Let $G$ be a finite abelian group, and let $n$ divide $|G|$. Let $m$ be the number of solutions of $x^n=1$. Prove that $n\mid m$. https://ift.tt/eA8V8J

Let $G$ be a finite abelian group, and let $n$ divide $|G|$. Let $m$ be the number of solutions of $x^n=1$. Prove that $n\mid m$.

My attempt

It's tempting to find a way to use Lagrange's theorem. Maybe something here is a subgroup of something else? We can fix $n$ and take the subgroup of $G$ of all elements which solve $x^n=1$. Proof that this is a subgroup: Inverses of solutions are always solutions. Because the group is abelian, products of solutions are solutions. QED.

Great, so it's a subgroup, so $m$ divides the order of $G$. So does $n$. I'm not sure that this really got me anywhere. It'd be nice if there were some relevant subgroup of order $n$.

Being finite and abelian then it has a representation as $G\cong C_{p_1^{n_1}}\times\dots\times C_{p_k^{n_k}}$, a product of cyclic groups of prime power order. The solutions are exactly the product of solutions "in each factor", i.e. solutions of the form $\langle e, \dots, e, x, e, \dots, e\rangle$ where $x\in C_{p_i^{k_i}}$ for some $i$. So perhaps something comes from thinking about the number of solutions to $x^n=1$ where $x$ is taken from $C_{p_i^{k_i}}$.

Again this is a subgroup so the number of solutions divides $p_i^{k_i}$, and $p_i^{k_i}$ divides $|G|$. And $n$ divides the order of $G$. But at this point I'm not sure whether I'm on a productive path, since these facts don't seem to be enough to show that $n|m$.

In fact the more that I think about how $n$ is so-to-speak missing factors from $|G|$ the more I think that finding numbers which divide $|G|$ just isn't a productive path.

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