Question #350791

2.5. If G is a finite group, show that there exists a positive integer m such that am = e for all a ∈ G.


1
Expert's answer
2022-06-20T17:54:11-0400

Let GG be finite group and 1aG1\ne a\in G.

Consider the set a,a2,a3,,ak,a,a^2,a^3,\dots,a^k,\dots. It is clear that aiai+1a^i\ne a^{i+1} for some integers from the beginning. Since GG is a finite group there exists ii and jj such that ai=aja^i=a^j implies aij=1a^{i-j}=1. Therefore every element has the finite order. That is the smallest positive integer kk satisfying ak=1a^k=1. (One may assume without loss of generality that i>ji>j). One can do this for each aGa\in G. The least common multiple mm of the order of all elements of GG satisfies am=1a^m=1 for all aGa\in G.


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