Heuristics for p-class towers of imaginary quadratic fields

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.

Appendix: On the nucleus of certain p-groups—by Jonathan Blackhurst

In this appendix we³ 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

$$\begin{aligned} 1\rightarrow K\rightarrow C\rightarrow G\rightarrow 1 \end{aligned}$$

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

$$\begin{aligned}1\rightarrow K\rightarrow C\rightarrow G\rightarrow 1\end{aligned}$$

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

$$\begin{aligned} 1\rightarrow R\rightarrow F\rightarrow G\rightarrow 1 \end{aligned}$$

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:

$$\begin{aligned} 1\rightarrow R^*\rightarrow F\rightarrow G^*\rightarrow 1, \end{aligned}$$

and

$$\begin{aligned} 1\rightarrow R/R^*\rightarrow G^*\rightarrow G\rightarrow 1. \end{aligned}$$

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

$$\begin{aligned} 1\rightarrow K\rightarrow C\rightarrow G\rightarrow 1. \end{aligned}$$

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:

$$\begin{aligned} 1\rightarrow M\rightarrow F\rightarrow C\rightarrow 1. \end{aligned}$$

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. $\square$