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.
If you are only looking at groups, then the name of the operation (addition, multiplication, etc.) does not matter...
If you write your group operation using "multiplication" ($x y$) then the definition of generator looks like "$x \in G$ is a generator for $G$ if every element in $G$ is of the form $x^n$ for some integer $n$."
If you write your group operation using "addition" ($x+y$), then the definition looks like "$x \in G$ is a generator for $G$ if every element of $G$ is of the form $nx$ for some integer $n$."
In complete generality, if the operation is expressed as $x*y$, then $x$ is a generator of $G$ if every element of $G$ is of the form
$$\underbrace{x*x*\cdots*x}_{n}$$
for some integer $n$.
Best Answer
For any ring $R$, it is standard to consider the set of units $$R^\times=\{a\in R:\textsf{there exists some }\,b\in R\,\textsf{ with }\,a\cdot b=b\cdot a=1\}$$ as a group under the multiplication operation of $R$. (Here's the relevant Wikipedia page.)
(In fact, if $R$ is not the trivial ring, then $R^\times$ is not closed under the addition operation of $R$, so it certainly will not be a group under that operation.)
You're told to look at the group of units of $R=\mathbb{Z}_{14}$, namely the set $$(\mathbb{Z}_{14})^\times=\{\overline{1},\overline{3},\overline{5},\overline{9},\overline{11},\overline{13}\}$$ under the multiplication operation of $\mathbb{Z}_{14}$, and you are looking for the elements $x\in(\mathbb{Z}_{14})^\times$ that generate the entire group, i.e., the elements with the property that $$(\mathbb{Z}_{14})^\times=\{x^n:n\in\mathbb{Z}\}$$ (Here's the relevant Wikipedia page.)