Let $U_p$ be the subgroup of $\mathbb{C}^*$ satisfying that: for any $x\in U_p$, there exists an integer $n$ such that $x^{p^n}=1$.
Let $K$ be a algebraic closed field with characteristic $0$. Denote $\mu_{p}$ be the subset of $K$ satisfying: for any $x\in\mu_p$, there exists an integer $n$ such that $x^{p^n}=1$.
Intuitively, $\mu_p$ and $U_p$ are isomorphic. Can anyone show me a clear and rigorous proof?
For any algebraically closed field $K$ of characteristic $0$, you propose to consider the subgroup $\mu_{p^{\infty}}$ of $p$-power roots of unity: equivalently, the $p$-primary torsion subgroup of $K^{\times}$. This group is the union of its finite subgroups $\mu_{p^n}$ of $p^n$th roots of unity, each of which is a cylic group of order $p^n$, since by the derivative criterion the polynomial $t^{p^n} -1$ has distinct roots in any algebraically closed field of characteristic different from $p$.
It is easy to see that there is up to isomorphism exactly one abelian group all of whose elements are $p$-power torsion and such that for each $n \in \mathbb{Z}^+$, the $p^n$-torsion subgroup is cyclic of order $p^n$. Indeed, we can chose a generator $\sigma_n$ of each order $p^n$-cyclic subgroup in such way so that $\sigma_{n+1}^p = \sigma_n$. Such a choice of generators induces an isomorphism from our group $\mu_{p^{\infty}}$ to $\varinjlim \mathbb{Z}/p^n \mathbb{Z}$, where the map from $\mathbb{Z}/p^n \mathbb{Z} \rightarrow \mathbb{Z}/p^{n+1} \mathbb{Z}$ sends $1$ to $p$.
If you don't know what a direct limit is, this is a good opportunity to learn, but on the other hand it is not necessary here. Mapping a "compatible sequence of generators" $\sigma_n$ in one such group to a compatible sequence of generators in another such group induces an isomorphism between them.
By the way, the unique abelian group in question is rather famous: it is called the Prufer p-group.
It is not necessary to assume that the characteristic is$~0$, just that it is different from $p$. Then the polynomials$~X^{p^n}-1$ are squarefree, so that all its $p^n$ roots in the algebraic closure are distinct. It is well known that any finite subgroup of the multiplicative group of a field is cyclic, so for every $n$ we've got a cyclic subgroup of order$~p^n$, and we are looking at the union of all these subgroups. Moreover a cyclic group has exactly one (cyclic) subgroup of every order dividing its own order, so there is no difficulty in finding which subset of the $p^{n+1}$-st roots of unity should be the set of $p^n$-th roots of unity.
In order to establish an isomorphism between the groups $U_p$ for two different fields, it suffices to choose for every $n$ an isomorphism between their cyclic subgroups of order $p^n$ in a way that is compatible (commutes) with the inclusions of these subgroups into each other. This can be done by choosing a specific generator for each such cyclic subgroup for both fields (the isomorphism will map the chosen generator to the chosen one for the corresponding subgroup), in such a way that if $g_n,g_{n+1}$ are generators for the subgroups of respective orders $p^n,p^{n+1}$ in the same field, then $g_{n+1}^p=g_n$. This is clearly possible successively while advancing the $n\in\mathbf N$.
