Let $G$ be a finite group with order $n$. For each $d$ dividing $n$, the number of subgroups of $G$ of order $d$ equals the number of subgroups of order $n/d$ if $G$ is abelian. More broadly, the lattice of subgroups of a finite abelian group looks the same if you flip it around by 180 degrees.

This is not at all obvious at the level at which the statement can first be understood, essentially because there is no natural way to construct subgroups of index $d$ from subgroups of order $d$ in a general finite abelian group with order divisible by $d$. It is not clear at a beginning level how the commutativity of the group leads to such conclusions.