Exercise 1.1.14 (Multiplicative orders of elements of $(\mathbb{Z}/36\mathbb{Z})^ imes$)

Question

Find the orders of the following elements of the multiplicative group $(\mathbb{Z}/36\mathbb{Z})^	imes$: $\overline{1}, \overline{-1},\overline{5}, \overline{13}, \overline{-13}, \overline{17}$.

Richard Ganaye · Accepted Answer

\begin{proof} 
For instance, 
\begin{align*}
\overline{5}^2 &= \overline{25} = -\overline{11},\
\overline{5^3} &= -\overline{55} = \overline{17},\
\overline{5^6} &= \overline{17}^2 = \overline{289} = \overline{ 8 \cdot 36 + 1} = \overline{1}.
\end{align*}
Since $\overline{5^6} = \overline{1}$ and $\overline{5^2} 
e \overline{1}, \ \overline{5}^3 
e \overline{1}$, we obtain $|\overline{5}| = 6$.

More generally,
\begin{center}
\begin{tabular}{c|cccccc}
$x$ & $\ \overline{1}$ & $\ \overline{-1}$ & $\ \overline{5}$ & $\ \overline{13}$ &$\ \overline{-13}$ &$\ \overline{17}$ \
\hline
$|x|$ & 1 & 2 & 6 & 3 & 6 &2 
\end{tabular}
\end{center}
\end{proof}
With Sagemath:
\begin{verbatim}
sage: l = [1,-1,5,13,-13,17]
sage: for k in l:
....:     print(k, '=>', G(k).multiplicative_order())
....:     
(1, '=>', 1)
(-1, '=>', 2)
(5, '=>', 6)
(13, '=>', 3)
(-13, '=>', 6)
(17, '=>', 2)
\end{verbatim}

$x$	$\bar{1}$	$\bar{- 1}$	$\bar{5}$	$\bar{13}$	$\bar{- 13}$	$\bar{17}$

$\| x \|$	1	2	6	3	6	2

Exercise 1.1.14 (Multiplicative orders of elements of $(\mathbb{Z}/36\mathbb{Z})^\times$)

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Exercise 1.1.14 (Multiplicative orders of elements of $(\mathbb{Z}/36\mathbb{Z})^\times$)

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