The Cohen-Lenstra paper says the probability that the odd part of a class group being cyclic is close to 0.98. So I was thinking: can we find any conditions on a finite abelian group so that it cannot be a class group of any number field?

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    $\begingroup$ related discussion math.stackexchange.com/questions/10949/… $\endgroup$ – Matthias Wendt Jul 16 at 17:23
  • $\begingroup$ @YCor I'd think "old part" probably should be "odd part"... $\endgroup$ – paul garrett Jul 16 at 18:43
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    $\begingroup$ Actually, the passage from Cohen--Lenstra that the question refers to concerns class groups of imaginary quadratic fields. The paper makes no statements about the statistics of class groups of all number fields. $\endgroup$ – Alex B. Jul 16 at 20:04

It follows from the Cohen-Lenstra heuristic that every finite abelian group is isomorphic to infinitely many class groups of real quadratic fields (even to a positive proportion of real quadratics, which is stronger), but nothing like this is known.

However, if you take the Galois action into account, then things get interesting: there are Galois modules that are not isomorphic to any class group of a Galois number field with the respective Galois group. See Corollary 4.8 and the bottom of page 17 in this paper of mine with Lenstra: https://arxiv.org/abs/1803.06903v2. You have to read a bit between the lines, but it is shown there that if $G$ is a cyclic group of order degree $58$, then there are finite $\mathbb{Q}[G]$-modules that cannot be realised as the class group of a $G$-extension (the "almost all" in the last line of page 17 can be strengthened to "all but one, namely the one for the field $\mathbb{Q}_{\zeta_{59}}$").

Of course, there are cheap ways of doing that, by demanding that the fixed submodule is something silly, contradicting the fact that the class group of $\mathbb{Q}$ is trivial, but that is not what is happening in our paper. For example our obstruction cannot be seen by looking at any particular $p$-Sylow of the class group.


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