# The group of positive rationals is isomorphic to its subgroup.

By this we know that the group of positive rationals under multiplication isomorphic to its subgroup consisting of rationals with odd numerators and denominators?

I tried to give a direct isomorphism between these two groups, Is my attempt right?

My attempt: Let $$2,3,5,7, \ldots, p_i,p_{i+1}, \ldots,$$ be enumeration of primes in increasing order. $$f(1)=1, f(2)=3, f(3)=5,\ldots, f(p_i)=p_{i+1}$$ and extend the map to any positive rational $$p/q$$ so that it is homomorphism.

I have one more similar question, We know that the group of non zero rationals is not isomorphic to a group of positive rationals since one has an element of order 2 and the other doesn't? Is there any other way to prove that these two groups are not isomorphic?

• $\mathbb{Q}^\times \cong \{ 1,-1\} \times \prod_p' p^\mathbb{Z}$ where $\prod'$ means restricted product that is all but finitely many terms must be $=1$, so $\mathbb{Q}^\times \cong C_2 \times \sum' \mathbb{Z}$ – reuns Jan 31 at 10:59

The first part is true. This is because the fundamental theorem of arithmetic guarantees each positive rational has unique representation as product of prime powers $$x=\prod p_i^{\alpha_i}$$, so $$f(x)=\prod f(p_i)^{\alpha_i}=\prod p_{i+1}^{\alpha_i}$$ because it's a homomorphism. So $$f(x)=1$$ implies the $$p_{i+1}$$-adic valuation of $$f(x)$$ is $$0$$ for each $$i$$, hence that $$\alpha_i=0$$ and $$x=1$$.
The second part comes down to showing each odd rational $$x$$ can be written as a product $$\prod p_i^{\alpha_i}$$ for primes at least $$3$$, which is obvious. We are done.