# 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 $n\mid m$.

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.

• I think this hinges on the fact that for finite abelian groups $G$, if $n$ divides $\#G$ then $G$ contains a subgroup of cardinality $n$ (proofs of which follow your ideas). Here, that subgroup will sit inside $\{x\in G\colon x^n=1\}$. Oct 31 '20 at 21:38

By the fundamental theorem of finite abelian groups we may choose a $$G$$ subgroup $$G_n$$ of size $$n$$. Lagrange's theorem gaurantees $$G_n\leq\ker(\varphi_n)$$ where $$\varphi_n$$ denotes the $$G$$ endomorphism $$\varphi_n:x\mapsto x^n$$ and $$\leq$$ denotes subgroup inclusion. Finally, by Lagrange's theorem once again, $$n=|G_n|\;\Big\vert\;|\ker(\varphi_n)|=m$$
• How does Lagrange's Theorem guarantee that $G_n\leq \ker(\varphi )$? Nov 1 '20 at 2:00
• @Addem Since $|G_n|=n$, by Lagrange's Theorem, $y^n=1$ for all $y\in G_n$ (because $|y|$ divides $|G_n|=n$). Nov 1 '20 at 2:55
Let $$p$$ be a prime and assume $$p^k\|n$$. If one of the factors in the product representation of $$G$$ is $$C_{p^r}$$ for some $$r\ge k$$, then this contains a subgroup isomorphic to $$C_{p^k}$$, which consists of solutions to $$x^n=1$$. If no such factor exists, then there exist several factors $$C_{p^r}$$ with $$r, which are completely solutions to $$x^n=1$$. But their product must have cardinality at least $$p^k$$. So at any rate, $$G$$ has a subgroup of order $$p^k$$ of solutions of $$x^n=1$$. Combining all relevant primes, we obtain a subgroup of order $$n$$ or - as we may have ignored a few factors - a multiple of $$n$$.