Let $f \in \mathbb{Z}[x]$ be an irreducible monic polynomial, with $f(0) \ne 0$, and $p$ a prime number. Also, let $a_1, \ldots, a_n \in \overline{\mathbb{Q}}_p$ be all the roots of $f$ taken in the algebraic closure of the $p$-adic numbers, and let $b_1, \ldots, b_n \in \mathbb{C}$ be all the roots of $f$ taken in the complex numbers. Finally, let $A$ be the multiplicative group generated by $a_1, \ldots, a_n$ and let $B$ be the multiplicative group generated by $b_1, \ldots, b_n$.
Is is true that there is a group isomorphism $\phi : A \to B$, such that $\phi(a_i) = b_{\pi(i)}$ for all $i \in \{1, \ldots, n\}$ and some permutation $\pi$ of $\{1, \ldots, n\}$? If so, how can I construct it?
This question is related to this other.
Although Mariano Suárez-Álvarez answer is good. I would like to make work mercio's idea.
Please, let me know is this is fine:
Inside $\overline{\mathbb{Q}_p}$ we can construct an algebraic closure of $\mathbb{Q}$ as $$\overline{\mathbb{Q}}^{\;\prime} = \{a \in \overline{\mathbb{Q}_p} : f(a) = 0 \text{ for some } f \in \mathbb{Q}[x]\}$$ (Exactly how $\overline{\mathbb{Q}}$ is usually constructed inside $\mathbb{C}$). Now $a_1, \ldots, a_n \in \overline{\mathbb{Q}}^{\;\prime}$ while $b_1, \ldots, b_n \in \overline{\mathbb{Q}}$. But the algebraic closure of $\mathbb{Q}$ is unique up to isomorphism, so there exists an isomorphism $\overline{\mathbb{Q}}^{\; \prime} \to \overline{\mathbb{Q}}$ that restricted to $A \to B$ does the job. Right?
Thank you again.
The field $\overline{\mathbb Q}_p$ is isomorphic to $\mathbb C$, so yes. Fix any isomorphism $\overline{\mathbb Q}_p\to\mathbb C$, show that it restricts to a bijection $A\to B$, and voilà.
(See this, for example, for a discussion of the isomorphism.)