# Theorems on the Order of an Element of a Group

__Theorem 1__**:** The order of every element of $$a$$ finite group is finite.

__Proof__**:** Let $$G$$ be a finite group and let $$a \in G$$, we consider all positive integral powers of $$a$$, i.e.

\[a,{a^2},{a^3},{a^4},…\]

Every one of these powers must be an element of $$G$$. But $$G$$ is of finite order. Hence these elements cannot all be different. We may therefore suppose that $${a^s} = {a^r},\,\,\,s > r$$

Now

\[\begin{gathered} {a^s} = {a^r} \Rightarrow {a^s}{a^{ – r}} = {a^r}{a^{ – r}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {a^{s – r}} = {a^0} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {a^{s – r}} = e \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {a^t} = e\,\,\,\left( {{\text{putting}}\,s – r = t} \right) \\ \end{gathered} \]

Since $$s > r,\,\,\,t$$ is a positive integer,

hence there exists a positive integer $$t$$ such that$${a^t} = e$$.

Now, we know that every set of positive integers has at least number. It follows that the set of all those positive integer $$t$$ such that $${a^t} = e$$ has a least member, say $$m$$, thus there exists a least positive integer $$m$$ such that $${a^m} = e$$, showing that the order of every element of a finite group is finite.

__Theorem 2__**: **The order of an element of a group is the same as that of its inverse $${a^{ – 1}}$$.

__Proof__**:** Let $$n$$ and $$m$$ be the orders of $$a$$ and $${a^{ – 1}}$$ respectively.

Then, $${a^n} = e$$ and $${\left( { – a} \right)^m} = e$$

Now

\[\begin{gathered} {a^n} = e \Rightarrow {\left( {{a^n}} \right)^{ – 1}} = {e^{ – 1}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {\left( {{a^{ – 1}}} \right)^n} = e \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow O\left( {{a^{ – 1}}} \right) \leqslant n \Rightarrow m \leqslant n \\ \end{gathered} \]

Also

\[\begin{gathered} O\left( {{a^{ – 1}}} \right) = m \Rightarrow {\left( {{a^{ – 1}}} \right)^m} = e \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {\left( {{a^m}} \right)^{ – 1}} = e \Rightarrow {a^m} = e \\ \end{gathered} \]

Because $${b^{ – 1}} = e \Rightarrow b = e$$

\[\Rightarrow O\left( a \right) \leqslant m \Rightarrow n \leqslant m\]

Now $$m \leqslant n$$ and $$n \leqslant m \Rightarrow m = n$$

If the order of $$a$$ is infinite, then the order of $${a^{ – 1}}$$ cannot be finite. Because \[O\left( {{a^{ – 1}}} \right) = m \Rightarrow O\left( a \right) \leqslant m \Rightarrow O\left( a \right)\] is finite. Therefore if the order of $$a$$ infinite, then the order of $${a^{ – 1}}$$ must also be infinite.

__Theorem 3__**:** The order of any integral power of an element $$a$$ cannot exceed the order of $$a$$.

__Proof__**: **Let $${a^k}$$ be any integral power of $$a$$. Let $$O\left( a \right) = n$$.

Now,

\[\begin{gathered} O\left( a \right) = n \Rightarrow {a^n} = e \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {\left( {{a^n}} \right)^k} = {e^k} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {a^{nk}} = e \Rightarrow {\left( {{a^k}} \right)^n} = e \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow O\left( {{a^k}} \right) \leqslant n \\ \end{gathered} \]

__Theorem 4__**:** If the element $$a$$ of a group $$G$$ is order $$n$$, then $${a^m} = e$$ if and only if $$n$$ is a divisor of $$m$$.

__Theorem 5__**:** The order of the elements $$a$$ and $${x^{ – 1}}ax$$ is the same where $$a,x$$ are any two elements of a group.

__Theorem 6__**:** If $$a$$ is an element of order $$n$$ and $$p$$ is prime to $$n$$, then $$ap$$ is also of order $$n$$.

__Corollary__**:** The order of $$ab$$ is the same as that of $$ba$$ where $$a$$ and $$b$$ are any elements of a group.