It is an exercise in Kargapolov to prove that a set G with a binary operation · is a group, if
(R+, · ) ≅ (R, +) via x ↦ ln x, but (by M. A. Armstrong's Groups and Symmetry (1988), pp.34–5) (Q, +) ≇ (Q+, ×), for were φ : Q → Q+ an isomorphism, then for some x in Q we'd have 2 = φ(x) = φ(x/2 + x/2) = φ(x/2)2, which is impossible because φ(x/2) is rational and √2 is not.
Three exercises (1.2.2–1.2.4):
Proposition. If a, b ∈ G then |ab| = |ba|.
Proof. If |ba| < |ab| then
(|ab| times) = a(ba)⋯(ba)b
(⩾ |ba| times) = (ab)⋯, contradicting the minimality of the order of |ab|. ▮
Alternate proof. One may note that 1 = (ab)|ab| ⇒ b−1a−1 = (ab)|ab|−1, so that (ba)|ab| = b(ab)|ab|−1a = bb−1a−1a = 1, whence |ba| ⩽ |ab|. Reversing the roles of a and b establishes the result. ▮