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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

|

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

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+Ak(h) -h), which one can see is infinite as one varies h (for k ∈ ℤ - 0 ).

1

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

|2 1|

|1 1|

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+Ak(h) -h), which one can see is infinite as one varies h (for k ∈ ℤ - 0 ).