aa r X i v : . [ m a t h . N T ] J u l MOMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OFQUADRATIC FIELDS
JACK KLYS
Abstract.
Let f ( K ) be the number of unramified extensions L/K of a quadraticnumber field K with Gal ( L/K ) = H and Gal ( L/ Q ) = G where G is a centralextension of F n by F . We find a function g ( K ) such that f /g has finite momentsand a distribution on its values. We show this distribution is a point mass when H isnon-abelian and the Cohen-Lenstra distribution when H is abelian, despite the factthat the set of values of f /g do not form a discrete set. We prove an explicit formulafor f as well as a refined counting function with local conditions. We also determinecorrelations of such counting functions for different groups G . Lastly we formulate aconjecture about moments and correlations for any pair of 2-groups ( G, H ). Introduction
Background.
The Cohen-Lenstra heuristics are a well known series of conjecturesand heuristics for the distributions of class groups of number fields. There has beenmuch recent interest and progress on these conjectures and their variants, includingnon-abelian versions. In this paper we make progress towards a potential (currentlynot yet formulated) non-abelian version of the Cohen-Lenstra conjectures for 2-groupsin the setting of quadratic fields by proving distributional results about unramified2-extensions of quadratic number fields.Let D ± X denote the set of positive (resp. negative) quadratic discriminants of absolutevalue less than X . For any function f ( K ) defined on the set of quadratic fields let E ± k ( f ) = lim X −→∞ P K,D K ∈D ± X f k ( K ) (cid:12)(cid:12) D ± X (cid:12)(cid:12) be the average of f k over positive (resp. negative) discriminants.For any pair of groups H ≤ G with [ G : H ] = 2 define f ( K ) = |{ L/K unramified | ∃ φ : G ( L/ Q ) ∼ = G, φ ( G ( L/K )) = H }| . We call any such extension as in the definition of f a ( G, H )-extension.The Cohen-Lenstra heuristics predict E ± k ( f ) for all k when H is an abelian p -group with p odd and G = H ⋊ C with C acting by inversion [CL84]. In thiscase f ( K ) = | Surj ( Cl K , H ) | · | Aut ( H ) | − . In fact it can be shown that the Cohen-Lenstra conjectures about the distribution of Cl K,p are equivalent to the statementthat E ± ( | Surj ( Cl K , H ) | ) = m ± for all finite abelian p -groups H , where m ± equals1 / Aut ( H ) and 1 in the postive and negative cases respectively.If we instead restrict to asking about the distribution of Cl K,p [ p ] this is equivalentto asking for a distribution of the values of the function | Surj ( Cl K , H ) | and in turn is equivalent to computing E ± k ( | Surj ( Cl K , Z /p Z ) | ) for all k ∈ Z ≥ . Thus computation ofthe moments E ± k ( f ) for any pair of groups ( G, H ) as above is a natural generalizationof the abelian Cohen-Lenstra conjectures.For any even group G Wood conjectured that E ± ( f ) is finite when there is a uniqueconjugacy class of order 2 elements not in H and infinite otherwise [Woo16]. Albertsand the author investigated the latter case for the pair ( G, H ) = ( Q ⋊ C , C ) bycomputing a finite value for E ± k (cid:0) f / ω ( D k ) (cid:1) for all k (here Q ⋊ C is the central productof D and C along C ) where f / ω ( D k ) is normalized precisely to make the momentsfinite [Alb16, AK16] ( ω ( n ) denotes the number of unique prime divisors of n ). Werefer the reader there for a more detailed exposition of all the above results.In this paper we consider the generalization of this to all 2-groups ( G, H ). In par-ticular when G is a central extension of F n by F we determine a value c such thatthe function f /c ω ( D k ) determines a distribution, and we compute this distribution andits moments. Furthermore we conjecture this value of c works for all pairs of finite2-groups ( G, H ).1.2.
Main results.
Let G be a finite 2-group with H ≤ G a subgroup such that[ G : H ] = 2. We will call the pair ( G, H ) admissible if there exists a (
G, H )-extension.This implies G can be generated by order 2 elements not contained in H . For exampleany pair ( G, H ) with G an abelian 2-group of exponent greater than 2 is not admissible.Let Φ ( G ) be the Frattini subgroup of G . Let T ⊂ G/ Φ ( G ) be the subset of allelements whose lifts to G have order 2 and are not contained in H . We will refer to T as the maximal admissible generating set. Let c = c T be the number of conjugacyclasses of order 2 elements in G which are not contained in H . This can be viewedas the number of orbits of the set of lifts of T acting on itself by conjugation. LetAut H ( G, Φ ( G )) be the subgroup of Aut ( G/ Φ ( G )) preserving the set H whose elementslift to Aut ( G ).Let µ ± CL ( H ) be the Cohen-Lenstra probability measure on finite abelian 2-groups H which is proportional to 1 / ( | Aut H |·| H | a ) where a = 0 , P ± CL ( i ) be the probability with respect to the Cohen-Lenstra measure that afinite abelian 2-group has rank i .Finally for any real valued function h defined on quadratic discriminants we willdenote by µ h the distribution on R (with the Borel σ -algebra) determined by its values(see Section 8.1 for a precise definition of µ h ).Our main result is the following. Theorem 1.1.
Let G be a central extension of F n by F with ( G, H ) is admissible. Let h ( d ) = f ( d ) /c ω ( d ) and a = | Aut H ( G, Φ( G )) | .If H is not abelian then the distribution of h ( d ) is µ h = δ P where δ P is the point-massat P ∈ Q and P = 0 .If H is abelian then h ( d ) has distribution µ h supported on the set { (2 i − /a | i ∈ Z ≥ } given by µ h (cid:0)(cid:0) i − (cid:1) / n − a (cid:1) = P ± CL ( i ) over the set of real (respectively imaginary) quadratic fields. OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 3
See (Theorem 8.2) for a precise formula for P . Considering the function f ( d ) /c ω ( d ) rather than f ( d ) is crucial. The latter has infinite moments of all orders and there isno distribution on its values.We note that even though in the case when H is abelian and µ h is supported on adiscrete subset of Q , the values of h themselves do not form a discrete subset. In fact h takes values not in the support of µ h with positive density in the set of quadraticfields. The proof of Theorem 1.1 shows that in this case h can be partially expressedin terms of | Cl K [4] /Cl K [2] | and this term is what controls the distribution of h andresults in the occurence of the distribution P ± CL ( i ).We can additionally deduce a result about moments of the above function as wellas correlations of functions for several pairs ( G i , H i ) . Let N ( k ) denote the number ofsubspaces of F k . Theorem 1.2.
Let ( G , H ) , . . . , ( G k , H k ) be a sequence of admissible pairs. Let T i, bethe maximal admissible generating set corresponding to ( G i , H i ) . Let f i be the functioncounting ( G i , H i ) -extensions. Let Y ⊂ [ k ] be the set of indices i for which H i is abelian.Then lim X −→∞ P d ∈D ± X Q ki =1 (cid:18) f i ( d ) c ω ( d ) Ti, (cid:19)P d ∈D ± X k Y j =1 q − j ! Y j ∈ [ k ] \ Y P j | Y | X i =0 ( − i (cid:18) | Y | i (cid:19) M ± ( | Y | − i ) where • q j = 2 n j − (cid:12)(cid:12) Aut H j ( G j , Φ ( G j )) (cid:12)(cid:12) and P j ∈ Q • M − ( j ) = N ( j ) and M + ( j ) = 2 − j ( N ( j + 1) − N ( j ))See (Theorem 9.4) for an explicit formula for P j . Note when ( G i , H i ) are all equalthis theorem gives the k th moment of f ( d ) /c ω ( d ) T .We have the following corollary of Theorem 1.1 on the density of quotients of theGalois group of the maximal unramified extension of quadratic fields. Corollary 1.3.
Let ( G , H , T , ) , . . . , ( G k , H k , T ,k ) be a set of admissible tuples. Let K ⊂ K be the set of all quadratic fields K whose maximal unramified extension K un / Q contains a ( G i , H i ) extension for all i .If H i is non-abelian for all i then K has density 1. Otherwise K has density − P CL (0) . The first step to proving the above results is obtaining an explicit formula for f expressed as a sum of Legendre symbols. In fact we obtain a formula for a refinedcounting function. Let T ⊂ G/ Φ ( G ) be any subset. We will call T admissible for( G, H ), or say that (
G, H, T ) is admissible if T lifts to a generating set of order 2elements of G not contained in H .For any L/K a (
G, H )-extension there is a canonical generating set U for G ( L/ Q ) / Φ ( G ( L/ Q )) which is not contained in G ( L/K ) and lifts to a generating setof order 2 elements ( U is given by projecting the inertia groups). We define f T ( K ) = |{ L/K unramified | ∃ φ : ( L/ Q ) ∼ = G, φ ( G ( L/K )) =
H, φ ( U ) = T }| JACK KLYS and call such an extension a (
G, H, T )-extension.Let Aut
H,T ( G, Φ ( G )) be the subgroup of Aut ( G/ Φ ( G )) preserving the sets H and T whose elements lift to Aut ( G ). Let c T denote the number of conjugacy classes oflifts of T . This can be interpreted as the number of orbits of the set of lifts of T actingon itself by conjugation. Theorem 1.4.
Suppose ( G, H, T ) is admissible. Let T = { t , . . . , t r } and let { x , . . . , x r } ⊂ G be any set of lifts. Let S i = { ≤ j ≤ r | [ x i , x j ] = 1 } . Then f T ( d ) = 12 n | Aut
H,T ( G, Φ ( G )) | X ( d ,...,d r ) r Y i =1 Y p | d i (cid:18) (cid:18) Q j ∈ S i d j p (cid:19)(cid:19) where the sum is over tuples of coprime fundamental discriminants such that d = Q ri =1 d i . In the proof we utilize the Embedding Theorem [MS91] which gives a condition forthe existence of the type of extensions we are interested in. This formula generalizesthe one in [AK16] used to count Q -extensions.We define a graph G ( T ) with vertex set T , and say u, v ∈ T are connected by anedge if their lifts do not commute. We show H being abelian is equivalent to G ( T )being complete and bipartite and split the proof up based on this property. We thenproceeds by determining the k th moments and hence the distribution of f T in thecase when G ( T ) is not complete bipartite, and building on this information to obtainthe distribution directly in the complete bipartite case by re-expressing f T in termsof the class group and other functions of the form f T and applying measure-theoreticarguments.The theorems of Fouvry and Kl¨uners on character sum cancellation [FK07] can beapplied to f T to show that the computation of the k th moments of f T depends onclassifying the maximal disconnected subgraphs of a graph G ∗ (cid:0) T k (cid:1) (see Section 4 forthe definition). In particular the sizes of these subgraphs determine the asymptoticsof f T , whereas their particular form is needed to compute the coefficient of the mainterm.While for k = 1 G ∗ ( T ) is closely related to G ( T ) above which is easy to handle,the structure of G ∗ (cid:0) T k (cid:1) is more complicated and necessitates additional tools. Since G ∗ (cid:0) T k (cid:1) depends on relations in G we encode these using a bilinear form in characteristic2 which can be used to study the graph by employing the theory of bilinear forms andlinear algebra in characteristic 2 (see Section 3).1.3. Related work.
There are some result known regarding the values E ± k ( f ) forvarious pairs of groups ( G, H ) and variants of f .The well known theorem of Davenport-Heilbronn gives E ± ( f ) for H ∼ = Z / Z [DH71].Bhargava computed E ± ( f ) for ( G, H ) = ( S n , A n ) for n = 3 , , p = 2) can be rephrased as computing E ± k (cid:0) f / ω ( D k ) (cid:1) for ( D , C ) for all k (see Section 10.3).The Embedding Theorem [MS91] used in the proof of Theorem 1.4 has been gener-alized by Alberts [Alb17] to a larger class of extensions, and consequently can be used OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 5 to obtain more general counting functions, the study of which we expect to form thebasis of future work.1.4.
Conjectures.
We state some conjectures which are suggested by our main theo-rems.
Conjecture 1.5.
Let G be a finite -group. Let H ≤ G be a subgroup with [ G : H ] = 2 such that ( G, H ) is admissible. Then(1) < E ± k (cid:18) fc ω ( D K ) (cid:19) < ∞ and these moments determine a distribution on the values of f .(2) There exists a constant C ( G, H ) such that X K, < ± D K Let ( G , H , T ) , . . . , ( G k , H k , T k ) be a sequence of admissible tuplesof finite 2-groups and generating sets. Then(1) For all positive integers k < E ± Q ki =1 f T i Q ki =1 c ω ( D K ) T i ! < ∞ . If ( G i , H i , T i ) = ( G, H, T ) for all i these moments determine a distribution onthe values of f T .(2) There exists a constant C ( G, H ) such that X K, < ± D K Let G be a central extension of F n by F and let ( G, H, T ) be admissible.Suppose G is not elementary abelian. Then Conjecture 1.5 (1) is true.If either of the following conditions is true:(1) T i = T i, is the maximal admissible generating set for all i = 1 , . . . , k JACK KLYS (2) k = 1 and T is any admissible generating set for ( G , H ) then Conjecture 1.6 (1) true. The proofs of our main theorems imply that in fact part (2) of Conjectures 1.5and 1.6 is true but for the function counting extensions unramified away from infinity(indeed throughout the proof we work with this counting function, and show that whennormalized it is asymptotic to f T /c ω ). This at least gives a bound for the quantity inpart (2) of the above conjectures.1.5. Plan of the proof. In Section 2 we prove a formula for f the number of ( G, H, T )-extensions of any quadratic field. In Sections 3 we prove necessary technical resultsabout quadratic forms in characteristic 2. We define the graphs we will use in ourcomputation of the moments of f in Section 4 and classify the disconnected subgraphsin Section 5.Applying these results we compute the asymptotics and moments of f in Sections 6and 7. Finally we compute the distributions and moments and correlations in Sections8 and 9. 2. Counting 2-extensions The main goal of this section is to obtain a formula for the function f T ( K ) whichgives the number of ( G, H, T )-extensions (defined below) of a quadratic field K for anyfixed admissible ( G, H, T ).2.1. Preliminaries. Let K = Q (cid:16) √ d (cid:17) . Let G be a central extension of F n by F andlet L be an unramified extension of K with φ : G ( L/ Q ) ∼ = G such that φ ( G ( L/K )) = H .Let L g = L ∩ K gen where K gen = Q (cid:0) √ q , . . . , √ q r (cid:1) and the q i are prime discriminantswith Q q i = d . It is easy to show that L g = Q (cid:16)p d ′ , . . . , p d ′ n (cid:17) where d ′ i | d and areindependent modulo ( Q ∗ ) .It is also clear that G ( L g / Q ) ∼ = C n and we can pick the isomorphism such that thestandard basis element e i ∈ C n projects nontrivially onto G i = G (cid:0) K ( p d ′ i ) / Q (cid:1) . Let y i be any chosen lift of e i in G ( L/ Q ). Let h a i be the distinguished central subgroupof G of order 2. Note this implies there are only two possibilities for y i and they differby a multiple of a . Lemma 2.1. The set of y i generates G ( L/ Q ) .Proof. The e i generate C n ∼ = G ( L/ Q ) / h a i so the y i generate at least 2 n elements in G ( L/ Q ) \ { a } and hence h y i i ni =1 ∪{ a } generates G ( L/ Q ). Since a ∈ Z ( G ) and G ( L/ Q )is not abelian there must exist y i and y j which don’t commute, so that [ y i , y j ] = a . (cid:3) For any such extension L/ Q let U ⊂ G ( L/ Q ) / Φ ( G ( L/ Q )) be the projection of thegenerators of all the inertia subgroups of G ( L/ Q ). Definition 2.2. We call L/K an unramified (resp. unramified away from infinity)( G, H, T )-extension if it is unramified everywhere (resp. away from infinity) and there OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 7 exists an isomorphism φ : G ( L/ Q ) ∼ = G such that φ ( G ( L/K )) = H and φ ( U ) = T .We call φ the associated isomorphism of the ( G, H, T )-extension (note φ is not uniquein general).Until Section 8 we will work with unramified away from infinity ( G, H, T )-extensions,and use f T to denote the number of these. Note this differs from the definition used upto this point. To prove our theorems for unramified everywhere ( G, H, T )-extensionswe will apply Lemma 8.10. Thus until further notice we will simply say ( G, H, T )-extensions to refer to those unramified away from infinity.2.2. Extensions with a fixed genus subfield. Throughout this subsection we willfix an unramified away from infinity extension L ′ /K such that G ( L ′ /K ) ∼ = C n − and G ( L ′ / Q ) ∼ = C n . We will count ( G, H, T )-extensions L/K with L g = L ′ . Fix anisomorphism φ ′ : Gal ( L ′ / Q ) ∼ = G/ Φ ( G ) such that φ ′ ( U ) = T and φ ′ (Gal ( L ′ /K )) = H .First we define a notion which will be helpful to this end, using ideas of Lemmermeyerfrom [Lem97]. The field L can be written as L = L g (cid:0) √ µ (cid:1) for some µ ∈ L g . Then since L/L g is a Kummer extension we see that µ e i e j = µ in L g . Let α ij = µ e i e j − . Then (cid:0) α ij (cid:1) e i e j = µ ( e i e j − e i e j +1) = 1. Hence α e i e j ij = ± 1. Define S ( µ ) ∈ F ( n ) by S ( µ ) ij = ( α e i e j ij = − 10 if α e i e j ij = +1 . Lemma 2.3. y i and y j commute in G if and only if S ( µ ) ij = 0 .Proof. Since the y i have order 2 we have [ y i , y j ] = ( y i y j ) . We have √ µ y i y j = α ij √ µ and hence (cid:0) a + b √ µ (cid:1) ( y i y j ) = a + α e i e j ij b √ µ for a, b ∈ L g . Thus y i y j has order 2 if andonly if α e i e j ij = 1 and the result follows. (cid:3) This lemma in particular shows that the group G is determined by the choice of e i ’salong with S ( µ ) (up to permutation of the entries) since by Lemma 2.1 G is generatedby the y i which by definition have order 2, and S ( µ ) encodes the relations betweenthem. Then it is clear that S ( µ ) = 0 in F ( n ) if and only if G = C n +12 . Lemma 2.4. Let L = L g (cid:0) √ µ (cid:1) and L = L g ( √ ν ) be ( G, H, T ) -extensions with L gi = L ′ and associated isomorphisms φ and φ . Assume φ = φ = φ ′ .Then µ = νδ for some δ ∈ Z a fundamental discriminant such that δ | d . Converselyfor any fundamental discriminant δ ∈ Z such that δ | d we have that L g (cid:0) √ δµ (cid:1) is a ( G, H, T ) -extension with associated isomorphism that projects to φ ′ .Proof. Let M = L L and let L be the third quadratic extension of L g containedin M , so L = L g (cid:0) √ µν (cid:1) . Let β ij = ν e i e j − . Then ( α ij β ij ) = ( µν ) e i e j − . Hence S ( µν ) ij = S ( µ ) ij · S ( ν ) ij .Let ϕ = φ − ◦ φ . Then ϕ : G ( L / Q ) ∼ = G ( L / Q ) such that the morphism ϕ induced on the quotient G ( L ′ / Q ) is the identity. From the existence of ϕ we see that S ( µ ) ij = S ( ν ) ij for all i, j , and thus G ( L / Q ) ∼ = C n +12 . This implies µν = δ ∈ Z and δ JACK KLYS can be chosen to be a fundamental discriminant. Since L and L are unramified over K = Q ( √ d ) so is L , and hence δ | d .Now suppose δ ∈ Z is a fundamental discriminant such that δ | d . From thedefinitions it is clear that S ( µ ) = S ( δµ ) and hence there is an isomorphism ϕ : G (cid:0) L g (cid:0) √ δµ (cid:1) / Q (cid:1) ∼ = G (cid:0) L g (cid:0) √ µ (cid:1) / Q (cid:1) such that ϕ = id on G ( L g / Q ). Additionally L g (cid:0) √ δµ (cid:1) /K is unramified since it is a subfield of the composite of L g (cid:0) √ µ (cid:1) /K and K (cid:16) √ δ (cid:17) /K , both of which are unramified (the latter because it is contained in K gen ).This implies L g (cid:0) √ δµ (cid:1) is a ( G, H, T )-extension with associated isomorphism φ ◦ ϕ . (cid:3) Proposition 2.5. Let L ′ = K ( p d ′ , . . . p d ′ n ) . If there exists a ( G, H, T ) -extension L/ Q such that L g = L ′ with associated isomorphism φ satisfying φ = φ ′ then there areexactly ω ( d ) − n such extensions.Proof. If L = K ( p d ′ , . . . p d ′ n ) (cid:0) √ µ (cid:1) is a ( G, H, T )-extension with associated isomor-phism φ ′ then by Lemma 2.4 every other such extension is of the form L g (cid:0) √ δµ (cid:1) for somefundamental discriminant δ | d . Define the F vector space V = (cid:10) q , . . . , q ω ( d ) (cid:11) / D q , . . . , q ω ( d ) E where the q i are the divisors of d which are prime fundamental discriminants or oneof {− , ± } . Then the number of such extensions which are distinct is the size of thevector space V / h d ′ , . . . , d ′ n i , which is clearly 2 ω ( d ) − n (recall the d ′ i are all independentmod Q ∗ ). (cid:3) For a, b ∈ Q let ( a, b ) denote the cyclic algebra in Br ( Q ) defined by( a, b ) = K h u, v i / (cid:10) u = a, v = b, uv = − vu (cid:11) . Recall that there is an injection inv : Br ( Q ) ֒ → Y v Br ( Q v ) ∼ = Y v Q / Z . Since ( a, b ) has order 2 we can view each factor of the image of inv as lying in Z / Z .Identify this group with h± i . Denote the composition of inv with the projection ontothe factor corresponding to v by inv v . Denote by ( a, b ) v the quadratic Hilbert symbolover the field Q v . Then the crucial property we will need is inv v (( a, b )) = ( a, b ) v . We will use the following theorem, stated as Theorem 1.2 in [MS91]. Theorem 2.6 (Embedding Criterion) . Let L ′ = F ( p d ′ , . . . , p d ′ n ) , where the d ′ i areelements of F ∗ independent modulo F ∗ . Let G = Gal ( L ′ /F ) , and consider a non-split central extension G ∗ of F by G . Let e , . . . , e n generate G , where e i p d ′ j =( − δ ( i,j ) p d ′ j and let x , . . . , x n be any set of preimages of e , . . . , e n in G ∗ . Define c ij = 1 if [ x i , x j ] = 1 and 0 otherwise for i = j and c ii = 1 if x i = 1 . OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 9 Then there exists a Galois extension L/F with L ′ ⊂ L such that Gal ( L/F ) ∼ = G ∗ and the surjection G ∗ −→ G is the natural surjection of Galois groups, if and only if Y i ≤ j (cid:0) d ′ i , d ′ j (cid:1) c ij = 1 ∈ Br ( F ) . Extensions of a fixed quadratic field. For the next proposition we define somenotation. Let H be the projection of H to G/ Φ ( G ). Let K T be the set of pairs ( L ′ , φ ′ )where L ′ /K is a subfield of K gen with isomorphism φ ′ : Gal ( L ′ / Q ) ∼ = G/ Φ ( G ) suchthat φ ′ ( U ) = T and φ ′ (Gal ( L ′ /K )) = H . Let D be the set of r -tuples ( d , . . . , d r ) ofcoprime fundamental discriminants with each d i = 1 and such that d = Q ri =1 d i . Define U i ⊂ [ n ] by the expression t i = P j ∈ U i e j for each i = 1 , . . . , r . Proposition 2.7. There is a bijective correspondence between K T and D .Proof. Suppose ( L ′ , φ ′ ) ∈ K T . Define d i to be the product of all prime fundamentaldiscriminants p (if p = 2 use ± ord d ) such that φ ′ ( I p ) = t i for some (equivalently all) p | p in L ′ . Then all the d i are coprime and d = Q ri =1 d i since each t i is in the image ofsome inertia group by definition.Conversely given a factorization d = Q ri =1 d i in D define d ′ j = Q i : j ∈ U i d i for each j = 1 , . . . , n . Let L ′ = Q ( p d ′ , . . . , p d ′ n ). For each j = 1 , . . . , n let φ ′ j : Gal ( L ′ / Q ) −→ Gal (cid:0) Q (cid:0)p d ′ j (cid:1)(cid:1) be the projection. Then φ ′ = ( φ ′ , . . . , φ ′ n ) is a homomorphism φ ′ :Gal ( L ′ / Q ) −→ G/ Φ ( G ) where G/ Φ ( G ) has basis { e , . . . , e n } .We will show φ ′ is an isomorphism. The relation d ′ j = Q i : j ∈ U i d i implies that if p | d i then p | d ′ j exactly when j ∈ U i . This implies that for any prime p | p in L ′ if I p is the inertia group then φ ′ j ( I p ) = 1 exactly when j ∈ U i . Thus φ ′ ( I p ) = t i . Sincethe { t , . . . , t n } generate G/ Φ ( G ) this implies φ ′ is surjective. Since Gal ( L ′ / Q ) is aquotient of F n this implies φ ′ is an isomorphism.Pick a basis { e ′ , . . . , e ′ n } for Gal ( L ′ / Q ) such that e ′ i p d ′ j = ( − δ ij p d ′ j . Then clearly φ ′ (cid:0) e ′ j (cid:1) = e j . Let t ′ i = P j ∈ U i e ′ i . It follows from the above argument that φ ′ ( U ) = T .Each t ′ i projects nontrivially to Gal ( K/ Q ) = F so Gal ( L ′ /K ) is equal to the trace0 subspace of Gal ( L ′ / Q ) with respect to the generating set { t ′ , . . . , t ′ n } . The same istrue for H ⊂ G/ Φ ( G ) with { t , . . . , t n } . Thus φ ′ (Gal ( L ′ /K )) = H .Next we will show the above maps are inverses of each other.Let ( L ′ , φ ′ ) ∈ K T . We can write L ′ = Q ( p d ′ , . . . , p d ′ n ) where Q ( p d ′ j ) is the subfieldcorresponding to φ ′− ( e j ). Let d = Q ri =1 d i be the corresponding factorization.We claim that d ′ j = Q i,j ∈ U i d i for each j = 1 , . . . , n . Fix p | d . Then p | d ′ j if andonly if I p projects non-trivially to Q ( p d ′ j ). This is equivalent to φ ′ ( I p ) = t i for some i satisfying j ∈ U i which by definition of the correspondence is equivalent to p | d i for some i satisfying j ∈ U i . Furthermore it is clear that φ ′ = ( φ ′ , . . . , φ ′ n ) as definedabove. This proves one direction.Let d = Q ri =1 d i in D . It is shown above that if p | d i then φ ′ ( I p ) = t i for any p | p in L ′ where ( L ′ , φ ′ ) is the corresponding pair. Thus the maps are inverse to each other,which completes the proof. (cid:3) Recall { e , . . . , e n } is a fixed basis for G/ Φ ( G ). For each i = 1 , . . . , n let x ′ i ∈ G beany lift of e i . Let S ′ i = (cid:8) ≤ j ≤ n | j = i, (cid:2) x ′ i , x ′ j (cid:3) = 1 (cid:9) ∪ { i | ord ( x ′ i ) = 4 } . Lemma 2.8. Let ( L ′ , φ ′ ) ∈ K T with L ′ = K ( p d ′ , . . . p d ′ n ) . There exists a ( G, H, T ) -extension L/L ′ / Q with isomorphism φ such that φ ′− ◦ φ is the natural surjection ofGalois groups if and only if Y p | d − χ ( p ) Q j ≤ i,j ∈ S ′ i (cid:0) d ′ j /p α j ( p ) (cid:1) α i ( p ) (cid:0) d ′ i /p α i ( p ) (cid:1) α j ( p ) p = 0 where χ ( p ) = ( p − P j ≤ i,j ∈ S ′ i α i,j ( p ) p odd P j ≤ i,j ∈ S ′ i ( d ′ j / αj (2) ) − ( d ′ i / αi (2) ) − p = 2 . Proof. We identify Gal ( L ′ / Q ) with G/ Φ ( G ) using φ ′ . Then by Theorem 2.6 such anextension exists if and only if Q j ≤ i (cid:0) d ′ i , d ′ j (cid:1) c ij = 1 ∈ Br ( Q ) (it is easy to show thatwhen F = Q and L ′ /K is unramified at the finite places, the extension L given byTheorem 2.6 is always also unramified above K at the finite places). Clearly c ij = 1 ifand only if j ∈ S ′ i .By the discussion preceding the theorem, since inv v is an injection such an extensionexists if and only if 1 = inv v Y j ≤ i,j ∈ S ′ i (cid:0) d ′ i , d ′ j (cid:1) = Y j ≤ i,j ∈ S ′ i (cid:0) inv v (cid:0) d ′ i , d ′ j (cid:1)(cid:1) = Y j ≤ i,j ∈ S ′ i (cid:0) d ′ i , d ′ j (cid:1) v . at all places v of Q . By the product formula for the Hilbert symbol this condition onlyneeds to be checked at the finite primes. This is trivially satisfied for v ∤ d .To simplify notation let α i ( p ) = ord p d ′ i and α i,j ( p ) = α i ( p ) α j ( p ). Note the propertyof the quadratic Hilbert symbol (cid:0) d ′ i , d ′ j (cid:1) v = (cid:0) d ′ j , d ′ i (cid:1) v . Then the condition at odd p becomes( − ( p − P j ≤ i,j ∈ S ′ i α i,j ( p ) Q j ≤ i,j ∈ S ′ i (cid:0) d ′ j /p α j ( p ) (cid:1) α i ( p ) (cid:0) d ′ i /p α i ( p ) (cid:1) α j ( p ) p = 1The condition at p = 2 becomes − P j ≤ i,j ∈ S ′ i (cid:18) d ′ j/ αj (2) (cid:19) − ( d ′ i/ αi (2) ) − + α i (2) (cid:18) d ′ j/ αj (2) (cid:19) − + α j (2) ( d ′ i/ αi (2) ) − = 1 OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 11 which can be rewritten as( − P j ≤ i,j ∈ S ′ i (cid:18) d ′ j/ αj (2) (cid:19) − ( d ′ i/ αi (2) ) − Q j ≤ i,j ∈ S ′ i (cid:0) d ′ j /p α j (2) (cid:1) α i (2) (cid:0) d ′ i /p α i (2) (cid:1) α j (2) . The result follows. (cid:3) For each i = 1 , . . . , r let x i ∈ G be any lift of t i . Let S i = { ≤ j ≤ r | [ x i , x j ] = 1 } . Lemma 2.9. Let ( L ′ , φ ′ ) ∈ K T and d = Q ri =1 d i ∈ D the corresponding factorization(see Proposition 2.7). There exists a ( G, H, T ) -extension L/ Q with isomorphism φ such that L ′ ⊂ L and φ ′− ◦ φ is the natural surjection of Galois groups if and only if (2.1) r Y i =1 Y p | d i (cid:18) Q j ∈ S i d j p (cid:19) ord p ( d ) ! = 0 . Proof. By Lemma 2.8 there exists a ( G, H, T )-extension L/ Q with isomorphism φ suchthat L ′ ⊂ L and φ ′− ◦ φ is the natural surjection of Galois groups if and only if(2.2) Y p | d − χ ( p ) Q j ≤ i,j ∈ S ′ i (cid:0) d ′ j /p α j ( p ) (cid:1) α i ( p ) (cid:0) d ′ i /p α i ( p ) (cid:1) α j ( p ) p = 0where χ ( p ) = ( p − P j ≤ i,j ∈ S ′ i α i,j ( p ) p odd P j ≤ i,j ∈ S ′ i ( d ′ j / αj (2) ) − ( d ′ i / αi (2) ) − p = 2 . Since each x i has order 2 this implies that the size of the set (cid:8) ( j , j ) | ≤ j ≤ n, j ≤ j , j ∈ S ′ j , j , j ∈ U i (cid:9) is even for all 1 ≤ i ≤ r . Note that for 1 ≤ j ≤ n and any p , α j ( p ) ≥ p | d i and j ∈ U i for some 1 ≤ i ≤ r .Hence for p odd χ ( p ) = ( p − X j ≤ j ,j ∈ S ′ j ; j ,j ∈ U i 1= ( p − (cid:12)(cid:12)(cid:8) ( j , j ) | ≤ j ≤ n, j ≤ j , j ∈ S ′ j , j , j ∈ U i (cid:9)(cid:12)(cid:12) for some i . Thus χ ( p ) is even for p odd. Since every odd fundamental discriminantis congruent to 1 mod4, ( d ′ i / αi (2) ) − is odd exactly when α i (2) ≥ 1. This similarlyimplies that χ (2) is even. Now we partition the product over p | d in (2 . d i . Fix 1 ≤ i ≤ r and p | d i .Recall this implies α j ( p ) ≥ j ∈ U i . Fix 1 ≤ j ≤ n . Then Y k ∈ S ′ j (cid:0) d ′ j /p α j ( p ) (cid:1) α k ( p ) (cid:0) d ′ k /p α k ( p ) (cid:1) α j ( p ) = Y k ∈ S ′ j ∩ U i (cid:0) d ′ j /p α j ( p ) (cid:1) α k ( p ) (cid:0) d ′ k /p α k ( p ) (cid:1) α j ( p ) × Y k ∈ S ′ j \ U i ( d ′ k ) α j ( p ) = (cid:0) d ′ j /p α j ( p ) (cid:1) ord p ( d ) | S ′ j ∩ U i | · a for some a . Hence considering the contribution from each j we see that for any1 ≤ l ≤ r with l = i the power of d l in the factor corresponding to d i in (2 . p ( d ) P j ∈ U l (cid:12)(cid:12) S ′ j ∩ U i (cid:12)(cid:12) and P j ∈ U l (cid:12)(cid:12) S ′ j ∩ U i (cid:12)(cid:12) is odd exactly when x l and x i donot commute. Note division by p only occurs in (2 . 2) when α j,k ( p ) ≥ Y p | d i (cid:18) Q j ∈ S i d j p (cid:19) ord p ( d ) ! . (cid:3) Let Aut H,T ( G/ Φ ( G )) be the subgroup of Aut( G/ Φ ( G )) preserving the set H andthe set T . Let Aut H,T ( G, Φ ( G )) be the subgroup of Aut H,T ( G/ Φ ( G )) consisting ofelements which lift to Aut ( G ). Theorem 2.10. Suppose ( G, H, T ) is admissible. Then f T ( d ) = 12 n | Aut H,T ( G, Φ ( G )) | X ( d ,...,d r ) ∈D r Y i =1 Y p | d i (cid:18) Q j ∈ S i d j p (cid:19) ord p ( d ) ! . (2.3) Proof. Let L ′ = Q ( p d ′ , . . . , p d ′ n ) and suppose ( L ′ , φ ′ ) ∈ K T with corresponding tuple( d , . . . , d r ) ∈ D given by Proposition 2.7. By Lemma 2.9 there exists a ( G, H, T )-extension L/ Q with L g = L ′ and associated isomorphism φ such that φ ′− ◦ φ is thenatural surjection of Galois groups (equivalently φ ′ = φ ) if and only if(2.4) r Y i =1 Y p | d i (cid:18) Q j ∈ S i d j p (cid:19) ord p ( d ) ! = 0 . This function takes on the values 0 or 2 ω ( d ) . By Proposition 2.5 if a ( G, H, T )-extension L/ Q with L g = L ′ and associated isomorphism φ such that φ ′ = φ exists,there are exactly 2 ω ( d ) − n such extensions. Hence the formula (2 . 4) divided by 2 n givesthe number of ( G, H, T )-extensions L/ Q with L g = L ′ and associated isomorphism φ such that φ ′ = φ .For fixed L ′ the group Aut H,T ( G/ Φ ( G )) acts freely transitively on the set I = { ( L ′ , φ ′ ) ∈ K T } by composition with φ ′ . The set of ( G, H, T )-extensions L/ Q with L g = L ′ and associated isomorphism φ such that φ ′ = φ is equal for every element in OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 13 the orbit of φ ′ under Aut H,T ( G, Φ ( G )). Hence summing over D overcounts by a factorof | Aut H,T ( G, Φ ( G )) | . This completes the proof. (cid:3) Remark . We will eventually look at the growth of the function 2.3 as d rangesover all fundamental discriminants. We can make some heuristic considerations to seewhat kind of behaviour to expect. If we expect the legendre symbol to take each value ± / Q i,S i = ∅ Q p | d i (cid:16) (cid:16) Q j ∈ Si d j p (cid:17)(cid:17) is12 ω ( Q i,Si = ∅ d i ) 2 ω ( Q i,Si = ∅ d i ) + (cid:18) − ω ( Q i,Si = ∅ d i ) (cid:19) . Let r = |{ i | S i = ∅ }| and r = r − r . Hence on average we expect2 n | Aut H,T ( G, Φ ( G )) | f T ( d ) = X d = d ··· d r ω ( Q i,Si = ∅ d i )= ω ( d ) X i =0 i (cid:18) ω ( d ) i (cid:19) r i r ω ( d ) − i = r ω ( d )1 ω ( d ) X i =0 (cid:18) r r (cid:19) i (cid:18) ω ( d ) i (cid:19) = r ω ( d )1 (cid:18) r r (cid:19) ω ( d ) = ( r + 2 r ) ω ( d ) . Thus we expect P d In this section we prove a result about maximal isotropic subsets of quadratic formsin characteristic 2. We will use this result to deduce statements about maximal dis-connected subgraphs of the graph G ∗ ( T ) on the generating set T (see 4) by encodingthe graph structure using a particular quadratic form defined in the next section.We collect some facts regarding quadratic forms in characteristic 2. Let Q : F n −→ F be a quadratic form and let B ( u, v ) = Q ( u + v )+ Q ( u )+ Q ( v ) be the correspondingbilinear form.For any subspace W ⊂ F n we define rad ( W ) = W ∩ W ⊥ (where W ⊥ is taken withrespect to B ). We can decompose F n = V ⊕ rad( F n ), and rad( F n ) = R ⊕ R such that Q ( R ) = 0. We say Q is non-degenerate if rad( F n ) = 0. Hence in the above notation Q is non-degenerate on V and it is easy to see that dim R = 0 or 1. We say any v ∈ F n is singular if v = 0 and F n . We say a subspace W ⊂ F n is totallysingular if Q ( v ) = 0 for all v ∈ W .The following result is Proposition 14.6 from [Gro02]. Lemma 3.1. If Q is nondegenerate, given any totally singular space there exists an-other one disjoint from it. We extend this result to the case when Q is degenerate and ker Q | F n ⊥ = 0. In theabove notation this means F n = V ⊕ R . Let p V , p R be the projections onto V and R . Proposition 3.2. Let Q be a quadratic form on F n and let R = rad( F n ) be non-trivial.Suppose ker( Q | R ) = 0 . Given any totally singular subspace W there exists anothertotally singular subspace W ′ of the same size such that W ∩ W ′ ⊂ h u i for some u ∈ F n with p R ( u ) = 0 .Proof. As noted above we have F n = V ⊕ R . Note the subspace W = W ∩ V has index ≤ W .We can apply Proposition 14.6 from [Gro02] to W to obtain a totally singularsubspace W ⊂ V of the same size such that W ∩ W = 0. Furthermore W has basis { u , . . . , u k } and W has basis { v , . . . , v k } and H i = h u i , v i i is hyperbolic, meaning B ( u i , v i ) = 1, and B ( H i , H j ) = 0 for i = j .If W = W then we let W ′ = W . Then W ′ satisfies the desired properties and weare done. Otherwise let u ∈ W \ W . Since Q is non-degenerate on V we can write V = W ⊕ W ⊕ W with W = ( W ⊕ W ) ⊥ (taken in V ).Write p V ( u ) in this basis and let v be the element obtained by applying to p V ( u )the transposition exchanging u i and v i for all i (note that u = v is possible).First we will show Q ( v ) = 0. Write u = w + w + w + r with w i ∈ W i and r ∈ R .Recall B ( u, v ) = Q ( u + v ) + Q ( u ) + Q ( v ). We have Q ( u ) = Q ( w + w + w + r )= Q ( w + w + w ) + Q ( r )= Q ( w + w ) + Q ( w ) + Q ( r )= B ( w , w ) + Q ( w ) + Q ( r )where we are using B ( w + w + w , r ) = 0, B ( w + w , w ) = 0 and the fact that W and W are totally singular. This implies in particular that Q ( u + w ′ ) = B ( w + w ′ , w )+ Q ( w ) + Q ( r ) for any w ′ ∈ W . Since Q ( u + w ′ ) = 0 for all w ′ ∈ W , by theproperties of B on hyperbolic subspaces listed above this implies w = 0. Thus Q ( u ) = Q ( w ) + Q ( r ). By symmetry also Q ( v ) = 0.Then also by symmetry v ∈ W ⊥ (taken in F n ) and hence W ′ = h W , v i is totallysingular. Clearly p R ( v ) = 0 and | W ′ | = | W | .It remains to show W ∩ W ′ ⊂ h u i . Suppose w + u = w + v with w i ∈ W i . Asshown above u is not supported on W and v is not supported on W . Thus we haveequality if and only if u = v and w i = 0. This completes the proof. (cid:3) We now prove some facts about maximal isotropic subsets of certain generating setsof F n . We start with the case when ker( Q | F n ⊥ ) = 0. OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 15 Lemma 3.3. Suppose ker( Q | F n ⊥ ) = 0 . Let H ⊂ F n be a codimension 2 subspace. Let S = (ker Q ) \ H and suppose S generates F n . Then for any subset T ⊂ S (cid:12)(cid:12) T ⊥ ∩ T (cid:12)(cid:12) ≤ | S | − | T | . Proof. Decompose F n = V ⊕ R , where R = F n ⊥ and ker( Q | R ) = 0. To simplify notationlet b T = T ⊥ ∩ T and r = | S | . Suppose T = { w , . . . , w r ′ } ⊂ S and { w , . . . , w m } = b T .We want to show m ≤ r − r ′ . Assume m ≥ B ( w i , w j ) = 0 and Q ( w i ) = 0 for all w i , w j ∈ b T . The set b T can beextended to a maximal totally singular subspace W ⊂ F n containing b T .If dim( V ⊕ R ) is even then R = 0. If dim( V ⊕ R ) is odd then R = h z i and Q ( z ) = 1.Let p R be the projection onto R . By Lemma 3.2 there exists a totally singular subspace W ′ ⊂ Y of the same size as W such that W ∩ W ′ ⊂ h w i for some w with p R ( w ) = 0.First suppose w / ∈ b T , so that W ∩ W ′ = { } . Let W ⊂ W ′ be the set of all elements w such that B ( w i , w ) = 1 for some i = 1 , . . . , m . Suppose towards a contradictionthat | W ∩ S | < m . Then since ˆ T ∩ W ′ = ∅ we see that W = D ( W ′ \ W ) ∪ b T E isa totally singular subspace containing b T which has strictly more elements not in H than W ′ . Since dim( W / ( W ∩ H )) = 1, that is half the elements in W are in H , thisimplies W is a totally singular space containing b T which is strictly larger than W ′ andhence W . Thus it must be that | W ∩ S | ≥ m .If w ∈ b T then by the same argument we get | W ∩ S | ≥ m − 1. Note B ( w, w i ) = 0for i = 1 , . . . , r ′ . Since ker( Q | R ) = 0 it follows that w / ∈ R . If B ( w, u ) = 0 for all u ∈ S then this would imply w ∈ R since S is a generating set for Y . Thus thereexists some w ′ ∈ S such that B ( w, w ′ ) = 1. Since w ∈ W ′ this implies w ′ / ∈ W ′ . Thus | ( W ∪ { w ′ } ) ∩ S | ≥ m .From the definition of W and w ′ it follows that (( W ∪ { w ′ } ) ∩ S ) ∩ { w , . . . , w r ′ } = ∅ . Thus m ≤ r − r ′ as desired. (cid:3) Remark . A simpler version of the above proof gives the same result when S =(ker Q ) \ { } and S generates F n . We will require this fact in the proof of the nextproposition. Proposition 3.5. Let H ⊂ F n be a codimension 2 subspace. Let S = (ker Q ) \ H andsuppose S generates F n . Then for any subset T ⊂ S satisfying T ∩ ker( Q | F n ⊥ ) = ∅ wehave (cid:12)(cid:12) T ⊥ ∩ T (cid:12)(cid:12) ≤ | S | − | T | . Proof. By Lemma 3.3 the result holds when ker( Q | F n ⊥ ) = 0, so assume ker( Q | F n ⊥ ) = 0.Decompose F n = V ⊕ R ⊕ R , where R ⊕ R = F n ⊥ and R = ker( Q | R ⊕ R ). Let Y = V ⊕ R . Let T z = T ∩ ( Y + z ) for any z ∈ R . By translating back each of these cosets weobtain the subset T ′ = ∪ z ∈ R ( T z + z ) of Y . Note T ′⊥ = T ⊥ . We have (cid:12)(cid:12) T ⊥ ∩ T (cid:12)(cid:12) = X z ∈ R (cid:12)(cid:12) T ⊥ ∩ T z (cid:12)(cid:12) ≤ X z ∈ R \ H (cid:12)(cid:12) T ′⊥ ∩ ( T ′ ∩ H ) (cid:12)(cid:12) + X z ∈ R ∩ H (cid:12)(cid:12) T ′⊥ ∩ ( T ′ \ H ) (cid:12)(cid:12) = | R | [ R : R ∩ H ] (cid:12)(cid:12) T ′⊥ ∩ T ′ (cid:12)(cid:12) . We consider two cases. Case 1: If R ⊂ H then since T z ∩ H = ∅ this implies T ′ ⊂ Y \ H so by Lemma 3.3 (cid:12)(cid:12) T ′⊥ ∩ T ′ (cid:12)(cid:12) ≤ | (ker Q | Y ) \ H | − | T ′ | . If x ∈ (ker Q | Y ) \ ( H ∪ T ′ ) then clearly x + z ∈ (ker Q | Y + z ) \ ( H ∪ T z ) for any z ∈ R .Hence | R | ( | (ker Q | Y ) \ H | − | T ′ | ) ≤ | R | min z ∈ R ( | (ker Q | Y + z ) \ H | − | T z | ) ≤ X z ∈ R | (ker Q | Y + z ) \ H | − | T z | = | S | − | T | . Case 2: If R * H then by Remark 3.4 (note that the condition T ∩ ker( Q | F n ⊥ ) = ∅ implies T ′ ⊂ (ker Q | Y ) \ { } ) we have (cid:12)(cid:12) T ′⊥ ∩ T ′ (cid:12)(cid:12) ≤ | (ker Q | Y ) \ { }| − | T ′ | . Then (cid:12)(cid:12) T ⊥ ∩ T (cid:12)(cid:12) ≤ | R | | (ker Q | Y ) \ { }| − | T ′ | ) . Let U = (ker( Q | Y ) \ ( { } ∪ T ′ )) ∩ H and U = ker( Q | Y ) \ ( { } ∪ T ′ ∪ H ). If x ∈ U then x + z ∈ (ker Q | Y + z ) \ ( H ∪ T z ) for z ∈ R \ H and similarly if x ∈ U and z ∈ R ∩ H .Thus | R | | (ker Q | Y ) \ { }| − | T ′ | ) = X z ∈ R \ H | U | + X z ∈ R ∩ H | U |≤ X z ∈ R | (ker Q | Y + z ) \ H | − | T z | = | S | − | T | . This completes the proof. (cid:3) The graph G ∗ In this section we define the graph whose maximal disconnected subgraphs we willbe interested in classifying.Recall we have a fixed isomorphism G/ Φ ( G ) ∼ = F n . Define the graph G ( F n ) withvertex set F n and let u, v ∈ F n be connected by an edge if their lifts in G do notcommute. OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 17 Define a set of elements F = { u w , v w | w ∈ F n } . We say u ∈ F is supported on w ∈ F n if u is of the form u w or v w . Let s : F −→ F n be the function sending anelement to its support. Additionally for any subsets F ⊂ F and V ⊂ F n we say F issupported on V if s ( F ) ⊂ V . We say that F has full support on V if s ( F ) = V . Forany subset T ⊂ F we let F ( T ) = { u ∈ F | s ( u ) ∈ T } .We define u, v ∈ F to be connected by an edge if u = u w and v = v z and ( w, z ) isan edge in G ( F n ). Define L ( u, v ) = 1 if u, v are connected and 0 if not. Definition 4.1. Let T , . . . , T k be subsets of F n and let b T = Q ki =1 T i . Define thegraph G ∗ ( b T ) with vertex set F ( b T ) = Q ki =1 F ( T i ) and an edge between u, v ∈ F ( b T ) if P ki =1 L ( u i , v i ) = 1.We explain the relationship G ∗ ( b T ) holds to the counting function (2.3).For each i = 1 , . . . , r let d i = D i − D i . Let U = [2 r ]. Note that the expression inthe summation in formula (2.3) can be expanded as X ( d ,...,d r ) ∈D r Y i =1 Y p | d i (cid:18) Q j ∈ S i d j p (cid:19) ord p ( d ) ! = X d = Q D u Y u,v ∈ U (cid:18) D u D v (cid:19) Φ( u,v ) where Φ ( u, v ) ∈ F . That is Φ ( u, v ) = 1 if the symbol ( D u D v ) appears in the expressionand 0 otherwise.Let U , . . . , U k be k sets with U i = [2 r i ]. For u, v ∈ Q ki =1 U i define Φ k ( u, v ) = P ki =1 Φ ( u i , v i ). Then we define u, v ∈ Q ki =1 U i to be unlinked if∆ k ( u, v ) = Φ k ( u, v ) + Φ k ( v, u ) = 0 . For any 1 ≤ i ≤ r , if x i = Q nj =1 x w j j define a bijective mapping ϕ = ( ϕ , . . . , ϕ k ) : Q ki =1 U i −→ F ( b T ) coordinate-wise with ϕ i given by2 i − u w i v w . Then it is easy to see that ∆ k = L k ◦ ϕ , and we will use this fact to view ∆ k as definedon G ∗ ( b T ).Let T ⊂ F n be an admissible generating set of G/ Φ ( G ). Define a quadratic form on F n by Q ( u ) = X i ≤ j a ij u i u j where a ij = 1 if [ x i , x j ] = 1 for i = j and a ii = 1 if ord x i = 4 and 0 otherwise. It hasan associated symmetric bilinear form B ( u, v ) = Q ( u + v ) + Q ( u ) + Q ( v ) on F n × F n .Alternately B ( u, v ) = u T Av where A is the symmetric matrix given by A ij = a ij asdefined above. Lemma 4.2. Let x = Q ni =1 x u i i be an element of G with u ∈ F n . Then x has order 2if and only if Q ( u ) = 0 . Proof. If x i and x j do not commute then [ x i , x j ] = a where a is the distinguishedcentral element of G . If ord x i = 4 then x i = a . Explicitly writing out x as a productand interchanging the basis elements gives x = n Y i =1 x u i i ! = n Y i =1 x u i i a P i Lemma 4.3. Let x = Q ni =1 x u i i and y = Q ni =1 x v i i . Then [ x, y ] = 1 if and only if B ( u, v ) = 0 .Proof. Let c ( x, y ) = 1 if [ x, y ] = 1 and 0 otherwise. By an argument analogous to theabove we have ( xy ) = x y [ x, y ]= a Q ( u )+ Q ( v )+ c ( x,y ) Note that, up to a multiple by a , xy equals Q ni =1 x u i + v i i . Since ord ( xy ) = ord ( xya ) wehave ( xy ) = a Q ( u + v ) . Thus B ( u, v ) = c ( x, y ) which proves the result. (cid:3) Maximal disconnected subgraphs Fix a set T ⊂ G/ Φ ( G ) given by T = { t , . . . t r } . Let { x , . . . , x r } be any choice oflifts of { t , . . . , t r } . Recall we defined S i = { ≤ j ≤ r | [ x i , x j ] = 1 } . This is the set ofindices j such that t j has an edge to t i in the graph G ( T ).Let C , . . . , C s be the connected components of G ( T ). Without loss of generality wecan assume i ∈ { , . . . , r } satisfies S i = ∅ .For a fixed pair ( G, H ) we will let T ⊂ G/ Φ ( G ) be the set of all elements which liftto order 2 elements in G \ H and call T the maximal admissible generating set. Thegraph G ( T ) determines ( G, H ) as follows. Lemma 5.1. The group H is abelian if and only if G ( T ) is a complete and bipartitegraph.Proof. Suppose H is abelian. Then G ∼ = H ⋊ C with C acting by inversion. Henceevery element of G can be written as xσ with x ∈ H and σ ∈ C . Every elementof G − H has order 2 since xσxσ = xx − σ = 1 since σ acts by inversion. Thus T = G − H ⊂ G/ Φ ( G ).If y ∈ G has order 4 then y = a where h a i = Φ ( G ) and a is central of order 2.Hence y − = ya . Let x σ, x σ ∈ G . Then x σx σ = x − σx − σ. OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 19 Thus x σ and x σ commute if and only if x i both have order 2, or both have order 4in H .Combining the above we see that G ( T ) = ( W , W ) is complete and bipartite where W = { xσ | ord ( x ) = 2 } and W = { xσ | ord ( x ) = 4 } .Now suppose G ( T ) = ( W , W ) is complete and bipartite. Since H has index 2 in G ,clearly H = n c t i t j | t i , t j ∈ T o where c t i t j denotes the lift from G/ Φ ( G ) to G . Suppose x = d t t and x = d s s with t i , s i ∈ T . By checking the cases t i ∈ W k , s j ∈ W l , with i, j, k, l = 1 , x and x commute, hence H is abelian. (cid:3) Throughout this paper we will primarily work with the sets T ⊂ G/ Φ ( G ) and hencerefer to the condition on G ( T ) rather than that on ( G, H ) from the above lemma. Thelemma will be applied at the end to translate the theorem statements to ones about( G, H ).For any T ⊂ G/ Φ ( G ) define c T to be the number of orbits of the set of lifts of T acting on itself by conjugation. If T is a generating set this is the number of conjugacyclasses of the lifts of T in G .We will write c = c T .The numbers c T will be crucial to determining the maximal disconnected subgraphsof G ∗ . Lemma 5.2. Let T ⊆ T . Then c T ≤ c . Furthermore if T is a generating set thenequality holds if and only if T = T .Proof. Note T ⊂ T = ker Q \ H by definition and Lemma 4.2.By definition of the quadratic form Q and its associated bilinear form B and Lemma4.3, T ⊥ ∩ T is the subset of T whose lifts have trivial action. There are two lifts of any t ∈ T , and hence they form one orbit if t / ∈ T ⊥ ∩ T and two distinct orbits if t ∈ T ⊥ ∩ T .Let K = ker( Q | F n ⊥ ). Let T ′ = T \ K and T ′′ = T ∩ K . Similarly let T ′ = T ∩ T ′ and T ′′ = T ∩ T ′′ . The above implies that the lifts of each element of T ′ form one orbit(under the action of T ) and the lifts of each element of T ′′ form two orbits.Then c T ≤ c − ( | T ′ | − | T ′ | ) − | T ′′ | − | T ′′ | ) + (cid:12)(cid:12) T ′⊥ ∩ T ′ (cid:12)(cid:12) . It follows from Proposition 3.5 that (cid:12)(cid:12) T ′⊥ ∩ T ′ (cid:12)(cid:12) ≤ | T ′ | − | T ′ | . Thus we have c T ≤ c for any T ⊆ T .Suppose c T = c and T is a generating set. If T = T then there is some x ∈ T whichis fixed by the conjugation action of T but not T . Since any set of lifts of T generates G this implies any lift x ′ ∈ Z ( G ). This is a contradiction since x is not fixed by T .Thus T = T . (cid:3) If T is a generating set then the proof of the above result does not require Proposition3.5. However the proofs in Section 5 will necessitate dealing with sets T which are notgenerating sets. Maximal unlinked sets. We will now define certain subsets of G ∗ ( b T ) called max-imal unlinked sets, which we will show are equal to maximal disconnected subgraphsof G ∗ ( b T ) of maximum size.As above fix a set T ⊂ G/ Φ ( G ) given by T = { t , . . . t r } . Throughout this sectionwe will use the identification made in Section 4 between the vertices of G ∗ ( T ) and U = { , . . . , r } . We partition the vertices of G ∗ ( T ) as follows.For 1 ≤ j ≤ s we let A j = { i − | ≤ i ≤ r , i ∈ C j } ,B j = { i | ≤ i ≤ r , i ∈ C j } Also we let C = { i − , i | i = r + 1 , . . . , r } . Let E j = A j ∪ B j so that U = ∪ sj =1 E j ∪ C .It is easy to verify the following description of G ∗ ( T ). If i ∈ C j and u = 2 i − u ∈ A j ) then u has an edge to every 2 i with i ∈ C j and i = i . It hasno edge to every other element of U . Similarly u = 2 i (that is u ∈ B j ) has an edgeto every 2 i − i ∈ C j and i = i and has no edge to everything else. Everyelement of C is disconnected.Consider the set of subsets of U (5.1) T = (cid:8) ∪ sj =1 W j ∪ C | W j = A j , B j (cid:9) . Let ( G, H ) = ( D , C ) and S ⊂ F be the maximal adimissible generating set forthis pair. For the purpose of the next subsection we will identify G ∗ (cid:0) S k (cid:1) with F k andrefer to disconnected subgraphs of F k .Suppose T is a generating set, such that T = ( W , W ) is a complete bipartite graph.Partition the vertices into four disjoint sets A = { u − | u ∈ W } ,A = { u − | u ∈ W } ,A = { u | u ∈ W } ,A = { u | u ∈ W } . Let T , . . . , T k be a sequence of such generating sets. Define a bijection between { A , A , A , A } k and elements of F k which acts coordinatewise by sending A i to thebinary representation of i in F . We will implicitly use this identification in whatfollows. For any q ∈ ( F ) k let R q = ( x ∈ k Y i =1 U | x i ∈ q i ) . Let D be the set of maximal disconnected subgraphs of G ∗ ( S ) with full supporton S (see Section 4 for the definition of the support). For any V ⊂ ( F ) k define R V = S q ∈ V R q . Then define OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 21 (5.2) R = { R V | V ∈ D} . Now we put together the above definitions of T and R . For generating sets T , . . . , T k let Y be the subset of indices i = 1 , . . . , k for which T i is a complete bipartite graph.Let T i correspond to T i for i / ∈ Y and let R correspond to the set of { T i } i ∈ Y as definedabove. Then let M ( b T ) = R × Y j / ∈ Y T j . We will denote maximal unlinked sets in R to be of type 1 and those in Q j T j of type 2 .5.2. Classification when T = T .Definition 5.3. For any b T = Q ki =1 T i let U ( b T ) be the set of largest maximal discon-nected subgraphs of G ∗ ( b T ) with full support on b T (see Section 4 for the definition ofthe support of b T ).We will show that when b T = T k we have the equality of sets M ( b T ) = U ( b T ).Throughout this section we will say u, v ∈ G ∗ ( F n ) are unlinked if there is no edgebetween them. Lemma 5.4. Let t ∈ M ( b T ) . Then | t | ≤ c k . Furthermore, if T is a product ofgenerating sets then equality holds if and only if T i = T for all i .Proof. Write t = R V × Q j / ∈ Y t j . Clearly | t j | = c T j and by Lemma 5.2 c T j ≤ c . For i ∈ Y write T i = ( W i, , W i, ) and without loss of generality suppose | W i, | ≥ | W i, | forall i . Note 2 | W i,j | = c W i,j ≤ c for all i, j . Since | R q | ≤ Q i ∈ Y | W i, | r for all q ∈ V wehave | R V | ≤ r Y i ∈ Y | W i, | r = Y i ∈ Y c W i, ≤ c r . Suppose | W i, | > | W i, | for some i . Since t has full support on T there is some q ∈ V such that for some i we have q i ∈ { (0 , , (1 , } , that is, for j = 1 or 3, every x ∈ R q satisfies x i ∈ A j . Hence | R q | ≤ | W i, | Y i ′ = i | W i ′ , | r < Y i ∈ Y | W i, | r . Thus in this case equality holds if and only if | W i, | = | W i, | (which implies | R q | =( c T i / r ) and c T i = c (which for generating sets only holds when T i = T ) for all i .Since | t | = | R V | · Q j / ∈ Y | t j | this completes the proof. (cid:3) Note that for any x ∈ R q and y ∈ R q the value of Φ r ( x, y ) and hence ∆ r ( x, y )is determined by q and q . This induces the function ∆ r : F r × F r −→ F . This isthe same function used to detect maximal disconnected subsets in the sense of [FK07,Section 5.2, p. 472] (which they refer to as maximal unlinked sets). In particularthere exists a non-degenerate quadratic form P defined on F r such that ∆ r ( u, v ) = P ( u + v ). It can be seen from the explicit formula for P given there that P is non-degenerate.We will show that when b T is a product of complete bipartite generating sets theelements of U ( b T ) are indexed by maximal unlinked sets in the sense of Fouvry-Kl¨uners.Recall in the previous subsection we identified G ∗ (cid:0) S k (cid:1) with F k . Lemma 5.5 ([FK07, Lemma 18, Section 5.5, p. 478]) . Let U ⊂ F r be a disconnectedsubgraph. Then |U | ≤ r and for any c ∈ F r , c + U is also a disconnected subgraph.If |U | = 2 r , then either U is a subspace of F r of dimension r or a coset of such asubspace. Lemma 5.6. For i = 1 , . . . , k let T i ⊆ T (not necessarily generating sets). Let T i , R , be as defined above. Let T ′ i ⊂ T i for each i and define similarly T ′ i , R ′ . Suppose c T ′ ,...,T ′ k = c T ,...,T k = c k .Let x ∈ t for some t ∈ M ( b T ) . Then for any t ′ ∈ M ( T ′ ) there exists some y ∈ t ′ such that x and y are unlinked.Proof. First note that since c T ′ i , c T i ≤ c for each i by Lemma 5.2, this implies that c = c T ′ i = c T i for each i . Let Y ′ be the set of indices for which T ′ i is a completebipartite graph. Let t = t Y ′ × t and similarly for t ′ . Let x be the projection of x ontothe coordinates in Y ′ . We construct an element y such that y is unlinked with x and y i is unlinked with x i for all i / ∈ Y ′ .First suppose i / ∈ Y ′ . If supp ( x i ) ∈ T ′ i then let y i be the element of t ′ i with supp ( y i ) = supp ( x i ). Now suppose supp ( x i ) / ∈ T ′ i . Since c T ′ i = c T i this implies there exists some u ∈ T ′ i which is a fixed point for the action of T ′ i but is not fixed for the action of T i .Hence 2 u − , u ∈ t ′ i so we let y i ∈ { u − , u } have the same pairity as x i .Now consider all i ∈ Y ′ . If T ′ i is strictly contained in T i with c T ′ i = c T i then T ′ i must contain at least one disconnected element, which contradicts that it is complete.Thus T ′ i = T i . Suppose x ∈ q . Since the function ∆ r is invariant under translation(simulatneously in both coordinates) by Lemma 5.5, we can assume t ′ Y ′ = R V where V is a subspace of F r . Thus we want to show that there exists q ′ ∈ V with ∆ r ( q , q ′ ) = 0.Since V is an unlinked set containing 0 we have P ( V ) = 0, that is V is a totallysingular space for P . By Lemma 3.1 there exists a subspace W ⊂ F r such that P ( W ) = 0 and W ∩ V = ∅ and | W | = | V | = 2 r . Thus V ⊕ W = F r . Write q = v + w for some v ∈ V and w ∈ W . Let q ′ = v . Then∆ r ( q , q ′ ) = P ( q + q ′ )= P ( w )= 0 . Thus we can let y ∈ R q ′ be any element and we get ∆ r ( x, y ) = 0. OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 23 Combining the above for all coordinates we construct an element with ∆ k ( x, y ) =0. (cid:3) We need one more graph-theoretic lemma. For any edge ( u, v ) ∈ E ( T ) consider theproperty P ( u, v ): there exists some w ∈ T such that either ( w, u ) , ( w, v ) / ∈ E ( T ) or( w, u ) , ( w, v ) ∈ E ( T ). Lemma 5.7. The condition P ( u, v ) is false for all ( u, v ) ∈ E ( T ) if and only if T is acomplete bipartite graph.Proof. Let ( u, v ) ∈ E ( T ). If P ( u, v ) is false then for every w ∈ T either ( w, u ) ∈ E ( T )or ( w, v ) ∈ E ( T ) but not both. Let W u = { w ∈ T | ( w, v ) ∈ E ( T ) } and similarly for W v . Then it is not hard to see that ( W u , W v ) is a complete partition of T . The otherdirection is trivial. (cid:3) We will make some notation which will be used in the following proofs. Given anysubset U ⊆ U k we can write it as a disjoint union U = S s +1 j =1 U j where we let U j = { u ∈ U | u k ∈ E j } for 1 ≤ j ≤ s, U s +1 = { u ∈ U | u k ∈ C } . Let p be the projection onto the first k − U j,i = { u ∈ U j | u k = i } .For simplicity we will write V i = p ( U j,i ). Lemma 5.8. Any disconnected subgraph U of G ∗ ( b T ) satisfies |U | ≤ c k .Proof. We proceed by induction on k . Let b T = Q ki =1 T i and U ∈ U ( b T ). Assumewithout loss of generality that T i is bipartite complete exactly for i = 1 , . . . , r . Thebase case k = r follows by Lemma 5.5 which says that the U ( b T ) = M ( b T ). Any elementof M ( b T ) has size bounded by c k by Lemma 5.4.Suppose the statement is true for k − 1. For any u ∈ T k , V u − ∪ V u is an unlinked set,hence by the induction hypothesis | V u − ∪ V u | ≤ c k − . Suppose { u − , u } ⊂ E j forsome 1 ≤ j ≤ s , so ( u, v ) ∈ E ( T k ) for some v ∈ T k and either V v − or V v is non-empty.This implies that V u − ∩ V u = ∅ . Hence |U j, u − | + |U j, u | = |U j, u − ∪ U j, u | ≤ c k − .If { u − , u } ⊂ C then clearly |U j, u − | , |U j, u | ≤ c k − .Thus summing |U j,i | over all i ∈ U k gives the desired result. (cid:3) Theorem 5.9. Let b T = Q ki =1 T i where T i are all generating sets. Then any U ∈ U ( b T ) satisfies |U | = c k if and only if T i = T for all i . In this case U ( b T ) = M ( b T ) .Proof. It is clear that if T i = T for all i then any t ∈ M ( b T ) is an unlinked set of size c k with full support. This gives one direction of the first part of the theorem statement.To prove the other direction and the second part we proceed by induction on k . Let U ∈ U ( b T ). Assume without loss of generality that T i is bipartite complete exactly for i = 1 , . . . , r . The base case is k = r . As above, by Lemmas 5.5 and 5.4 U ( b T ) = M ( b T )and its elements have size bounded by c r with equality if and only if T i = T for all i .Suppose the statement is true for k − 1. Suppose U is an unlinked set with fullsupport on T such that |U | = c k . We will show that either p ( U j,i ) = ∅ exactly for all i ∈ A j or for all i ∈ B j . From the coordinatewise definition of ∆ k it is clear that forany ( u, v ) ∈ E ( T k ) the elements of the set X = { V u − ∪ V u , V u − ∪ V v − , V u ∪ V v , V v − ∪ V v } are unlinked sets. Furthermore every element of V u − is linked with every element of V v , and similarly for V u and V v − . Since U has full support, either V u − or V u isnon-empty. Also note that V u − ∩ V u = ∅ . Lemma 5.8 implies |U j, u − ∪ U j, u | ≤ c k − .By Lemma 5.7 there exists ( u, v ) ∈ E ( T k ) such that the condition P ( u, v ) is true.We will use this to rule out the case when V u − , V u , V v − , V v are all non-empty.Suppose towards a contradiction that this is true. This implies V u − , V u , V v − , V v are all disjoint.Lemma 5.8 implies |U u ∪ U v | ≤ c k − . Let w ∈ T k verify P ( u, v ). First suppose that( w, u ) , ( w, v ) ∈ E ( T k ) and without loss of generality that V w = ∅ .Let T ′ = Q T ′ i be the support of the coordinates of the unlinked set V u − ∪ V v − .By the induction hypothesis, if | V u − ∪ V v − | = c k − then T ′ i = T i = T and hence c T ′ i = c T i = c for all 1 ≤ i ≤ k − 1. Any element of V w is linked with every element of V u − ∪ V v − and hence by Lemma 5.6 | V u − ∪ V v − | < c k − . Since V u − ∩ V v − = ∅ this implies |U u − ∪ U v − | < c k − . Thus |U u − ∪ U v − ∪ U u ∪ U v | < c k − which implies |U j | < | C j | c k − .Now suppose that ( w, u ) , ( w, v ) / ∈ E ( T k ) and again without loss of generality that V w = ∅ . Then any element of V w is unlinked with every set in X . Since V u − ∪ V u and V v − ∪ V v are disjoint this implies they cannot both be maximal so as above weget |U j | < | C j | c k − .Thus V u − , V u , V v − , V v cannot all be non-empty. Suppose without loss of gener-ality that V u − = ∅ which implies V u = ∅ .If V v − = ∅ and x ∈ V v − then every element of V u is connected to x . We willapply Lemma 5.6 to show that V u cannot be a maximal diconnected subgraph. Let T ′ = Q T ′ i be the support of the coordinates of the unlinked set V u . If | V u | = c k − bythe same argument as above by Lemma 5.6 there is an element of V u which is unlinkedwith x , a contradiction. Thus | V u | < c k − . If in addition V v = ∅ then this implies |U j | < | C j | c k − .Thus we have shown that if U ∈ U ( b T ) such that |U | = c k then U ∈ M ( b T ).Thus we have shown that either p ( U j,i ) = ∅ exactly for all i ∈ A j or exactly for all i ∈ B j as claimed. Hence ∪ j,i p ( U j,i ) is an unlinked set, which implies U j has maximalsize when p ( U j,i ) is equal to a unique maximal unlinked set fully supported on Q k − i =1 T i for all i and j . (cid:3) Asymptotic analysis Given a product of generating sets b T = Q i T i we define f b T = Q i f T i . We will computethe averages of functions of the form f b T , and use them to obtain the k th moments of f which will in turn determine a distribution. OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 25 Notational preliminaries. We define some notation for the next theorem. LetAut H, b T ( G, Φ( G )) = Q i Aut H,T i ( G, Φ( G )). Define N ± ( T i ) be the set of subsets N ⊂ T i of even (respectively odd) cardinality. Let N ± ( b T ) = Q i N ± ( T i ).We define a condition on tuples of congruence classes which will be used in the nexttheorem. Let r ( j ) = | T j | for j = 1 , . . . , k . Let Υ ∈ Q kj =1 [ r ( j )], α ∈ { , , } . Let b U = Q ki =1 U i with U i = [2 r ( j )] (recall from Section 4 there is a bijection b U −→ G ∗ ( b T )).Let ( h u ) u ∈ b U be a tuple of integers. For each 1 ≤ j ≤ k and i ∈ [ r ( j )] define thefollowing conditions on ( h u ) modulo 4(6.1) Y u,v h u h v ≡ − i ∈ N j , i = Υ j i / ∈ N j , i = Υ j i ∈ N j , i = Υ j , α = 2 ± i ∈ N j , i = Υ j , α = 3where the product is over all u, v ∈ b U with u j = 2 i − , v j = 2 i . For any positiveinteger m with 4 | m letΛ m ( N, Υ) = n ( h u ) ∈ ( Z /m Z ) b U | ( h u mod4) u satisfies (6 . o , Λ m ( U , N, Υ) = (cid:8) ( h u ) u ∈ b U ∈ Λ m ( N, Υ) | h u = 1 if u / ∈ U (cid:9) . We will use Λ m ( N, Υ) to restrict the congruence classes of certain factorizations of thediscriminant in the upcoming theorem.Next we define some functions which will be used in the description of the coefficientof the main term of f b T . For N ∈ N ± ( T ) and u ∈ G ∗ ( b T ) let λ N ( u ) = ( u = 2 i, i ≤ r , | N ∩ S i | odd0 else.Also recall we defined Φ k ( u, v ) = P ki =1 Φ ( u i , v i ) for u, v ∈ G ∗ ( T ), where the functionΦ is defined at the start of Section 4. Let Γ ⊆ [ k ]. Then let λ Nk ( u ) = k X i =1 λ N i ( u i ) , (6.2) γ Υ , Γ ( u ) = X i ∈ Γ Φ ( u i , i ) ,ψ Υ ( u ) = k X i =1 Φ (2Υ i , u i ) . Finally for ( h u ) ∈ Λ m ( N, Υ) and U ⊆ b U define χ (( h u ) , U ) = Y { u,v }⊂U ( − Φ k ( u,v ) hu − hv − , (6.3) χ (( h u ) , U ) = Y u ∈U ( − λ Nk ( u ) hu − ,χ (( h u ) , U ) = Y u ∈U ( − h u − ( γ Υ , Γ ( u )+ ψ Υ ( u ) α ) . Let D ± X,α to be the set of positive (resp. negative) fundamental discriminants divisibleby 2 α .6.2. Proof of the theorem.Theorem 6.1. Let b T = Q i T i be a product of generating sets and let c = max U∈ U ( b T ) |U | .With the above notation, for any positive constant a ∈ R and α ∈ { , , } the sum of f b T ( d ) /a ω ( d ) over D ± X,α is X d ∈D ± X,α f b T ( d ) a ω ( d ) = Γ U ( b T ) c kn (cid:12)(cid:12)(cid:12) Aut H, b T ( G, Φ( G )) (cid:12)(cid:12)(cid:12) π X (log X ) ( c /a ) − + o (cid:0) X (log X ) ( c /a ) − (cid:1) where (6.4) Γ U ( b T ) = X N ∈N ± ( b T ) X Γ ⊂ [ k ] X Υ ∈ Q j [ r ( j )] S N, Γ , Υ ( b T ) , and S N, Γ , Υ = X U∈ U ( b T ) |U| = c X ( h u ) ∈ Λ ( U ,N, Υ) 3 Y i =1 χ i (( h u ) , U ) . (6.5) Proof. Let d ∈ D ± X,α . For any T which is one of the factors of b T we expand the formulafrom Theorem (2.10) as f T ( d ) = 12 n (cid:12)(cid:12)(cid:12) Aut H, b T ( G, Φ( G )) (cid:12)(cid:12)(cid:12) X N ∈N ± ( T ) r X Υ=1 X d =2 α Q D u Y u,v (cid:18) D u D v (cid:19) Φ( u,v ) Y u (cid:18) − D u (cid:19) λ N ( u ) × Y u (cid:18) α D u (cid:19) ψ Υ ( u ) Y u (cid:18) D u (cid:19) γ Υ , ( u ) ! . where the second sum only appears when α = 0 and the third sum is over factorizationsof d into positive integers D u such that ( D u ) ∈ Λ ( N, l ) and the functions appearing inthe exponents are defined in (6.2) with k = 1. Note the D u were allowed to be negativeand divisible by 2 previously, hence we introduce λ N to keep track of the negative signsand γ Υ , , ψ Υ to keep track of the factor 2 α .For b T = Q ki =1 T i we obtain, using the same procedure as in [FK07, p.470-472] OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 27 f b T ( d ) = 12 kn (cid:12)(cid:12)(cid:12) Aut H, b T ( G, Φ( G )) (cid:12)(cid:12)(cid:12) X Γ ⊂{ ,...,k } X N ∈N ± ( b T ) X Υ ∈ Q j [ r ( j )] X d =2 α Q D u Y u,v (cid:18) D u D v (cid:19) Φ k ( u,v ) × Y u (cid:18) − D u (cid:19) λ Nk ( u ) (cid:18) D u (cid:19) γ Υ , Γ ( u ) (cid:18) α D u (cid:19) ψ Υ ( u ) where ( D u ) ∈ Λ ( N, Υ) and λ Nk , γ Υ , Γ , ψ Υ are defined in the theorem statement. Thesum over Γ ⊂ [ k ] comes from expanding the product Q kj =1 (1 + Q u (cid:0) D u (cid:1) γ Υ ( u ) ).An analysis of character sums and their cancellation, and a removal of the congruenceconditions on the D u as in [FK07, p.473-478] gives X d ∈D ± X,α f b T ( d ) a ω ( d ) = 12 kn c (cid:12)(cid:12)(cid:12) Aut H, b T ( G, Φ( G )) (cid:12)(cid:12)(cid:12) " X N, Γ , Υ S N, Γ , Υ ( D u ) µ (2 Q D u ) a ω ( Q D u ) (6.6) + o (cid:0) X (log X ) ( c /a ) − (cid:1) where the range of the last sum is over factorizations into c -tuples of coprime positiveintegers up to X and S N, Γ , Υ is as given by (6.5) in the statement. The factor of 4 − c comes from removing the congruence conditions on the remaining c variables in thelast sum in (6.6) as in the argument from [FK07, p.480-482] (Lemma 19 is modifiedto give a factor of 1 / / x < X as a product of c integers is c ω ( x ) we have X ( D u ) µ (2 Q D u ) a ω ( Q D u ) = "X x Throughout this section we will prove results about Γ M ( b T ) (defined as in Theorem 6.1with M ( b T ) in place of U ( b T )) when b T = Q ki =1 T i with all T i not complete bipartite. Wewill apply them in the case when U ( b T ) = M ( b T ). Let c b T = Q ki =1 c T i be the cardinalityof any element of M ( b T ).7.1. The congruence conditions. We will first reduce to a simpler set of congruenceconditions. Let b T = Q ki =1 T i where all T i are not complete bipartite. Let N ∈ N ± ( b T )and U ∈ M ( b T ). Let Λ ( U , N ) be the set of tuples of congruence classes modulo 4 such that for each 1 ≤ j ≤ k and i ∈ [ r ( j )](7.1) Y u,v h u h v ≡ ( − i ∈ N j i / ∈ N j where the product is over all u, v ∈ b U with u j = 2 i − , v j = 2 i . Lemma 7.1. Suppose U ( b T ) = M ( b T ) . In the notation of Theorem 6.1 we have Γ M ( b T ) = 2 c b T (cid:12)(cid:12)(cid:12) b T (cid:12)(cid:12)(cid:12) X N ∈N ± ( b T ) X U∈M ( b T ) X ( h u ) ∈ Λ( U ,N ) χ (( h u ) , U ) χ (( h u ) , U ) where χ i is defined in (6.3).Proof. Since χ and χ in (6.5) only depend on the class of h u modulo 4, we can let( h ′ u ) denote the reduction of ( h u ) modulo 4 and factor the sum X Γ ⊂ [ k ] X Υ ∈ Q j [ r ( j )] S N, Γ , Υ = X U∈M ( b T ) X ( h ′ u ) χ (( h ′ u ) , U ) χ (( h ′ u ) , U ) × X Γ ⊂ [ k ] X Υ ∈ Q j [ r ( j )] X ( h u ) χ (( h u ) , U ) . Note in the above sum ( h u ) depends on ( h ′ u ) since h u ≡ h ′ u mod4 and hence there aretwo possible choices for the class of h u modulo 8. If h ′ u ≡ h u ≡ h u ≡ h u − is even, and if h u ≡ h u − is odd. Hence both pairitiesare possible. Similarly when h ′ u ≡ κ (Υ) = X Γ ⊂ [ k ] X ( h u ) Y u ∈U ( − h u − ( γ Υ , Γ ( u )+ ψ Υ ( u ) α ) . Using the above observation we can replace the sum over ( h u ) as follows κ (Υ) = X Γ ⊂ [ k ] Y u ∈U (cid:16) − ( γ Υ , Γ ( u )+ ψ Υ ( u ) α ) (cid:17) . Recall the definitions γ Υ , Γ ( u ) = P i ∈ Γ Φ ( u i , i ) and ψ Υ ( u ) = P ki =1 Φ (2Υ i , u i ).It follows from the definition of Φ and M ( b T ) for b T not complete bipartite that thereis only one Γ in the above sum for which γ Υ , Γ ( u ) + ψ Υ ( u ) α is even for all u ∈ U . Inparticular if α is even γ Υ , Γ ( u ) is even for all u ∈ U only when Γ = ∅ . If α is odd then γ Υ , Γ ( u ) + ψ Υ ( u ) is even for all u ∈ U only when Γ = { i ∈ [ k ] | u i is even for all u ∈ U } .In both cases we get κ (Υ) = 2 c b T for all Υ. Finally note that (cid:12)(cid:12)(cid:12)Q kj =1 r ( j ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) b T (cid:12)(cid:12)(cid:12) . Thiscompletes the proof. (cid:3) Next we rewrite the congruence conditions (6.1) in linear algebraic terms.Let U ∈ M ( b T ). Let ( h u ) ∈ Λ ( U , N ). These congruence conditions can be encodedas vectors lying in a certain coset of a subspace of F |U| . For any ( h u ) define the OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 29 corresponding element x ∈ F |U| by(7.2) x u = ( h u = − 10 if h u = 1 . Let r k = P kj =1 r ( j ). Then there is a r k × |U | matrix M b T over F and vector h ∈ F r k such that the set of elements x ∈ F |U| corresponding to ( h u ) ∈ Λ ( U , N ) is equal to theset of solutions of the system M b T x = h (that is, equal to a coset of ker M T ).7.2. Factoring the main term. We begin by investigating how the congruence con-ditions (6.1) behave under products of maximal unlinked sets.Factor b T = b T × b T . Suppose U ∈ M ( b T ) factors as U × U . We can correspondinglyfactor N = N × N . Define a map λ : Λ ( U , N ) −→ Λ ( U , N ) by[ λ (( h u ))] u = Y u ∈U ×U u =( u ,u ) h u for each u ∈ U . It follows from the conditions (6.1) that this map is well-defined. Let p be the projection onto the coordinates in [ k ] corresponding to b T . If we let x ∈ F r represent λ (( h u )) then M b T x = p ( h ). Similarly define λ : Λ ( U , N ) −→ Λ ( U , N ). Lemma 7.2. In the above notation the map λ = λ × λ : Λ ( U , N ) −→ Λ ( U , N ) × Λ ( U , N ) is a |U ||U |−|U |−|U | +1 to 1 function.Proof. Under the correspondence (7.2) the map λ becomes[ λ ( x )] u = X u ∈U ×U u =( u ,u ) x u . Put an arbitrary ordering on the elements of U and U , and the lexicographic orderingon U × U . Then it is easy to see that λ is given by a matrix M of size |U | × |U | of full rank. Similarly λ is given by M of size |U | × |U | and full rank. Thus λ is thematrix M λ = (cid:2) M T , M T (cid:3) T which has size ( |U | + |U | ) × |U | and rank |U | + |U | − 1. Itfollows that dim ker M λ = |U | |U | − |U | − |U | + 1.Finally we need to show that Λ ( U , N ) + ker M λ ⊂ Λ ( U , N ) which is equivalent toker M λ ⊂ ker M b T by definition of Λ ( U , N ) as a coset of ker M b T . We have Λ ( U i , N i ) = p i ( h ) + ker M b T i . This implies that M b T i M i x = p i ( h ) = p i (cid:0) M b T x (cid:1) for any x ∈ F |U| .Let M ′ be the block diagonal matrix constructed from M b T and M b T . Then for any x ∈ Λ ( U , N ) and y ∈ ker M λ we have M b T y = M ′ M λ y = 0. Hence ker M λ ⊂ ker M b T asrequired. This completes the proof. (cid:3) By defintion each U ∈ M ( b T ) factors into U = U × U ∈ M ( b T ). We can factor N = N × N and let x ∈ Λ ( U , N ). Suppose χ ( U , x ) is a function satisfying(7.3) χ ( U , x ) = χ ( U , λ ( x )) χ ( U , λ ( x )) for all U ∈ M ( b T ). Define(7.4) γ ( U , N ) = X x ∈ Λ( U ,N ) χ ( U , x )and γ M ( b T ) ,N = P U∈M ( b T ) γ ( U , N ). Then Γ M ( b T ) = 2 c k (cid:12)(cid:12)(cid:12) b T (cid:12)(cid:12)(cid:12) P N ∈N ± ( b T ) γ M ( b T ) ,N . Lemma 7.3. Let b T = b T × b T and N = N × N . Let c b T i be the cardinality of anyelement of M ( b T i ) . Then γ M ( b T ) ,N = 2 c b T c b T − c b T − c b T +1 γ M ( b T ) ,N γ M ( b T ) ,N . Proof. Write U = U × U . We have |U | = c b T and |U | = c b T . Let θ = c b T c b T − c b T − c b T + 1. Hence by Lemma 7.2 we can factor the resulting sum γ M ( b T ) ,N = X U∈M ( b T ) X x ∈ Λ( U ,N ) χ ( U , x )= 2 θ X U = U ×U X ( x ,x ) ∈ Λ( U ,N ) × Λ( U ,N ) χ ( U , x ) χ ( U , x )= 2 θ X U = U ×U X x ∈ Λ( U ,N ) χ ( U , x ) X x ∈ Λ( U ,N ) χ ( U , x ) = 2 θ γ M ( b T ) ,N γ M ( b T ) ,N . (cid:3) Computing the main term. Let s ( j ) = s ( j ) + s ( j ) be the number of con-nected components of T j where s ( j ) is the number of components of size 1. Recall b T = Q ki =1 T i and M ( b T ) = Q kj =1 T j . We recall some notation from previous sections.For j ∈ [ k ] we denote by C j the set of fixed points for the action of T j on itself. Hence | C j | = s ( j ).Let C j,i be the i th connected component of T j . For i = 1 . . . . , s ( j ) we let A j,i , B j,i be the i -th connected component of A j , B j respectively.For any U ∈ M ( b T ) and x ∈ Λ ( U , N ) define χ ( U , x ) = χ ( U , x ) χ ( U , x ) (see 6.3).Then χ satisfies (7.3).We want to compute Γ M ( b T ) = 2 c k (cid:12)(cid:12)(cid:12) b T (cid:12)(cid:12)(cid:12) P N ∈N ± ( b T ) γ M ( b T ) ,N where γ M ( b T ) ,N = X U∈M ( b T ) X x ∈ Λ( U ,N ) χ ( U , x ) . Lemma 7.4. Let ω = c b T − P j ∈ [ k ] c T j + P j ∈ [ k ] s ( j ) + ( k − . For any N let m j,i = | N j ∩ supp B j,i | . Then we have γ M ( b T ) ,N = ( ω if m j,i ≡ , mod4 for all i, j else . OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 31 Proof. For the remainder of the proof we will write b T = Q kj =1 T j , and let b T i = Q j ≥ i T j .Any U ∈ M ( b T ) factors as U = Q kj =1 W j with each W j maximal unlinked with fullsupport on T j . Hence by repeated application of Proposition 7.3 we have γ M ( b T ) ,N = 2 c b T − P j c Tj +( k − k Y j =1 X W j X x j ∈ Λ( W j ,N j ) χ ( W j , x j )(7.5)where we are using that k − X i =1 c b T i − c b T i +1 − c T i + 1 = c b T − X j c T j + ( k − . Note that | Λ ( W j , N j ) | = 2 s ( j ) but the value of e ( W j , · ) + f N j ( W j , · ) is constant onΛ ( W j , N j ) (since Φ and λ N j are zero on C j ).Let W ′ j = W j \ C j . Then (cid:12)(cid:12) Λ (cid:0) W ′ j , N j (cid:1)(cid:12)(cid:12) = 1. Hence (7.5) becomes(7.6) 2 ω k Y j =1 X W ′ j χ (cid:0) W ′ j , x j (cid:1) where ω = 2 c b T − P j c Tj +( k − P j s ( j ) .Note that for every j , if u ∈ A j,i then Φ ( v, u ) = 0 for all v ∈ W j . Furthemore forevery u, v ∈ W j we have Φ ( u, v ) = Φ ( v, u ). Hence it follows from the definition of thesets W j = n(cid:16) ∪ s ( j ) i =1 W i (cid:17) ∪ C j | W i = A j,i or B j,i o that (7.6) factors as2 ω k Y j =1 s ( j ) Y i =1 (1 + χ ( B j,i , x j ))where by abuse of notation we are denoting by x j its projection onto the coordinatescorresponding to the i th connected component C j,i .We now turn to computing χ ( B j,i , x j ) which is determined by the pairity of X { u,v }⊂ B j,i Φ ( u, v ) x u x v + X u ∈ B j,i λ N j ( u ) x u . Note that Φ ( u, v ) = 1 for all u, v ∈ B j if u = v . It follows from ( ?? ) that for any u = 2 l ∈ B j,i x u = ( l ∈ N j i / ∈ N j . It follows from this that P { u,v }⊂ B j,i Φ ( u, v ) x u x v = (cid:0) m j,i (cid:1) which is even when m j,i ≡ , u = 2 l ∈ B j,i we have λ N j ( u ) = ( | N j ∩ S l | odd0 else and hence X u ∈ B j,i λ N j ( u ) x u = ( m j,i if m j,i even0 elsewhich is always even. Thus if m j,i ≡ , i and j then γ M ( b T ) ,N = 2 ω k Y j =1 s ( j ) Y i =1 (1 + χ ( B j,i , x j ))= 2 ω + P j s ( j ) where ω = 2 c b T − P j c Tj +( k − P j s ( j ) . If m j,i = 0 , i or j then γ M ( b T ) ,N =0. Since s ( j ) = s ( j ) + s ( j ) for all j , this completes the proof. (cid:3) Recall for the next proposition that we defined C j,i be the i th connected componentof T j . Proposition 7.5. Let a = 0 , in the case of positive (respectively negative) discrimi-nants. We have (7.7) Γ M ( b T ) = 2 ω (cid:12)(cid:12)(cid:12) b T (cid:12)(cid:12)(cid:12) k Y j =1 r ( j ) X m ≡ a (2) X m i ≡ , P s ( j ) i =1 m i = m s ( j ) Y i =1 (cid:18) |C j,i | m i (cid:19) where ω = 2 c b T − P j ∈ [ k ] c T j + P j ∈ [ k ] s ( j ) + ( k − .Proof. We begin with the formula from Lemma 7.4 and sum over N ∈ N ± ( b T ) satisfyingthe appropriate conditions. For any N let m j,i = | N j ∩ supp B j,i | . Then we have P N γ M ( b T ) ,N = 2 ω X ′ N N satisfying | N j | ≡ a (2) and m j,i ≡ , Q ± j = |{ N j ⊂ T j | | N j | ≡ a (2) , m i,j ≡ , }| thenwe have X ′ N Q kj =1 Q ± j . One can further compute(7.8) Q ± j = r ( j ) X m ≡ a (2) X m i ≡ , P s ( j ) i =1 m i = m s ( j ) Y i =1 (cid:18) |C i,j | m i (cid:19) . The outer summation corresponds to choices of possible sizes of the set N j and theinner sum to the choices of elements of T j contained in N j . (cid:3) The k th moment. We can combine the results of the previous two subsections. Proposition 7.6. Let b T = Q ki =1 T i be a product of generating sets and suppose T i isnot complete bipartite for all i . Suppose U ( b T ) = M ( b T ) . Then for all positive integers OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 33 k lim X −→∞ P d ∈D ± X,α f b T ( d ) c ω ( d ) b T P d ∈D ± X,α k + P j ∈ [ k ] ( s ( j ) − c Tj ) − kn (cid:12)(cid:12)(cid:12) b T (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Aut H, b T ( G/ Φ( G )) (cid:12)(cid:12)(cid:12) k Y j =1 Q ± j where • Q ± j is given by (7.8), • s ( j ) is the number of connected components of T j .Furthermore if a > c b T then lim X −→∞ P d By Theorem 6.1 we have X d Preliminaries for distributions of functions on quadratic fields. We startby recalling some preliminaries on measures and distributions. In particular we specifywhat we mean by the distribution of a function on the set of quadratic fields.Let K be the set of real or imaginary quadratic fields. Let Π be the σ -algebra on K consisting of all subsets and define the measurable space M = ( K , Π).For any positive X ∈ R let µ X be the measure on M with mass supported uniformlyon { K ∈ K | | D K | < X } . Then for any measurable space I = ( X , Σ) and function g : M −→ I we have a measure µ g,X = g ∗ µ X on I (note any function is measurablesince Π is the power set of K ). Thus for any measurable set U ∈ Σ, µ g,X ( U ) is theproportion of quadratic fields in K ∈ K with | D K | < X such that g ( K ) ∈ U .Now suppose X is also a metric space. Then for any function g : K −→ X we define µ g = lim X −→∞ µ g,X to be weak limit of measures, if it exists. If X has the discrete topology this is equivalentto strong convergence: µ g,X ( A ) −→ µ g ( A ) for all A , but this equivalence is not truein general. Note also the limit µ g may not be a probability measure (that is escape ofmass can occur). We call µ g the distribution of g if µ g is a probability measure on I . Example. Let X be the set of finite abelian p -groups, and let Σ be the power set of X . Let g : K −→ X be defined by g ( K ) = Cl K,p (note squaring only has an effectwhen p = 2).Then the Cohen-Lenstra conjectures state that µ g exists and describe what it shouldbe equal to, for any p . Example. Let X be the set of finite abelian p -torsion groups, and let Σ be the powerset of X . In this case any A ∈ X is determined by its order, and hence we can identify X with the set { p i | i ∈ Z ≥ } . Let g : K −→ X be defined by g ( K ) = | Cl K [ p ] | . Fouvryand Kl¨uners computed µ g in the case when p = 2.Since in this case X is discrete we have strong convergence of the µ g,X . Since X ⊂ Q it is easy to see that this also implies convergence when the measures in question areviewed on Q with its usual metric as follows. Let B be the Borel σ -algebra on Q (the σ -algebra generated by the open subsets of Q ). Then we can equivalently consider µ g onthe measurable space I ′ = ( Q , B ). Letting µ ′ g be the new measure on I ′ the connectionis given by µ ′ g = P ∞ i =0 δ { p i } µ g ( p i ) where δ { p i } is the points-mass with respect to B .In our case the functions f are rational valued, however taking the discrete metricon the set of values we find that lim X −→∞ µ f,X ( x ) = 0 for all x ∈ Q , that is µ f existsbut there is escape of mass. This issue is corrected if we instead view the set of valueslying inside Q with its usual metric, as in the second example above. We will see thatin this case µ f will be a probability measure.We prove a lemma which will be needed throughout the next two sections. Lemma 8.1. Let M , I be measurable spaces with I = ( R , B ) . Let g, h : M −→ I measurable functions such that µ g exists and µ h = δ a and let t a be translation by a .Then µ g + h = ( t a ) ∗ µ g . Proof. For any measure µ let F µ denote its cumulative distribution function. It is afact that for a sequence of measures µ n −→ µ weakly if and only if F µ n ( x ) −→ F µ ( x )at all points of continuity of F n , F .First note that F µ g ( x − a ) = F ( t a ) ∗ µ g ( x ), since for any U ∈ B , ( t a ) ∗ µ g ( U ) = µ g ( U − a ). Hence we need to show thatlim X −→∞ F µ g + h,X ( x ) = F µ g ( x − a ) . By definition we have F µ g + h,X ( x ) = µ g + h,X (( −∞ , x ])= µ X (cid:0) ( g + h ) − (( −∞ , x ]) (cid:1) . Fix ǫ > 0. Let ǫ > (cid:12)(cid:12) F µ g ( x − a ) − F µ g ( x − a ± ǫ ) (cid:12)(cid:12) < ǫ (this is possible since F ( t a ) ∗ µ g is continuous at x which implies F µ g is continuous at OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 35 x − a ) and there exists X large enough such that for W = h − ( a − ǫ , a + ǫ ) we have µ X ( W ) ≥ − ǫ .We have the set inclusions( g ) − (( −∞ , x − a − ǫ ]) ∩ W ⊂ ( h + g ) − (( −∞ , x ]) ∩ W, ( h + g ) − (( −∞ , x ]) ∩ W ⊂ ( g ) − (( −∞ , x − a + ǫ ]) ∩ W. By choice of X we have | µ X ( U ∩ W ) − µ X ( U ) | < ǫ for any U ∈ B . Hence (cid:12)(cid:12) F µ g,X ( x − a ± ǫ ) − µ X (( g ) − (( −∞ , x − a ± ǫ ]) ∩ W ) (cid:12)(cid:12) < ǫ, (cid:12)(cid:12) F µ g + h,X ( x ) − µ X (( h + g ) − (( −∞ , x ]) ∩ W ) (cid:12)(cid:12) < ǫ. Furthermore since F µ g,X −→ F µ g we have (cid:12)(cid:12) F µ g,X ( x − a ± ǫ ) − F µ g,X ( x − a ) (cid:12)(cid:12) < ǫ. Putting the above together we obtain that for any ǫ > X large enoughsuch that F µ g,X ( x − a ) − ǫ ≤ F µ g + h,X ( x ) ≤ F µ g,X ( x − a ) + ǫ for some ǫ a multiple of ǫ . Thuslim X −→∞ F µ g + h,X ( x ) = lim X −→∞ F µ g,X ( x − a )= F µ g ( x − a ) . As stated at the beginning F µ g ( x − a ) = F ( t a ) ∗ µ g ( x ) which implies µ g + h,X −→ ( t a ) ∗ µ g weakly. This completes the proof. (cid:3) We are now interested in computing the distribution µ f of the function f = P T i f T i where the sum is over generating sets T i ⊂ T .We first consider the case when T is not complete bipartite and later apply this tohandle the complete bipartite case.8.2. The non complete bipartite case.Theorem 8.2. Let a = | Aut H ( G/ Φ( G )) | . Suppose T is not complete bipartite. Forall positive integers k lim X −→∞ P d ∈D ± X,α (cid:16) f ( d ) c ω ( d ) (cid:17) k P d ∈D ± X,α ( (2 c − s + n − a ) − k · ( | T | Q ± ) k if T i = T for all i elsewhere Q ± is defined in (7.8).Proof. Since f = P T i f T i , taking the k th power gives f k = P b T f b T where b T = Q ki =1 T i .Fix b T and let c = max U∈ U ( b T ) |U | . First suppose b T = Q ki =1 T . By Theorem 5.9 c < c k . Then the second case of the statement follows by Theorem 6.1 with a = c k .Now suppose T i = T for all i . Then by Theorem 5.9 c = c k and U ( b T ) = M ( b T ).Then the first case of the statement follows by Proposition 7.6. (cid:3) Corollary 8.3. Let a = | Aut H ( G/ Φ( G )) | . Let q = (2 c − s + n − a ) − k · ( Q ± ) k . Suppose T is not complete bipartite.Then the distribution of h ( d ) = f ( d ) c ω ( d ) is µ h = δ q where δ q is the point-mass at q .Proof. In the case where T i = T and T is complete bipartite the k th moment of f ( d ) / ( c ) ω ( d ) is the k th power of some non-zero rational number by Theorem 8.2. Thusthe distribution determined is a point-mass. (cid:3) The complete bipartite case. Suppose T is complete bipartite. Fix a qua-dratic discriminant d of the quadratic field K . Let(8.1) g ( d ) = X ( d ,...,d r ) ∈D r Y i =1 Y p | d i (cid:18) Q j ∈ S i d j p (cid:19) ord p ( d ) ! . Note g ( d ) = 2 n | Aut H ( G/ Φ ( G )) | f T ( d ). Let b D = { ( d , . . . , d r ) | Q ri =1 d i = d } . Recallwe defined D ⊂ b D to be the subset such that d i = 1 for all i . In the following we willwrite D T and b D T to keep track of tuples corresponding to any subset T ⊂ T and g D T for the function in (8.1) with D = D T . We have g b D T ( d ) = X T ⊂ T g D T ( d )where we sum over all subsets T ⊂ T (not necessarily generating sets).Note g D T /c ω ( d ) = 2 n | Aut H ( G/ Φ ( G )) | f T ( d ) /c ω ( d ) is the function whose distribu-tion we want to compute. We can write(8.2) g D T ( d ) c ω ( d ) = g b D T ( d ) c ω ( d ) − X T ⊂ T g D T ( d ) c ω ( d ) . Now we consider the distribution of each of the functions on the right-hand side. Westart with g b D T ( d ) /c ω ( d ) .For any d ′ , d ′ dividing d define P d ′ ( d ′ ) = Y p | d i (cid:18) d ′ p (cid:19) ord p ( d ) ! . Now define the function C ( d ) = 12 ω ( d ) X ( d ′ ,d ′ ) d ′ d ′ = d P d ′ ( d ′ ) P d ′ ( d ′ ) . We will use this function to relate g b D T /c ω ( d ) to the 4-torsion in Cl K . Lemma 8.4. For any quadratic field K with discriminant dg b D T ( d ) c ω ( d ) = C ( d ) . OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 37 Proof. Since T is bipartite we have T = ( W , W ). Let T ,i = T ∩ W i . Then we have g b D T ( d ) c ω ( d ) = 1 c ω ( d ) X ( d ,...,d r ) ∈D r Y i =1 Y p | d i (cid:18) Q j ∈ S i d j p (cid:19) ord p ( d ) ! = 1 c ω ( d ) X ( d ′ ,d ′ ) d ′ d ′ = d X ( d ,...,d r/ ) ∈ b D T , ( d ′ ) P d ′ ( d ′ ) X ( d r/ ,...,d r ) ∈ b D T , ( d ′ ) P d ′ ( d ′ ) . Noting that the terms in the inner summations depend only on the factors d ′ and d ′ we can write this as1 c ω ( d ) X ( d ′ ,d ′ ) d ′ d ′ = d P d ′ ( d ′ ) P d ′ ( d ′ ) X ( d ,...,d r/ ) ∈ b D T , ( d ′ ) 1 X ( d r/ ,...,d r ) ∈ b D T , ( d ′ ) 1 . Now (cid:12)(cid:12)(cid:12) b D T ,i ( d ′ i ) (cid:12)(cid:12)(cid:12) = ( r/ ω ( d ′ i ) = c ω ( d ′ i ) T ,i since this is the number of ways of factoring d ′ i into r/ c T ,i = c T / 2. Thus we get1 c ω ( d ) X ( d ′ ,d ′ ) d ′ d ′ = d P d ′ ( d ′ ) P d ′ ( d ′ ) c ω ( d ′ ) T , c ω ( d ′ ) T , = 12 ω ( d ) X ( d ′ ,d ′ ) d ′ d ′ = d P d ′ ( d ′ ) P d ′ ( d ′ ) . This completes the proof. (cid:3) Lemma 8.5. Suppose d is the discriminant of a quadratic field K . Then g b D T ( d ) c ω ( d ) = 2 | Cl K [4] /Cl K [2] | . Proof. By Lemma 8.4 g b D T ( d ) /c ω ( d ) = C ( d ).Let G = D ∼ = C ⋊ C , H = C and T be the two non-trivial elements of G/ Φ ( G ) ∼ = C × C not in the projection of H . Then by Theorem 2.10 C ( d ) = 2 n | Aut H,T ( G/ Φ ( G )) | ω ( d ) f T ( d ) + 2= 12 ω ( d ) (cid:0) f T ( d ) + 2 · ω ( d ) (cid:1) (8.3)where n = 2 and | Aut H,T ( G/ Φ ( G )) | = 2.Now f T is the number of ( G, H, T )-extensions of K . This is also the number of( G, H )-extensions since this is the unique T for the pair ( G, H ) = ( D , C ). We claim that f T is in fact equal to the number of unramified C -extensions of K .Clearly any ( G, H )-extension is an unramified C -extension of K . Conversely suppose L/K is unramified and α : Gal ( L/K ) −→ C is an isomorphism. Since C is abelian L is a subfield of the Hilbert class field of K hence Galois over Q . Then since G ∼ = C ⋊ C and Gal ( L/ Q ) ∼ = Gal ( L/K ) ⋊ Gal ( K/ Q ) we can lift α to α ′ : Gal ( L/ Q ) −→ G . Hence L/K is a ( G, H )-extension.For any such extension there are exactly 2 surjections from Cl K to C . Hence2 · f T = | Surj ( Cl K , C ) | .Note | Hom ( Cl K , C m ) | = | Cl K [2 m ] | . Furthermore | Hom ( Cl K , C ) | = | Surj ( Cl K , C ) | + | Hom ( Cl K , C ) | . It is well known that | Cl K [2] | = 2 ω ( d ) − . Combining the above facts with (8.3) we get C ( d ) = 12 ω ( d ) (cid:0) | Surj ( Cl K , C ) | + 2 · ω ( d ) (cid:1) = 42 ω ( d ) (cid:0) | Surj ( Cl K , C ) | + 2 ω ( d ) − (cid:1) = 2 | Cl K [4] | / | Cl K [2] | . (cid:3) Let µ ± CL ( H ) be the Cohen-Lenstra probability measure on finite abelian 2-groups H which is proportional to 1 / ( | Aut H |·| H | a ) where a = 0 , P ± CL ( i ) be the probability with respect to the Cohen-Lenstra measure that afinite abelian 2-group has rank i .The distribution of the funtion | Cl K [4] /Cl K [2] | over both imaginary and real qua-dratic fields was determined by Fouvry and Kl¨uners in [FK07] and [FK06]. Thus weconclude the following. Lemma 8.6. The function h ( d ) = g b D T ( d ) /c ω ( d ) has distribution µ h supported on theset { i | i ∈ Z ≥ } given by µ h (cid:0) i (cid:1) = P ± CL ( i − over all real (respectively imaginary) quadratic fields.Proof. Follows from Lemma 8.5 and Theorem 3 from [FK07]. (cid:3) Next we consider the distribution of h T ( d ) = g D T ( d ) /c ω ( d ) for T ⊂ T . We will dothis by computing the k th moments of h T . Lemma 8.7. If T ∈ { W , W } then µ h T = δ . If T / ∈ { W , W , T } then µ h T = δ .Proof. Let c = max U∈ U ( T k ) |U | . Since T ⊂ T and T is complete bipartite it is easy tosee that c T ≤ c T with equality if and only if T ∈ { W , W , T } .Let G ′ be the group generated by T , H ′ = H ∩ G and let T ′ be the correspondingmaximal generating set which lifts to order 2 elements in G ′ − H ′ . By Theorem 5.9applied to T ′ we see that c ≤ c kT ′ with equality if and only if T = T ′ . By Lemma 5.8 c T ′ ≤ c T . OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 39 If T = T ′ then c = c kT = c kT ′ . Putting these facts together we see that c ≤ c kT withequality if and only if T ∈ { W , W , T } .Hence if T / ∈ { W , W , T } by Theorem 6.1lim X −→∞ P d Suppose T is complete bipartite. Let q = 2 n − | Aut H ( G/ Φ ( G )) | . Then h ( d ) = f ( d ) /c ω ( d ) has distribution µ h supported on the set { (2 i − /q | i ∈ Z ≥ } givenby µ h (cid:0)(cid:0) i − (cid:1) /q (cid:1) = P ± CL ( i ) over the set of real (respectively imaginary) quadratic fields.Proof. Recall we have the relations(8.4) f = X T ⊂ T f T where the sum is over generating sets, g D T ( d ) = g b D T ( d ) − X T $ T g D T ( d )where the sum is over all strict subsets of T , including non-generating sets, and g D T ( d ) = 2 n | Aut H ( G/ Φ ( G )) | f T ( d ).By an argument similar to Lemma 8.7 we can show that when T = T the k thmoments of f T /c ω ( d ) are all equal to 0 since in this case T is a generating set andhence max U∈ U ( T k ) |U | < c kT . Thus µ f T /c ω = δ . Combined with (8.4) this implies that µ h = µ f T /c ω .Let h g ( d ) = g D T /c ω ( d ) . By Lemmas 8.6 and 8.7 µ h g is supported on the set { i − | i ∈ Z ≥ } and µ h g (cid:0) i − (cid:1) = P ± CL ( i − µ h is supported on { (2 i − /q | i ∈ Z ≥ } given by µ h (cid:0)(cid:0) i − (cid:1) /q (cid:1) = P ± CL ( i ) . (cid:3) Remark . From the above proof we see that f /c ω ( d ) = 12 n | Aut H ( G/ Φ ( G )) | | Cl K [4] /Cl K [2] | − X T $ T g D T ( d ) /c ω ( d ) + X ′ T $ T f T /c ω ( d ) = 12 n | Aut H ( G/ Φ ( G )) | (cid:18) | Cl K [4] /Cl K [2] | − X ′′ T $ T g D T ( d ) /c ω ( d ) (cid:19) (8.5)where the prime indicates a sum over subsets T which are generating sets, and thedouble-prime indicates a sum is over subsets T which are not generating sets.It is evident from this that when T is complete bipartite the distribution of f ( d ) /c ω ( d ) is controlled by the 4-torsion of the class group. The values of the function howeverdepend on additional information, which comes from the other unramified 2-extensionsof the quadratic field K .Up until now we have been working with the formula for the function f T which counts( G, H, T )-extensions unramified away from infinity. For the remainder of the paper weswitch notation as follows. Let f ′ T denote the function counting ( G, H, T )-extensionsunramified away from infinity and let f T denote that counting ( G, H, T )-extensionsunramified everywhere. Similarly define f ′ and f . We now show that the distributionsand moments of these two functions are the same. Lemma 8.10. Let f ′ T ( K ) and f T ( K ) denote the number of ( G, H, T ) -extensions of K which are unramified away from infinity (resp. unramified everywhere). Then µ f T /c ω = µ f ′ T /c ω and E k ( f T /c ω ) = E k (cid:0) f ′ T /c ω (cid:1) for all positive integers k .Proof. By the same proofs as in Propositions 11 and Lemma 12 of [AK16] we see that ifwe define g = f T /c ω − f ′ T /c ω then E k ( g ) = 0 for all k . This implies µ g = δ and henceby Lemma 8.1 µ f T /c ω = µ f ′ T /c ω . Then also clearly E k ( f T /c ω ) = E k (cid:0) f ′ T /c ω (cid:1) . (cid:3) Moments and Correlations Using results from the previous section we can determine the moments of f for afixed pair ( G, H ) as well as correlations of such functions for several different pairs( G, H ).Using almost the same proof as in Lemma 8.1 we can also prove Lemma 9.1. Let M , I be measurable spaces with I = ( R , B ) . Let g, h : M −→ I measurable functions such that µ g exists and µ h = δ a and let m a be multiplication by a . Then µ g · h = ( m a ) ∗ µ g . We first need a lemma on correlations of the type of functions we will encounter.For any measure µ let E k ( µ ) denote the k th moment defined by R x k dµ . OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 41 Lemma 9.2. Let M , I be measurable spaces with I = ( R , B ) . Let g, h : M −→I be positive measurable functions. Suppose g has countable well-ordered image and µ g,X −→ µ g strongly and µ h = δ a . Furthemore suppose E k ( µ g,X ) −→ E k ( µ g ) and E k ( µ h,X ) −→ E k ( µ h ) for k = 1 , . Then E ( µ g · h,X ) −→ E ( µ g · h ) .Proof. Let { y i } ∞ i =1 be the image of g in increasing order. Then for each k = 1 , E k ( µ g ) = ∞ X i =0 µ g ( y i ) y ki . Let ǫ > 0. Since the above series converges there exists N large such that R x>N x k dµ g <ǫ for k = 1 , 2. Since E k ( µ g,X ) −→ E k ( µ g ) and µ g,X −→ µ g strongly we can make X large enough so that R x>N x k dµ g,X < ǫ for k = 1 , ǫ , δ > I = ( a − δ, a + δ ) and let J = R \ I . Since µ h = δ a we can find X large enough such that µ h,X ( I ) > − ǫ and µ h,X ( J ) < ǫ . Since E k ( µ h,X ) −→ E k ( µ h ) = a k we can also find X large enough such that (cid:12)(cid:12) a k − R x k dµ h,X (cid:12)(cid:12) < ǫ for k = 1 , 2. Hence (cid:12)(cid:12)(cid:12)(cid:12)Z J x k dµ h,X (cid:12)(cid:12)(cid:12)(cid:12) < ǫ + (cid:12)(cid:12)(cid:12)(cid:12) a k − Z I x k dµ h,X (cid:12)(cid:12)(cid:12)(cid:12) (9.1)and the right side goes to 0 with ǫ and δ . Also choose ǫ such that ǫ N is small. Sowe can choose ǫ , ǫ , δ arbitrarily small and X large enough such that all of the aboveholds.Now fix such an X and let W = h − ( I ) ∩ D X and W c = D X \ W . Note µ X ( W ) = µ h,X ( I ) > − ǫ and µ X ( W c ) = µ h,X ( J ) < ǫ . By definition E ( µ g · h,X ) = Z D X ( g · h ) ( x ) dµ X = Z W ( g · h ) ( x ) dµ X + Z W c ( g · h ) ( x ) dµ X For k = 1 , Z W c g ( x ) k dµ X = Z W c ∩ g − ([0 ,N ]) g ( x ) k dµ X + Z W c ∩ g − ( N, ∞ ) g ( x ) k dµ X (9.2) ≤ ǫ N k + Z x>N x k dµ g,X ≤ ǫ N + ǫ . Setting k = 1 this implies that (cid:12)(cid:12)(cid:12) E ( µ g ) − R W g ( x ) dµ X (cid:12)(cid:12)(cid:12) < | E ( µ g ) − E ( µ g,X ) | + ǫ N + ǫ . Then it follows from the definition of W and the previous fact that( a − δ ) Z W g ( x ) dµ X ≤ Z W ( g · h ) ( x ) dµ X ≤ ( a + δ ) Z W g ( x ) dµ X so R W ( g · h ) ( x ) dµ X can be made arbitrarily close to aE ( µ g ) = E ( µ g · h ) (see Lemma9.1).Finally it follows from Cauchy-Schwarz that Z W c ( g · h ) ( x ) dµ X ≤ Z W c g ( x ) dµ X ! / Z W c h ( x ) dµ X ! / . Note R W c h ( x ) dµ X = R J x dµ h,X . Hence by (9.1) and (9.2) with k = 2 we see that R W c ( g · h ) ( x ) dµ X goes to 0 with ǫ , ǫ and δ . (cid:3) Remark . A similar easier proof can be used to show Lemma 9.2 holds when g satisfies the same conditions as h and µ g = δ b for some non-zero b ∈ R .Considering the explicit form of the function f when T is complete bipartite weobtain the following result on correlations. Theorem 9.4. Let ( G , H ) , . . . , ( G k , H k ) be a sequence of admissible pairs. Let T i, bethe maximal admissible generating set corresponding to ( G i , H i ) . Let f i be the functioncounting ( G i , H i ) -extensions. Let Y ⊂ [ k ] be the set of indices for which T i, is completebipartite. Then lim X −→∞ P d ∈D ± X Q ki =1 (cid:18) f i ( d ) c ω ( d ) Ti, (cid:19)P d ∈D ± X k Y j =1 q − j ! Y j ∈ [ k ] \ Y P ± j | Y | X i =0 ( − i (cid:18) | Y | i (cid:19) M ± ( | Y | − i ) where • q j = 2 n j − (cid:12)(cid:12) Aut H j ( G j , Φ ( G j )) (cid:12)(cid:12) • M − ( j ) = N ( j ) and M + ( j ) = 2 − j ( N ( j + 1) − N ( j )) • P ± j = 2 c Tj, − s j + n j − | T j, | Q ± j where s j is the number of connected components of T j, and Q ± j is defined in (7.8).Proof. From Remark 8.9 we see that if T i, is complete bipartite then f i ( d ) /c ω ( d ) T i, canbe written as a sum of ( | Cl K [4] /Cl K [2] | − /q i and functions g i,j satisfying µ g i,j = δ and E k (cid:0) µ g i,j ,X (cid:1) −→ E k (cid:0) µ g i,j (cid:1) . Let f ( K ) = | Cl K [4] /Cl K [2] | − 1. Note f k ( K ) = P ki =0 ( − i (cid:0) ki (cid:1) | Cl K [4] /Cl K [2] | i . By Fouvry-Kl¨unerslim X −→∞ P K ∈D ± X ( | Cl K [4] /Cl K [2] | ) k P K ∈D ± X M ± ( k ) . Hence lim X −→∞ P d ∈D ± X ( f ( K )) k P d ∈D ± X k X i =0 ( − i (cid:18) ki (cid:19) M ± ( k − i )and µ f (2 i − 1) = P ± CL ( i ) for i ∈ Z ≥ . Note f ( K ) k satisfies the conditions of Lemma9.2. OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 43 By Theorem 8.2 and Corollary 8.3 we know the limiting distribution and momentsof f i ( d ) /c ω ( d ) T i, for each i / ∈ Y . Let g ( d ) = Q i ∈ [ k ] \ Y (cid:16) f i ( d ) /c ω ( d ) T i, (cid:17) . By applying Remark9.3 iteratively to products of powers of the functions f i ( d ) /c ω ( d ) T i, for i ∈ [ k ] \ Y weobtain that µ g is a point-mass supported at Q j ∈ [ k ] \ Y P ± j /q j andlim X −→∞ P d ∈D ± X g ( d ) P d ∈D ± X Y j ∈ [ k ] \ Y P ± j /q j . Let h ( d ) = Q i ∈ Y ( f i ( d ) /c ω ( d ) T i, ). Then similarly by Remark 9.3 applied to the functions g i,j for i ∈ Y and Lemma 9.2 to products of f and the g i,j we seelim X −→∞ P d ∈D ± X h ( d ) P d ∈D ± X k X i =0 ( − i (cid:18) ki (cid:19) M ± ( k − i )and µ h (2 i − 1) = P ± CL ( i ). Finally applying Lemma 9.2 to g and h once more completesthe proof. (cid:3) Corollary 9.5. Let ( G , H , T , ) , . . . , ( G k , H k , T ,k ) be a set of admissible tuples. Let K ⊂ K be the set of all quadratic fields K whose maximal unramified extension K un / Q contains a ( G i , H i ) extension for all i .If H i is non-abelian for all i then K has density 1. Otherwise K has density − P CL (0) .Proof. Let f i and c i be the counting function and normalizing constant correspondingto ( G i , H i , T ,i ). Since in the first case for each i the function h i = f i /c ωi is distributedas a point mass at some non-zero value this implies that h i evaluates to be non-zeroat one hundred percent of quadratic fields. Clearly the field K un contains the desiredcompositum if Q ki =1 h i is non-zero at K .On the other hand if H i is abelian for at least one i then Q ki =1 h i will be zero withdensity P CL (0) (since for such i the functions h i are maximally correlated). (cid:3) Additional remarks Malle-Bhargava heuristics. As further evidence for the refined Conjecture1.6 we obtain an asymptotic for P K, < ± D K With the above notation m ′ p = 1 + c T p . Proof. Note that since D/ h x i has to be cyclic and x cannot be a square in G (if x isa square then x ∈ Φ ( G )) D has to be of the form C × C m where m ≥ c and m let r p ( c, m ) = X x ∈ c X h x i≤ D ≤ G,D ∼ = C × C m (cid:12)(cid:12) Aut h x i ( D ) (cid:12)(cid:12) |{ F/ Q p | Gal ( F/ Q p ) ∼ = D }| so that by switching summations the first term in (10 . 1) becomes1 | G | X c X m ≥ r p · p − . We now compute the values of r p ( c, m ). Note that (cid:12)(cid:12) Aut h x i ( D ) (cid:12)(cid:12) = 2 m . Also, for anyodd prime there is exactly 1 ramified extension F/ Q p with Gal ( F/ Q p ) ∼ = C × C m for m > m = 0.If D ∼ = C (that is m = 0) then r p ( c, 0) = X x ∈ c 2= 2 | c | . Now suppose m ≥ 1. Then r p ( c, m ) = 2 m X x ∈ c X h x i≤ D ≤ G,D ∼ = C × C m OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 45 and X h x i≤ D ≤ G,D ∼ = C × C m | C G ( x ) [2 m ] | − | C G ( x ) [2 m − ] | − δ ( m ) (cid:12)(cid:12) Aut h x i ( D ) (cid:12)(cid:12) where δ ( m ) = 1 if m = 1 and 0 otherwise. We subtract δ ( m ) to account for theelement x ∈ C G ( x ) [2] which does not contribute to the count of D ∼ = C × C in thecase m = 1.Thus for any m ≥ r p ( c, m ) = X x ∈ c | C G ( x ) [2 m ] | − (cid:12)(cid:12) C G ( x ) (cid:2) m − (cid:3)(cid:12)(cid:12) = | c | (cid:0) | C G ( x ) [2 m ] | − (cid:12)(cid:12) C G ( x ) (cid:2) m − (cid:3) − δ ( m ) (cid:12)(cid:12)(cid:1) . Noting that | C G ( x ) [2 m ] | = | C G ( x ) | for large enough m and | C G ( x ) [2 ] | + δ (1) = 2 weget 1 | G | X c X m ≥ r p ( c, m ) = 1 | G | X c | c | | C G ( x ) | = X c c T . In the second equality we are using the orbit stabilizer theorem which says | G | = | c | | C G ( x ) | .By a similar argument one can show that X D ≤ G (cid:12)(cid:12) Aut h x i ( D ) (cid:12)(cid:12) |{ F/ Q p | Gal ( F/ Q p ) ∼ = D }| = | G | where the sum is over all cyclic subgroups (recall this term corresponds to the unram-ified extensions).Thus the local mass at p is 1 + c T p . This translates to an asymptotic of X (log X ) c T − for the global count, in agreementwith Conjecture 1.6. (cid:3) Examples. We list all the examples of admissible pairs ( G, H ) along with theirgraphs and asymptotics for n = 2 and 3.When n = 2 the only example is ( G, H ) = ( D , C ) and T = { t , t } which form anedge in G ( T ). Hence c = 2 . The following example shows that f ( G,H ) can have different asymptotics (that isdifferent values c T ) for different subgroups H of the same group G .Let n = 3. There are two possibilities for G which is an admissible central extensionof F by F . Let G be the central product of D and C (the quotient of D ⊕ C obtained byidentifying their Frattini subgroups). It is isomorphic to both Q ⋊ C and ( C × C ) ⋊ C with the appropriate actions.If we let H = Q then T = { t , t , t } such that G ( T ) is complete on its vertices.Hence c = 3.If we let H = C × C then T = { t , t , t , t t t } such that G ( T ) is completebipartite with partitions W = { t , t } and W = { t , t t t } . Hence c = 4.Let G = D × C . The only possibility is H = D in which case T = { t , t , t } suchthat ( t , t ) ∈ G ( T ) and t is disconnected. Hence c = 4.10.3. Comparison with Fouvry-Kl¨uners. In the case ( G, H ) = ( D , C ) the for-mula for our counting function f in Theorem 2.10 closely resembles the expressioncomputed by Fouvry and Kl¨uners for | Cl K [4] /Cl K [2] | . They showed that the function h ( d ) = | Cl K [4] /Cl K [2] | has distribution µ h (2 i ) = P ± CL ( i ) for all i ∈ Z ≥ . We makeexplicit the relationship.In this case the only admissible generating set T is the maximal one consisting oftwo elements, hence f = f T and c T = 2. Hence (using the notation of Section 8.3) wehave g D T ( d ) = g b D T ( d ) − X T $ T g D T ( d )= g b D T ( d ) − · ω ( d ) where the sum in the first line is over all strict subsets of T , including non-generatingsets. Furthermore in this case g D T ( d ) = 2 n | Aut H ( G/ Φ ( G )) | f T ( d )= 8 f T ( d ) . By Lemma 8.5 g b D T ( d )2 ω ( d ) = 2 | Cl K [4] /Cl K [2] | so by the above it follows that f ( d )2 ω ( d ) = 18 (2 | Cl K [4] /Cl K [2] | − | Cl K [4] /Cl K [2] | − . Thus in this case the function f is supported exactly on the same set as its distribution µ f/ ω which by Theorem 8.8 (or directly from the above result of Fouvry-Kl¨uners) is µ h (cid:0)(cid:0) i − (cid:1) / (cid:1) = P ± CL ( i )for all i ∈ Z ≥ (we reiterate that for any pair ( G, H ) = ( D , C ) f will be supportedaway from this set with positive density). OMENTS OF UNRAMIFIED 2-GROUP EXTENSIONS OF QUADRATIC FIELDS 47 References [AK16] Brandon Alberts and Jack Klys. The distribution of h -extensions of quadratic fields. 2016.Preprint https://arxiv.org/abs/1611.05595 .[Alb16] Brandon Alberts. Cohen-Lenstra Moments for Some Nonabelian Groups, 2016. Preprint https://arxiv.org/abs/1606.07867 .[Alb17] Brandon Alberts. Certain Unramified Metabelian Extensions Using Lemmermeyer Factor-izations, 2017. Preprint https://arxiv.org/abs/1710.00900 .[Bha07] Manjul Bhargava. Mass formulae for extensions of local fields, and conjectures on the densityof number field discriminants. Int. Math. Res. 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