Saturday 7 December 2019

group theory - Is there an operational isomorphism from $(mathbb{Z},+)$ to $(mathbb{Q}^{+},cdot)$?



Let $\left(\mathbb{Z},+\right)$ and $\left(\mathbb{Q}^{+},\cdot\right)$ be groups (let the integers be a group with respect to addition and the positive rationals be a group with respect to multiplication). Is there a function $\phi\colon\mathbb{Z}\mapsto\mathbb{Q}^{+}$ such that:





  • $\phi(a)=\phi(b) \implies a=b$ (injection)

  • $\forall p\in\mathbb{Q}^{+} : \exists a\in\mathbb{Z} : \phi(a)=p$ (surjection)

  • $\phi(a+b) = \phi(a)\cdot\phi(b)$ (homomorphism)



? If so, provide an example. If not, disprove.


Answer



If such an isomorphism existed it would of course be onto so there would exist some $n \in \mathbb{Z}$ such that $\phi(n) = \frac{1}{2}$ for example. But then, $$\phi(n) = \phi(1+\cdots + 1) = \phi(1)\cdots \phi(1) = \phi(1)^n = \frac{1}{2}.$$ This implies $n=1$ since otherwise $\left(\frac{1}{2}\right)^{1/n} \notin \mathbb{Q}$. So, $\phi(1) = \frac{1}{2}$. Thus, for any $n \in \mathbb{Z}$, $\phi(n) = \phi(1)^n = \frac{1}{2^n}$, so clearly $\phi$ is not onto since we only achieve powers of two in the image. So such an isomorphism cannot exist.


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