HW1 Problem 13

### Problem 13

Example 1.15 asserts that there is an isomorphism of $U_8$ with $\mathbb{Z}_8$ in which $\zeta = e^{i(\pi/4)} \leftrightarrow 5$ and $\zeta^{2} \leftrightarrow 2$. Find the element of $\mathbb{Z}_8$ that corresponds to each of the remaining six elements $\zeta^{m}$ must correspond for $m = 0, 2, 3, 4, 5,$ and $6$.

Solution

(1)
\begin{align} \zeta^{0} \leftrightarrow 0 \end{align}
(2)
\begin{align} \zeta = e^{i(\pi/4)} \leftrightarrow 5 \end{align}
(3)
\begin{align} \zeta^{2} \leftrightarrow 2 \end{align}
(4)
\begin{align} \zeta^{3} = \zeta^{2} \cdot \zeta \leftrightarrow 2 + _{8}5 = 7 \end{align}
(5)
\begin{align} \zeta^{4} = \zeta^{2} \cdot \zeta^{2} \leftrightarrow 2 + _{8}2 = 4 \end{align}
(6)
\begin{align} \zeta^{5} = \zeta^{3} \cdot \zeta^{2} \leftrightarrow 7 + _{8}2 = 9 - 8 = 1 \end{align}
(7)
\begin{align} \zeta^{6} = \zeta^{3} \cdot \zeta^{3} \leftrightarrow 7 + _{8}7 = 14 - 8 = 6 \end{align}
(8)
\begin{align} \zeta^{7} = \zeta^{4} \cdot \zeta^{3} \leftrightarrow 4 + _{8}7 = 11 - 8 = 3 \end{align}