# The Order of an Element of a Group

If $G$ is a group and $a$ is an element of group $G$, the order (or period) of $a$ is the least positive integer $n$, such that
${a^n} = e$

If there exists no such integer, we say that $a$ is a finite order or zero order. We shall use the notation $O\left( a \right)$ for the order of $a$.

Note that the only element of order one in a  group is the identity element $e$.

Important Note: If there exists a positive integer $m$ such that ${a^m} = e$, then the order of $a$ is definitely finite. Also we must have $O\left( a \right) \leqslant m$. When ${a^m} = e$, then the question of order of a being greater than $m$ does not arise. At the most it can be equal to $m$. If $m$ itself is the least positive such that ${a^m} = e$, then we will have $O\left( a \right) = m$.

Example:
Find the order of each element of the multiplicative group $G$, where $G = \left\{ {1, – 1,i, – i} \right\}$
Since 1 is the identity element, its order is 1.

Now
${\left( { – 1} \right)^1} = – 1,\,\,{\left( { – 1} \right)^2} = \left( { – 1} \right)\left( { – 1} \right) = 1$

Hence the order of -1 is 2.

Again
${i^1} = i,\,\,\,{i^2} = – 1,\,\,\,{i^3} = – i,\,\,\,{i^4} = 1$

Therefore the order of $i$ is 4.

Similarly,
${\left( { – i} \right)^1} = – i,\,\,\,{\left( { – i} \right)^2} = {i^2} = – 1$
${\left( { – i} \right)^3} = i,\,\,\,{\left( { – i} \right)^4} = 1$

Hence the order of -1 is 4.