[Math] If $a$ is a member of a finite group, and $e$ is the identity, how can I prove that $a^n=e$ for some $n$

group-theory

I'm doing some self-teaching on abstract algebra. One of the chapter's exercises is to prove that $a^n = e$ for some $n>0$. I can't figure out how to do this. I know that if $a,b$ are in a group $G$, then $ax=b$ must have a single solution. But I can't translate that into a proof of this hypothesis. Any hints?

Reference: Fraleigh p. 49 Question 4.34 or p. 58 Question 5.49 in A First Course in Abstract Algebra

Best Answer

I'll assume that you mean that the relevant group $G$ is finite, because otherwise the claim is very false (the integers, or the non-zero reals under multiplication give such examples). Suppose that there were no such $N$; then can you prove that all the elements of $\{a^1, a^2, a^3, \dots\}$ are distinct? And then use this to contradict the finiteness assumption?

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