# 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.**