Determine whether the following groups are isomorphic.

Question

Let $G=\lbrace \frac{p}{q}:p,q\in\mathbb{N} \rbrace$, and let $G'=\lbrace \frac{p}{q}:p,q\in\mathbb{N},p$ and $q$ both are odd $ \rbrace$. It is clear that both $G$ and $G'$ are groups under common multiplication.

Now the question : Is $G$ and $G'$ isomorphic as groups ?

I clearly don't have any idea how to prove/disprove it. Please give me some hints.

Answer 1

Hints:

Each positive rational number $x$ has a unique factorization as $$x = 2^{e_2}\cdot 3^{e_3} \cdot 5^{e_5}\cdot 7^{e_7} \cdots$$ where $e_2,e_3,e_5,e_7, ...\;$are integers.

Define $f:G \to G'$ by $$ f(2^{e_2}\cdot 3^{e_3} \cdot 5^{e_5}\cdot 7^{e_7} \cdots) = 3^{e_2}\cdot 5^{e_3} \cdot 7^{e_5}\cdot 11^{e_7} \cdots $$

Verify that $f$ is an isomorphism.

Answer 2

Consider the map $r:\mathbb{N}\rightarrow\mathbb{N}$ that sends $1$ to $1$, $p_{i}$ to $p_{i+1}$ for all $i\geq 1$ ($p_{i}$ denoting the $i$th prime, so that in particular, $p_{1}=2$), and for $n\in\mathbb{N},$ $n=p_{j_{1}}^{k_{1}}\cdots p_{j_{\ell}}^{k_{\ell}},$ $r(n):=p_{j_{1}+1}^{k_{1}}\cdots p_{j_{\ell}+1}^{k_{\ell}}.$ If we define $R:G\rightarrow G'$ by $R(p/q)=r(p)/r(q),$ this clearly maps into $G',$ and by our definition of $r$ and $R$, $$R\left(\frac{p}{q}\cdot\frac{p'}{q'}\right)=R\left(\frac{pp'}{qq'}\right)=\frac{r(pp')}{r(qq')}=\frac{r(p)r(p')}{r(q)r(q')}=R\left(\frac{p}{q}\right)R\left(\frac{p'}{q'}\right),$$ so this is a group homomorphism (you'll need to check that it's well-defined, since $p/q=p'/q'$ is possible for $p\neq p',q\neq q',$ but this is easy). Moreover, it is injective, since $R(p/q)=1$ implies $r(p)=r(q),$ which implies $p=q.$ To show that it is surjective, suppose $p$ and $q$ are odd natural numbers. Then their prime expansions do not have any powers of $2$, so if $p=p_{j_{1}}^{k_{1}}\cdots p_{j_{\ell}}^{k_{\ell}},$ then all $j_{i}\geq 2$, and therefore $r^{-1}(p)=p_{j_{1}-1}^{k_{1}}\cdots p_{j_{\ell}-1}^{k_{\ell}}$ is well-defined, and similarly for $r^{-1}(q).$ Since $R(r^{-1}(p)/r^{-1}(q))=p/q$, $R$ is surjective, and the obvious definition of $R^{-1}(p/q)$, $p/q\in G$, is $r^{-1}(p)/r^{-1}(q).$ A bijective group homomorphism is an isomorphism, so we are done.