Cyclic Groups - $a^k = e \text{ iff } n|k$

Question

I saw this proof in the book on Abstract Algebra. Here is part of it:

Let $G$ be a cyclic group of order $n$ and $a$ is the generator of $G$. Then $a^k = e \iff n|k$

Proof: Suppose $a^k=e$. By the division algorithm, $\exists q,r \text{ such that } k=nq+r$ where $0\le r < n$. So $e=a^k = a^{nq+r} = a^{nq}a^r = ea^r = a^r$

My question is: is this circular, because we are already assuming that $a^{nq} = e$?

Answer 1

Exponent laws still work in groups, so we can prove that $a^{nq}=e$ by noting: $$a^{nq}=\left(a^n\right)^q=e^q=e$$ where the first two expressions both expand to $$\underbrace{a\cdot a\cdot a\cdot \ldots\cdot a \cdot a}_{nq\text{ times}}$$ and the next follows from replacement with $a^n=e$.

Moreover, notice that the statement $a^{nq}=e$ is exactly the same as

$n|k$ implies $a^k=e$

whereas the proof you present aims to show the converse:

$a^k=e$ implies $n|k$

So it's not actually circular - the premise it is taking is distinct from what it is proving and can be proven separately.