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.

  • 3
    $\begingroup$ 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 $\{x\in G\colon x^n=1\}$. $\endgroup$ 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, $$n=|G_n|\;\Big\vert\;|\ker(\varphi_n)|=m$$

  • $\begingroup$ How does Lagrange's Theorem guarantee that $G_n\leq \ker(\varphi ) $? $\endgroup$
    – Addem
    Nov 1 '20 at 2:00
  • 2
    $\begingroup$ @Addem Since $|G_n|=n$, by Lagrange's Theorem, $y^n=1$ for all $y\in G_n$ (because $|y|$ divides $|G_n|=n$). $\endgroup$
    – Alan Wang
    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 $r<k$, 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 $n$.


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