I’m trying to understand the meaning of Let $G = \{z\in \mathbb{C} \mid z^n=1$ for some $n\in \mathbb{Z}^+\}$ I’m trying to understand the meaning of Let $G = \{z\in \mathbb{C} \mid z^n=1$ for some $n\in \mathbb{Z}^+\}$
I need to prove:
A) that $G$ is a group under multiplication
B) that $G$ is not a group under addition
However, I’m sure I don’t quite know what the group $G$ is... or $z$ for that matter. I feel like $z$ must be $\sqrt{-1}$ and $n$ are even integers, but am I missing something?
If this is correct it would be a group under multiplication because basically $1*1 =1$ which is still in the group and not under addition because $1+1=2$ which is not in the group.
 A: I assume that $n$ is not fixed as it is defined inside the set definition.
You have properly showed that B) is true: $G$ is not even closed under addition. To show A), i.e. that $(G, \cdot)$ is a group, you need to prove 4 things:


*

*Closure under $\cdot$ (multiplication). If $x \in G$ and $y \in G$, then there are $n, m \in \mathbb Z^+$ such that $x^n=1$ and $y^m=1$. Then $(xy)^{nm}=x^{nm}y^{nm}=(x^n)^m(y^m)^n=1^m1^n=1$, so $xy \in G$.

*The associativity of the operation $\cdot$. This is inherited from $\mathbb C$.

*Presence of the identity element. Obviously, $1 \in G$.

*Closure under taking an inverse element. Indeed, if $z \in G$, then $z^n=1$ for some $n \in \mathbb Z^+$. Hence, $(z^{-1})^n=z^{-n}=(z^n)^{-1}=1^{-1}=1$.

A: $G$ consists of all $n$-roots of unity for some positive integer $n$. Explicitly, any element of $G$ can be uniquely written as $exp({\frac{2i\pi k}{n}})$ for some $k\in \{ 0,\ldots ,n-1\}$ that is prime with $n$.  
It is a multiplicative group because $\frac{2i\pi k}{n}+\frac{2i\pi k'}{n'} = \frac{2i\pi (kn'+k'n)}{nn'}$, so that the multiplication of an $n$-root with an $n'$-root is an $nn'$-root of unity. Your example proves that it is not an additive group.
A: It need not be $\sqrt{-1}$.
Suppose we fix $n$, then all $z$ that satisfies $z^n=1$ would be $\exp\left( \frac{2k\pi i}{n}\right)$ where $k=0, \ldots, n-1$.  It is easy to check that they are indeed solutions of $$\exp\left( \frac{2k\pi i}{n}\right)^n=\exp\left( \frac{2\pi ki}{n}\cdot n\right)=\exp(2\pi k i)=1$$
Also they are distinct for all those values of $k$ since we can plot it on the Argand diagram and see that they are spreaded out evenly.
You might like to read up on root of unity.
