How do we interpret the group structure of the ideal class group? Given a number field $K/\mathbb{Q}$, let $\text{Cl}(K)$ denote the ideal class group of $K$.
My question is: how do we interpret the group structure of $\text{Cl}(K)$? To be clear, I understand how to interpret the order of the class group: if the class group of $K$ has order $1$, then the number ring $\mathcal{O}_K$ has unique factorization; if the class group has order bigger than $1$, then $\mathcal{O}_K$ doesn't have unique factorization. But that is only part of the picture: what if you have two number fields whose class groups have the same order, but different group structures? (Say, for example: a class group of $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ versus a class group of $\mathbb{Z}/4\mathbb{Z}$.) What is an intuitive way to interpret this difference? Does it have any meaning?
I have some vague sense that the structure of $\text{Cl}(K)$ measures "the complexity of the arithmetic" in the number ring, but I don't really know what that means. Any of your insights would be greatly appreciated!
 A: When the class group consists only of cycles having order 2, like the Klein four-group, then we can profitably augment the ring with "square-root" elements and then turn a nonunique factorization into a unique one with the square roots.
To illustrate, consider the number $66$ as it factors in $\mathbb Z[\sqrt{-17}]$ and in $\mathbb Z[\sqrt{-30}]$, both of which have class number $4$:
$\mathbb Z[\sqrt{-17}]: 66=2×3×11=(7+\sqrt{-17})(7-\sqrt{-17})$
$\mathbb Z[\sqrt{-30}]: 66=2×3×11=(6+\sqrt{-30})(6-\sqrt{-30})$
I now claim that in the latter case, where the class group is $\mathbb Z/2\mathbb Z×\mathbb Z/2\mathbb Z$ rather than $\mathbb Z/4\mathbb Z$, I can use square-root augmentation to render a unique factorization of $66$, from which the two factorizations $2×3×11$ and $(6+\sqrt{-30})(6-\sqrt{-30})$ can both be derived.
The number $1+2\sqrt{-30}$ has a perfect square Euclidean norm, but we can't reduce it in $\mathbb Z[\sqrt{-30}]$ per se. But, if we bring elements of the form $a\sqrt{6}+b\sqrt{-5}$, then we can render
$1+2\sqrt{-30}=(\sqrt{6}+\sqrt{-5})^2.$
Similarly, using two other square root extensions with integer radicands, we have
$7+2\sqrt{-30}=(\sqrt{10}+\sqrt{-3})^2,$
$13+2\sqrt{-30}=(\sqrt{15}+\sqrt{-2})^2.$
These then represent the square-root augmentation of the Klein four-group class domain $\mathbb Z[\sqrt{-30}]$ with which a number like $66$ may be uniquely factored. To wit, using all three of the square root extensions:
$66 = (\sqrt{-2})^2(\sqrt{-3})^2(\sqrt{6}+\sqrt{-5})(\sqrt{6}-\sqrt{-5}).$
And from this one augmented-domain factorization, allowing also the unit factor $-1$, we get both:
$66=[(-1)(\sqrt{-2})^2][(-1)(\sqrt{-3})^2][(\sqrt{6}+\sqrt{-5})(\sqrt{6}-\sqrt{-5})]=2×3×11$
$66=[(-1)\sqrt{-2}\sqrt{-3}(\sqrt{6}+\sqrt{-5})][(-1)\sqrt{-2}\sqrt{-3}(\sqrt{6}-\sqrt{-5})]=(6+\sqrt{-30})(6-\sqrt{-30}).$
None of this works with $\mathbb Z[\sqrt{-17}]$ where the class group $\mathbb Z/4\mathbb Z$ has a cycle of order greater than $2$, at least not using square roots. We can identify a square root of (let us say) $\pm(8+\sqrt{-17})$, but its terms will not augment the domain in a way that maps both $2×3×11$ and $(7+\sqrt{-17})(7-\sqrt{-17})$ into a unique factorization of $66$.
