Exercise 3.3.8 (The group of $p$-power roots of $1$ is isomorphic to a proper quotient of itself)

Question

Let $p$ be a prime and let $G$ be the group of $p$-power roots of $1$ in $\mathbb{C}$ (cf. Exercise 18, Section 2.4). Prove that the map $z \mapsto z^p$ is a surjective homomorphism. Deduce that $G$ is isomorphic to a proper quotient of itself.

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 $p$ be a prime and let
$$G = \{z \in \mathbb{C} \mid \exists n \in  \mathbb{Z}^+,\ z^{p^n}= 1\}.$$
Consider the map
$$\varphi\ 
\left\{
\begin{array}{ccl}
G & 	o &G\
z & 	o & z^p.
\end{array}
ight.
$$
Then
\begin{itemize}
\item  $\varphi$ is a homomorphism: for all $z, z' \in G$,
$$\varphi(zz') = (zz')^p = z^p z'^p = \varphi(z) \varphi(z').$$

\item $\varphi$ is surjective: Let $t$ be any element of $G$. Then $t^{p^n} = 1$ for some positive integer $n$.

Therefore $t = e^{2i\pi k/p^n}$ for some $k \in [\![0, p^n[\![$. Put $z = e^{2i\pi k/p^{n+1}}$. Then $z^{p^{n+1}}  = e^{2i\pi k} = 1$, so $z \in G$. Moreover $z^p = e^{2i\pi k/p^n} = t$, so $t =\varphi(z)$. This shows that $\varphi$ is surjective, so
$$\varphi(G) = G.$$

\item Kernel of $\varphi$:
$$\ker(\varphi) = \{z \in \C \mid z^p = 1\} = \{1, e^{2i\pi/p}, e^{4i\pi/p}, \ldots, e^{2(p-1)i\pi/p} \} = \langle e^{2i\pi/p} angle = H_1$$
 is a cyclic group of order $p$.
\end{itemize}
By the First Isomorphism Theorem, $ \varphi(G) \simeq G/\ker(\varphi) $, so
$$G \simeq G/H_1,$$
where $\{1\} 
e H_1 
e  G$.

\end{proof}

Exercise 3.3.8 (The group of $p$-power roots of $1$ is isomorphic to a proper quotient of itself)

Answers

Comments

Exercise 3.3.8 (The group of $p$-power roots of $1$ is isomorphic to a proper quotient of itself)

Answers

Comments

Add answer