langle a \mid a \rangle$

$\Bbb Z^1 = \langle a, b \mid b \rangle$

$\Bbb Z^2 = \langle a, b \mid aba^{-1}b^{-1} \rangle$

but is it possible for $\Bbb Z^3$ to be represented with just one relator? I can’t think of a way in which it would be possible but I am also unsure how to prove it would be impossible.

cede
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  • An obvious presentation with three relators is $\langle a,b,c \mid aba^{-1}b^{-1}, aca^{-1}c^{-1}, bcb^{-1}c^{-1} \rangle$ (specifying that the generators commute with each other). Could the three relators be combined into one? – Geoffrey Trang Jul 04 '23 at 01:02
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    Are you familiar with the Freiheitssatz? – James Jul 04 '23 at 01:49
  • @James This is my first time reading about it. Applying this theorem, does this mean any group with at least three generators and one relator must contain a free group of rank 2? And this couldn't be contained in $\Bbb Z^3$? Or am I not understanding it correctly. – cede Jul 04 '23 at 07:38
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    @GeoffreyTrang Asking if the three relators can be combined into one is not immediately relevant. For example, the group with presentation $\langle a, t\mid t^{-1}a^2ta^{-3}, [t^{-1}at, a]\rangle$ is a one-relator group but not on the generators $(a, t)$, and so the two given relators cannot be "combined into one". – user1729 Jul 04 '23 at 11:58
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    @cede Yes, any presentation must have at least three generators, and all three must be involved in a single relator, lest the group be a non-trivial free product which is not abelian. Therefore, the Freiheitssatz implies it contains a free group of rank two, which is not abelian. – James Jul 04 '23 at 20:24

2 Answers2

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There is a super-pretty, and very poweful, theorem which you can use here. Its proof is not so hard (probably master's level). Reference is: Baumslag, Benjamin; Pride, Stephen J. Groups with two more generators than relators. J. London Math. Soc. (2) 17 (1978), no.3, 425–426.

Theorem (B.Baumslag-Pride). If $G$ admits a presentation with $2$ more generators than relators, then $G$ contains a finite index subgroup which surjects onto the free group of rank $2$, $F_2$.

In modern parlance, $G$ is large.

To apply this to your problem, note that by considering abelianisations of one-relator groups, if $\mathbb{Z}^3$ is an $n$-generator one-relator group then $n\geq3$. As $3-1\geq2$, the Baumslag-Pride theorem then says that $\mathbb{Z}^3$ contains a finite index subgroup $H$ which surjects onto $F_2$. As $H$ is necessarily abelian but $F_2$ is not, we have a contradiction. Hence, $\mathbb{Z}^3$ is not a one-relator group.

In fact, using more powerful tools we can prove the following.

Theorem. The only possible abelian subgroups of a one-relator group are cyclic groups and $\mathbb{Z}^2$.

One way to prove this would be to first use (virtual) cohomological dimension to eliminate the possibility of $\mathbb{Z}^n$ for $n\geq3$ - one relator groups have $(v)cd\leq2$, and hence so do all their subgroups, while $cd(\mathbb{Z}^n)>2$. [This argument also works for the problem in the question.]

We therefore just need to eliminate $\mathbb{Z}\times\mathbb{Z}_n$ and $\mathbb{Z}_m\times\mathbb{Z}_n$ as possibilities. To do this, first apply a theorem of Pride (Pride, Stephen J. The two-generator subgroups of one-relator groups with torsion. Trans. Amer. Math. Soc.234(1977), no.2, 483–496) which says that if a subgroup of a one-relator group is two-generated and contains elements of finite order, then the subgroup is itself a one-relator group, and moreover (via another deep result...) the relator is a proper power. So then:

  • $\mathbb{Z}_m\times\mathbb{Z}_n$ is eliminated by considering abelianisations.
  • For $\mathbb{Z}\times\mathbb{Z}_n$, we can apply a beefed-up version of the Baumslag-Pride result, due to Gromov, which states that if $G$ admits a presentation with one more generator than relator and where at least one relator is a proper power, then $G$ contains a finite index subgroup which surjects onto $F_2$. If $\mathbb{Z}\times\mathbb{Z}_n$ is a subgroup of a one-relator groups, then as noted above it a one-relator group where the relator is a proper power. It is also non-cyclic, and so Gromov's result applies to this one-relator presentation. However, no subgroup of $\mathbb{Z}\times\mathbb{Z}_n$ surjects onto $F_2$.

Phew.

Arturo Magidin
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user1729
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Here is a proof of the impossibility of such a presentation that converts it into a problem on finite groups and the ranks of their Schur Multiplier.

If ${\mathbb Z}^3$ had a presentation with $r$ generators and $1$ relator, then $C_2^3$ would have a presentation with $r$ generators and $r+1$ relators, and hence would have deficiency at least $-1$.

This would imply that its Schur Multiplier had rank at most $1$, but it has rank $3$, contradiction.

Derek Holt
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