Profinite genus of fundamental groups of compact flat manifolds with the cyclic holonomy group of square-free order
aa r X i v : . [ m a t h . G R ] F e b Profinite genus of fundamental groups of compactflat manifolds with the cyclic holonomy group ofsquare-free order
Genildo de Jesus Nery ∗ February 4, 2021
Abstract
In this article we study the extent to which an n -dimensional compact flat man-ifold with the cyclic holonomy group of square-free order may be distinguished bythe finite quotients of its fundamental group. In particular, we display a formulafor the cardinality of profinite genus of the fundamental group of an n -dimensionalcompact flat manifold with the cyclic holonomy group of square-free order. Keywords:
Profinite genus; Bieberbach group; compact flat manifold.
Mathematics Subject Classification (2010):
The study of what structural properties of a manifold can be detected by the set offinite quotients of the fundamental group of a manifold attracted a lot of attention in21-st centure (see [2, 3, 4, 23, 24, 25, 27]).The n -dimensional compact flat manifolds were well described by Bieberbach whocharacterized such manifolds as isometrically covered by a flat torus such that theirfundamental groups Γ are torsion-free with a maximal abelian normal subgroup M (subgroup of all translations) of finite index (see [7]). Consequently, the group Γ iscalled an n -dimensional Bieberbach group; the quotient G = Γ /M , called the holonomygroup, is a finite group acting faithfully on M . Note that Γ d , the full inverse image ofa subgroup H of G of order d under the quotient map Γ → Γ /M , is an n -dimensionalBieberbach group with maximal abelian normal subgroup M d and holonomy group H .In this paper we investigate the extent to which a Bieberbach group may be distin-guished from each other by its set of finite quotient groups. Since there is no completeclassification of Bieberbach groups in all dimensions, it becomes more difficult to inves-tigate this problem in the general case. Thus, it makes sense to consider this problemfor some families. For example, in the previous study [15] we give a complete answer tothis problem in the case that the holonomy group is cyclic of prime order. In this paperwe consider the Bieberbach groups Γ with cyclic holonomy group G = C p × · · · × C p k , ∗ The author held CNPq scholarship during the preparation of this article. C p i is a cyclic group of prime order p i . Thus, this paper can be considered as anatural continuation of the previous work.Following [11], we define the genus g (Γ) as the set of isomorphism classes of finitelygenerated residually finite groups with the profinite completion isomorphic to the profi-nite completion b Γ of Γ . In this paper we find a formula for the genus of an n -dimensionalBieberbach group with the cyclic holonomy group of square-free order.Fix a faithful Z G -lattice M for a finite group G . We define the crystal class ( G, M ) as the set of all the free-torsion extensions Γ of G by M . We say that two crystal classes ( G, M ) and ( G ′ , M ′ ) are arithmetically equivalent if G and G ′ are conjugate subgroupsof GL( n, Z ) . The resulting equivalence classes are the arithmetic crystal classes. Similarto the definition of g (Γ) , let C ( M ) denote the set of isomorphism classes of Z G -lattice N such that the b Z G -modules b N and c M are isomorphic.Note that to calculate the cardinality | g (Γ) | of the genus we have to calculate thenumber of isomorphism classes of Bieberbach groups in the crystal class ( G, M ) and |C ( M ) | . We first calculate |C ( M ) | . For this purpose, we will divide our study into twocases: special and not special, according to the following Definition 1.1.
Let Γ be an n -dimensional Bieberbach group with the cyclic holonomygroup G of square-free order. Let D be the set of prime divisors of | G | such that for each p ∈ D the n -dimensional Bieberbach subgroups Γ p of Γ with the cyclic holonomy group C p of order p , have corresponding Z C p -modules M p with all indecomposable summandsof Z -rank p − except one trivial summand of Z -rank . We sasy that Γ is special , if D = ∅ .Let | G | = δ = p p · · · p k be the decomposition of δ into distinct primes. Denote by Q ( ζ δ ) the cyclotomic field generated by a primitive δ -th root of unity ζ δ , by H ( Q ( ζ δ )) its class group and by Gal( ζ δ ) its Galois group, the latter acts naturally on H ( Q ( ζ δ )) .Using that Gal( ζ δ ) ∼ = Gal( ζ p ) × · · · × Gal( ζ p k ) (1)(see [10, Chapter 14, Corollary 27]) and that Gal( ζ p i ) ∼ = C p i − contains the uniquesubgroup C or order 2, for a set D of prime divisors of δ we can define a subgroup H D of Gal( ζ δ ) that has C instead of Gal( ζ p ) as a factor in the direct product in Gal( ζ p ) × · · · × Gal( ζ p k ) for each p i ∈ D .With these notations, we obtain the following formula for |C ( M ) | : Theorem 1.2.
Let Γ be a n -dimensional Bieberbach group with maximal abelian normalsubgroup M and cyclic holonomy group G of square-free order δ . If Γ is special, then |C ( M ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) H D \ Y d | δ H ( Q ( ζ d )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Otherwise, |C ( M ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Gal( ζ δ ) \ Y d | δ H ( Q ( ζ d )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Let M be a Z C p -module (for a cyclic group C p = h x i of prime order p ). Then M can be written in the form M = M ⊕ M , where M is the largest direct summand2f M on which C p acts trivially. For an element n ∈ N Aut( M ) ( C p ) denote by ˜ n the itsnatural image in F ∗ p = F p \ { } under an identification Aut( C p ) ∼ = F ∗ p .Now let "bar" denotes the reduction modulo p . The normalizer N Aut( M ) ( C p ) acts ina natural way on the set of all fixed elements M C p under the action of C p on M . Thisinduces the action of N Aut( M ) ( C p ) on ¯ M C p , and hence N Aut( M ) ( C p ) acts on ¯ M C p / ∆ · ¯ M ∼ =¯ M , where ∆ = 1 + x + · · · + x p − (see [7, p. 168] for more details).Define a new action “ • ” of the normalizer N Aut( M ) ( C p ) on ¯ M by n • m := n · e nm, ( m ∈ ¯ M ) (2)where “ · ” denotes the action of N Aut( M ) ( C p ) on ¯ M described in the preceding para-graph. Since for each prime p dividing | G | the normalizer N Aut( M ) ( G ) is a subgroup of N Aut( M ) ( C p ) , the normalizer N Aut( M ) ( G ) also acts on M /pM , and hence N Aut( M ) ( G ) acts on ( M /pM ) G/C p = ( M /pM ) G for each prime p dividing | G | .We denote ¯ M ∗ = ¯ M − { } and state the following Theorem 1.3.
Let Γ be an n -dimensional Bieberbach group with maximal abelian nor-mal subgroup M and cyclic holonomy group G of square-free order δ . Then, | g (Γ) | = X M ∈ T Y p | δ |N Aut( M ) ( G ) \ ( ¯ M ∗ ,p ) G | , where T is a set of representatives for the isomorphism classes of Z G -lattices in C ( M ) and M ,p is the largest direct summand of M on which C p acts trivially. Corollary 1.4. | g (Γ) | ≤ | H ( Q ( ζ δ )) | a (cid:0) max {| ( ¯ M ∗ ,p ) G | : p | δ } (cid:1) b , where a is the number of divisors of δ and b is the number of prime divisors of δ . Corollary 1.5. If Γ is a special Bieberbach group, then | g (Γ) | = X M ∈ T Y p ∈ D |N Aut( M ) ( G ) \ ¯ M ,p | Y q | δq / ∈ D |N Aut( M ) ( G ) \ ( ¯ M ∗ ,q ) G | We also deduce as a corollary the main result of [15].
Corollary 1.6.
Let Γ be an n -dimensional Bieberbach group with maximal abeliannormal subgroup M and cyclic holonomy group G of square-free order. If | G | is aprime number, then | g (Γ) | = |C ( M ) | . The outline of this paper is as follows. In Section 2, we summarize without proofsthe Oppenheim’s classification of integral representations of a cyclic group of square-free order – important result used in the proof of Theorem 1.2 – and presents somepreliminaries. Section 3 contains a special case of Theorem 1.3 which generalizes theresults of [15] (see Theorem 3.9) and the proofs of our main results. We also give anexample of use of the formula of Theorem 1.3.3 otation δ = square-free positive integer. a | b = a divides b (for a, b ∈ Z ). ζ d = primitive d -th root of unity (for d ∈ Z ). Φ d ( x ) = d -th cyclotomic polynomial (for d ∈ Z ). Q ( ζ d ) = cyclotomic field generated by ζ d ; Z [ ζ d ] = ring of integers of Q ( ζ d ) (for d ∈ Z ). Gal( ζ d ) = Galois group of Q ( ζ d ) over Q . G = cyclic group of order δ . b Γ = profinite completion of Γ (see [18]). H \ X = set of orbits of the action of the group H on a set X . | X | = cardinality of the set X . For the convenience of the reader we repeat the relevant results of the Oppenheim’s clas-sification of integral representations of a cyclic group of square-free order [17] withoutproofs.Let D denote the set of all n ∈ Z that divides δ such that there is only an evennumber of distinct primes in its decomposition. Similarly, let D denote the set of all n ∈ Z that divides δ such that there is only an odd number of distinct primes in itsdecomposition.In Z [ x ] set s ( x ) = Y d ∈ D Φ d ( x ) , s ( x ) = Y d ∈ D Φ d ( x ) and let s i = s i ( g ) , i = 0 , . Definition 2.1. A Z G -lattice is a Z G -module which is finitely generated and free as a Z -module. Lemma 2.2 ([17], Lemma 3.4) . Let M be a Z G -lattice. Then, M = { m ∈ M : s m = 0 } and M = M/M are Z G -lattices. Note that the module M is an extension of M by M , i.e., an exact sequence → M → M → M → of Z G -modules. It is well known that there is a bijection between the set of equivalenceclasses of extensions of M by M and Ext Z G ( M , M ) (see [19] for details). Thus, tocharacterize a Z G -lattice M we must determine not only the structure of M and M but also the group Ext Z G ( M , M ) . Proposition 2.3 ([17], p. 17) . Let M and M be as in Lemma 2.2. Then, M i ∼ = M d ∈ D i M d where M d := t d M i with t d := s i / Φ d ( g ) , i = 0 , . s M = 0 = s M . Thus, Φ d ( g ) M d = 0 for all d | δ . Hence, M d is a Z G/ Φ d ( g ) Z G -module.Let ζ d be a primitive d -th root of unity. We have a ring isomorphism Z G Φ d ( g ) Z G ∼ = Z [ ζ d ] given by g ζ d . Thus, we may then turn M d into a Z [ ζ d ] -module by setting ζ d · m := gm, m ∈ M d . Since Z [ ζ d ] is a Dedekind domain, it follows that we may write M d ∼ = I ,d ⊕ I ,d ⊕ · · · ⊕ I r ( d ) ,d where the { I j,d } are ideals in Z [ ζ d ] (see [9, Theorem 4.13]). Moreover, the isomorphisminvariants of M d are its rank r ( d ) and the ideal class of the product I ,d I ,d · · · I r ( d ) ,d . Proposition 2.4 ([17], Theorem 4.1 and Corollary 4.8) . Let M and M be as inLemma 2.2. Then, Ext Z G ( M , M ) ∼ = M ( s,t ) ∈ D × D Λ( s, t ) where Λ( s, t ) is the Z [ ζ s ] / Φ t ( ζ s ) Z [ ζ s ] -module of r ( s ) × r ( t ) matrices with entries in Z [ ζ s ] / Φ t ( ζ s ) Z [ ζ s ] . Let L ( s, t ) denote Z [ ζ s ] / Φ t ( ζ s ) Z [ ζ s ] . Suppose now that M is an extension of M by M corresponding to an element λ ∈ Ext Z G ( M , M ) . According to Proposition 2.4 wecan write λ = ( λ ( s , t ) , · · · , λ ( s l , t l )) where ( s i , t i ) ∈ D × D and λ ( s i , t i ) is an r ( s i ) × r ( t i ) matrix with entries in L ( s i , t i ) , i =1 , · · · , l . Lemma 2.5 ([17], Lemma 4.4) . Let s | δ, t | δ and t > s . Then, L ( s, t ) is a trivial ringunless there is a prime p such that t = ps . If t = ps , then L ( s, t ) = F p, ⊕ · · · ⊕ F p,v ,where F p,i is a field of characteristic p , i = 1 , · · · , v . Thus, the entries a ij ∈ L ( s, t ) of the matrix λ ( s, t ) can be written as a ij = ( α ij , · · · , α vij ) where α kij ∈ F p,k . Then to λ ( s, t ) corresponds the v -tuple of matrices (( α ij ) , · · · , ( α vij )) .Set ρ k ( λ ( s, t )) = rank( α kij ) .We need a little more notation to state the classification of the Z G -lattices. Let D ∗ = { ( s, t ) : s | δ, t | δ and s/t is a prime power } and let D ∗ = { ( s, t ) : ( s, t ) ∈ D ∗ , s ∈ D } . Proposition 2.6 ([17], Theorem 4.13) . Let M be a Z G -lattice. A full set of isomor-phism invariants of M consists of:(i) The Z G -rank of M d , r ( d ) , for each M d and d | δ .(ii) The ideal class of product I ,d I ,d · · · I r ( d ) ,d associated with M d , for each d | δ . iii) { ρ k ( λ ( s, t )) : ( λ ( s, t )) ∈ Ext Z G ( M , M ) , ( s, t ) ∈ D ∗ , k = 1 , · · · , v } . In what follows, r ( d, M ) and λ ( s, t, M ) denote r ( d ) and λ ( s, t ) on the module M ,respectively. Moreover, to shorten notation, let [ I M d ] denote the ideal class of product I ,d I ,d · · · I r ( d ) ,d associated with M d , for each d | δ .By [17, Theorem 7.3] and [9, Proposition 31.15] we have the following profiniteversion of Proposition 2.6. Proposition 2.7.
Let M and N be Z G -lattices. Then, c M ∼ = b N as b Z G -modules if andonly if(i) r ( d, M ) = r ( d, N ) for each d | δ .(ii) ρ k ( λ ( s, t, M )) = ρ k ( λ ( s, t, N )) for each k = 1 , · · · , v and ( s, t ) ∈ D ∗ . The next result will be useful to us, later on.
Proposition 2.8 ([26], Proposition 5.1) . Let C n be a cyclic group of order n ( n square-free) and let M, N, and L be Z C n -lattices. Then, M ⊕ L ∼ = N ⊕ L implies M ∼ = N ifand only if n = 6 , , , or n is prime. The next result tells us that the direct-sum cancellation holds for b Z G -modules. Proposition 2.9.
Let M , M , N and N be Z G -lattices. Then, c M ⊕ c M ∼ = b N ⊕ b N if and only if r ( d, M ) + r ( d, M ) = r ( d, N ) + r ( d, N ) and ρ k ( λ ( s, t, M )) + ρ k ( λ ( s, t, M )) = ρ k ( λ ( s, t, N )) + ρ k ( λ ( s, t, N )) for each d | δ and k = 1 , , · · · , v .Proof. The result follows from [17, Theorem 7.4] and [9, Proposition 31.15].
Let ζ m be a primitive m -th root of unity. It is well known that the Galois group Gal( ζ m ) of the cyclotomic field Q ( ζ m ) is isomorphic to ( Z /m Z ) × (the multiplicative group ofunits of the ring Z /m Z ). Proposition 2.10 ([10], Chapter 14, Corollary 27) . Let m = p a p a · · · p a k k be the de-composition of the positive integer m into distinct prime powers. Then, Gal( ζ m ) ∼ = Gal( ζ p a ) × · · · × Gal( ζ p akk ) . The next lemma shows that there is an action of the Galois group
Gal( ζ m ) on theideal class group H ( Q ( ζ m )) . Lemma 2.11 ([7], Chapter IV, Exercise 6.2) . Let A and B be ideals of Z [ ζ ] . If A and B are in the same ideal class, then σ ( A ) and σ ( B ) are in the same ideal class for any σ ∈ Gal( ζ m ) . emark 2.12. (i) Given any positive square-free integer δ suppose that d | δ . Then,there is u ∈ Z such that δ = du . Since ζ u is a primitive d -th root of unity, the field Q ( ζ d ) is a subfield of Q ( ζ δ ) . Thus, for any σ in Gal( ζ δ ) we have σ | Q ( ζ d ) ∈ Gal( ζ d ) .Therefore, Gal( ζ δ ) acts via automorphisms on H ( Q ( ζ d )) for each d | δ .(ii) From (i) it follows that Gal( ζ δ ) acts via automorphisms on the direct product Q d | δ H ( Q ( ζ d )) . We let Gal( ζ δ ) \ Q d | δ H ( Q ( ζ d )) be the orbit set. Definition 2.13.
Let M and N be Z G -modules. A semi-linear homomorphism from M to N is a pair ( f, ϕ ) where f : M → N is an abelian group homomorphism and ϕ isan automorphism of G such that f ( x · m ) = ϕ ( x ) · f ( m ) for x ∈ G and m ∈ M .Let ϕ be an automorphism of a finite group G and let M be a Z G -module. Let ( M ) ϕ denote the Z G -module M given by the action g ⋆ m := ϕ ( g ) · m for g ∈ G and m ∈ M . Proposition 2.14.
Let M and N be Z G -lattices. Then, M will be semi-linearly iso-morphic to N if and only if(i) r ( d, M ) = r ( d, N ) for each d | δ .(ii) ρ k ( λ ( s, t, M )) = ρ k ( λ ( s, t, N )) for each k = 1 , · · · , v and ( s, t ) ∈ D ∗ .(iii) σ · [ I M d ] = [ J N d ] for each d | δ and for some σ ∈ Gal( ζ δ ) .Proof. Let M and N be Z G -lattices. Note that M and N are semi-linearly isomorphicif and only if M ∼ = ( N ) ϕ as Z G -modules for some ϕ ∈ Aut( G ) . As G is a cyclic groupof square-free order δ , by Proposition 2.6 we have(i) r ( d, M ) = r ( d, N ) for each d | δ .(ii) ρ k ( λ ( s, t, M )) = ρ k ( λ ( s, t, N )) for each k = 1 , · · · , v and ( s, t ) ∈ D ∗ .(iii) I M d ∼ = ( J N d ) ϕ as Z [ ζ d ] -modules for each d | δ and for some ϕ ∈ Aut( G ) (see [8,Lemma 22.2] ).For each d | δ , we will denote by C d the subgroup of G of order d . Recall that Gal( ζ δ ) ∼ =( Z /δ Z ) × ∼ = Aut( G ) . So to finish the proof it suffices to show that ( J N d ) ϕ ∼ = ϕ − · J N d as Z C d -modules. Suppose ϕ ( c ) = c l where c ∈ C d and l is a positive integer such that ( l, δ ) = 1 . Consider the map η : ( J N d ) ϕ → ϕ − · J N d defined by u ϕ − ( u ) . It is clearthat η is a group isomorphism, so we have to show that η is a C d -homomorphism. Notethat η ( c ⋆ u ) = η ( ϕ ( c ) · u )= ϕ − ( c l · u )= ϕ − ( ζ ld u )= ϕ − ( ζ ld ) ϕ − ( u )= ζ d ϕ − ( u )= c · η ( u ) where c ∈ C d . This finishes the proof of Proposition 2.14.7 .3 Bieberbach groups and Cohomology of Groups If Γ is any Bieberbach group, then Γ satisfies an exact sequence → M → Γ → H → , (3)where M is a maximal abelian subgroup (subgroup of all translations) of Γ and H is afinite group. Proposition 2.15 ([7], Chapter I, Proposition 4.1) . Let Γ be an n -dimensional Bieber-bach group. Then, the translation subgroup M is the unique normal, maximal abeliansubgroup of Γ . Note that the exact sequence (3) gives a natural structure of Z H -module on M .Since M is maximal abelian, it follows that M is a faithful Z H -module. Recall thata crystal class ( H, M ) is the set all n -dimensional Bieberbach groups Γ that appearsas an extension of H by M and that two crystal classes ( H, M ) and ( H ′ , M ′ ) arearithmetically equivalent if there exist isomorphisms ϕ : H → H ′ and f : M → M ′ such that f ( h · m ) = ϕ ( h ) · f ( m ) for all h ∈ H and m ∈ M , i.e., if H and H ′ areconjugate subgroups of GL( n, Z ) . The resulting equivalence classes are the arithmeticcrystal classes. Lemma 2.16.
Isomorphic Bieberbach groups determine the same arithmetic crystalclass.Proof.
Let Γ and Γ be n -dimensional Bieberbach groups with the holonomy groups H , H and maximal abelian normal subgroups M , M , respectively. Suppose that F : Γ → Γ is an isomorphism. By Proposition 2.15, F ( M ) = M . Hence, F inducesan isomorphism ϕ : H → H . Let π : Γ → H and π : Γ → H be the naturalprojections. Let γ ∈ Γ and suppose π ( γ ) = h ∈ H . By the definition of ϕ , wehave π ( F ( γ )) = ϕ ( h ) . Hence, if f denotes the restriction of F to M , then f ( h · m ) = F ( γ mγ − ) = F ( γ ) F ( m ) F ( γ ) − = ϕ ( h ) · f ( m ) , where m ∈ M . Therefore, Γ and Γ determine the same arithmetic crystal class. Lemma 2.17.
Let ( H, M ) and ( H ′ , M ′ ) be belong to the same arithmetic crystal class.Then, for each Bieberbach group Γ in ( H, M ) there is an isomorphic group Γ ′ in ( H ′ , M ′ ) .Proof. Let Γ be a Bieberbach group of the crystal class ( H, M ) . Let ξ : H × H → M be a -cocycle corresponding to the extension → M → Γ → H → , and consider Γ to be M × H with the multiplication ( m , h )( m , h ) = ( m + h m + ξ ( h , h ) , h h ) , where m , m ∈ M and h , h ∈ H . Since ( H, M ) and ( H ′ , M ′ ) are in the samearithmetic crystal class, there exist isomorphisms ϕ : H → H ′ and f : M → M ′ suchthat f ( h · m ) = ϕ ( h ) · f ( m ) h ∈ H and m ∈ M . Now define a map F : M × H → M ′ × H ′ by F ( m, h ) = ( f ( m ) , ϕ ( h )) for m ∈ M and h ∈ H . Clearly, F is a bijective and F [( m , h )( m , h )] = F ( m + h m + ξ ( h , h ) , h h )= ( f ( m ) + ϕ ( h ) f ( m ) + f ( ξ ( h , h )) , ϕ ( h ) ϕ ( h ))= ( f ( m ) + ϕ ( h ) f ( m ) + f ( ξ ( f − f h f − f, f − f h f − f )) , ϕ ( h ) ϕ ( h ))= ( f ( m ) + ϕ ( h ) f ( m ) + f ( ξ ( f − ϕ ( h ) f, f − ϕ ( h ) f )) , ϕ ( h ) ϕ ( h ))= ( f ( m ) + ϕ ( h ) f ( m ) + ξ ′ ( ϕ ( h ) , ϕ ( h )) , ϕ ( h ) ϕ ( h ))= ( f ( m ) , ϕ ( h ))( f ( m ) , ϕ ( h )) where ξ ′ ( ϕ ( h ) , ϕ ( h )) = f ( ξ ( f − ϕ ( h ) f, f − ϕ ( h ) f )) is a -cocycle from H ′ × H ′ to M ′ . Therefore, F (Γ) = Γ ′ is a Bieberbach group of the crystal class ( H ′ , M ′ ) . Remark 2.18.
It follows from Lemmas 2.16 and 2.17 that to find all possible Bieberbachgroups (up to isomorphism) of an arithmetic crystal class it is sufficient to find allpossible Bieberbach groups of only one representative of the class.
Recall that the groups Γ that satisfies an exact sequence (3) can be classified byelements of the second cohomology group H ( H, M ) of H with coefficients in M . Itis known that if H is a finite group and M is a finitely generated Z H -module, then H ( H, M ) is a finite group (see [19, Corollary 9.41]). Proposition 2.19 ([7], Chapter III, Theorem 2.1) . Let Γ be the extension correspondingto α ∈ H ( H, M ) . Then, Γ is torsion-free if and only if, for any cyclic subgroup C p of H of prime order p , the image of the restriction homomorphism res ( α ) ∈ H ( C p , M ) is not zero. Definition 2.20.
An element α ∈ H ( H, M ) is special, if it defines a Bieberbach group.Let X ( H, M ) denote the subset of H ( H, M ) formed by all the special elements.We need a little more notation to characterize all possible Bieberbach groups in thecrystal class. The normalizer N Aut( M ) ( H ) of H in Aut( M ) acts on H ( H, M ) by theformula [ φ ⊛ c ]( h , h ) = φ ( c ( φ − h φ, φ − h φ )) , (4)where c : H × H → M is a -cocycle, φ ∈ N Aut( M ) ( H ) , and h , h ∈ H (see [7, p. 168]). Remark 2.21.
Let M be a Z C p -module for a cyclic group C p = h x i of prime order p . Then, M can be written in the form M = M ⊕ M , where M is the largest directsummand of M on which C p acts trivially. Thus, H ( C p , M ) ∼ = ¯ M (see [7, Chap. IV,Theorem 5.1]). Therefore, we can define an action of N Aut( M ) ( C p ) on H ( C p , M ) bythe formula (2) . We claim that this action is equal to the action ˝ ⊛ defined in (4) .Indeed, let φ ∈ N Aut( M ) ( C p ) and define e φ ∈ Aut( C p ) by e φ ( x ) = φ − xφ . Note that if c : C p × C p → M is a -cocycle, then [ e φ ∗ ( c )]( x, y ) = c ( e φ ( x ) , e φ ( y )) nd [ φ ∗ ( c )]( x, y ) = φ ( c ( x, y )) , for x, y ∈ C p , define homomorphisms e φ ∗ : H ( C p , M ) → H ( C p , M ) and φ ∗ : H ( C p , M ) → H ( C p , M ) , respectively. Thus, the action ˝ ⊛ defined in (4) can be rewritten as φ ⊛ α = φ ∗ ( e φ ∗ ( α )) , (5) where α ∈ H ( C p , M ) . On the other hand, since H ( C p , M ) ∼ = ¯ M , the action ˝ • definedin (2) can be rewritten as φ • m = φ · e φm = φ ∗ ( e φ ∗ ( m )) , (6) where m ∈ ¯ M . Therefore, from (5) and (6) we see that the action ˝ ⊛ defined in (4) is equal to the action ˝ • defined in (2) in this case. Proposition 2.22 ([13], Theorem 5.2) . There exists a one-to-one correspondence be-tween the isomorphism classes of Bieberbach groups in the crystal class ( H, M ) and theorbits of the action of N Aut( M ) ( H ) on X ( H, M ) . Lemma 2.23.
Let C p be a cyclic group of prime order p . Then, N Aut( M ) ( C p ) actstransitively on H ( C p , M ) ∗ . To prove this, we will need the following definition:
Definition 2.24.
Let C p be a cyclic group of prime order p . A finitely generated Z C p -module M is exceptional if all indecomposable summands of M have Z -rank p − except one trivial summand of Z -rank . Proof of Lemma 2.23.
Suppose that M is an exceptional Z C p -module, i.e., M = L ri =1 A i ⊕ Z where A i are ideals of Z [ ζ p ] . For each ϕ ∈ Gal( ζ p ) , consider the map φ : L ri =1 A i ⊕ Z → L ri =1 ϕ ( A i ) ⊕ Z defined by a ⊕ · · · ⊕ a r ⊕ u ϕ ( a ) ⊕ · · · ⊕ ϕ ( a r ) ⊕ u, for a i ∈ A i and u ∈ Z . Recall that Gal( ζ p ) ∼ = Aut( C p ) ; every ϕ ∈ Aut( C p ) is of theform ϕ ( x ) = x k where x is a generator of C p and k is some integer between and p .Thus, ϕ ( x · a i ) = ϕ ( ζ p a i ) = ϕ ( ζ p ) ϕ ( a i ) = ϕ ( x ) · ϕ ( a i ) , for i = 1 , · · · , r . Since Z is a trivial Z C p -module, we have φ ( x · m ) = ϕ ( x ) · φ ( m ) foreach m ∈ M . Hence, φ ∈ N Aut( M ) ( C p ) . As H ( C p , M ) = H ( C p , Z ) ∼ = ¯ Z , we have φ • m = e φm = km, where m ∈ ¯ Z and e φ = k . Therefore, by construction we have N Aut( M ) ( C p ) acts transi-tively on H ( C p , M ) ∗ in this case.Now, if M is a non-exceptional Z C p -module, then N Aut( M ) ( C p ) also acts transitivelyon H ( C p , M ) ∗ by [7, Theorem 6.1, p. 140].Let p be a prime number. Let H p denote the p -primary component of the group H .10 roposition 2.25 ([6], p. 84) . Let H be a finite group and P a normal p -Sylowsubgroup of H . Then,(i) H n ( H, M ) = L p || H | H n ( H, M ) p for all n > .(ii) H n ( H, M ) p ∼ = H n ( P, M ) G/P for all n > . In particular, if G is a cyclic group of square-free order, then X ( G, M ) = Q p || G | ( ¯ M ∗ ,p ) G/C p = Q p || G | ( ¯ M ∗ ,p ) G where M ,p is the largest direct summand of M on which C p acts trivially(see Proposition 2.19). Lemma 2.26.
Let M be a Z G -lattice for a cyclic group G of square-free order δ . Then, |N Aut( M ) ( G ) \ X ( G, M ) | = Y p | δ |N Aut( M ) ( G ) \ ( ¯ M ∗ ,p ) G | , (7) where M ,p is the largest direct summand of M on which C p acts trivially.Proof. Since N Aut( M ) ( G ) is a subgroup of N Aut( M ) ( C p ) for each p | δ , the result followsimmediately from Proposition 2.25. In this section we shall determine a full set of isomorphism invariants for the profinitecompletion of an n -dimensional Bieberbach group with the cyclic holonomy group ofsquare-free order and also for an arithmetic crystal class. Lemma 3.1.
Let Γ be an n -dimensional Bieberbach group with cyclic holonomy group G of square-free order δ and M its maximal abelian normal subgroup. Then, M = M n − ⊕ Z admits a Z G -decomposition, where Z is trivial module generated by the δ -thpower of some element c of Γ and Γ = M n − ⋊ C with C = h c i .Proof. By [20, Theorem 2] we have the following exact sequence → M n − → Γ → Z → . Since Z is free, this sequence splits as a semidirect product Γ = M n − ⋊ C . Lemma 3.2.
Let Γ and Γ be n -dimensional Bieberbach groups with the cyclic holon-omy group G of square-free order δ and maximal abelian normal subgroups M and N ,respectively. Then, Γ and Γ are in the same arithmetic crystal class if and only if(i) r ( d, M ) = r ( d, N ) for each d | δ .(ii) ρ k ( λ ( s, t, M )) = ρ k ( λ ( s, t, N )) for each k = 1 , · · · , v and ( s, t ) ∈ D ∗ .(iii) σ · [ I M d ] = [ J N d ] for each d | δ and for some σ ∈ Gal( ζ δ ) .Proof. This is a consequence of Proposition 2.14.11 roposition 3.3.
Let Γ and Γ be n -dimensional Bieberbach groups with the cyclicholonomy group G of order δ = 6 , , , or δ is prime and maximal abelian normalsubgroups M and N , respectively. Then, Γ ∼ = Γ if and only if(i) r ( d, M ) = r ( d, N ) for each d | δ .(ii) ρ k ( λ ( s, t, M )) = ρ k ( λ ( s, t, N )) for each k = 1 , · · · , v and ( s, t ) ∈ D ∗ .(iii) σ · [ I M d ] = [ J N d ] for each d | δ and for some σ ∈ Gal( ζ δ ) .Proof. If Γ ∼ = Γ , then M is semi-linearly isomorphic to N by Lemma 2.16. Therefore,it follows from Proposition 2.14 that the conditions (i)–(iii) are fulfilled.Conversely, let M and N be Z G -lattices satisfying the conditions (i)–(iii). ByLemma 3.1, M = M n − ⊕ Z and N = N n − ⊕ Z ′ such that Γ = M n − ⋊ C and Γ = N n − ⋊ C where C , C contain Z and Z ′ as subgroups of index δ and acts on M n − , N n − as G . Since the conditions (i)–(iii) are satisfied for M n − ⊕ Z and N n − ⊕ Z ′ ,it follows from Proposition 2.14 that M n − ⊕ Z is semi-linearly isomorphic to N n − ⊕ Z ′ .Hence, there is a semi-linearly isomorphism ( f, ϕ ) from M n − to N n − by Proposition2.8. Let Θ : C → C be the isomorphism induced by ϕ : G → G . Now, consider themap F : M n − ⋊ C → N n − ⋊ C given by F ( m, c ) = ( f ( m ) , Θ( c )) , for m ∈ M n − and c ∈ C . We have F is a group homomorphism, because ( f, ϕ ) is asemi-linearly homomorphism from M n − to N n − . Since F is clearly bijective, it is anisomorphism, as we wanted. Lemma 3.4 ([15], Lemma 3.2) . Let Γ and Γ be n -dimensional Bieberbach groupswith maximal abelian normal subgroups M and M and holonomy groups G and G ,respectively. If ψ : b Γ → b Γ is an isomorphism, then there are isomorphisms φ : c M → c M and ϕ : G → G such that the following diagram commutes: −−−→ c M −−−→ b Γ −−−→ G −−−→ y φ y ψ y ϕ −−−→ c M −−−→ b Γ −−−→ G −−−→ . Proposition 3.5.
Let Γ and Γ be n -dimensional Bieberbach groups with the cyclicholonomy group G of square-free order δ and maximal abelian normal subgroups M and N , respectively. Then, b Γ ∼ = b Γ if and only if(i) r ( d, M ) = r ( d, N ) for each d | δ .(ii) ρ k ( λ ( s, t, M )) = ρ k ( λ ( s, t, N )) for each k = 1 , · · · , v and ( s, t ) ∈ D ∗ .Proof. Suppose b Γ ∼ = b Γ . By Lemma 3.4, there are isomorphisms φ : c M → c M and ϕ : G → G such that the following diagram is commutative −−−→ c M −−−→ b Γ −−−→ G −−−→ y φ y ψ y ϕ −−−→ c M −−−→ b Γ −−−→ G −−−→ . φ : c M → c M is a b Z G -module isomorphism. Now, the statements (i) and (ii)follow immediately from Proposition 2.7.Conversely assume that M and N are maximal abelian normal subgroups of Γ and Γ , respectively, satisfying the conditions (i) and (ii). Since G is a cyclic group of square-free order δ , it follows from Lemma 3.1 that M = M n − ⊕ Z and N = N n − ⊕ Z ′ suchthat Γ = M n − ⋊ C and Γ = N n − ⋊ C where C , C contain Z and Z ′ as subgroupsof index δ and acts on M n − , N n − as G . Hence, c M = c M n − ⊕ b Z , b N = b N n − ⊕ b Z ′ and b Γ = c M n − ⋊ b C , b Γ = b N n − ⋊ b C , where b C and b C act on c M n − and b N n − as G . ByProposition 2.7, c M n − ⊕ b Z ∼ = b N n − ⊕ b Z ′ as b Z G -modules and hence r ( d, M n − ) + r ( d, Z ) = r ( d, N n − ) + r ( d, Z ′ ) and ρ k ( λ ( s, t, M n − )) + ρ k ( λ ( s, t, Z )) = ρ k ( λ ( s, t, N n − )) + ρ k ( λ ( s, t, Z ′ )) for each d | δ and k = 1 , , · · · , v by Proposition 2.9. As Z ∼ = Z ′ as Z G -modules, byProposition 2.6 we have r ( d, Z ) = r ( d, Z ′ ) and ρ k ( λ ( s, t, Z )) = ρ k ( λ ( s, t, Z ′ )) for each d | δ and k = 1 , , · · · , v . Hence, r ( d, M n − ) = r ( d, N n − ) and ρ k ( λ ( s, t, M n − )) = ρ k ( λ ( s, t, N n − )) for each k = 1 , , · · · , v and d | δ , so that by Proposition 2.7, c M n − ∼ = b N n − as b Z G -modules. Hence, b Γ ∼ = b Γ , and the proposition is proved.We finish this section with the Charlap’s classification of Bieberbach groups withthe holonomy group of prime order. Definition 3.6.
A Bieberbach group Γ with prime order holonomy group C p is excep-tional if its maximal abelian normal subgroup M is an exceptional Z C p -module. Proposition 3.7 ([7], Chapter IV, Theorem 6.3) . There is a one-to-one correspondencebetween the isomorphism classes of non-exceptional Bieberbach groups whose holonomygroup has prime order p and -tuples ( a, b, c ; θ ) where a, b, c ∈ Z with a > , b ≥ , c ≥ , ( a, c ) = (1 , , ( b, c ) = (0 , and θ ∈ Gal( ζ p ) \ H ( Q ( ζ p )) . Note that
Gal( ζ p ) is cyclic of even order if p > ; hence Gal( ζ p ) has a subgroup C of order if p > . Proposition 3.8 ([7], Chapter IV, Theorem 6.4) . There is a one-to-one correspondencebetween the isomorphism classes of exceptional Bieberbach groups whose holonomy grouphas prime order p and pairs ( b, θ ) where b > , and θ ∈ C \ H ( Q ( ζ p )) . .2 Proofs of main results Proof of Theorem 1.2.
Let Γ and Γ be n -dimensional Bieberbach groups with thecyclic holonomy group of square-free order and maximal abelian normal subgroups M and N , respectively. Suppose b Γ ∼ = b Γ . By Lemma 3.4, we can assume that Γ and Γ have the same cyclic holonomy group G of square-free order δ .For each d | δ , let C d denote the subgroup of G of order d and let Γ i,d denote the n -dimensional Bieberbach subgroup of Γ i with the holonomy group C d , for i = 1 , .Combining Propositions 3.5 and 2.7, we have b Γ ∼ = b Γ ⇔ c M ∼ = b N as b Z G -modules ⇔ for all d | δ, c M ∼ = b N as b Z C d -modules (8) ⇔ for all d | δ, b Γ ,d ∼ = b Γ ,d . Suppose that there are prime numbers p and q dividing δ such that the Bieberbachgroups Γ ,p and Γ ,q are exceptional. Since b Γ ∼ = b Γ , it follows from (8) that b Γ ,d ∼ = b Γ ,d for each d | δ . In particular, b Γ ,p ∼ = b Γ ,p and b Γ ,q ∼ = b Γ ,q . This implies that, if p = q , thenthe Bieberbach groups Γ i,p and Γ i,q are exceptional, for i = 1 , . Thus, as δ = p p · · · p k where p l ( l = 1 , · · · , k ) are distinct prime numbers, we can assume that there is apositive integer r with ≤ r ≤ k such that D = { q , · · · , q r } is a subset of { p , · · · , p k } with r distinct elements, so that the Bieberbach groups Γ i,q are exceptional, for each q ∈ D and i = 1 , .We conclude from Proposition 2.10 that the Galois group Gal( ζ δ ) has a subgroup H D that has Z / Z instead of Gal( ζ q ) as a factor in the direct product for all q ∈ D .Since Gal( ζ δ ) ∼ = ( Z /δ Z ) × ∼ = Aut( G ) , we have Aut( G ) contains an isomorphic copy of H D , which we will also denote by H D .Now, if [( G, M )] denotes the arithmetic crystal class of Γ , then, by Lemma 3.2 andProposition 2.14, we have Γ , Γ ∈ [( G, M )] ⇔ M ∼ = ( N ) ϕ as Z G -modules for some ϕ ∈ H D ⇔ for all d | δ, M ∼ = ( N ) ϕ as Z C d -modules for some ϕ ∈ H D ⇔ for all d | δ, Γ ,d , Γ ,d ∈ [( C d , M )] . Since the Bieberbach groups Γ i,q ( q ∈ D, i = 1 , are exceptional, it follows fromPropositions 3.5 and 3.8 together with Lemma 3.2 that |C ( M ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) H D \ Y d | δ H ( Q ( ζ d )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . On the contrary, if for each prime p dividing δ the Bieberbach groups Γ i,p ( i = 1 , are not exceptional, then by Propositions 3.5 and 3.7 together with Lemma 3.2, we have |C ( M ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Gal( ζ δ ) \ Y d | δ H ( Q ( ζ d )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , and therefore the Theorem 1.2 is proved. 14n particular, using Proposition 3.3 and similar arguments as in the proof of Theorem1.2, we get the following generalization of Theorem 1.1 of [15]. Theorem 3.9.
Let Γ , Γ d and H D be as in Theorem 1.2. If δ = 6 , , , or δ is prime,then | g (Γ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) H D \ Y d | δ H ( Q ( ζ d )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , for the special case. Otherwise, | g (Γ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Gal( ζ δ ) \ Y d | δ H ( Q ( ζ d )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Since the action of the Galois group on ideal class group is transitive if and only ifthe class group is trivial, we have
Corollary 3.10.
Let Γ be an n -dimensional Bieberbach group with the cyclic holonomygroup of order δ = 6 , , , or δ is prime. Then, | g (Γ) | = 1 if and only if for eachinteger d dividing δ the class group H ( Q ( ζ d )) is trivial. We can now prove our main theorem.
Proof of Theorem 1.3.
Let Γ be an n -dimensional Bieberbach group with maximalabelian normal subgroup M and cyclic holonomy group G of square-free order δ . Itfollows from Remark 2.18 that to find all the isomorphism classes of Bieberbach groupsof the arithmetic crystal classes that corresponds to the Z G -lattices in C ( M ) , it is suf-ficient to consider a set T of representatives for the isomorphism classes of Z G -latticesin C ( M ) . Theorem 1.2 gives us a formula for the cardinality of T . Now, applying firstProposition 2.22 and then Lemma 2.26, we have | g (Γ) | = X M ∈ T |N Aut( M ) ( G ) \ X ( G, M ) | = X M ∈ T Y p | δ |N Aut( M ) ( G ) \ ( ¯ M ∗ ,p ) G | , where M ,p is the largest direct summand of M on which C p acts trivially. Proof of Corollary 1.4.
To simplify notation, we let max {| ( ¯ M ∗ ,p ) G |} stand for max {| ( ¯ M ∗ ,p ) G | : p | δ } . By Theorem 1.3, | g (Γ) | = X M ∈ T Y p | δ (cid:12)(cid:12) N Aut( M ) ( G ) \ (( ¯ M ,p ) ∗ ) G (cid:12)(cid:12) ≤ X M ∈ T Y p | δ max {| ( ¯ M ∗ ,p ) G |} = X M ∈ T (cid:0) max {| ( ¯ M ∗ ,p ) G |} (cid:1) b = |C ( M ) | (cid:0) max {| ( ¯ M ∗ ,p ) G |} (cid:1) b ≤ ( H ( Q ( ζ δ ))) a (cid:0) max {| ( ¯ M ∗ ,p ) G |} (cid:1) b , ( by Theorem 1.2 ) a is the number of divisors of δ and b is the number of prime divisors of δ . Proof of Corollary 1.5.
This follows from Theorem 1.3 together with Lemma 3.1.
Proof of Corollary 1.6.
This is a consequence of Theorem 1.3 together with Lemma2.23.
Suppose that G is a cyclic group of square-free order δ > . Let h a, b | R i be apresentation of G , i.e., G ∼ = F / e R where F is the free group on { a, b } and e R is thenormal closure of R in F . This defines an exact sequence → e R → F → G → . (9)Hence, if [ e R, e R ] denotes the commutator subgroup of e R , then (9) induces the exactsequence → M → Γ → G → , where M = e R/ [ e R, e R ] and Γ = F / [ e R, e R ] . Note that M is a free abelian group whoserank is given by the Schrier’s formula (rank( F ) − δ + 1 = δ + 1 (see [14, Chapter6, Proposition 2]). Moreover, we claim that M is maximal abelian in Γ . Suppose theclaim were false. Then we could find x ∈ F such that the image of x in G is non-trivial,so that [ x, r ] ∈ [ e R, e R ] for all r ∈ e R . Thus, since G is a cyclic group, we have M is inthe center of Γ . Hence, Γ is abelian and [ F , F ] = [ e R, e R ] . As M has finite index in Γ , we have rank( e R ) = rank( F ) , and hence | G | = 1 , a contradiction. Then, Γ is an ( δ + 1) -dimensional Bieberbach group whose holonomy group is G . By [21, Corollary1], M = Z G ⊕ Z where Z is a trivial Z G -module.Thus, by Theorem 1.3, | g (Γ) | = X M ∈ T Y p | δ |N Aut( M ) ( G ) \ F ∗ p | , since Z is the largest direct summand of M on which C p acts trivially for each prime p | δ . Since Z G is a permutation Z G -module (i.e., the Z -basis of Z G is fixed underthe action of G ), we have G ≤ Sym( G ) ≤ GL( δ + 1 , Z ) ∼ = Aut( M ) , where Sym( G ) isthe group of all permutations of the elements of G . Hence, N Aut( M ) ( G ) contains theholomorph of G by [12, Theorem 6.3.2], i.e., the group Hol( G ) = G ⋊ Aut( G ) . As Aut( C p ) ≤ Aut( G ) for each prime p | δ , we have N Aut( M ) ( G ) acts transitively on F ∗ p foreach prime p | δ . Thus, | g (Γ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Gal( ζ d ) \ Y d | δ H ( Q ( ζ d )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , by Theorem 1.2.In particular, if δ = 6 , , , , or δ is prime ≤ , for example, then | g (Γ) | = 1 because for each d | δ the class group H ( Q ( ζ d )) is trivial (see [1, Table 10]).Now, if δ = 46 , , or δ is prime > , for example, then | g (Γ) | > because | and | H ( Q ( ζ ) | ≥ ; | H ( Q ( ζ ) | ≥ ; | H ( Q ( ζ ) | ≥ ; and | H ( Q ( ζ p ) | > forall prime p > (see [22, Theorem 11.1 and p. 353]).16 emark 3.11. In general, calculating the normalizer of a finite subgroup in
GL( n, Z ) isnot easy. We refer the interested reader to [5], which presents a method for calculatingsuch normalizers. Acknowledgements
The author wishes to express his thanks to Prof. Dr. Pavel Zalesskii for his manyvaluable advices and several discussions.
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E-mail address : [email protected]@mat.unb.br