Problem
Find the orders of the following elements of the multiplicative group \((\mathbb{Z}/12\mathbb{Z})^\times\)
- \(\overline{1}\)
- \(\overline{-1}\)
- \(\overline{5}\)
- \(\overline{7}\)
- \(\overline{-7}\)
- \(\overline{13}\)
Solution
\(\overline{1}\)
\(1^1 = 1\). Therefore, \(|\overline{1}| = 1\).
\(\overline{-1}\)
Note \(\overline{-1} = \overline{11}\).
- \(11^1 = 11 \notin \overline{1}\)
- \(11^2 = 121 = 12 \times 10 + 1 \in \overline{1}\)
Therefore, \(|\overline{-1}| = 2\).
\(\overline{5}\)
- \(5^1 = 5 \notin \overline{1}\)
- \(5^2 = 25 = 12 \times 2 + 1 \in \overline{1}\)
Therefore, \(|\overline{5}| = 2\).
\(\overline{7}\)
- \(7^1 = 7 \notin \overline{1}\)
- \(7^2 = 49 = 12 \times 4 + 1 \in \overline{1}\)
Therefore, \(|\overline{7}| = 2\).
\(\overline{-7}\)
Note \(\overline{-7} = \overline{5}\). Therefore, using what’s shown above, \(|\overline{-7}| = |\overline{5}| = 2\).
\(\overline{13}\)
Note \(\overline{13} = \overline{1}\). Therefore, using what’s shown above, \(|\overline{13}| = |\overline{1}| = 1\).