# Q+ not isomorphic to Q

1. Oct 10, 2011

### Essnov

I'm taking a first course in algebra, and in my textbook, there is the following problem:

a) Show that, for every natural # n, there is a subgroup A of Q+ such that |Q+/H| = n.

b) Suppose that B is a proper subgroup of Q. Show that |Q / B| = ∞.

c) Conclude that Q+ ≠ Q.

I did parts a and b just fine, but I'm not sure if I see how to (properly) conclude from them that Q+ ≠ Q.

I'm thinking that if there is to be a bijection from Q+ → Q, then there should be a bijection from Q+ / A → Q / B, but that's not possible since Q+ / A and Q / B should at least have the same order.

So basically if P : Q+ → Q is a bijection, the following are also bijections:
For g in Q+, h in Q,
R taking g to gA
S taking h to hB

So the composition S o P o R-1 should be a bijection from Q+ / A → Q / B, which is (I think) a contradiction, so there should be no such P.

Am I doing this properly or am I missing something more obvious?

2. Oct 10, 2011

### micromass

Staff Emeritus
Let $f:\mathbb{Q}^+\rightarrow \mathbb{Q}$ be an isomorphism. What can you say about f(A)? (where A is the subgroup with $|\mathbb{Q}^+/A|=3$ or another number).

3. Oct 10, 2011

### Essnov

I cannot think of anything helpful. f(A) will be a subgroup of Q? I feel like we should have | Q / f(A) | = n but I don't know how to show this.

4. Oct 10, 2011

### micromass

Staff Emeritus
Show that

$$\mathbb{Q}^+/A \rightarrow \mathbb{Q}/f(A): x+A\rightarrow f(x)+f(A)$$

is a bijection. So show

1) it is well-defined
2) it is injective
3) it is surjective

5. Oct 10, 2011

### Essnov

Thank you very much for your help :)