# 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.