Does such a subgroup exist?

Question

I am looking for a certain masa in a $II_1$ factor which is singular and has nontrivial Takesaki invariant. For this I am looking for an example of an inclusion of groups $H\subset G$ such that:

• $G$ is a countable icc (infinite conjugacy class) group
• $H$ is abelian
• $\forall g\in G-H,\{ hgh^{-1} |h\in H \}$ is infinite
• $ |H\backslash G/H| \geq 3$
• there exists $g\in G-H$ and $h_1\neq h_2\in H$ with $h_1 g=gh_2$.

Does such an example exist?

Answer 1

[generalization of Agol's answer]

Take $H$ a group and let $K$ act on $H$ by isomorphisms (write the action as $\sigma$) and consider $G=H\rtimes_\sigma K$. Then

• condition 2 is satisfied if $H$ is abelian
• condition 4 is satisfied if $K$ contains at least 3 elements
• condition 5 is satisfied if $K$ acts non-trivially
• condition 3 is satisfied if $\{h^{-1}\sigma_k(h) : h\in H\}$ is infinite for all $k \in K$
• condition 1 is satisfied if $K$ acts with infinite orbits on $H$ and condition 3 is satisfied.

Answer 2

I think a lattice in the rank 3 solvable Lie group Sol works. For any 2x2 matrix A ∈ SL₂ ℤ with tr(A) > 3, take the extension G of H=ℤ² by ℤ, where 1 acts by A on ℤ². We may write elements of G as (k, h), k ∈ ℤ, h ∈ ℤ². The subgroups (k,0) and (0,h) are additive in the coordinates, and (k,0)(0,h)=(k,h). We have the relation (0,h)(1,0) =(1,0)(0, A(h)) (so the 5th condition holds). For example, the matrix

$$\begin{pmatrix} 2 & 1 \\\\ 1& 1\end{pmatrix}$$

gives rise to the fundamental group of 0-framed surgery on the figure 8 knot complement. G is countable icc, and H=ℤ² is a normal subgroup, G/H = ℤ, so the 2nd and 4th conditions are satisfied. The 3rd condition is satisfied, since for h ∈ H= ℤ², (0,h)(k,g)(0,-h) = (k, g+A^k(h) -h), which one can see is infinite as one varies h (for k ∈ ℤ - 0 ).

Answer 3

This is off the cuff, and I'm very much a dilettante when it comes to group theory, so I hope there isn't an error in what follows. Corrections welcome, of course.

[EDIT: it has been pointed out below that the group given below doesn't quite work. I'm leaving the bulk of this "answer" here, in case it suggests a correct solution or warns people off the same mistake I made.]

I think the group $G$ with presentation $\langle g, h | hg =gh^n \rangle$, where $n\geq 2$, will do the job, with $H$ being the group generated by $h$. [Conditions 2,5]

Elements of this group have a normal form with all the $g$s on the left and all the $h$s on the right. [EDIT: this is not quite right, one has to take care over negative powers of $g$.] Multiplying on the left or on the right by an element of $h$ should, once we bring to normal form, not change the index of $g$ in the normal form, and so there are infinitely many double $H$-cosets, taking care of Condition 4.

Also, given an element of the form $g^ah^b$ where $a\neq 0$, then some back-of-the-envelope scribbling indicates that repeated conjugation by $h$ ought to increase the absolute value of the index of $h$ in the resulting normal form, so that conjugation by $h$ cannot be an operation of finite order. That would take care of Condition 3. [EDIT: this is incorrect/insufficient, see comments below.]

Finally, I think Condition 1 should follow from some further case-by-case analysis (given a non-identity element in $H$, conjugate repeatedly by $g$; and all the elements in $G-H$ are taken care of by condition 3).

(The group $G$ is an example of a Baumslag-Solitar group, and these beasts have been quite well studied over the years, I'm told. I don't know if you can do similar games with other B-S groups.)

Answer 4

A more singular example:

Take an infinite index inclusion of abelian groups $K\subset H$. Let a non-trivial group $L$ act on $K$ by automorphisms. Then the amalgamated free product $G=H\underset{K}{\ast} (K\rtimes L)$ satisfies the conditions. Moreover, $L(H)\subset L(G)$ $\,$is a singular masa. On can use the results of Ioana, Peterson and Popa for this, but maybe there are more elementary ways to see this.