Exercise 1.1.8 (Group of roots of unity in $\mathbb{C}$)

Question

Let $G = \{z \in \mathbb{C} \mid z^n = 1 	ext{ for some } n \in \mathbb{Z}^+\}$.
\begin{enumerate}
\item[{\bf(a)}] Prove that $G$ is a group under multiplication (called the group of {\bf roots of unity} in $\mathbb{C}$)
\item[{\bf(b)}] Prove that $G$ is not a group under addition.
\end{enumerate}

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

\begin{proof} 
Let
$$G =  \{z \in \mathbb{C} \mid\exists n \in \Z^+,\  z^n = 1 \}.$$
\begin{enumerate}
\item [{\bf(a)}]
We show that $G$ is a subgroup of $(\C^*, 	imes)$ (see Section 2.1).
\begin{itemize}
\item Since $1^1 = 1$, then $1 \in G$, so $G 
e \varnothing$.
\item let $z,t \in G$. Then there are integers $n , m\in \Z^+$ such that $z ^n = 1$ and $t^m = 1$. Then
$$(zt^{-1})^{nm} = (z^n)^m /(t^m)^n = 1,$$
where $nm \in \Z^+$, so  $zt^{-1} \in G$.
\end{itemize}
Therefore $G$ is a subgroup of $(\C^*, 	imes)$.

\item [{\bf(b)}] $1 \in G$, but $2 = 1 + 1 
ot \in G$, since $2^n 
e 1$ for all $n \geq 1$. Therefore $G$ is not a group under addition.
\end{enumerate}
\end{proof}

Exercise 1.1.8 (Group of roots of unity in $\mathbb{C}$)

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Exercise 1.1.8 (Group of roots of unity in $\mathbb{C}$)

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