Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.13
Find the orders of the following elements of the additive group $\mathbb{Z}/(36)$: $\overline{1}$, $\overline{2}$, $\overline{6}$, $\overline{9}$, $\overline{10}$, $\overline{12}$, $\overline{-1}$, $\overline{-10}$, $\overline{-18}$.
Solution:
| $\overline{n}$ | Reasoning | $\overline{n}$ |
|---|---|---|
| $\overline{1}$ | 36 is the smallest multiple of 1 that is congruent to 0 mod 36. | 36 |
| $\overline{2}$ | 36 is the smallest multiple of 2 that is congruent to 0 mod 36. | 18 |
| $\overline{6}$ | Multiples of $\overline{6}$ are $\overline{6}$, $\overline{12}$, $\overline{18}$, $\overline{24}$, $\overline{30}$, $\overline{36} = \overline{0}$ | 6 |
| $\overline{9}$ | Multiples of $\overline{9}$ are $\overline{9}$, $\overline{18}$, $\overline{27}$, $\overline{36} = \overline{0}$ | 4 |
| $\overline{10}$ | Multiples of $\overline{10}$ are $\overline{10}$, $\overline{20}$, $\overline{30}$, $\overline{4}$, $\overline{14}$, $\overline{24}$, $\overline{34}$, $\overline{8}$, $\overline{18}$, $\overline{28}$, $\overline{2}$, $\overline{12}$, $\overline{22}$, $\overline{32}$, $\overline{6}$, $\overline{16}$, $\overline{26}$, $\overline{36} = \overline{0}$ | 18 |
| $\overline{12}$ | Multiples of $\overline{12}$ are $\overline{12}$, $\overline{24}$, $\overline{36} = \overline{0}$ | 3 |
| $\overline{-1}$ | 36 is the smallest multiple of $-1$ that is congruent to 0 mod 36. | 36 |
| $\overline{-10}$ | Multiples of $\overline{-10}$ are $\overline{-10} = \overline{26}$, $\overline{16}$, $\overline{6}$, $\overline{32}$, $\overline{22}$, $\overline{12}$, $\overline{2}$, $\overline{28}$, $\overline{18}$, $\overline{8}$, $\overline{34}$, $\overline{24}$, $\overline{14}$, $\overline{4}$, $\overline{30}$, $\overline{20}$, $\overline{10}$, $\overline{0}$ | 18 |
| $\overline{-18}$ | $\overline{-18} + \overline{-18} = \overline{0}$ | 2 |
