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?
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4$\begingroup$ I can't make any sense of this question. Is $G$ an additive group or a multiplicative group? If it is multiplicative and $a \in G$ then trivially $aG=G$. If it is an additive group, then what does $aG$ mean? Or did you mean to take $a \in {\mathbb Z}$? $\endgroup$– Derek HoltCommented Jan 29, 2022 at 18:59
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$\begingroup$ Sorry, I forgot to specify that $G$ is a subset of the complex numbers, and I'm considering it's additive structure. An example of such a $G$ is the ring of integers of any number field (it's additive group is free of rank equal to the degree of the number field). By $a^n G$ in this case I'm denoting the set of all elements of the form $a^n g$ for $g\in G$. $\endgroup$– MarkG99Commented Jan 29, 2022 at 21:03
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