Why is it assuming in both sides that $a^{n}=e$?

Question

Let $G$ be a cyclic group of order $n$ generated by $a$. Then $a^{k} = e$ if and only if $n | k$.

Proof:

I have a question about the above proof. Why is it assuming in both sides that $a^{n}=e$?

Answer 1

When you say a finite cyclic group is generated by $a$ that means powers of $a$ will each be distinct elements of that group until all the elements are exhausted. Since the first $a^{n-1}$ powers are all distinct from $a^0=e$ by considering the order of the group so this forces $a^n=e$. At this point the cycle begins again, which is the motivation for calling them cyclic groups. Note that since all powers of $a$ commute with all other powers of $a$ that this group is abelian as well so it has a lot of additional structure because of how it's constructed.

Answer 2

By definition $G$ is given by the presentation $$\langle a\mid a^n=e\rangle.$$