Is this inverse limit of groups of roots of unity the Prüfer group?

Question

Let $p$ be a prime number and for each $n\geq 0$, let $\mu_{p^{n}}$ denote the group of $p^{n}$-th complex roots of $1$.

Let us say that $n\leq m$ if $n$ divides $m$. If this is the case, we have a homomorphism \begin{equation} \varphi_{mn} \colon\mu_{p^{m}}\to\mu_{p^{n}},\qquad\qquad \zeta\mapsto \zeta^{\frac{m}{n}} \end{equation} which satisfies $\varphi_{ca}=\varphi_{ba}\circ\varphi_{cb}$ whenever $a\leq b\leq c$ and $\varphi_{ii}=1$ for any $i\geq 0$.

In other words, this forms an inverse system.

Considering each finite group $\mu_{p^{n}}$ as a discrete space, we then can take the inverse limit.

My question is:

Is this inverse limit the Prüfer group?

I suspect it is, but I didn't found any characterization of the latter as this limit, so I'm not sure.

Thanks! Any tips on how to prove this will be appreciated.

Answer 1

Here’s a rather more advanced take on your question than is necessary to answer it.

It all has an origin in the field $\Bbb Q_p$ of $p$-adic numbers. Just considering the additive structure, $\Bbb Q_p$ has an important compact subgroup $\Bbb Z_p$, and the quotient $\Bbb Q_p/\Bbb Z_p$ can be seen to be countable, indeed the union of the subgroups $\left(\frac1{p^n}\Bbb Z_p\right)\big/\Bbb Z_p$. each of the limiting groups is cyclic, of order $p^n$, and that limit is the Prüfer group.

The subgroup $\Bbb Z_p\subset\Bbb Q_p$ can be looked at as the inverse limit of the groups $\Bbb Z_p/p^n\Bbb Z_p\cong\Bbb Z/p^n\Bbb Z$. This is your group.

Answer 2

The direct limit of the groups $\mu_{p^n} \simeq \mathbb{Z} /p^n \mathbb {Z}$ (which is just the union of the $\mu_{p^n} $) is the Prüfer. But the inverse limit you're considering is the group $\mathbb{Z}_p$ of $p$-adic integers. It is an uncountable group in which $\mathbb{Z}$ can be embedded (two reasons why it can't be isomorphic to the Prüfer group).