# Prove (Q+, *) is isomorphic to a proper subgroup of itself

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1. Dec 9, 2016

### gummz

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.

2. Dec 9, 2016

### Staff: Mentor

Looks fine, beside some minor issues on the notation (the missing definition of G, sqrt cannot be defined, it should be $\phi^{-1}$ (preimage) instead, and the equation under "subgroup" is a bit short, i.e. doesn't introduce a,b, injectivity could be a little more explicitly, i.e. why does x=y follow, resp. what properties of $\mathbb{Q}$ do you use).

3. Dec 9, 2016

### gummz

Thank you so much again fresh_42!