Abstract Algebra – Cyclic Properties of Multiplicative Group G of Complex $2^n$ Roots of Unity

Question

Consider the multiplicative group G of all the complex $2^n$ roots of unity, $n=0,1,2,\ldots$

I am asked to verify whether $G$ is a cyclic group and whether it has a finite set of generators.

The answer to both these questions is coming as yes as

$$G=\left\{e^{\frac{2\pi ik}{2^n}}\bigg|k=0,1,2,\ldots ,n-1\right\}.$$

and it has only one generator i.e.$e^{\frac{2\pi i}{2^n}}$

But the answer to both these questions is given as no. Any help?

Answer 1

It is false because cyclic groups means that every element can be written as a power of one single element. However, your candidate $e^{2\pi i/2^n}$ cannot generate, for example $e^{2\pi i/2^{n+1}}$. The number of generators is necessarily infinite, since if we have a finite set of candidates

$$\left\lbrace\exp\left({2\pi i k_j\over 2^{n_j}}\right)\right\rbrace_{j=1}^m$$

Then note that this cannot generate $\exp\left({2\pi i\over 2^{N+1}}\right)$ with $N=\max\{n_j\}$, so no finite set of generators can work. In particular, it cannot be cyclic, since cyclic means $1$ single generator, which is a finite number.

Now how does $G$ look? Well, if we denote the $\left(2^{n}\right)^{th}$ roots of unity for fixed $n$ as $\mu_{2^n}$ then our group looks like their union, i.e.

$$G=\bigcup_n\mu_{2^n}$$

We see that a set of generators is $\left\lbrace\exp\left({2\pi i\over 2^n}\right)\right\rbrace_{n=1}^\infty$, and that it is isomorphic as a group to

$$\Bbb Z\left[{1\over 2}\right]/\Bbb Z$$

because the big group is $\left\{{a\over 2^n} \big| a\in\Bbb Z\right\}$ and modding out by $\Bbb Z$ tells us that $1\equiv 0$, which corresponds to the property that $\exp\left(2\pi i\right)=1$ is the identity of $G$. This is another way to tell that $G$ is not finitely generated--if you know a little number theory and abstract algebra together you know two important facts:

• If $G$ is not finitely generated, neither is $G/\langle x\rangle$ for any $x\in G$ (here we are using abelian so that $\langle x\rangle$ is a normal subgroup)
• $\alpha\in \Bbb C$ satisfies a monic, irreducible polynomial with coefficients in $\Bbb Z$ iff $\Bbb Z[\alpha]$ is finitely generated. (This is not hard to prove)

By the second fact, since $2x-1$ is the minimal polynomial over $\Bbb Z$ for $\alpha={1\over 2}$ and this is not monic, we know that $\Bbb Z\left[{1\over 2}\right]$ is not finitely generated, and since we mod out only by one generator, i.e. $1$, the quotient is not finitely generated.