Hw1 Problem 15

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Problem 15

Why can there be no isomorphism of $U_{6}$ with $\mathbb{Z}_{6}$ in which $\zeta = e^{i(\pi/3)}$ corresponds to 4?

Solution

(1)

\begin{align} \zeta = e^{i(\pi/3)} \leftrightarrow 4 \end{align}

(2)

\begin{align} \zeta^{2} = \zeta \cdot \zeta \leftrightarrow 4 + _{6}4 = 8 - 6 = 2 \end{align}

(3)

\begin{align} \zeta^{3} = \zeta^{2} \cdot \zeta \leftrightarrow 2 + _{6}4 = 6 - 6 = 0 \end{align}

(4)

\begin{align} \zeta^{4} = \zeta^{2} \cdot \zeta^{2} \leftrightarrow 2 + _{6}2 = 4 \end{align}

(5)

\begin{align} \zeta^{5} = \zeta^{3} \cdot \zeta^{2} \leftrightarrow 3 + _{6}2 = 5 \end{align}

(6)

\begin{align} \zeta^{0} = \zeta^{6} = \zeta^{3} \cdot \zeta^{3} \leftrightarrow 0 + _{6}0 = 0 \end{align}

$\zeta$ and $\zeta^{4}$ both correspond to $4$ meaning there is no one-to-one correspondence of $U_{6}$ with $\mathbb{Z}_{6}$.