Hw1 Problem 14

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Problem 1

(+) There is an isomorphism of $U_{7}$ with $\mathbb{Z}_{7}$ in which $\zeta=e^{i(2\pi/7)}\leftrightarrow 4$. Find the element in $\mathbb{Z}_{7}$ to which $\zeta^m$ must correspond for $m=0,2,3,4,5,$ and $6$.

Solution

(1)

\begin{align} \zeta^0=\zeta^7=\zeta \cdot \zeta \cdot \zeta \cdot \zeta \cdot \zeta \cdot \zeta \cdot \zeta \leftrightarrow 4+_{7}4+_{7}4+_{7}4+_{7}4+_{7}4+_{7}4=28-28=0 \end{align}

(2)

\begin{align} \zeta^2=\zeta \cdot \zeta \leftrightarrow 4+_{7}4=8-7=1 \end{align}

(3)

\begin{align} \zeta^3=\zeta^2 \cdot \zeta \leftrightarrow 1+_{7}4=5 \end{align}

(4)

\begin{align} \zeta^4=\zeta^3 \cdot \zeta \leftrightarrow 5+_{7}4=9-7=2 \end{align}

(5)

\begin{align} \zeta^5=\zeta^4 \cdot \zeta \leftrightarrow 2+_{7}4=6 \end{align}

(6)

\begin{align} \zeta^6=\zeta^5 \cdot \zeta \leftrightarrow 6+_{7}4=10-7=3 \end{align}