Prove that Q+, the group of positive rational numbers under multiplication, is isomorphic to a proper subgroup of itself.
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+
phi(x) = phi(y)
x2 = y2
x, y > 0
x = y
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.
We have phi(x*y) = (xy)2 = x^2y2
phi(x)phi(y) = x2y2
Therefore, phi is an isomorphism between Q+ and a proper subgroup of itself.