The Order of Elements in a Group

The Order of Elements in a Group

Definition: Let $(G, *)$ be a group and let $a \in G$. The Order of $a$ is the least positive integer $m$ such that $a^m = \underbrace{a * a * … * a}_{m : \mathrm{many \: factors}} = e$ where $e$ is the identity element of $G$. If no such $m$ exists then $a$ is said to have order $\infty$.

If the operation is multiplicative in nature then we usually define the order of $a \in G$ as above. If the operation is instead additive in nature then we define the order of $a \in G$ as the smallest positive integer $m$ such that $ma = \underbrace{(a + a + … + a)}_{m \: \mathrm{many}} = e$ or $\infty$ is no such positive integer $m$ exists.

For example, consider the group $(\mathbb{Z}_5, +)$ where $+$ is defined for all $x, y \in \mathbb{Z}_5 = \{ 0, 1, 2, 3, 4 \}$ to be:

(1)
\begin{align} \quad x + y = (x + y) \mod 5 \end{align}

The order of $e = 0$ is trivially $1$. The order of $1$ is $5$ since:

(2)
\begin{align} \quad 1^5 = (1 + 1 + 1 + 1 + 1) = 0 \mod 5 \end{align}

The order of $2$ is also $5$. In fact, the orders of $3$ and $4$ are also $5$.

Now instead consider the group $(\mathbb{Z}_6, +)$ where $+$ is defined for all $x, y \in \mathbb{Z}_6 = \{ 0, 1, 2, 3, 4, 5 \}$ to be:

(3)
\begin{align} \quad x + y = (x + y) \mod 6 \end{align}

You should verify that the order of $0$ is $1$, the order of $1$ is $6$, the order of $2$ is $3$, the order of $3$ is $2$, the order of $4$ is $3$, and the order of $5$ is $6$

For a third set of examples, consider the set of real numbers under the operation of standard addition, $(\mathbb{R}, +)$. The order of every nonzero real number is $\infty$ since for $m$ as a positive integer, the equation below equals $0$ if and only if $x = 0$:

(4)
\begin{equation} mx = 0 \end{equation}
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