Is there a one relator group with property (T)?

Question

Is there a one-relator group with property (T)?

That is, is there an $n > 2$, and some $x \in F_n$ (the free group on $n$ generators) such that the quotient of $F_n$ by the normal subgroup generated by $x$ has Kazhdan's property $\mathrm{(T)}$ ?

Answer 1

No, the abelianization of any such quotient will be infinite (the abelianization of $F_n$ is $\mathbb{Z}^n$, which does not have a cyclic subgroup of finite index), but Kazhdan groups always have finite abelianization.

Answer 2

In fact, much more is true: if $H$ is a finitely generated subgroup of a one-relator group and $H$ has property (T) then $H$ is finite. Indeed, a classical theorem of Brodskii and Howie asserts that torsion-free one-relator groups are locally indicable, meaning that every non-trivial finitely generated subgroup surjects $\mathbb{Z}$, and in particular can’t have property (T). It’s also true that one-relator groups with torsion have subgroups of finite index that are locally indicable.

[Thanks to Giles Gardam for pointing out the correct attribution for this fact. The papers of Brodskii and Howie proving that one-relator groups without torsion are locally indicable are:

"Equations over groups and groups with one defining relation." (Russian) Uspekhi Mat. Nauk 35 (1980), no. 4(214), 183.

Howie, James "On locally indicable groups." Math. Z. 180 (1982), no. 4, 445–461. ]

By popular request, I'll leave in a few words about how to deduce this from recent work of Helfer--Wise and Louder--W.

Corollary 5 of this paper of Louder and myself should state:

Let $G$ be a torsion-free one-relator group. If $H < G$ is finitely generated and non-trivial then $b_2(H)\leq b_1(H)−1$.

Here, $b_i(H)$ is the rank of the $i$th homology group of $H$ with coefficients in $\mathbb{Q}$. In particular, if $H$ is non-trivial then $b_1(H)\geq 1$, so the abelianisation of $H$ is infinite and in particular $H$ certainly does not have property (T).

Remarks:

1. I notice that the hypothesis that $H$ should be non-trivial was accidentally omitted from the paper, and unfortunately this typo made it all the way to the published version!

2. The main theorem, from which this follows, was proved independently by Helfer--Wise, though I don't think they stated this corollary.

3. The case of one-relator groups with torsion follows quickly. This paper of Louder and myself explains how to apply the above techniques in that case, and a similar result follows.

Answer 3

Property (T) implies Property FA: every action on a tree has a global fixpoint. The Magnus–Moldavanskii hierarchy expresses every one-relator group as (a subgroup of) an HNN extension of a "simpler" one-relator group (over a free subgroup, but that doesn't matter here). The hierarchy terminates in finitely many steps with a free group or finite cyclic group. Putting this together, any group with (T) that embeds in a one-relator group fixes a vertex of the Bass–Serre tree, so is a subgroup of a simpler one-relator group, and acts on the Bass–Serre tree of that one-relator group, etc until we conclude that, since it can't be a non-trivial subgroup of a free group (infinite abelianization), it must be finite.