Order of a quotient of a free abelian group

Question

Let $G\subseteq \mathbb{C}$ be a free abelian group of rank $n$ and let $p$ be a prime. Then, we know that $$|G/pG|=p^n$$ and in fact for this we don't even need $p$ to be a prime. Suppose now that there exists $a\in G$ so that $pG=a^n G$. Then, again, we trivially know that $|G/a^nG|=p^n$ since $a^nG=pG$. However, is this sufficient to conclude that $$|G/aG|=p \,?$$ I tried with some particular examples of $G$ and it seems to be true; also, I apparently never used the fact that $p$ is a prime. So, is this true in general, and is that hypothesis redundant? And if so, is there an easy proof of this that apparently went completely over my head?