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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?

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  • $\begingroup$ $\mathbb{Q}^\times \cong \{ 1,-1\} \times \prod_p' p^\mathbb{Z}$ where $\prod'$ means restricted product that is all but finitely many terms must be $=1$, so $\mathbb{Q}^\times \cong C_2 \times \sum' \mathbb{Z}$ $\endgroup$ – reuns Jan 31 at 10:59
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You provided an explicit map, defined it to be a homomorphism, but never actually showed it's an isomorphism. To show that your map, is an isomorphism, we need to show it has trivial kernel as well as that it is a surjection. (You also need to check that it's well-defined, but that's easy.)

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.

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