# Subgroups of the group of all roots of unity.

Let $$G=\mathbb{C}^*$$ and let $$\mu$$ be the subgroup of roots of unity in $$\mathbb{C}^*$$. Show that any finitely generated subgroup of $$\mu$$ is cyclic. Show that $$\mu$$ is not finitely generated and find a non-trivial subgroup of $$\mu$$ which is not finitely generated.

I can see that a subgroup of $$\mu$$ is cyclic due to the nature of complex numbers and De-Moivre's theorem. The second part of this question confuses me though since, by the same logic, should follow a similar procedure. Perhaps it has something to do with the "finite" nature of the subgroup in comparison to $$\mu$$. Any assistance would be helpful ! Thank you in advance.

• Yes, there's a big difference between "finitely generated" subgroups of $\mu$, and $\mu$ itself or any nontrivial subgroup of $\mu$ that is NOT finitely generated. Mar 26, 2013 at 0:02
• See $\mu$ as $\mathbb Q / \mathbb Z$.
– lhf
Mar 26, 2013 at 17:38

Edit: after reading your comments, I would just like to point out that $$\mu = \{z : \exists n \in \mathbb{N}, z^n = 1\}$$. This union of roots of the equations $$z^n-1 = 0$$ for each $$n$$, not for a particular one.

For the first part, suppose we have a subgroup $$H$$ generated by $$z_1,z_2,\dots,z_m$$, where $$z_k$$ has order $$n_k$$.(So that e.g. $$n_3$$ is the least integer $$j$$ such that $$z_3^j = 1$$). Then it follows that $$H$$ is also generated by $$e^{\frac{2\pi i}{n_1}},e^{\frac{2\pi i}{n_2}},\dots,e^{\frac{2\pi i}{n_m}}$$. (Why?)

Now it is relatively easy to show there is an element of $$H$$ with order $$l =\operatorname{lcm}(n_1,\dots,n_m)$$, and that it generates the group. (Assuming the fundamental theorem of arithmetic and writing the $$n_k$$ as products of primes.) Hence $$H$$ is cyclic (and finite).

The reason that $$\mu$$ is not finitely generated is that it is infinite, and the above proof shows that any finitely generated group is finite. You can also prove that any finitely generated subgroup of $$\mu$$ is finite more easily from first principles as follows, using the (important) fact that every element of $$\mu$$ has finite order: If $$H$$ is a subgroup generated by $$z_1,\dots,z_m$$ where the order of $$z_k$$ is $$n_k$$ as above, $$H$$ has at most $$n_1\times n_2\times\dots \times n_m$$ elements.

To find a non-trivial subgroup which is not finitely generated, from the above, we only need to show that it is infinite. We could also try and come up with one from first principles, if it is not finitely generated, then it must have elements of arbitrarily large orders. (Since there are only a finite number of elements of $$\mu$$ with order less than or equal to $$n$$ for any natural number $$n$$.)

• -1 because the statement "..any finitely generated group is finite" is not true as written, since e.g. $\Bbb{Z}$ is generated by 1. Perhaps you meant any finitely generated subgroup of this particular group..? Dec 27, 2019 at 19:41

The first part should probably include "Show that any f.g. subgroup of $\,\mu\,$ is cyclic finite...". From here it follows at once that $\,\mu\,$ cannot be f.g. as it isn't finite.

For a non-trivial non f.g. subgroup think of the roots of unit of order $\,p^n\,\,,\,\,p\,$ a prime, and running exponent $\,n\in\Bbb N\,$ ...

• Gosh, I'm a little confused. So we're saying that $\mu$ is not finite. But if the roots of unity are the solutions to $z^n=1$, this means that solving $z^n-1=0$ will have at most $n$ complex roots...which suggests that $\mu$ is finite. :'( Mar 26, 2013 at 10:39
• You don't need that a finitely generated subgroup is finite. You just need to notice that $\mu$ isn't cyclic. Oct 24, 2016 at 19:44