Open Questions
We collect here the open questions mentioned in the book. We have main- tained the numeration of the original question so that the reader may consult the context in which the question is posed. The wording of the questions are sometimes modified slightly to make them self-contained. We also mention here those open problems that appeared in the first edition of this book which have been solved or advanced in the meantime; in these cases we provide some relevant information.
Open Question 3.5.3 (Inverse problem of Galois Theory) Is every finite group a continuous homomorphic image of the absolute Galois group GQ¯ /Q of the field Q of rational numbers?
Question in the First Edition of this Book Let F be a free profinite (or, more generally, pro - C) group on a profinite space X. Is there a canonical way of constructing a basis converging to 1 for F ? NOTE: J-P. Serre has given a negative answer to this question; see Theorem 3.5.13.
Open Question 3.7.2 What pro - C groups are pro - C completions of finitely generated abstract groups?
Question in the First Edition of this Book Let G be a finitely gener- ated profinite group. Is every subgroup of finite index in G necessarily open? NOTE: This question has been answered positively by N. Nikolov and D. Segal (see Theorem 4.2.2; for a proof see Nikolov and Segal [2007a, 2007b]).
Question in the First Edition of this Book Let G be a finitely generated prosolvable group. Are the terms (other than [G, G]) of the derived series of G closed? NOTE: This question has a negative answer; in fact V.A. Roman’kov had already provided a counterexample in 1982 for a more general setting; see Roman’kov [1982].
Open Question 4.8.4 Let G be a finitely generated profinite group and let n be a natural number. Let Gn = xn | x ∈ G be the abstract subgroup of G generated by the n-th powers of its elements. Is Gn closed?
L. Ribes, P. Zalesskii, Profinite Groups, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 40, DOI 10.1007/978-3-642-01642-4, c Springer-Verlag Berlin Heidelberg 2010 394 Open Questions
Open Question 4.8.5b Is a torsion profinite group necessarily of finite exponent?
Open Question 6.12.1 Let G be a solvable pro-p group such that Hn(G, Z/pZ) is finite for every n. Is G polycyclic?
Open Question 7.10.1 For what finite p-groups G does one have rr(G)= arr(G)? [rr = relation rank as a profinite group; arr = relation rank as an abstract group]
Open Question 7.10.4 Let G be a finitely generated pro-p group such that cd(G) > 2 and dim H2(G, Z/pZ)=1,(i.e., relation rank rr(G) is 1). Does G admit a presentation with a single defining relator of the form up?
Open Question 7.10.5 Study finitely generated pro-p groups with the fol- lowing property: every closed subgroup of infinite index is free pro-p.
Open Question 7.10.6 Let F be a free pro-p group of finite rank. Is vcd(Aut(F )) finite?
Question in the First Edition of this Book Does the Grushko-Neumann theorem hold for free profinite products of profinite groups, that is, if G = G1 G2 is the free profinite product of two profinite groups G1 and G2, is d(G)=d(G1)+d(G2)? NOTE: The answer to this is negative. It was answered by A. Lucchini; see Lucchini [2001a, 2001b].
Open Question 9.1.21 Let F be a free pro-p group and let H and K be closed finitely generated subgroups of F . Is there a bound on the rank of H ∩ K intermsoftheranksof H and K?
Open Question 9.5.2 Is a general inverse limit of a surjective inverse sys- tem of free profinite groups of finite rank necessarily a free profinite group?
Question in the First Edition of this Book For which extension closed varieties C of finite groups is it always true that whenever we are given G1,G2 ∈C, then there is a group G ∈Csuch that G1,G2 ≤ G, G = G1,G2 and d(G)=d(G1)+d(G2)? NOTE: For the class C of all finite solvable groups the answer is negative, see Kov´acs and Sim [1991]; for the class C of all finite groups the answer is negative, see Lucchini [2001a, 2001b].
Question in the First Edition of this Book Do all profinite Frobenius groups of the form Zπ C (C is finite cyclic, p |C| for all p ∈ π and C acts fixed-point-free on Zπ) appear as subgroups of free profinite products A B? NOTE: A positive answer is provided in Guralnick and Haran [2010]. Open Questions 395
Open Question 9.5.5 Give (verifiable) sufficient conditions for H to have ∗ bounded generation, where H = H1 H0 H2.
Open Question 9.5.6 Give (verifiable) sufficient conditions for Hpˆ to have ∗ bounded generation, where H = H1 H0 H2.
Open Question 9.5.7 Let G be a limit pro-p group. Does G satisfy the Howson property? In other words, if H1 and H2 are finitely generated closed subgroups of G, is H1 ∩ H2 a finitely generated pro-p group?
Open Question 9.5.8 Are limit pro-p groups residually free pro-p?
Open Question 9.5.9 Let H be a finitely generated subgroup of a limit pro-p group G. Is then H a virtual retract of G, i.e., is there an open subgroup M of G and a normal closed subgroup K of M so that M = K H?
Open Problem C.3.2 Let G be a finitely generated profinite (respectively, ≥ ≥ pro-p) group with finite def (G) 2(respectively, def p(G) 2). Does G containanopensubgroupU such that there exists a continuous epimorphism U → F onto a free profinite (respectively, pro-p) group F of rank at least 2? Appendix A: Spectral Sequences
A.1 Spectral Sequences
r,s A bigraded abelian group E is a family E =(E )r,s∈Z of abelian groups. A differential d of E of bidegree (p, q) is a family of homomorphisms
d : Er,s → Er+p,s+q such that dd =0.
s • • 2,4 •••• E3
• • • ••••d3
• ••••5,2 • E3
• 1,1 • 3,1 ••• E3 E3
• • • • • • • r
A spectral sequence consists of a sequence E = {E1, E2, E3,...} of bi- r,s −→ graded abelian groups Et =(Et )r,s∈Z, with differentials dt : Et Et of bidegree (t, −t + 1), such that
r,s ∼ r,s −→dt r+t,s−t+1 r−t,s+t−1 −→dt r,s Et+1 = Ker(Et Et )/Im(Et Et ). (1)
L. Ribes, P. Zalesskii, Profinite Groups, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics 40, DOI 10.1007/978-3-642-01642-4, c Springer-Verlag Berlin Heidelberg 2010 398 A: Spectral Sequences
To simplify the notation, from now on we assume that the isomorphism in (1) is in fact an equality. The bigraded abelian group E2 is called the initial term of the spectral sequence. ∈ r,s Lemma A.1.1 For each r, s Z there exists a series of subgroups of E2 r,s ≤ r,s ≤ r,s ≤···≤ r,s ≤ r,s ≤ r,s r,s 0=B2 B3 B4 C4 C3 C2 = E2 such that r,s r,s r,s ≥ Et = Ct /Bt (t 2). r,s r,s r,s r,s r,s r,s Proof. Set B2 =0andC2 = E2 ;thenE2 = C2 /B2 . Define induc- tively
r,s r,s r−t,s+t−1 r−t,s+t−1 r−t,s+t−1 →dt r,s r,s r,s Bt+1/Bt =Im(Et = Ct /Bt Et = Ct /Bt ), and
r,s r,s r,s r,s r,s →dt r+t,s−t+1 r+t,s−t+1 r+s,s−t+1 Ct+1/Bt =Ker(Et = Ct /Bt E = Ct /Bt ). Hence r,s ≤ r,s ≤ r,s ≤ r,s ≤ r,s ≤ r,s B2 Bt Bt+1 Ct+1 Ct C2 , and r,s r,s r,s r,s r,s r,s r,s Et+1 =(Ct+1/Bt )/(Bt+1/Bt )=Ct+1/Bt+1.
r,s r,s Let Ct , Bt be as in Lemma A.1.1. Define r,s r,s r,s r,s C∞ = Ct ,B∞ = Bt t t and r,s r,s r,s E∞ = C∞ /B∞ . r,s The bigraded abelian group E∞ =(E∞ )r,s∈Z, is completely determined by the spectral sequence. We think of the terms Et of the spectral sequence as approximating E∞. A filtered abelian group with filtration F consists of an abelian group A together with a family of subgroups F n(A)ofA,(n ∈ Z), such that
A ≥···≥F n(A) ≥ F n+1(A) ≥···.
We always assume that a filtration satisfies the additional condition: F r(A)=A and F r(A)=0. (2) r r To each filtered abelian group A we associate a grading in the following manner A.2 Positive Spectral Sequences 399
Gr(A)=F r(A)/F r+1(A)(r ∈ Z). A filtered graded abelian group with filtration F , consists of a family H = n n (H )n∈Z, of filtered groups H . A spectral sequence E =(Et)issaidtoconverge to the filtered graded abelian group H =(Hn) with filtration F if
r,s ∼ r r+s r r+s r+1 r+s E∞ = G (H )=F (H )/F (H ). r,s ⇒ n ⇒ We indicate this situation by E2 = H or by E = H.
A.2 Positive Spectral Sequences
r,s We say that a spectral sequence E is positive or first quadrant if E2 =0, whenever r<0ors<0. It is clear that if E is a positive spectral sequence r,s ≥ then Et =0fort 2andr<0ors<0. From now on we assume that all spectral sequences are positive. Proposition A.2.1 Let E be a positive spectral sequence converging to H. Then
r,s r,s (a) Et = E∞ if t>max(r, s +1), n (b) H =0 if n<0, 0 if r>n, (c) F r(Hn)= Hn if r ≤ 0.
Proof. (a) Note that
r−t,s+t−1 −→dt r,s −→dt r+t,s−t+1 Et Et Et . r−t,s+t−1 r+t,s−t+1 If t>r,thenEt =0;ift>s+1,then Et =0.So,if r,s r,s ··· r,s r,s ··· t>max(r, s + 1), then Ct = Cr+1 = ,andBt = Br+1 = ; hence, by Lemma A.1.1, r,s r,s ··· r,s Et = Et+1 = = E∞ . (b) If r + s = n<0, then either r<0ors<0; so F r(Hn)/F r+1(Hn)= r,s r n r+1 n r n E ∞ = 0; therefore F (H )=F H , for all r ∈ Z;thusF (H )=0(since r n n r n r F H = 0). This implies that H = r F (H )=0. r,s ∼ r n r+1 n (c) Let r + s = n.ThenE∞ = F (H )/F (H ). Now, if r<0or r,s r n r+1 n s<0, then E∞ =0;soF (H )=F (H ). Hence, ···= F −2(Hn)=F −1(Hn)=F 0(Hn) and F n+1(Hn)=F n+2(Hn)=F n+2(Hn)=···. Thus, it follows from condition (2)thatHr = F 0(Hn)ifr ≤ 0, and F r(Hn)= 0ifr>n. 400 A: Spectral Sequences
Proposition A.2.2 For each n thereisasequence
n,0 ι n π 0,n E∞ −→ H −→ E∞ , where ι is an injection,πa surjection and πι =0. The sequence is exact if n =1.
Proof. One has the following composition of maps
∼ n,0 = n n n n 1 n = 0,n E∞ −→ F (H ) → H −→ H /F (H ) −→ E∞ , and so, n,0 ι n π 0,n E∞ −→ H −→ E∞ . Note that Im(ι)=F n(Hn) ≤ F 1(Hn)=Ker(π); hence πι =0.Ifn =1, Im(ι)=Ker(π)=F 1(Hn), so the sequence is exact.
The Base Terms
r,0 The terms of the form Et are called the base terms of the spectral sequence. Proposition A.2.3 For each r there exist epimorphisms
∼ r,0 −→ r,0 −→ · · · −→ r,0 −→= r,0 E2 E3 Er+1 E∞ .
r,0 ∼ Proof. The last arrow is an isomorphism by Proposition A.2.1.SinceE3 = r,0 r,0 −→ r,0 Ker(d2)/Im(d2)=E2 /Im(d2), we have a surjection E2 E3 .One obtains the other maps in a similar way.
r,0 −→ r,0 −→ι r Each of the maps of Proposition A.2.3 and the map E2 E∞ H obtained from the maps of Propositions A.2.2 and A.2.3, are called edge homomorphisms on the base, and will be denoted by eB.
The Fiber Terms
0,s The terms of the form Et are called the fiber terms of the spectral sequence. Proposition A.2.4 For each s, there exist monomorphisms
∼ 0,s ←− 0,s ←− · · · ←− 0,s ←−= 0,s E2 E3 Es+2 E∞ .
0,s ∼ Proof. The last arrow is an isomorphism by Proposition A.2.1.SinceE3 = 0,s −→ 0,s Ker(d2)/Im(d2)=Ker(d2), we have an injection E3 E2 . The other injections are obtained similarly. A.2 Positive Spectral Sequences 401
Each of the maps of the Proposition A.2.4, and the map
s −→π 0,s −→ 0,s H E∞ E2 obtained by composing the maps of Propositions A.2.2 and A.2.4, are called edge homomorphisms on the fiber, and will be denoted by eF . ≥ 0,n −→ n+1,0 For n 1, the homomorphism dn+1 : En+1 En+1 is called a trans- gression.
Condition *(n). For a fixed n ≥ 1, we will say that the spectral sequence E satisfies condition ∗(n)if r,s ≤ ≤ − ≤ ≤ − E2 = 0 whenever 1 s n 1andr + s = n, and whenever 1 s n 1 and r + s = n +1. Note that condition ∗(1) is vacuous. Proposition A.2.5 Assume condition ∗(n) holds for a positive spectral se- quence E.Then 0,n −→ 0,n (a) the monomorphism eF : En+1 E2 is an isomorphism; n+1,0 −→ n+1,0 (b) the epimorphism eB : E2 En+1 is an isomorphism. t,n−t+1 0,n −→ t,n−t+1 0,n Proof. (a) Et =0ift = n +1.SoKer(dt : Et Et )=Et 0,n ∼ 0,n ∼ ···∼ 0,n if t = n + 1. Therefore, E2 = E3 = = En+1. n−t+1,t−1 n−t+1,t−1 −→ n+1,0 (b) Et =0ift = n +1.SoIm(dt : Et Et )=0. n+1,0 ∼ n+1,0 ∼ ···∼ n+1,0 Therefore, E2 = E3 = = En+1 . By the proposition above we can define a map
− − e 1 e 1 0,n −→F 0,n d−→n+1 n+1,0 −→B n+1,0 E2 En+1 En+1 E2 if condition ∗(n) is satisfied. This homomorphism will also be called a trans- gression and denoted tr. r,s Theorem A.2.6 Let E =(Et ) be a positive spectral sequence converging to n r,s ≤ ≤ − H =(H ). Assume that E2 =0for 1 s n 1(for n =1this condition is vacuous). Then there exists a five term exact sequence
−→ n,0 −→eB n −→eF 0,n −→tr n+1,0 −→eB n+1 0 E2 H E2 E2 H . Proof. First notice that r,0 −→eB r,0 r−t,t−1 −→dt r,0 Ker Et Et+1 =Im Et Et (3) 0,s −→eF 0,s 0,s −→dt t,s−t+1 Im Et+1 Et =Ker Et Et . (4) 402 A: Spectral Sequences
We shall prove exactness at each term. n,0 n,0 −→ n,0 Exactness at E2 : It is enough to prove that each Et Et+1 is n−t,t−1 an injection (r =2,...,n). But this follows from (3)sinceEt =0 (t =2,...,n). Exactness at Hn: Since condition ∗(n) holds, it follows then from Propo- sitions A.2.2 and A.2.5 that n,0 −→ n n,0 −→ι n n n Im(eB)=Im E2 H =Im E∞ H = F (H ) and n −→ 0,n n −→π 0,n 1 n Ker(eF )=Ker H E2 =Ker H E∞ = F (H ).
r,s Now, by hypothesis, if n = r + s and 1 ≤ r ≤ n − 1, then 0 = E∞ = F r(Hn)/F r+1(Hn); so F r(Hn)=F r+1(Hn). Hence F 1(Hn)=F n(Hn). Thus Im(eB)=Ker(eF ). 0,n Exactness at E2 : By Proposition A.2.5 and the definition of tr we have n −→ 0,n 0,n −→ 0,n Im(eF )=Im H En+1 =Im En+2 En+1 , and 0,n −→ n+1,0 Ker(tr)=Ker En+1 En+1 .
Thus Im(eF )=Ker(tr). n+1,0 Exactness at E2 : Analogously, 0,n −→ n+1,0 Im(tr)=Im En+1 En+1 , and n+1,0 −→ n+1 n+1,0 −→ n+1,0 Ker(eB)=Ker En+1 H =Ker En+1 En+2 .
Therefore Im(tr)=Ker(eB). Corollary A.2.7 There exists a five term exact sequence
−→ 1,0 −→eB 1 −→eF 0,1 −→tr 2,0 −→eB 2 0 E2 H B2 E2 H . Proof. This is a special case of the theorem since for n = 1 the hypothesis is vacuous.
A.3 Spectral Sequence of a Filtered Complex
In this section we study a canonical way of constructing spectral sequences. Given a complex with a suitable filtration, we define a spectral sequence that A.3 Spectral Sequence of a Filtered Complex 403 converges to the filtered graded abelian group consisting of the homology groups of that complex. Let
X =(X,∂)=···−→Xn−1 −→∂ Xn −→ Xn+1 −→··· be a complex of abelian groups. We say that X is filtered if each Xn has a filtration F compatible with ∂, i.e., for each r and each n, ∂Fr(Xn) ≤ F r(Xn+1). Assume that X is a filtered complex: . . . ··· . ↓↓ Xn−1 ≥···≥ F r(Xn−1) ≥··· ↓↓ Xn ≥···≥ F r(Xn) ≥··· ↓↓ Xn+1 ≥···≥ F r(Xn+1) ≥··· ↓↓ ......
Then the sequence of homology groups H = {Hn(X)} of this complex can be thought of as a single graded abelian group with a filtration inherited from the filtration of the complex X; namely, F r(Hn(X)) is the image of Hn(F r(X)) under the injection F r(X) → X. Next we begin the construction of a spectral sequence associated to X. Let r + s = n and r ∈ Z.Set r,s { ∈ r n | ∈ r+t n+1 } Zt = a F (X ) ∂(a) F (X ) , r,s r−t+1,s+t−2 r−t+1 n−1 ∩ r n Bt = ∂Zt−1 = ∂(F (X )) F (X ), and r,s r,s r,s r+1,s−1 Et = Zt /(Bt + Zt−1 ). (5) Since r,s ≤ r+t,s−t+1 ∂Zt Zt , and r,s r+1,s−1 r+1,s−1 r+t,s−t+1 ∂(Bt + Zt−1 )=∂Zt−1 = Bt , we have that the map ∂ induces a homomorphism r,s −→ r+t,s−t+1 dt : Et Et , (6) with dtdt = 0. Moreover, one checks that r,s −→dt r+t,s−t+1 r,s r+1,s−1 r,s r+1,s−1 Ker Et Et = Zt+1 + Zt−1 Bt + Zt−1 , 404 A: Spectral Sequences and r−t,s+t−1 −→dt r,s r,s r+1,s−1 r,s r+1,s−1 Im Et Et = Bt+1 + Zt−1 Bt + Zt−1 .
Hence ∼ r,s r+1,s−1 r,s r+1,s−1 Ker(dt)/Im(dt) = Zt+1 + Zt−1 Bt+1 + Zt−1 ∼ r,s r,s r+1,s−1 r,s = Zt+1 Bt+1 + Zt−1 = Et+1.
Observe that this is valid for every t ∈ Z. Thus we have proved the first part of the following Theorem A.3.1 Let (X,∂) be a filtered complex. Then r,s (a) There exists a spectral sequence E, where Et is given by (5). (b) Assume, in addition, that the filtration F of (X,∂) is bounded, i.e., for each n there are integers u = u(n) r,s r n Z∞ = {a ∈ F (X ) | ∂(a)=0}, and r,s n−1 r n B∞ = ∂(X ) ∩ F (X )(r + s = n). Then, r n ∼ r,s n−1 n−1 F (H (X)) = Z∞ + ∂X ∂X . So, r n r+1 n ∼ r,s n−1 r+1,s−1 n−1 F (H (X))/F (H (X)) = Z∞ + ∂X Z∞ + ∂X ∼ r,s r+1,s−1 n−1 r,s = Z∞ Z∞ + ∂X ∩ Z∞ ∼ r,s r+1,s−1 r,s = Z∞ Z∞ + B∞ . Since the filtration of (X,∂) is bounded, it is clear that r,s ∼ r,s r,s ∼ r,s Zu = Z∞ and Bu = B∞ for u large enough. Hence r n r+1 n ∼ r,s F (H (X))/F (H (X)) = Eu for u large enough. Finally, it is immediate that the boundedness of the filtration of (X,∂) r,s ∼ r,s ⇒ implies that Eu = E∞ for u large enough. Thus E = H(X). A.4 Spectral Sequences of a Double Complex 405 A.4 Spectral Sequences of a Double Complex r,s A double complex is a family K =(K )r,s∈Z of abelian groups together with differentials ∂ : Kr,s → Kr+1,s,∂ : Kr,s → Kr,s+1 such that ∂∂ =0,∂∂ =0and∂∂ + ∂∂ =0. Using the double complex K we define a complex (X,∂)=X = Tot(K), the total complex of K,by Xn = Kr,s, r+s=n and where ∂ : Xn −→ Xn+1 is ∂ = ∂ + ∂. Note that (X,∂) is a complex, for ∂∂ = ∂∂ + ∂∂ + ∂∂ + ∂∂ =0. Now we construct in a canonical way two filtrations of its total complex X. The first filtration F of X is given by F r(Xn)= Kα,β. α+β=n α≥r The second filtration F of X is defined by F s(Xn)= Kα,β. α+β=n β≥s For each of these filtrations we can construct corresponding spectral se- r,s r,s quences E =(Et )and E =( Et ), called the first and second spectral sequence of the double complex K (see the construction in Section A.3). Now assume that the double complex K is positive, i.e., Kr,s =0ifr<0ors<0. Then both the first and second filtrations are bounded. In fact Xn = F 0(Xn) ≥ F 1(Xn) ≥···≥ F n+1(Xn)=0 and Xn = F 0(Xn) ≥ F 1(Xn) ≥···≥ F n+1(Xn)=0. So, according to Theorem A.3.1, there exist corresponding spectral sequences r,s r,s E =(Et )and E =( Et ) (the first and second spectral sequences of K) converging both of them to H(X) with the induced filtrations. Next we calculate the initial terms E2 and E2 of these two spectral sequences. In order to do this we compute first the terms E1 and E1.We start with the first spectral sequence. We have r,s { ∈ r n | ∈ r+1 n+1 } Z1 = a F (X ) ∂(a) F (X ) ∼ r,s ∂ r,s+1 r+1 n = Ker(K −→ K ) ⊕ F (X ); 406 A: Spectral Sequences and r,s r+1,s−1 ∼ r n−1 r+1 n B1 + Z−1 = ∂ F (X )+ F (X ) ∼ r,s−1 ∂ r,s r+1 n = Im(K −→ K ) ⊕ F (X ). Hence