5
$\begingroup$

Let $\mu_n$ denote the group of $n$-th roots of unity. In $\mathbb{C}$, this group has exactly $n$ elements. For a positive integer $n$ and a prime number $p$, using the canonical isomorphisms $\mu_{p^n} \cong \mathbb{Z}/p^n\mathbb{Z}$ for all $n$ as a direct system of factor groups, and the multiplication-by-$p$ homomorphisms $$\mathbb{Z}/p^n\mathbb{Z} \rightarrow \mathbb{Z}/p^{n+1}\mathbb{Z},$$ one sees that the direct limit of this system is the Prüfer group $$\mathbb{Z}(p^\infty) = \{x \in \mathbb{C}^\times: x^{p^n} = 1\,\, \mathrm{for\,\,some} \,\,n\}.$$

Similarly, as seen in this question Union of all finite cyclic groups, we can consider a direct system of cyclic group $\mathbb{Z}/n\mathbb{Z}$ and homomorphisms $\mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z}$ if $n|m$. The resulting direct limit is seen to be $\mathbb{Q}/\mathbb{Z}$, the set of rational numbers under addition modulo $1$. We have $$\mathbb{Q}/\mathbb{Z} = \bigoplus_p \mathbb{Z}(p^\infty).$$

Question. Can we view $\mathbb{Q}/\mathbb{Z}$ as being isomorphic to $\mu_\infty = \{x \in \mathbb{C}^\times:x^n = 1\,\,\mathrm{for\,\,some}\,\, n\in \mathbb{Z}^+\}$, i.e., the subgroup of all roots of unity? If so, is there any way to define the isomorphism?

$\endgroup$

1 Answer 1

8
$\begingroup$

Yes, send $a/b \in \mathbb{Q}/\mathbb{Z}$ to $\zeta_{b}^{a}$, where $\zeta_{b}$ is a primitive $b$-th root of unity.

$\endgroup$
2
  • $\begingroup$ Indeed... should've given some more thought before posting... Thanks! $\endgroup$ Jun 27, 2021 at 5:31
  • 2
    $\begingroup$ This is $\frac ab\mapsto e^{2\pi i\frac ab}$. $\endgroup$
    – robjohn
    Jun 27, 2021 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.