Theorems on 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.

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

Now

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

Hence there exist a positive integer $t$ such that${a^t} = e$.

Now, we know that every set of positive integer 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 exist 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

Also

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

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

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,

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

Theorem 5: The order of the elements $a$ and ${x^{ - 1}}ax$ are 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: Order of $ab$ is the same as that of $ba$ where $a$ and $b$ are any elements of a group.