Q:

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, $(z^{-1})^n=z^{-n}=(z^n)^{-1}=1^{-1}=1$.

