# Is there a nice explanation for this curious fact about cyclic subgroups?

Here's something that I noticed that quite surprised me.

Let $G$ be a finite abelian group. Consider the following expression. $$\nu(G) = \sum_{\substack{H \leq G \\ H \text{ is cyclic}}} |H|$$ It is easy to see that for cyclic groups, we have that $\nu(G) = \sigma_1(|G|)$. What is significantly more surprising is the following.

Theorem For every finite abelian group, we have that $$\nu(G) = \sigma_1(|G|)$$ That is, this only depends on the order of the group.

Now, one can see pretty quickly that this is a multiplicative function, and so proving this reduces to studying Abelian $p$-groups. But even there it isn't obvious: consider the groups $\mathbb{Z}/p \times \mathbb{Z}/p$ and $\mathbb{Z}/p^2$. The first of these has lots of small cyclic subgroups, which the second one just has one large one, and amazingly enough these work out to contribute the same amount.

So is there a nice explanation for this? This definitely surprised me, and the only way I can prove this is with not-so-pretty computations that don't enlighten me much.

• What is $\sigma_1(G)$? Should that be $\lvert G \rvert$? Jun 3 '16 at 9:56
• what is $\sigma_1$? Jun 3 '16 at 10:29
• Sum of divisors, I suppose. Jun 3 '16 at 10:30

(Essentially the same answer as Neil Strickland's:) Since a cyclic group of order $n$ has $\varphi(n)$ generators, your sum equals $$\DeclareMathOperator{\ord}{ord} \nu(G) = \sum_{g\in G} \frac{ \ord(g) }{ \varphi(\ord(g)) } .$$ For elements $g$ of $p$-power order, we get $\ord(g)/ \varphi( \ord(g) ) = p/(p-1)$ and thus if $G$ is a $p$-group, $\nu(G) = 1 + (|G|-1)p/(p-1)$ depends only on the order of $G$ and not on its structure, and so $\nu(G) =\nu(C_{|G|})$. So your equality holds for nilpotent groups, as was remarked in Nilpotency of a group by looking at orders of elements .

This may well be the proof that you have already.

Reduce to the case of $p$-groups as in the question. Put \begin{align*} n(x) &= |\langle x\rangle| \\ m(x) &= |\{x' : \langle x'\rangle = \langle x\rangle\}| \end{align*} Then $$\nu(G) = \sum_{H \text{cyclic}} |H| = \sum_{x\in G} n(x)/m(x).$$ Now, if $x=1$ then $m(x)=1$. On the other hand, if $x\neq 1$ then $n(x)=p^k$ for some $k>0$ and $m(x)=p^k-p^{k-1}=(1-1/p)n(x)$. This gives \begin{align*} \nu(G) &= 1 + \sum_{x\neq 1} \frac{1}{1-1/p} = 1 + \frac{|G|-1}{1-1/p} \end{align*} If $|G|=p^r$, this becomes $$\nu(G) = 1 + \frac{p^r-1}{1-1/p} = \frac{p^r-1/p}{1-1/p} = \frac{p^{r+1}-1}{p-1} = \sum_{i=0}^r p^i = \sigma_1(p^r).$$

• This proof works just as well for non-abelian $p$-groups! Jun 3 '16 at 11:35
• I wonder if $\nu(G)<\sigma_1(G)$ for all non-nilpotent finite $G$. Jun 3 '16 at 11:43
• @TomGoodwillie To the best of my knowledge, this is still open. The second paper linked in my answer shows a number of similar inequalities, but not this specific one (cf. the remark one p. 9 before Question~14). Jun 3 '16 at 12:17

Another proof of Strickland's result.

Let $G$ be a group of order $p^n$ and $\nu(G)$ be the sum of orders of its cyclic subgroups. To prove that $\nu(G)=\sigma_1(G)$, we proceed by induction on $|G|$. If $G$ is cyclic, then, obviously, $\nu(G)=\sigma_1(G)$. Therefore assume that $G$ is noncyclic. If $H<G$, then, by induction, $\nu(H)=\sigma_1(H)$ depends only on $|H|$. Therefore assume that $G$ is noncyclic. Then it has a normal subgroup $T$ such that $G/T$ is elementary abelian of order $p^2$. In that case, as all maximal subgroups of $G$ have the same order, we get $$\nu(G)=\sum_{T<H<G}\nu(H)-p\nu(T)=(p+1)\nu(H)-p\nu(T)$$ hence, by induction, $$\nu(G)=(p+1)\sigma_1(p^{n-1})-p\sigma_1(p^{n-2})$$ $$=(p+1)(1+p+\dots+p^{n-1})-p(1+p+\dots+p^{n-2})=1+p+\dots+p^n=\sigma_1(p^n),$$ completing the proof.

Above we applied a partial case of Hall's enumeration principle. Applying that principle, it is easy to compute, for not very complicated $p$-groups, the sum of orders of their subgroups. About Hall's enumeration principle, see Kap. III in Huppert's `Endliche Gruppen'.