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.



Is there a nice explanation for this curious fact about cyclic subgroups?
(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](https://mathoverflow.net/q/55071/10266) .  

See also the papers <http://arxiv.org/abs/1207.1020> and
<http://arxiv.org/abs/1503.00355>.


