Answer to Question #350791 in Abstract Algebra for eliz

Question

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

Answer 1

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

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