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Consider the group of all complex roots of unity, $\mathbb{Q}/\mathbb{Z}$ (where both groups are additive groups). I was wondering what its automorphism group is ?

Boccherini
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  • Maybe it helps to first consider the automorphism group of the rationals, https://math.stackexchange.com/questions/511796/automorphism-from-bbb-q-to-bbb-q – Gerry Myerson May 25 '22 at 13:09
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    Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Shaun May 25 '22 at 13:42
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    $\Bbb Q/\Bbb Z$ is a direct sum of Prüfer $p$-groups and and each of those has at least one automorphism for each unit of the p-adic integers, see this post. In fact, ${\rm Aut}(\Bbb Q/\Bbb Z)\cong \prod_p \Bbb Z_p^{\times}$. – Dietrich Burde May 25 '22 at 13:47

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Instead of looking just at automorphisms of this abelian group, let’s look at endomorphisms to get a ring and then the automorphisms are the units of this ring.

The torsion group $G = \mathbf Q/\mathbf Z$ is the direct sum of its subgroups $G_p$ of $p$-power elements. That is a fancy way of saying nontrivial roots of unity are unique products of finitely many nontrivial roots of unity of $p$-power order for different primes $p$. An endomorphism of $G$ induces an endomorphism of each $G_p$ and a set of choices $(f_p)_p$ of endomorphisms of each $G_p$ as $p$ runs over the primes defines an endomorphism of $G$. This bijection between elements of $\prod_p {\rm End}(G_p)$ and elements of $G$ shows the endomorphisms of $G$ are the direct product of the endomorphisms of each $G_p$.

What are the endomorphisms of the group $G_p$, which is isomorphic to the group of $p$-power roots of unity? They are the $p$-adic integers (each one acting as an exponent on all of $G_p$). Therefore the endomorphism ring of $G$ is $\prod_p \mathbf Z_p$, and passing to units shows the answer to your question (what are the automorphisms of $G$?) is $\prod_p \mathbf Z_p^\times$.

This answer is perhaps much more sophisticated than you expected (if you never heard of $p$-adic integers), because the description of the endomorphisms of the additive group $\mathbf Q$ is much simpler: they are the multiplication maps $r \mapsto sr$ for different rationals $s$, so the automorphism group of $\mathbf Q$ is $\mathbf Q^\times$. The lesson here is that the (additive) quotient group $\mathbf Q/\mathbf Z$ is a more complicated object than the additive group $\mathbf Q$: the former is an infinite torsion group while the latter is an infinite uniquely divisible group.

KCd
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  • Is this group isomorphic to the group of units of the profinite integers $\widehat{\mathbb{Z}}$? – Boccherini May 25 '22 at 14:50
  • @Boccherini yes, but I didn't want to use that terminology. The endomophism ring is the profinite integers. – KCd May 25 '22 at 15:10
  • I think your third sentence should stress distinct primes $p$ (to contrast with unique factorization in $\Bbb Z_{>0}$ where repetition of primes (in a somewhat different role) is allowed. (Also there are minor matters as either ignoring or imposing the order in which factors are taken into account, not considering $1$ a candidate $p$-power order element, and allowing an empty product.) – Marc van Leeuwen May 26 '22 at 08:23
  • @MarcvanLeeuwen I made some edits. – KCd May 26 '22 at 13:02