Abstract
Cohen and Lenstra have given a heuristic which, for a fixed odd prime p, leads to many interesting predictions about the distribution of p-class groups of imaginary quadratic fields. We extend the Cohen-Lenstra heuristic to a non-abelian setting by considering, for each imaginary quadratic field K, the Galois group of the p-class tower of K, i.e. \({G}_K:=\mathrm {Gal}(K_\infty /K)\) where \(K_\infty \) is the maximal unramified p-extension of K. By class field theory, the maximal abelian quotient of \({G}_K\) is isomorphic to the p-class group of K. For integers \(c\ge 1\), we give a heuristic of Cohen-Lenstra type for the maximal p-class c quotient of \({G}_K\) and thereby give a conjectural formula for how frequently a given p-group of p-class c occurs in this manner. In particular, we predict that every finite Schur \(\sigma \)-group occurs as \(G_K\) for infinitely many fields K. We present numerical data in support of these conjectures.
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Notes
See the earlier footnote to Remark 1.4.
Jonathan Blackhurst, 1037 E Millbrook Way, Bountiful, UT 84010, USA. E-mail: jblackhurst@gmail.com.
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Acknowledgments
We acknowledge useful correspondence and conversations with Bettina Eick, Jordan Ellenberg, John Labute, Daniel Mayer, Cam McLeman, Eamonn O’Brien, and Melanie Matchett Wood. We are grateful to Jonathan Blackhurst for providing the Appendix. We would also like to thank Joann Boston for drawing the figure in Sect. 2.
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Dedicated to Helmut Koch.
With an Appendix by Jonathan Blackhurst.
The research of the first author was supported by National Security Agency Grant MSN115460. The second author received support from several Lenfest Summer Research Grants (an internal college grant).
Appendix: On the nucleus of certain p-groups—by Jonathan Blackhurst
Appendix: On the nucleus of certain p-groups—by Jonathan Blackhurst
In this appendix weFootnote 3 prove the proposition that if the Schur multiplier of a finite non-cyclic p-group G is trivial, then the nucleus of G is trivial. Our proof of the proposition will use the facts that a p-group has trivial nucleus if and only if it has no immediate descendants and that a finite group has trivial Schur multiplier if and only if it has no non-trivial stem extensions, so we will begin by recalling a few definitions. For the definition of the lower p-central series and p-class of a group, we refer to Sect. 2 of the article.
Definition 6.1
Let G be a finite p-group with minimal number of generators \(d=d(G)\) and presentation F / R where F is the free pro-p group on d generators. The p -covering group \(G^*\) of G is \(F/R^*\) where \(R^*\) is the topological closure of \(R^p[F,R]\), and the nucleus of G is \(P_c(G^*)\) where c is the p-class of G. The p -multiplicator of G is defined to be the subgroup \(R/R^*\) of \(G^*\). The Schur multiplier \(\mathcal {M}(G)\) of G is defined to be \((R\cap [F,F])/[F,R]\). A group C is a stem extension of G if there is an exact sequence
where K is contained in the intersection of the center and derived subgroups of C.
We will need to recall some basic properties of Schur multipliers and p-covering groups. First, for a finite group G, the largest stem extension of G has size \(|G||\mathcal {M}(G)|\). Hence, the Schur multiplier of a finite group G is trivial if and only if G admits no non-trivial stem extensions. Second, every elementary abelian central extension of G is a quotient of \(G^*\). By this we mean that if H is a d-generated p-group with elementary abelian subgroup Z contained in the center of H such that H / Z is isomorphic to G, then H is a quotient of \(G^*\). Every immediate descendant of G is an elementary abelian central extension of G, hence is a quotient of \(G^*\). A subgroup M of the p-multiplicator of G is said to supplement the nucleus if M and the nucleus together generate the p-multiplicator, that is \(MP_c(G^*)=R/R^*\). The immediate descendants of G can be put in one-to-one correspondence with equivalence classes of proper subgroups M of the p-multiplicator of G that supplement the nucleus. The equivalence relation comes from the action of the outer automorphism group of \(G^*\), so M and N are equivalent if there is an outer automorphism \(\sigma \) of \(G^*\) such that \(\sigma (M)=N\). The reader is referred to O’Brien [27] for more details.
With these preliminaries in place, we can show that the non-cyclic hypothesis in our proposition is necessary by considering the finite cyclic p-group \(G=\mathbb {Z}/p^c\mathbb {Z}\). The Schur multiplier is trivial since in this case \(F=\mathbb {Z}\) so [F, F] is trivial. On the other hand, the nucleus is non-trivial since in this case \(F=\mathbb {Z}_p\) and \(R=p^c\mathbb {Z}_p\) so \(R^*=p^{c+1}\mathbb {Z}_p\) and \(G^*=F/R^*=\mathbb {Z}/p^{c+1}\mathbb {Z}\) which implies that \(P_c(G^*)=p^cG^*\) is non-trivial.
Proposition 6.2
Let G be a finite non-cyclic p-group. If the Schur multiplier of G is trivial, then the nucleus of G is trivial.
Proof
We will prove the following equivalent assertion: if the nucleus of G is non-trivial, then G has a non-trivial stem extension. We divide the problem into two cases depending on whether the abelianization of G has stabilized; that is, whether the abelianization of an immediate descendant of G can have larger order than the abelianization \(G^{ab}\) of G. We will see that this is equivalent to whether or not \(G^{ab}\simeq (G/P_{c-1}(G))^{ab}\) where G has p-class c.
CASE 1: Suppose that \(G^{ab}\simeq (G/P_{c-1}(G))^{ab}\) and that the nucleus of G is non-trivial. Since the nucleus is non-trivial, G has an immediate descendant C and we have the following diagram
where \(K=P_c(C)\). Note that since \(C/P_{k}(C)\simeq G/P_{k}(G)\) for \(k\le c\), we have that \((C/P_{c-1}(C))^{ab}\simeq (C/K)^{ab}\). If \(P_{c-1}(C)\) were not contained within the derived subgroup \(C'\) of C, then its image \(\overline{P_{c-1}}(C)\) in \(C/C'\) would be non-trivial. Since \(K=P_{c-1}(C)^p[C,P_{c-1}(C)]\), the image \(\overline{K}\) of K would be \(\overline{P_{c-1}}(C)^p\) and thus would be strictly smaller than \(\overline{P_{c-1}}(C)\). Now \((C/H)^{ab}\simeq (C/C')/\overline{H}\) for any \(H\triangleleft C\), so, replacing H with K and \(P_{c-1}(C)\), we see that \((C/P_{c-1}(C))^{ab}\) would be smaller than \((C/K)^{ab}\), contradicting that they are isomorphic. Thus \(P_{c-1}(C)<C'\), hence \(K<C'\), so C is a stem extension of G. Since G has a non-trivial stem extension, its Schur multiplier is non-trivial.
CASE 2: Suppose that \(G^{ab} \not \simeq (G/P_{c-1}(G))^{ab}\). Let
be a presentation of G where F is free pro-p group on d generators and d is the minimal number of generators of G. Induction and the argument in the preceding case shows that \((G/P_k(G))^{ab}\) is strictly smaller than \((G/P_{k+1}(G))^{ab}\) for any \(k<c\). Furthermore, since the image \(\overline{P_{k+1}}(G)\) of \(P_{k+1}(G)\) in \(G/G'\) is \(\overline{P_k}(G)^p\), there must be a generator b of F such that the image of \(b^{p^{c-1}}\) in G lies outside \(G'\). Now consider \(R^*=R^p[F,R]\) and let \(G^*=F/R^*\) be the p-covering group of G. We have the following diagrams:
and
We now show that the image of \(b^{p^c}\) in \(P_c(G^*)\) is non-trivial so G has non-trivial nucleus. Let G have abelianization isomorphic to \(\mathbb {Z}/p^{n_1}\mathbb {Z}\times \cdots \times \mathbb {Z}/p^{n_d}\mathbb {Z}\). Consider the topological closure S of \(R\cup [F,F]\). Then F / S is isomorphic to \(G^{ab}\). The group \(\mathbb {Z}/p^{n_1+1}\mathbb {Z}\times \cdots \times \mathbb {Z}/p^{n_d+1} \mathbb {Z}\) is an elementary abelian central extension of F / S. This implies that \(b^{p^c}\) lies outside \(S^*=S^p[F,S]\). Since \(R\subset S\), we have that \(R^*\subset S^*\). Hence \(b^{p^c}\) lies outside \(R^*\) so it has non-trivial image in \(G^*\). Since its image lies inside \(P_c(G^*)\), this group is non-trivial.
We have shown that G has non-trivial nucleus. Now let a be a generator of F independent of b—i.e., one that doesn’t map to the same element as b in the elementary abelianization of F—and let \(\overline{M}\) be a proper subgroup of \(R/R^*\) that contains the image of \(b^{p^c}[a,b^{p^{c-1}}]\) and that supplements the subgroup of \(R/R^*\) generated by the image of \(b^{p^c}\) (so \(\overline{M}\) and the image of \(b^{p^c}\) generate \(R/R^*\)). Now consider \(C=G^*/\overline{M}\). Letting \(K=(R/R^*)/\overline{M}\), we have the following diagram
Since \(G^*\) is a central extension of G and C is a quotient of \(G^*\), C is also a central extension of G. Furthermore, \(|K|=p\). Now let M be the subgroup of F corresponding to \(\overline{M}\) under the lattice isomorphism theorem. Then we have the following diagram:
Since M does not contain \(b^{p^c}\), its image in C is non-trivial. Since G has p-class c, the image of \(b^{p^c}\) is trivial in G. Also since \(|K|=p\), the image of the powers of \(b^{p^c}\) constitute K. Since M does contain \(b^{p^c}[a,b^{p^{c-1}}]\), the image of \(b^{p^c}\) in C equals the image of \([b^{p^{c-1}},a]\), hence K lies in the derived subgroup of C, so C is a non-trivial stem extension of G. Consequently, the Schur multiplier of G is non-trivial. \(\square \)
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Boston, N., Bush, M.R. & Hajir, F. Heuristics for p-class towers of imaginary quadratic fields. Math. Ann. 368, 633–669 (2017). https://doi.org/10.1007/s00208-016-1449-3
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DOI: https://doi.org/10.1007/s00208-016-1449-3