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


\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 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


\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}


\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.


\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 aof a group Gis 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.