Automorphism group of $\mathbb{Q}/\mathbb{Z}$

Question

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 ?

Answer 1

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.