Question #143114
if a is an element of a group of finite order, prove that a^m ≠ a^n whenever m ≠ n
1
Expert's answer
2020-11-09T20:07:09-0500

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

a0=e,a,a2,...ak,...a^0=e, a, a^2,...a^k,...

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


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



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