Algebra Solutions: The complex roots of unity form a multiplicative group

Let $G = \{ z \in \mathbb{C} \ |\ z^n = 1 \mathrm{for\ some} n \in \mathbb{Z}^+ \}$ .

Prove that $G$ is a group under multiplication (called the group of roots of unity in $\mathbb{C}$ ).
Prove that $G$ is not a group under addition.

- First we show that $G$ is closed under multiplication. Suppose $x,y \in G$ . Then there exist $n,m \in \mathbb{Z}^+$ such that $x^n = y^m = 1$ . Then $(xy)^{mn} = (x^n)^m \cdot (y^n)^m = 1 \cdot 1 = 1$ , so that $xy \in G$ .
- Multiplication on $G$ is associative since multiplication on $\mathbb{C}$ is associative.
- Note that $1^1 = 1$ so that $1 \in G$ , and that for all $x \in G$ we have $1 \cdot x = x \cdot 1 = x$ . So $1$ is an identity in $G$ under multiplication.
- Suppose $x \in G$ with $x^n = 1$ . Then $(x^{-1})^n = (x^n)^{-1} = 1^{-1} = 1$ , so that $x^{-1} \in G$ . Moreover, $x \cdot x^{-1} = x^{-1} \cdot x = 1$ . So every element of $G$ has a multiplicative inverse in $G$ .
Thus is a group under multiplication.
Note that $(-1)^2 = 1$ , so that $\pm 1 \in G$ . However $-1 + 1 = 0$ , and $0^k = 0 \neq 1$ for all $k \in \mathbb{Z}$ . So $G$ is not closed under addition and thus not a group.

Algebra Solutions