0. The facts

For simplicity, let's write $\mathbb{Z}_ m$ for $\mathbb{Z}/ m \mathbb{Z}$ with $m > 0$. Let's be casual about omitting brackets, writing $3 \in \mathbb{Z}_ {10}$ instead of $[3] _ {10}$. Also, $p$ will always refer to a prime.

(a) For $a \in \mathbb{Z}_ m$, if $a^n = 1$ with $n \geq 1$ then $a (a ^ {n-1}) = 1$, so $a$ is invertible (i.e., $a$ is a unit).

(b) For $a \in \mathbb{Z}_ m$, the list of powers $1, a, a^2, a^3, \dots$ must start cycling at some point:

• If $a$ is a unit, then the first repeat is back to $1$. The first $n > 0$ such that $a^n = 1$ is called the order of $a$.

  Example: For $3 \in \mathbb{Z}_ {10}$ the list of powers is $1, 3, 9, 7, 1, 3, 9, 7, \dots$ and the order of 3 is 4.

  Example: In a finite field, every nonzero element is a unit and so has an order.

• If $a$ is not a unit, then the first repeat is not back to $1$, but the powers do get stuck in a cycle sooner or later.

  Example: For $2 \in \mathbb{Z}_ {24}$, the list of powers is $1, 2, 4, 8, 16, 8, 16, 8, 16, \dots$.

(c) If $a$ is a unit in $\mathbb{Z}_ m$, with order $n$, then the powers equal to 1 are precisely $a ^0, a ^ n, a ^ {2n}, a ^ {3n}, \dots$.

In other words, $a^i = 1 \Leftrightarrow n \vert i$.

(d) (i) If $p$ is prime and $a$ is a nonzero element of $\mathbb{Z}_ p$, then Fermat's Little Theorem says $a^{p-1} = 1$. Therefore the order of $a$ divides $p-1$.

(ii) More generally, for any $m$, if $a$ is a unit of $\mathbb{Z}_ m$, then Euler's Theorem says $a^{\phi(m)} = 1$. Therefore the order of $a$ divides $\phi(m)$.

(iii) Even more generally, if $R$ is any commutative finite ring and $a$ is a unit of $R$, then the order of $a$ divides the number of units in $R$.

Example: In a finite field with $q$ elements, the order of each nonzero element divides $q - 1$. (As you know, $q$ has to be a prime power.)

(iv) Still more generally, if $G$ is any finite group, abelian (commutative) or not, then the order of each element divides the size of the group¹.

(e) If $a$ has order $n$ and if $k$ and $n$ are coprime, then $a^k$ also has order $n$.

More generally, if $a$ has order $n$ then for any $k \geq 0$, $a^k$ has order ${\frac{\displaystyle n}{\displaystyle \mbox{gcd}(n,k)}}$.

Example: In $\mathbb{Z}_ {11}$, $6$ has order $10$. Then $6^2$ has order 5, and so do $6^4$, $6^6$, and $6^8$, since all these exponents have 2 as their gcd with 10.

(f) (i) The units of a given finite ring might have a generator or primitive element, meaning an element $g$ for which the powers $1, g, g^2,\dots$ are all the units. An equivalent statement is that the order of $g$ is the same as the number of units.

Example: The units of $\mathbb{Z}_ {10}$ are $1, 3, 7, 9$, with generator $3$.

The units of $\mathbb{Z}_ 8$ are $1, 3, 5, 7$; there is no generator since each of these elements has square = 1.

(ii) In a finite field, where all nonzero elements are units, there is always a generator. In fact, there are $\phi(p)$ generators of $\mathbb{Z}_ p$ for $p$ prime.

(g) (i) In a commutative ring, if $a$ and $b$ are units and $a$ has order $m$ and $b$ has order $n$, where $m$ and $n$ are coprime, then $ab$ has order $mn$.

(ii) If a commutative ring $R$ has a unit $a$ of order $r$ and a unit $b$ of order $s$, then $R$ has a unit of order $d$ where $d =$   lcm$(r,s)$.

Note: This element is not necessarily $ab$, since for example if $b = a^{-1}$ then $a$ and $b$ have the same order $r$ and lcm$(r,r) = r$, but $ab = 1$, which is an element of order 1.

(h) If $m$ and $n$ are coprime, then the order of an element $a$ of $\mathbb{Z}/mn\mathbb{Z}$ is the lcm of the orders of the images of $a$ in $\mathbb{Z}/ m \mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$.

In other words, if $a \leftrightarrow (a _ 1, a _ 2)$ under the isomophism of $\mathbb{Z}/mn\mathbb{Z}$ with $\mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ according to the Chinese Remainder Theorem, then the order of $a$ is the lcm of the orders of $a _ 1$ and $a _ 2$.