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The complex roots of unity form a multiplicative group


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.8

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.


Solution:

(1) 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 $G$ is a group under multiplication.

(2) 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.

Linearity

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