Given $a\in G$, $\langle a\rangle$ is defined to be the smallest subgroup of $G$ containing $a$ as an element. Knowing that $\langle a\rangle=\langle b\rangle$ isn't enough to conclude that $a=b$, though. For example, we always have $\langle a\rangle=\langle a^{-1}\rangle,$ no matter the order of $a$.
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.
Best Answer
It's isomorphic to $\Bbb Z_6$, so has $\varphi(6)=2$ generators: $e^{2\pi ki/6},\, k=1,5$. These are the so-called primitive roots.
Since $U_6$ is cyclic, it has a unique subgroup of each order dividing $6$. For instance, $\{1,-1\}$ is the one of order $2$, generated by $-1$.