1. The problem statement, all variables and given/known data Prove that Q+, the group of positive rational numbers under multiplication, is isomorphic to a proper subgroup of itself. 2. Relevant equations None 3. The attempt at a solution Not at all sure if this is legit. Let phi: Q+ --> G phi(x) = x2, x is in Q+ We will demonstrate that G c Q+ It is a subgroup: 1=e is in G, and ab-1 = x2 y-2 = (xy-1)2 is in G It is a proper subgroup: 2 is in Q+, but sqrt(2) is not in G and indeed not in Q+ One-to-one: phi(x) = phi(y) x2 = y2 x, y > 0 x = y Onto: Take some g in G. We have that sqrt(g) satisfies phi(sqrt(g)) = sqrt(g)2 = g. Therefore, there is an element in Q+ such that phi(x)=g. Operation preservation: We have phi(x*y) = (xy)2 = x^2y2 phi(x)phi(y) = x2y2 So phi(x*y)=phi(x)*phi(y) Therefore, phi is an isomorphism between Q+ and a proper subgroup of itself.