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Contains a free group as a finite index subgroup). Moreover, the conformal dimension of zero is attained if and only if G is virtually cyclic. 42 3. GROMOV HYPERBOLIC SPACES AND BOUNDARIES In fact, to show that G is virtually free, it would suffice to assume that Cdim ∂∞ G < 1, since this implies that the boundary is totally disconnected. Therefore as a corollary we can conclude that Cdim ∂∞ G ∈ {0} ∪ [1, ∞) for any hyperbolic group G. We will see later on that no metric space can have conformal dimension in the interval (0, 1).

Thus, for p, q ∈ Kn−1 × Im(K), we define p ◦ q by the identity τq ◦ τp = τp◦q . 5) p ◦ q = (x, t) ◦ (y, s) = (x + y, t + s + 2 Im x, y ). 4. Let K = R. The model space is Rn−1 , equipped with the abelian (Euclidean) group law. 5. Let K = C. The model space is Cn−1 ×Im(C) which we identify with Cn−1 × R. Equipped with the group law in the preceding paragraph, this space is the (n − 1)st Heisenberg group. It is a sub-Riemannian stratified Lie (Carnot) group of step two, with a (2n − 2)-dimensional horizontal space and one dimensional center.

17 implies that 1 1 1 H∞ (Z) ≥ H∞ (f (Z)) = H∞ ([0, +∞)) = L1 ([0, +∞)) = ∞. Here we used the fact that f (Z) is an unbounded interval in the nonnegative real axis containing 0, hence is equal to [0, ∞). Now suppose that Z is bounded. Let > 0 and fix z0 ∈ Z so that sup d(z, z0 ) ≥ diam Z − . z∈Z Let f be as above. Again, f (Z) is an interval in the nonnegative real axis containing 0, with L1 (f (Z)) ≥ sup |f (z) − f (z0 )| ≥ diam Z − . z∈Z As before we find 1 1 H∞ (Z) ≥ H∞ (f (Z)) = L1 (f (Z)) ≥ diam Z − .