Answer to Question #143114 in Abstract Algebra for nan

Question

if a is an element of a group of finite order, prove that a^m ≠ a^n whenever m ≠ n

Answer 1

Let G be a group of finite order and $a$ be arbitrary element of G. Then G contains finite number of elements. Consider the following powers of $a:$

$a^0=e, a, a^2,...a^k,...$

Since G is a group, it contains all powers of $a$. Taking into account that exponenets of the degrees of the element $a$ are infinitely many and group G contains only finite number of elements, we conclude that $a^n=a^m$ for some $n\ne m.$

Consequently, it is not true that $a^m ≠ a^n$ whenever $m ≠ n.$