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Free profinite products of prosoluble groups
Journal of Algebra 529 (2019) 54–64
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Journal of Algebra www.elsevier.com/locate/jalgebra
Free profinite products of prosoluble groups ✩ Kıvanç Ersoy a,b,∗ , Wolfgang Herfort c a
Department of Mathematics, Mimar Sinan Fine Arts University, Istanbul, Turkey Department of Mathematics and Computer Science, Freie Universität Berlin, Germany c University of Technology at Vienna, Austria b
a r t i c l e
i n f o
Article history: Received 28 February 2018 Available online 26 March 2019 Communicated by E.I. Khukhro MSC: primary 20E06 secondary 20E18, 20E08
a b s t r a c t Let G := i∈I Gi be the restricted free profinite product of prosoluble groups Gi . We contribute to the description of prosoluble subgroups of such G. In particular, the question of existence of a prosoluble retract for the projection onto the maximal prosoluble factor group is discussed. © 2019 Elsevier Inc. All rights reserved.
Keywords: Free profinite products modulo their F-residuals
1. Introduction and main results Let G be the restricted free profinite product G = i∈I Gi of profinite groups Gi over a set I, as introduced by J. Neukirch in [7], see [10, D.3.1.] and [9, Example 5.9.1]. In [2,3,6] D. Haran, W. Herfort & L. Ribes, and, O.V. Melnikov independently describe pro-p subgroups of such G. In [11,12] profinite analogues of the Kurosh subgroup theorem are provided by P.A. Zalesskiˇi for open subgroups of G and also for closed nor✩ Wolfgang Herfort would like to thank for the warm hospitality during his stay at Mimar Sinan Fine Arts University in February 2012 and appreciates travel support by TU Vienna. Kıvanç Ersoy was partially supported by Mimar Sinan Fine Arts University Research Project Unit with project number 2917/21. * Corresponding author. E-mail addresses: [email protected] (K. Ersoy), [email protected] (W. Herfort).
https://doi.org/10.1016/j.jalgebra.2019.02.042 0021-8693/© 2019 Elsevier Inc. All rights reserved.
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mal subgroups. A closer analysis of closed subgroups having particular properties leads O.V. Melnikov and P.A. Zalesskii in [13] to the description of all soluble subgroups of groups acting on profinite trees. It should be said that any profinite product naturally acts on the so-called standard tree (see e.g. [9, Section 3.5]) so that these results contain, as a special case, the classification of soluble subgroups of free profinite products. Relaxing the condition of a closed subgroup being soluble, in [8, Theorem 1] F. Pop provides necessary conditions for a prosoluble subgroup F to be the closed subgroup of a free product of profinite groups. Theorem 1.1 ([8]). Let G = i Gi be the free product of profinite groups Gi and F a closed prosoluble subgroup of G. One of the following statements holds: (i) There is a prime p such that F ∩ Ggi is pro-p for every i ∈ I and g ∈ G, (ii) All non-trivial intersections F ∩ Ggi for i ∈ I and g ∈ G are finite and conjugate in F . (iii) A conjugate of F is subgroup of Gi for some i ∈ I. See also for conditions on a prosoluble closed subgroup of a free product to be conjugate into one of the free factors in [5]. In the same paper Pop asks whether or not the situation described in (ii) of Theorem 1.1 does occur in some free profinite product. Guralnick and Haran in [1] gave an affirmative answer to this question and described all profinite Frobenius subgroups of a free profinite product of finite groups A and B. They not only generalised an earlier result by Herfort and Ribes [4] where only the free prosoluble free product is treated, but provided a much simpler proof of this fact as well. Here we shall be concerned with the following Problem 1.2. Under which conditions does G = i∈I Gi have a prosoluble retract that is the free prosoluble product of the prosoluble groups Gi for i ∈ I. We shall denote by i∈I,s Gi the restricted free prosoluble product, i.e., the restricted free product for the class of all soluble groups (see [10, D.3]). Positive answers to Problem 1.2 are available under restrictions on the free factors Gi . If, for example, all factors have cohomological dimension ≤ 1, i.e., are closed subgroups of some free profinite group then H := i∈I,s Gi has cohomological dimension ≤ 1 and is a factor group of G. Then by [10, Proposition 7.5.4] there is a retract isomorphic to i∈I,s Gi . The factors in the retract, in general, are not conjugates of the original ones, as one may guess from Pop’s Theorem 1.1. Proposition 1.3. Let G = i∈I Gi be the restricted free product of nontrivial prosoluble groups and I contain at least 2 elements. If G has a prosoluble retract H = i∈I,s Hi and Hi = Gxi i for some elements xi ∈ G then for some prime p and all i ∈ I the free factor Gi is a pro-p group.
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Proof. For the proof we need [8, Claim 1 in the proof of Theorem 1]: Claim 1. Suppose that F ≤ G satisfies the following conditions: There exist prime numbers p = q, indices i = j in I, and x, y in G such that Gxi ∩ F has a nontrivial Sylow p-subgroup and Gyj ∩ F has a nontrivial Sylow q-subgroup. Then F cannot be prosoluble. x
Fix i ∈ I and let j ∈ I − {i} be arbitrary. Then F := Gxi i s Gj j is a prosoluble subgroup of H and hence of G. From Claim 1 by letting x := xi and y := xj deduce that there is a single prime p for which Gi and Gj are p-groups. Since j was arbitrary it follows that all free factors in G are p-groups. 2 The prosoluble residual of a profinite group G is Rs (G) =
{N : N c G : G/N is prosoluble}.
(Compare [10, 3.4, p. 96].) The condition on the prosoluble factors Gi for the existence of a prosoluble retract in i∈I Gi from Proposition 1.3 on the factors is indeed sufficient. Theorem 1.4. Suppose that G = i∈I Gi is the restricted free profinite product of pro-p groups. Then G = Rs (G) H where H is an internal restricted free prosoluble product of conjugates of the factors Gi . Theorem 1.1 hints to obstructions when trying to find a prosoluble retract for G = A F when A is a finite p-group and F is free profinite: Proposition 1.5. Let G = A Z where A is a finite soluble group with π(A) ≥ 2 and ˆ Then G does not have a prosoluble retract. Z∼ = Z. Proof. Fix a generator z of Z and a prime p. We let N denote the normal closure of z p in G and R the prosoluble residual of G. Then G/R ∼ = A s Z. Assume now the existence of a retraction map j : G/R → G. Set H := j(G/R) = A s Z1 ˆ is generated by some element z1 . Let U be the unique open subgroup where Z1 ∼ =Z∼ =Z of index p of H containing A and z1p . Then, by the Kurosh Subgroup Theorem (see [10, Theorem 9.1.9, p. 359] and observe that in our case G = H, n = 2, D = U , [G : D] = p, G1 = A, G2 = Z1 , and the rank of F equals 1 + (2 − 1)p − (p + 1) = 0) for a generator x of Z we have, setting Ip := {0, 1, . . . , p − 1}, ⎛ U =⎝
j∈Ip ,s
⎞ Ax ⎠ s xp . j
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j Thus F := j∈Ip ,s Ax is a prosoluble subgroup of the free product A Z. Theorem 1.1 shows that this is impossible. 2 The condition A being a finite p-group is sufficient. The following will be an immediate consequence of Theorem 4.4: ∼ Z. ˆ Theorem 1.6. Let G = A Z be the free profinite product with A pro-p and Z = Then there is a procyclic subgroup D such that Γ := A, D is prosoluble and DRs (G) = ZRs (G). Moreover, Γ ∼ = A s Z. 2. Preliminaries The set GX of all functions on a set X with values in a group G has a natural group structure, i.e., for f, g ∈ GX we let f g be the function defined as (f g)(x) = f (x)g(x) and f −1 is defined as f −1 (x) := f (x)−1 . The space of closed subgroups Subgr(G) of a profinite group G will be equipped with the Vietoris topology. The latter has a base consisting of sets W (U, F ) := {C ∈ Subgr(G) | CU = U F } where F is a finite set and U an open subgroup of G. It is well known that Subgr(G) is compact as it agrees with the inverse limit of the finite sets Subgr(G/N ) for N running through the directed set the filter base N of open normal subgroups of G. We note for later that a net (Xν )ν of subgroups in Subgr(G) and ν running through some upwards directed poset, converges to an element X ∈ Subgr(G) if, and only if, for every fixed open normal subgroup N of G there exists ν0 such that Xν N = Xν0 N holds for all ν ≥ ν0 . The following observation will turn out to be useful: Lemma 2.1. Let R be a normal subgroup of the profinite group G and N a net of normal subgroups of G converging to 1. Suppose that for every N ∈ N there is a subgroup HN of G and the following properties hold: (a) HN /HN ∩ N is prosoluble; (b) HN R = G. Let H be a clusterpoint of the net (HN )N ∈N . Then (i) H is prosoluble; (ii) HR = G. Proof. There is no loss of generality to assume that HN → H as N → 1. Fix any open normal subgroup M of G. Due to convergence there exists N0 ∈ N such that HN M = HM and N ≤ M holds for all N ≤ N0 .
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(i) As HN /HN ∩ N ∼ = HN N/N is prosoluble, so is HN M/M = HM/M – in fact the latter group is indeed soluble as M is open in G. Since M was chosen arbitrarily conclude that H/H ∩ M is soluble and thus H = ← lim H/H ∩ M is prosoluble. −−M (ii) Since HN R = G implies HN M R = G and, as HN M = HM , we conclude that HM R = G holds for every clopen normal subgroup M of G. Let U be an arbitrary open subgroup containing the closed subgroup HR. Then there is MU ∈ N contained in U and therefore HRMU ⊆ HRU . Therefore, letting U denote the set of all open subgroups U containing HR, [10, Proposition 2.1.4](d) implies G=
M
as claimed.
HM R ⊆
U ∈U
MU (HR) ⊆
U (HR) = HR
U ∈U
2
3. Splitting a free product over its prosoluble residual For a profinite group G we shall consider a convergent family of subgroups (Gi )i∈I in the sense of [10, D.3]: Every fixed open subgroup U of G contains Gi for almost all indices i ∈ I. A preliminary form of Theorem 1.4 reads as follows. Theorem 3.1. Let p be a fixed prime and the profinite group G be generated by a convergent family (Gi )i∈I of pro-p-groups. Then there exists a function g : I → G such that g(i) H(g) := Gi | i ∈ I is prosoluble and G = H(g)Rs (G). Proof. We prove the Theorem first under the additional assumption Rs (G) being finite. Also we can assume that Rs (G) = {1}, else there is nothing to prove. By contradiction assume that G is a counter-example with Rs (G) of minimal cardinality. We establish the following facts about G: (A) If, for some g : I → G, the equality H(g)Rs (G) = G holds (and hence H(g) is not prosoluble) then H(g) = G. (B) Rs (G) is a minimal normal subgroup of G and is isomorphic to a direct product i Si of groups Si isomorphic to some finite nonabelian simple group. (C) Rs (G) does not contain p-elements. (A) If H(g) = G there is nothing to prove. Suppose next that H(g) < G and thus Rs (H(g)) < Rs (G).
(1)
Then, by the minimality assumption, there is h : I → H(g) such that H(g)(h)Rs (H(g)) = H(g) and H(g)(h) = H(gh) is soluble. But, as Rs (H(g)) ≤ Rs (G), also H(gh)Rs (G) =
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H(g)Rs (G) = G holds, i.e., G is not a counter-example, in contrast to the assumption. Hence Rs (G) = Rs (H(g)) must hold, contradicting Eq. (1). But then H(g) = H(g)Rs (G) = G. (B) Let N be a minimal normal subgroup of G contained in Rs (G). Then N = 1 so that |Rs (G)/N | < |Rs (G)|. Then G/N satisfies the assumptions of our theorem for (Gi N/N )i∈I . Hence there is g : I → G such that H(g)Rs (G) = G and H(g)N/N is soluble. It shows that Rs (H(g)) is contained in N . On the other hand, by (A), Rs (H(g)) = Rs (G) and thus N ≥ Rs (G). Hence Rs (G) = N is a minimal normal subgroup. Since G is not prosoluble, Rs (G) has the claimed structure. (C) For every i ∈ I, we choose a Sylow p-subgroup Pi of G containing Gi . Fix i0 ∈ I. Since, for every i ∈ I, the intersection Pi ∩ Rs (G) is a Sylow p-subgroup of Rs (G), and all Sylow p-subgroups of Rs (G) are conjugate in Rs (G), there exists r : I → Rs (G) with r(i)
Pi0 ∩ Rs (G) = (Pi ∩ Rs (G))r(i) = Pi r(i)
∩ Rs (G).
r(i)
Hence the subgroup Gi of Pi normalises Pi0 ∩ Rs (G) for every i ∈ I and thus also H(r) normalises it. Observe that r(i)
H(r)Rs (G) = Gi
| i ∈ I Rs (G) = G
and therefore, by (A), it follows that H(r) = G. Thus the p-subgroup Pi0 ∩ Rs (G) is normal in G and contained in Rs (G). We cannot have Pi0 ∩ Rs (G) = Rs (G), since Rs (G) is not prosoluble. By (B), Pi0 ∩ Rs (G) = {1}. We come to finishing the proof for finite Rs (G). Fix i0 ∈ I and let Gp be a Sylow p-subgroup of G containing Gi0 . Let q ∈ π(Rs (G)) and pick a Sylow q-subgroup Q of G. By (C) we have Gp Rs (G)/Rs (G) ∼ = Gp and therefore making use of the Frattini argument, G = NG (Q)Rs (G), and the fact that Rs (G) does not contain p-elements by (C), we deduce that NG (Q)p ∼ = Gp . Thus, there is no loss of generality to assume that Gp ≤ NG (Q). There is k : I → G k(i) such that Gi ≤ Gp for all i ∈ I. Since, according to the Frattini argument, k(i) decomposes as k(i) = ν(i)r(i) with ν(i) ∈ NG (Q) and r(i) ∈ Rs (G) it follows that r(i)−1
Gi
≤ Gpν(i) r(i)−1
and since Gp ≤ NG (Q), we deduce from this that Gi −1
≤ NG (Q). r(i)−1
Gi | i ∈
I ≤ NG (Q) and Collecting facts, we have achieved that H(r ) = certainly H(r−1 )Rs (G) = G since for every i ∈ I we have r(i) ∈ Rs (G). Thus, by (A), H(r−1 ) = G and hence Q is a minimal normal subgroup of G properly contained in Rs (G). This contradicts (B).
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Thus we have established the Theorem for finite Rs (G) and turn to the general case. We let N denote the filter base of normal subgroups N of G open in Rs (G). This filter base is a directed set. Now, for fixed N ∈ N , apply the first part of the proof to G/N , and manufacture functions gN : I → G and subgroups HN := H(gN ) such that HN Rs (G) = G and HN N/N is a finite soluble group. The space GI × Subgr(G) is boolean and (gN , H(gN ))N ∈N is a net in this space. Making use of the compactness of the space GI × Subgr(G) allows us to pass to a cofinal convergent subnet and achieve that (1) the net (gN )N converges to some g : I → G; and (2) the subgroups HN := H(gN ) converge to a subgroup of the form H(g). Using Lemma 2.1 with R := Rs (G) implies that H(g) is prosoluble and G = H(g)Rs (G) as wanted. 2 We now can establish our first result. Proof of Theorem 1.4. Since G = i∈I Gi is a restricted free product the family (Gi )i∈I is convergent. In Theorem 3.1 we use this family of subgroups. Therefore there is a sequence g := (gi )i∈I ∈ GI such that H(g) := Ggi i | i ∈ I is prosoluble and G =
Rs (G)H(g). Define a map α : i∈I Gi → H(g) by sending every g ∈ Gi to g gi and observe that the family of closed subgroups (Ggi i )i∈I is convergent. Using the universal property of this convergent family (see [10, D.3, p. 425]), we can extend α to an epimorphism from the restricted free prosoluble product i∈I,s Gi to H(g). Then we have the following diagram of maps i∈I,s
Gi
α
H(g)
H(g)/H(g) ∩ Rs (G) ∼ = G/Rs (G) ∼ = i∈I,s Gi .
The composition of these maps constitutes an epimorphism of i∈I,s Gi onto itself which
is a bijection when restricted to i∈I Gi . The cited universal property of the free prosoluble product implies that this composition is an isomorphism of groups. In particular, H(g) is the internal free prosoluble product of the factors Gi . The canonically induced map from G onto H(g) has thus precisely the kernel Rs (G) and hence G is the semidirect product G = Rs (G) H(g). 2 Question 3.2. Let G = A B be the free profinite product of pro-p groups A and B. Is every prosoluble subgroup of G isomorphic to a closed subgroup of A s B?
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ˆ over the prosoluble residual 4. Splitting A Z We turn to proving Theorem 1.6 which states that for any pro-p group A the free ˆ retracts to the prosoluble free product A s Z ˆ over its prosoluble residual. product A Z Lemma 4.1. Let the profinite group G = R P be a semidirect product where P is a Sylow p-subgroup of G. Then, for p = q ∈ π(R) the normaliser NG (Q) of any Sylow q-subgroup Q of R contains a Sylow p-subgroup of G. / π(NG (Q) ∩ R). Therefore Proof. Observe that NG (Q)/NG (Q) ∩ R is a p-group and p ∈ the Schur-Zassenhaus theorem (see [10, Theorem 2.3.15]) implies the existence of a complement P1 of NG (Q) ∩ R in NG (Q), i.e., NG (Q) = (NG (Q) ∩ R) P1 . Then P1 is a p-subgroup of G and G = P R = NG (Q)R = P1 (NG (Q) ∩ R)R = P1 R. Since P1 ∩ R = P1 ∩ NG (Q) ∩ R = {1} deduce G = R P1 , i.e., P1 is another complement of R in G. By [10, Theorem 2.3.15] the complements P1 and P are conjugate in G. Therefore P1 is a Sylow p-subgroup of G contained in NG (Q). 2 Lemma 4.2. Let G be a profinite group and A ≤ G a p-subgroup. Suppose the prosoluble residual Rs (G) is not trivial. Then there exists a closed subgroup A ≤ X < G such that XRs (G) = G and X normalises a Sylow q-subgroup of R(G) for some q ∈ π(R(G)). Moreover X/Rs (G) ∩ X is prosoluble. Proof. Set R := Rs (G) and let P be a Sylow p-subgroup of G containing A. It will be enough to manufacture X < G with the claimed normaliser property and, in addition, containing P . Suppose first that a Sylow p-subgroup of R is not normal. Replacing the Sylow p-subgroup of R by a suitable conjugate we can assume that it agrees with P ∩ R. By the Frattini argument we have G = NG (P ∩ R)R. Setting X := NG (P ∩ R) this equality reads G = XR and since P ∩ R is normal in P we must have P ≤ X as well. Since P ∩ R is not normal in R it cannot be normal in G and hence X is a proper subgroup of G. Assume now P ∩ R R. Then P ∩ R is characteristic in R and thus normal in G. We claim that P ∩ R = {1} can be assumed. For proving the claim observe first that G/P ∩ R cannot be prosoluble because it is the extension of the nonprosoluble group R/P ∩ R by the prosoluble group G/R ∼ = (G/P ∩ R)/(R/P ∩ R). Thus there is Y < G/P ∩ R containing P/P ∩ R and satisfying Y Rs (G/P ∩ R) = G/P ∩ R. Letting φ denote the canonical epimorphism from G onto
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G/P ∩R, and, taking R/P ∩R = Rs (G/P ∩R) into account, it turns out that X := φ−1 (Y ) serves the purpose. Hence during the remainder of the proof we may assume P ∩ R = {1}. Since R cannot be pronilpotent there must be q ∈ π(R) such that the Sylow q-subgroup, say Q, of R is not normal in R. Then L := RP = R P is a semidirect product to which we may apply the Frattini argument R P = NL (Q)R. Lemma 4.1 implies that NL (Q) contains a Sylow p-subgroup of G and, conjugating Q, if necessary, we can achieve P ≤ NL (Q). Then X = NG (Q) contains P and is a proper subgroup of G since Q is not normal in G. Moreover, X/X ∩ R ∼ = XR/R is prosoluble, since G/R = G/Rs (G) is prosoluble. 2 Proposition 4.3. Let G be a profinite group and A a pro-p subgroup. Then G contains a prosoluble supplement A ≤ X of Rs (G), i.e., G = Rs (G)X. Proof. Denote, during this proof, by R the prosoluble residual Rs (G). Let S denote the set of all supplements Y of R in G containing A. Introduce a partial order on S by setting S > S if, and only if, S is a subgroup of S . We intend to apply Zorn’s Lemma and note first that S cannot be the empty set, as G itself belongs there. Let (Xi )i∈I be a chain in S, i.e., I is linearly ordered and Xi ≤ Xi if, and only if, i ≥ i . Then Xi R = G holds for all i ∈ I. Applying Proposition 2.1.4 in [10] to the decreasing collection (Xi R)i∈I yields
(Xi R) = (
i∈I
Xi )R
i∈I
so that i∈I Xi belongs to S, showing that the (arbitrarily chosen) chain (Xi )i∈I has an upper bound. By Zorn’s Lemma S has a maximal element, say X, and we need to show that X is prosoluble. Indeed, if X is not prosoluble, then applying Lemma 4.2 to X yields a proper subgroup Y of X containing A and X = Y Rs (X). Then, as Rs (X) ≤ R = Rs (G), G = XR = Y Rs (X)R = Y R. Hence Y > X contrasts the minimality of X.
2
Recall from [10] that for a finitely generated free pro-C group G its rank is the minimal number of topological generators. Theorem 4.4. Let G = A F be the free product of a finite p-group A and a finitely generated free profinite group F . Then G has a retract L ∼ = A s Fs with Fs the free prosoluble group of rank equal to the rank of F .
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Proof. The canonical epimorphism from F to F/Rs (F ) = Fs admits a prosoluble retract, i.e., F = Rs (F ) Fs . By [10, Corollary 9.1.7] the subgroup H := A, Fs is isomorphic to H∼ = A Fs . Suppose X is a prosoluble retract for H, i.e., H = Rs (H) X. Then G = (Rs (F ))G H = (Rs (F ))G (Rs (H) X) and hence, setting R := (Rs (F ))G Rs (H) it turns out that G = R X. Since X is prosoluble, we must have Rs (G) ≤ R. On the other hand, factoring Rs (G), yields an epimorphism φ from G/Rs (G) ∼ = X onto X = A s Fs and since X is finitely generated it is Hopfian. Therefore φ is an isomorphism and thus X would be a prosoluble retract of G as well. Hence there is no loss of generality to assume G = A Fs , where the rank of Fs is equal to the rank of F . By Proposition 4.3 there is a prosoluble subgroup X of G containing A with G = XRs (G). The canonical epimorphism ρ : G → A s Fs restricts to an epimorphism from X to A s Fs . Let T be a set of free generators of Fs and lift it to a set S. Our candidate for L will be the subgroup L := A, S . Note first that S ∼ = T ∼ = Fs . Furthermore ρ induces an epimorphism from L onto A s Fs that sends A to A and S to Fs . By the universal property of the free prosoluble product there is an epimorphism χ : A s Fs → L sending A to A and Fs to S . Since L is finitely generated it is Hopfian (see [10, Proposition 2.5.2]) and thus χ is an isomorphism. Thus G = Rs (G) L. 2 We have not been able to extend these findings. Problem 4.5. Let A be a pro-p group and F a projective prosoluble group. Does the free profinite product G := A F admit a prosoluble retract isomorphic to A s Fs ? Examples of prosoluble groups with a single conjugacy class of maximal finite subgroups are A s F with F torsion free and A a finite soluble group, as well as Frobenius groups K F with K the kernel and F a complement. Acknowledgment We would like to thank the referee for the interest in our work and careful reading. The reports led to significant improvements.
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Contents lists available at ScienceDirect
Journal of Algebra www.elsevier.com/locate/jalgebra
Free profinite products of prosoluble groups ✩ Kıvanç Ersoy a,b,∗ , Wolfgang Herfort c a
Department of Mathematics, Mimar Sinan Fine Arts University, Istanbul, Turkey Department of Mathematics and Computer Science, Freie Universität Berlin, Germany c University of Technology at Vienna, Austria b
a r t i c l e
i n f o
Article history: Received 28 February 2018 Available online 26 March 2019 Communicated by E.I. Khukhro MSC: primary 20E06 secondary 20E18, 20E08
a b s t r a c t Let G := i∈I Gi be the restricted free profinite product of prosoluble groups Gi . We contribute to the description of prosoluble subgroups of such G. In particular, the question of existence of a prosoluble retract for the projection onto the maximal prosoluble factor group is discussed. © 2019 Elsevier Inc. All rights reserved.
Keywords: Free profinite products modulo their F-residuals
1. Introduction and main results Let G be the restricted free profinite product G = i∈I Gi of profinite groups Gi over a set I, as introduced by J. Neukirch in [7], see [10, D.3.1.] and [9, Example 5.9.1]. In [2,3,6] D. Haran, W. Herfort & L. Ribes, and, O.V. Melnikov independently describe pro-p subgroups of such G. In [11,12] profinite analogues of the Kurosh subgroup theorem are provided by P.A. Zalesskiˇi for open subgroups of G and also for closed nor✩ Wolfgang Herfort would like to thank for the warm hospitality during his stay at Mimar Sinan Fine Arts University in February 2012 and appreciates travel support by TU Vienna. Kıvanç Ersoy was partially supported by Mimar Sinan Fine Arts University Research Project Unit with project number 2917/21. * Corresponding author. E-mail addresses: [email protected] (K. Ersoy), [email protected] (W. Herfort).
https://doi.org/10.1016/j.jalgebra.2019.02.042 0021-8693/© 2019 Elsevier Inc. All rights reserved.
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mal subgroups. A closer analysis of closed subgroups having particular properties leads O.V. Melnikov and P.A. Zalesskii in [13] to the description of all soluble subgroups of groups acting on profinite trees. It should be said that any profinite product naturally acts on the so-called standard tree (see e.g. [9, Section 3.5]) so that these results contain, as a special case, the classification of soluble subgroups of free profinite products. Relaxing the condition of a closed subgroup being soluble, in [8, Theorem 1] F. Pop provides necessary conditions for a prosoluble subgroup F to be the closed subgroup of a free product of profinite groups. Theorem 1.1 ([8]). Let G = i Gi be the free product of profinite groups Gi and F a closed prosoluble subgroup of G. One of the following statements holds: (i) There is a prime p such that F ∩ Ggi is pro-p for every i ∈ I and g ∈ G, (ii) All non-trivial intersections F ∩ Ggi for i ∈ I and g ∈ G are finite and conjugate in F . (iii) A conjugate of F is subgroup of Gi for some i ∈ I. See also for conditions on a prosoluble closed subgroup of a free product to be conjugate into one of the free factors in [5]. In the same paper Pop asks whether or not the situation described in (ii) of Theorem 1.1 does occur in some free profinite product. Guralnick and Haran in [1] gave an affirmative answer to this question and described all profinite Frobenius subgroups of a free profinite product of finite groups A and B. They not only generalised an earlier result by Herfort and Ribes [4] where only the free prosoluble free product is treated, but provided a much simpler proof of this fact as well. Here we shall be concerned with the following Problem 1.2. Under which conditions does G = i∈I Gi have a prosoluble retract that is the free prosoluble product of the prosoluble groups Gi for i ∈ I. We shall denote by i∈I,s Gi the restricted free prosoluble product, i.e., the restricted free product for the class of all soluble groups (see [10, D.3]). Positive answers to Problem 1.2 are available under restrictions on the free factors Gi . If, for example, all factors have cohomological dimension ≤ 1, i.e., are closed subgroups of some free profinite group then H := i∈I,s Gi has cohomological dimension ≤ 1 and is a factor group of G. Then by [10, Proposition 7.5.4] there is a retract isomorphic to i∈I,s Gi . The factors in the retract, in general, are not conjugates of the original ones, as one may guess from Pop’s Theorem 1.1. Proposition 1.3. Let G = i∈I Gi be the restricted free product of nontrivial prosoluble groups and I contain at least 2 elements. If G has a prosoluble retract H = i∈I,s Hi and Hi = Gxi i for some elements xi ∈ G then for some prime p and all i ∈ I the free factor Gi is a pro-p group.
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Proof. For the proof we need [8, Claim 1 in the proof of Theorem 1]: Claim 1. Suppose that F ≤ G satisfies the following conditions: There exist prime numbers p = q, indices i = j in I, and x, y in G such that Gxi ∩ F has a nontrivial Sylow p-subgroup and Gyj ∩ F has a nontrivial Sylow q-subgroup. Then F cannot be prosoluble. x
Fix i ∈ I and let j ∈ I − {i} be arbitrary. Then F := Gxi i s Gj j is a prosoluble subgroup of H and hence of G. From Claim 1 by letting x := xi and y := xj deduce that there is a single prime p for which Gi and Gj are p-groups. Since j was arbitrary it follows that all free factors in G are p-groups. 2 The prosoluble residual of a profinite group G is Rs (G) =
{N : N c G : G/N is prosoluble}.
(Compare [10, 3.4, p. 96].) The condition on the prosoluble factors Gi for the existence of a prosoluble retract in i∈I Gi from Proposition 1.3 on the factors is indeed sufficient. Theorem 1.4. Suppose that G = i∈I Gi is the restricted free profinite product of pro-p groups. Then G = Rs (G) H where H is an internal restricted free prosoluble product of conjugates of the factors Gi . Theorem 1.1 hints to obstructions when trying to find a prosoluble retract for G = A F when A is a finite p-group and F is free profinite: Proposition 1.5. Let G = A Z where A is a finite soluble group with π(A) ≥ 2 and ˆ Then G does not have a prosoluble retract. Z∼ = Z. Proof. Fix a generator z of Z and a prime p. We let N denote the normal closure of z p in G and R the prosoluble residual of G. Then G/R ∼ = A s Z. Assume now the existence of a retraction map j : G/R → G. Set H := j(G/R) = A s Z1 ˆ is generated by some element z1 . Let U be the unique open subgroup where Z1 ∼ =Z∼ =Z of index p of H containing A and z1p . Then, by the Kurosh Subgroup Theorem (see [10, Theorem 9.1.9, p. 359] and observe that in our case G = H, n = 2, D = U , [G : D] = p, G1 = A, G2 = Z1 , and the rank of F equals 1 + (2 − 1)p − (p + 1) = 0) for a generator x of Z we have, setting Ip := {0, 1, . . . , p − 1}, ⎛ U =⎝
j∈Ip ,s
⎞ Ax ⎠ s xp . j
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j Thus F := j∈Ip ,s Ax is a prosoluble subgroup of the free product A Z. Theorem 1.1 shows that this is impossible. 2 The condition A being a finite p-group is sufficient. The following will be an immediate consequence of Theorem 4.4: ∼ Z. ˆ Theorem 1.6. Let G = A Z be the free profinite product with A pro-p and Z = Then there is a procyclic subgroup D such that Γ := A, D is prosoluble and DRs (G) = ZRs (G). Moreover, Γ ∼ = A s Z. 2. Preliminaries The set GX of all functions on a set X with values in a group G has a natural group structure, i.e., for f, g ∈ GX we let f g be the function defined as (f g)(x) = f (x)g(x) and f −1 is defined as f −1 (x) := f (x)−1 . The space of closed subgroups Subgr(G) of a profinite group G will be equipped with the Vietoris topology. The latter has a base consisting of sets W (U, F ) := {C ∈ Subgr(G) | CU = U F } where F is a finite set and U an open subgroup of G. It is well known that Subgr(G) is compact as it agrees with the inverse limit of the finite sets Subgr(G/N ) for N running through the directed set the filter base N of open normal subgroups of G. We note for later that a net (Xν )ν of subgroups in Subgr(G) and ν running through some upwards directed poset, converges to an element X ∈ Subgr(G) if, and only if, for every fixed open normal subgroup N of G there exists ν0 such that Xν N = Xν0 N holds for all ν ≥ ν0 . The following observation will turn out to be useful: Lemma 2.1. Let R be a normal subgroup of the profinite group G and N a net of normal subgroups of G converging to 1. Suppose that for every N ∈ N there is a subgroup HN of G and the following properties hold: (a) HN /HN ∩ N is prosoluble; (b) HN R = G. Let H be a clusterpoint of the net (HN )N ∈N . Then (i) H is prosoluble; (ii) HR = G. Proof. There is no loss of generality to assume that HN → H as N → 1. Fix any open normal subgroup M of G. Due to convergence there exists N0 ∈ N such that HN M = HM and N ≤ M holds for all N ≤ N0 .
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(i) As HN /HN ∩ N ∼ = HN N/N is prosoluble, so is HN M/M = HM/M – in fact the latter group is indeed soluble as M is open in G. Since M was chosen arbitrarily conclude that H/H ∩ M is soluble and thus H = ← lim H/H ∩ M is prosoluble. −−M (ii) Since HN R = G implies HN M R = G and, as HN M = HM , we conclude that HM R = G holds for every clopen normal subgroup M of G. Let U be an arbitrary open subgroup containing the closed subgroup HR. Then there is MU ∈ N contained in U and therefore HRMU ⊆ HRU . Therefore, letting U denote the set of all open subgroups U containing HR, [10, Proposition 2.1.4](d) implies G=
M
as claimed.
HM R ⊆
U ∈U
MU (HR) ⊆
U (HR) = HR
U ∈U
2
3. Splitting a free product over its prosoluble residual For a profinite group G we shall consider a convergent family of subgroups (Gi )i∈I in the sense of [10, D.3]: Every fixed open subgroup U of G contains Gi for almost all indices i ∈ I. A preliminary form of Theorem 1.4 reads as follows. Theorem 3.1. Let p be a fixed prime and the profinite group G be generated by a convergent family (Gi )i∈I of pro-p-groups. Then there exists a function g : I → G such that g(i) H(g) := Gi | i ∈ I is prosoluble and G = H(g)Rs (G). Proof. We prove the Theorem first under the additional assumption Rs (G) being finite. Also we can assume that Rs (G) = {1}, else there is nothing to prove. By contradiction assume that G is a counter-example with Rs (G) of minimal cardinality. We establish the following facts about G: (A) If, for some g : I → G, the equality H(g)Rs (G) = G holds (and hence H(g) is not prosoluble) then H(g) = G. (B) Rs (G) is a minimal normal subgroup of G and is isomorphic to a direct product i Si of groups Si isomorphic to some finite nonabelian simple group. (C) Rs (G) does not contain p-elements. (A) If H(g) = G there is nothing to prove. Suppose next that H(g) < G and thus Rs (H(g)) < Rs (G).
(1)
Then, by the minimality assumption, there is h : I → H(g) such that H(g)(h)Rs (H(g)) = H(g) and H(g)(h) = H(gh) is soluble. But, as Rs (H(g)) ≤ Rs (G), also H(gh)Rs (G) =
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H(g)Rs (G) = G holds, i.e., G is not a counter-example, in contrast to the assumption. Hence Rs (G) = Rs (H(g)) must hold, contradicting Eq. (1). But then H(g) = H(g)Rs (G) = G. (B) Let N be a minimal normal subgroup of G contained in Rs (G). Then N = 1 so that |Rs (G)/N | < |Rs (G)|. Then G/N satisfies the assumptions of our theorem for (Gi N/N )i∈I . Hence there is g : I → G such that H(g)Rs (G) = G and H(g)N/N is soluble. It shows that Rs (H(g)) is contained in N . On the other hand, by (A), Rs (H(g)) = Rs (G) and thus N ≥ Rs (G). Hence Rs (G) = N is a minimal normal subgroup. Since G is not prosoluble, Rs (G) has the claimed structure. (C) For every i ∈ I, we choose a Sylow p-subgroup Pi of G containing Gi . Fix i0 ∈ I. Since, for every i ∈ I, the intersection Pi ∩ Rs (G) is a Sylow p-subgroup of Rs (G), and all Sylow p-subgroups of Rs (G) are conjugate in Rs (G), there exists r : I → Rs (G) with r(i)
Pi0 ∩ Rs (G) = (Pi ∩ Rs (G))r(i) = Pi r(i)
∩ Rs (G).
r(i)
Hence the subgroup Gi of Pi normalises Pi0 ∩ Rs (G) for every i ∈ I and thus also H(r) normalises it. Observe that r(i)
H(r)Rs (G) = Gi
| i ∈ I Rs (G) = G
and therefore, by (A), it follows that H(r) = G. Thus the p-subgroup Pi0 ∩ Rs (G) is normal in G and contained in Rs (G). We cannot have Pi0 ∩ Rs (G) = Rs (G), since Rs (G) is not prosoluble. By (B), Pi0 ∩ Rs (G) = {1}. We come to finishing the proof for finite Rs (G). Fix i0 ∈ I and let Gp be a Sylow p-subgroup of G containing Gi0 . Let q ∈ π(Rs (G)) and pick a Sylow q-subgroup Q of G. By (C) we have Gp Rs (G)/Rs (G) ∼ = Gp and therefore making use of the Frattini argument, G = NG (Q)Rs (G), and the fact that Rs (G) does not contain p-elements by (C), we deduce that NG (Q)p ∼ = Gp . Thus, there is no loss of generality to assume that Gp ≤ NG (Q). There is k : I → G k(i) such that Gi ≤ Gp for all i ∈ I. Since, according to the Frattini argument, k(i) decomposes as k(i) = ν(i)r(i) with ν(i) ∈ NG (Q) and r(i) ∈ Rs (G) it follows that r(i)−1
Gi
≤ Gpν(i) r(i)−1
and since Gp ≤ NG (Q), we deduce from this that Gi −1
≤ NG (Q). r(i)−1
Gi | i ∈
I ≤ NG (Q) and Collecting facts, we have achieved that H(r ) = certainly H(r−1 )Rs (G) = G since for every i ∈ I we have r(i) ∈ Rs (G). Thus, by (A), H(r−1 ) = G and hence Q is a minimal normal subgroup of G properly contained in Rs (G). This contradicts (B).
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Thus we have established the Theorem for finite Rs (G) and turn to the general case. We let N denote the filter base of normal subgroups N of G open in Rs (G). This filter base is a directed set. Now, for fixed N ∈ N , apply the first part of the proof to G/N , and manufacture functions gN : I → G and subgroups HN := H(gN ) such that HN Rs (G) = G and HN N/N is a finite soluble group. The space GI × Subgr(G) is boolean and (gN , H(gN ))N ∈N is a net in this space. Making use of the compactness of the space GI × Subgr(G) allows us to pass to a cofinal convergent subnet and achieve that (1) the net (gN )N converges to some g : I → G; and (2) the subgroups HN := H(gN ) converge to a subgroup of the form H(g). Using Lemma 2.1 with R := Rs (G) implies that H(g) is prosoluble and G = H(g)Rs (G) as wanted. 2 We now can establish our first result. Proof of Theorem 1.4. Since G = i∈I Gi is a restricted free product the family (Gi )i∈I is convergent. In Theorem 3.1 we use this family of subgroups. Therefore there is a sequence g := (gi )i∈I ∈ GI such that H(g) := Ggi i | i ∈ I is prosoluble and G =
Rs (G)H(g). Define a map α : i∈I Gi → H(g) by sending every g ∈ Gi to g gi and observe that the family of closed subgroups (Ggi i )i∈I is convergent. Using the universal property of this convergent family (see [10, D.3, p. 425]), we can extend α to an epimorphism from the restricted free prosoluble product i∈I,s Gi to H(g). Then we have the following diagram of maps i∈I,s
Gi
α
H(g)
H(g)/H(g) ∩ Rs (G) ∼ = G/Rs (G) ∼ = i∈I,s Gi .
The composition of these maps constitutes an epimorphism of i∈I,s Gi onto itself which
is a bijection when restricted to i∈I Gi . The cited universal property of the free prosoluble product implies that this composition is an isomorphism of groups. In particular, H(g) is the internal free prosoluble product of the factors Gi . The canonically induced map from G onto H(g) has thus precisely the kernel Rs (G) and hence G is the semidirect product G = Rs (G) H(g). 2 Question 3.2. Let G = A B be the free profinite product of pro-p groups A and B. Is every prosoluble subgroup of G isomorphic to a closed subgroup of A s B?
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ˆ over the prosoluble residual 4. Splitting A Z We turn to proving Theorem 1.6 which states that for any pro-p group A the free ˆ retracts to the prosoluble free product A s Z ˆ over its prosoluble residual. product A Z Lemma 4.1. Let the profinite group G = R P be a semidirect product where P is a Sylow p-subgroup of G. Then, for p = q ∈ π(R) the normaliser NG (Q) of any Sylow q-subgroup Q of R contains a Sylow p-subgroup of G. / π(NG (Q) ∩ R). Therefore Proof. Observe that NG (Q)/NG (Q) ∩ R is a p-group and p ∈ the Schur-Zassenhaus theorem (see [10, Theorem 2.3.15]) implies the existence of a complement P1 of NG (Q) ∩ R in NG (Q), i.e., NG (Q) = (NG (Q) ∩ R) P1 . Then P1 is a p-subgroup of G and G = P R = NG (Q)R = P1 (NG (Q) ∩ R)R = P1 R. Since P1 ∩ R = P1 ∩ NG (Q) ∩ R = {1} deduce G = R P1 , i.e., P1 is another complement of R in G. By [10, Theorem 2.3.15] the complements P1 and P are conjugate in G. Therefore P1 is a Sylow p-subgroup of G contained in NG (Q). 2 Lemma 4.2. Let G be a profinite group and A ≤ G a p-subgroup. Suppose the prosoluble residual Rs (G) is not trivial. Then there exists a closed subgroup A ≤ X < G such that XRs (G) = G and X normalises a Sylow q-subgroup of R(G) for some q ∈ π(R(G)). Moreover X/Rs (G) ∩ X is prosoluble. Proof. Set R := Rs (G) and let P be a Sylow p-subgroup of G containing A. It will be enough to manufacture X < G with the claimed normaliser property and, in addition, containing P . Suppose first that a Sylow p-subgroup of R is not normal. Replacing the Sylow p-subgroup of R by a suitable conjugate we can assume that it agrees with P ∩ R. By the Frattini argument we have G = NG (P ∩ R)R. Setting X := NG (P ∩ R) this equality reads G = XR and since P ∩ R is normal in P we must have P ≤ X as well. Since P ∩ R is not normal in R it cannot be normal in G and hence X is a proper subgroup of G. Assume now P ∩ R R. Then P ∩ R is characteristic in R and thus normal in G. We claim that P ∩ R = {1} can be assumed. For proving the claim observe first that G/P ∩ R cannot be prosoluble because it is the extension of the nonprosoluble group R/P ∩ R by the prosoluble group G/R ∼ = (G/P ∩ R)/(R/P ∩ R). Thus there is Y < G/P ∩ R containing P/P ∩ R and satisfying Y Rs (G/P ∩ R) = G/P ∩ R. Letting φ denote the canonical epimorphism from G onto
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G/P ∩R, and, taking R/P ∩R = Rs (G/P ∩R) into account, it turns out that X := φ−1 (Y ) serves the purpose. Hence during the remainder of the proof we may assume P ∩ R = {1}. Since R cannot be pronilpotent there must be q ∈ π(R) such that the Sylow q-subgroup, say Q, of R is not normal in R. Then L := RP = R P is a semidirect product to which we may apply the Frattini argument R P = NL (Q)R. Lemma 4.1 implies that NL (Q) contains a Sylow p-subgroup of G and, conjugating Q, if necessary, we can achieve P ≤ NL (Q). Then X = NG (Q) contains P and is a proper subgroup of G since Q is not normal in G. Moreover, X/X ∩ R ∼ = XR/R is prosoluble, since G/R = G/Rs (G) is prosoluble. 2 Proposition 4.3. Let G be a profinite group and A a pro-p subgroup. Then G contains a prosoluble supplement A ≤ X of Rs (G), i.e., G = Rs (G)X. Proof. Denote, during this proof, by R the prosoluble residual Rs (G). Let S denote the set of all supplements Y of R in G containing A. Introduce a partial order on S by setting S > S if, and only if, S is a subgroup of S . We intend to apply Zorn’s Lemma and note first that S cannot be the empty set, as G itself belongs there. Let (Xi )i∈I be a chain in S, i.e., I is linearly ordered and Xi ≤ Xi if, and only if, i ≥ i . Then Xi R = G holds for all i ∈ I. Applying Proposition 2.1.4 in [10] to the decreasing collection (Xi R)i∈I yields
(Xi R) = (
i∈I
Xi )R
i∈I
so that i∈I Xi belongs to S, showing that the (arbitrarily chosen) chain (Xi )i∈I has an upper bound. By Zorn’s Lemma S has a maximal element, say X, and we need to show that X is prosoluble. Indeed, if X is not prosoluble, then applying Lemma 4.2 to X yields a proper subgroup Y of X containing A and X = Y Rs (X). Then, as Rs (X) ≤ R = Rs (G), G = XR = Y Rs (X)R = Y R. Hence Y > X contrasts the minimality of X.
2
Recall from [10] that for a finitely generated free pro-C group G its rank is the minimal number of topological generators. Theorem 4.4. Let G = A F be the free product of a finite p-group A and a finitely generated free profinite group F . Then G has a retract L ∼ = A s Fs with Fs the free prosoluble group of rank equal to the rank of F .
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Proof. The canonical epimorphism from F to F/Rs (F ) = Fs admits a prosoluble retract, i.e., F = Rs (F ) Fs . By [10, Corollary 9.1.7] the subgroup H := A, Fs is isomorphic to H∼ = A Fs . Suppose X is a prosoluble retract for H, i.e., H = Rs (H) X. Then G = (Rs (F ))G H = (Rs (F ))G (Rs (H) X) and hence, setting R := (Rs (F ))G Rs (H) it turns out that G = R X. Since X is prosoluble, we must have Rs (G) ≤ R. On the other hand, factoring Rs (G), yields an epimorphism φ from G/Rs (G) ∼ = X onto X = A s Fs and since X is finitely generated it is Hopfian. Therefore φ is an isomorphism and thus X would be a prosoluble retract of G as well. Hence there is no loss of generality to assume G = A Fs , where the rank of Fs is equal to the rank of F . By Proposition 4.3 there is a prosoluble subgroup X of G containing A with G = XRs (G). The canonical epimorphism ρ : G → A s Fs restricts to an epimorphism from X to A s Fs . Let T be a set of free generators of Fs and lift it to a set S. Our candidate for L will be the subgroup L := A, S . Note first that S ∼ = T ∼ = Fs . Furthermore ρ induces an epimorphism from L onto A s Fs that sends A to A and S to Fs . By the universal property of the free prosoluble product there is an epimorphism χ : A s Fs → L sending A to A and Fs to S . Since L is finitely generated it is Hopfian (see [10, Proposition 2.5.2]) and thus χ is an isomorphism. Thus G = Rs (G) L. 2 We have not been able to extend these findings. Problem 4.5. Let A be a pro-p group and F a projective prosoluble group. Does the free profinite product G := A F admit a prosoluble retract isomorphic to A s Fs ? Examples of prosoluble groups with a single conjugacy class of maximal finite subgroups are A s F with F torsion free and A a finite soluble group, as well as Frobenius groups K F with K the kernel and F a complement. Acknowledgment We would like to thank the referee for the interest in our work and careful reading. The reports led to significant improvements.
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