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Here are a few (simple) questions on roots of unity I'm a little confused about.

Is the automorphism group of the group $\mu_{\infty}$ of all roots of unity $\widehat{\mathbf{Z}}^{\times}$, and what are its orbits?

For the first question, the answer should be yes because the automorphism group of a finite cyclic group of order $n$ is $(\mathbf{Z}/n)^{\times}$.

Suppose I define $G\subset\mu_{\infty}$ to be the union (not the subgroup generated by) of the subgroups of the form $\mu_{p^n-1}$ for $p$ prime and $n\ge 1$. Do we have $G=\mu_{\infty}$? Does $G$ generate $\mu_{\infty}$? Is $G$ dense in the circle group $T\subset\mathbf{C}^{\times}$ with respect to the topology induced by the complex topology?

I would expect the answer to the first question to be no, but I don't know about the second and third.

Tim
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    For the second question: take any $m$th root of unity, then it belongs to $G$, because by Dirichlet's theorem on arithmetic progressions there is a prime of the form $p=am+1$. – danneks Oct 31 '23 at 05:28
  • (And if $p$ is fixed, then we don't get any $p^k$-th roots of unity.) – Amateur_Algebraist Oct 31 '23 at 06:15
  • For the 1st question. The map $r\mapsto e^{2i\pi r}$ gives an iso between ${\bf Q}/{\bf Z}$ and $\mu_\infty$. Hence the answer is yes by https://math.stackexchange.com/questions/4458363/automorphism-group-of-mathbbq-mathbbz – Damian Rössler Oct 31 '23 at 14:31

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