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 exist 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 exist 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.

For 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 order of -1 is 2.
Again

{i^1}  = i,\,\,\,{i^2} = - 1,\,\,\,{i^3} = - i,\,\,\,{i^4} = 1


Therefore 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 order of -1 is 4.