Prosoluble subgroups of free profinite products

Abstract

We describe finitely generated and second countable prosoluble subgroups of free profinite products. We also give a description of relatively projective prosoluble groups.

1 Introduction

The result that describes the algebraic structure of subgroups of free products obtained by Kurosh [11] in 1934 is regarded as one of the fundamental results in combinatorial group theory. Unfortunately, the profinite analogue of the Kurosh Subgroup Theorem does not hold for free products of profinite groups, and so the algebraic structure of closed subgroups of free profinite products is not clear. Therefore, it is reasonable to study the structure of the most important classes of subgroups of free profinite products. One of the most natural and important class of profinite groups is the class of prosoluble groups, as they play the same role in the profinite group theory as soluble subgroups in the theory of finite groups.

The study of prosoluble subgroups of a free profinite product was initiated in middle 1990s by Jarden and Pop in [8] and [14], respectively; the interest in the question has origin in Galois theory (cf. [14], [5], [8]). In fact, the positive solution of [9, Problem 18], stating that a free profinite product of absolute Galois groups is an absolute Galois group strengthens this motivation (see [10] for a description, the statement and the history of the result).

Pop established the following necessary conditions for a prosoluble subgroup to be the closed subgroup of a free product of profinite groups.

Theorem 1. Let H be a prosoluble subgroup of a free profinite product $G=\coprod _{i\in I} G_i$ . Then one of the following statements holds:

1. (i) There is a prime p such that $H \cap g G_i g^{-1}$ is pro-p for every $i \in I$ and $g \in G$ ;

2. (ii) All nontrivial intersections $H \cap g G_i g^{-1}$ for $i \in I$ and $g \in G$ are finite and conjugate in H;

3. (iii) A conjugate of H is a subgroup of $G_i$ for some $i \in I$ .

In the same paper, Pop asks whether or not the situation described in (ii) of Theorem 1 does occur. Guralnick and Haran in [3] give an affirmative answer to this question. Namely, they prove that profinite Frobenius groups $\widehat {\mathbb {Z}}_\pi \rtimes C$ , where $\pi $ is a set of primes and C is cyclic finite, can be realized as a subgroup of a free profinite product.

Our first objective is to show that, in fact, profinite Frobenius groups $\widehat {\mathbb {Z}}_\pi \rtimes C$ are the only possibility in Theorem 1(ii).

Theorem 1.1. Let $G=\coprod _{t\in T} G_t$ be a free profinite product of profinite groups over a profinite space T and H a prosoluble subgroup of G that is not contained in any conjugate $G_t^g$ of a free factor $G_t$ . If the intersections $H \cap G_t^g, t\in T, g\in G$ are not all pro-p, then $H\cong \widehat {\mathbb {Z}}_\pi \rtimes C$ is a profinite Frobenius group (i.e., C is cyclic, $|C|$ divides $p-1$ for any $p\in \pi $ and $[c,z]\neq 1$ for all $1\neq c\in C, 0\neq z\in \widehat {\mathbb {Z}}_\pi $ ).

This gives the complete description of prosoluble subgroups whose intersections with the conjugates of the free factors involve two different primes.

The case when all such intersections are pro-p was studied by Ersoy and Herfort [2]. They showed [2, Theorem 1.4] that a free profinite product of pro-p groups contains a free prosoluble product of conjugates of these pro-p groups.

Ersoy and Herfort posed in the same paper the following questions.

Question 1.2. [2, Question 3.2]. Let $G = A \amalg B$ be the free profinite product of pro-p groups A and B. Is every prosoluble subgroup H of G isomorphic to a closed subgroup of the free prosoluble product $A \amalg ^s B$ ?

Question 1.3. [2, Problem 4.5]. Let A be a pro-p group and P a projective prosoluble group. Does the free profinite product $G := A \amalg P$ admit a prosoluble retract isomorphic to a free prosoluble product $A \amalg _s P$ ?

The second objective of the paper is to answer these questions. Namely, in Theorem 5.2, we answer positively Question 1.3. Theorem 5.3 answers positively Question 1.2 if H is second countable, in particular, if H is finitely generated. This practically completes a description of prosoluble subgroups of a free profinite product. Indeed, the answers to the above questions show that for a subgroup H of a free profinite product $\coprod _{x\in X} G_x$ , either H is contained in a conjugate of a free factor or it is a Frobenius group or it is a subgroup of a free prosoluble product of pro-p groups $H\cap G_i^{g_i}$ with no restrictions.

Theorem 1.4. Let $G=\coprod _{x \in X} G_x$ be a free profinite product over a profinite space X. A finitely generated prosoluble group H is isomorphic to a subgroup of G if and only if one of the following holds:

1. (i) H is isomorphic to a subgroup of some free prosoluble product $H_s=\coprod _{i=1}^n{}^s\ H_i$ of finitely generated pro-p groups $H_i$ such that each $H_i$ is isomorphic to a subgroup of some $G_x$ ;

2. (ii) H is isomorphic to a profinite Frobenius group $\widehat {\mathbb {Z}}_\pi \rtimes C$ , with C a finite cyclic subgroup of $G_x$ for some $x\in X$ ;

3. (iii) H isomorphic to a subgroup of a free factor $G_x$ .

In fact, Theorem 1.4 is valid also for a second countable subgroup H (see Theorem 4.15), and the hypothesis of second countability is due to the open question posed by Haran [4, Problem 5.6 (a)]. In this paper, Haran introduced the notion of relatively projective profinite groups and proved that a subgroup H of a free profinite product $\coprod _{x \in X} G_x$ is projective relative to the family $\{H\cap G_x^g\mid x\in X, g\in G\}$ . Haran [4, Problem 5.6 (a)] asks whether a profinite group which is projective relative to the family of subgroups $\{G_t\mid t\in T\}$ embeds into a free profinite product of groups isomorphic to $G_t, t\in T$ . This question was solved in [6] for relatively projective pro-p groups, but for profinite and prosoluble relatively projective groups, it is still open.

The next theorem is concerned with prosoluble relatively projective groups.

Theorem 1.5. Let G be a prosoluble group and $\{ G_t\mid t\in T\}$ a continuous family of its subgroups closed for conjugation such that $G_t\neq G_{t'}$ if $t\neq t'\in T_0=\{ t\in T\mid G_t\neq 1\}$ . Then G is projective relative to $\{ G_t\mid t\in T\}$ if and only if one of the following holds:

1. (i) there is a prime p such that all $G_t, t\in T$ are pro-p and G is $\mathcal {S}$ -projective relative to $\{ G_t\mid t\in T\}$ , where $\mathcal {S}$ is the class of all finite soluble groups;

2. (ii) $G= \widehat {\mathbb {Z}}_\pi \rtimes C$ is a profinite Frobenius group (i.e., C is cyclic, $|C|$ divides $p-1$ for any $p\in \pi $ and $[c,z]\neq 1$ for all $1\neq c\in C, 0\neq z\in \widehat {\mathbb {Z}}_\pi $ ), $T_0$ is closed in T, $\{ G_t\mid t\in T_0\}=\{C^z\mid z\in \widehat {\mathbb {Z}}_\pi \}$ and $|T_0/G|=1$ .

3. (iii) $|T_0|=1$ and $G_t=G$ for the unique $t\in T_0$ .

Corollary 1.6. If G is finitely generated with minimal number of generators $d=d(G)$ , then in (i) of Theorem 1.5, in addition one has $T_0=\{ t\in T\mid G_t\neq 1\}$ is closed in T and $|T_0/G|\leq d$ . Moreover, $\sum _t d(G_t)\leq d(G)$ , where t runs through representatives of G-orbits in $T_0$ .

From Theorem 1.5, we also deduce the corollary that generalizes [2, Theorems 1.4 and 4.4].

Corollary 1.7. Let $G=\coprod _{x\in X} G_x \amalg P$ and $G=\coprod _{x\in X}{}^s G_x \amalg ^s P$ be free profinite and free prosoluble products of pro-p groups $G_x$ over a profinite space X and a prosoluble projective group P. Then the natural epimorphism $\alpha :G\longrightarrow G_s$ that sends identically the factors of the free profinite product to the factors of the free prosoluble product admits a retraction $r:G_s\longrightarrow G$ such that $r(G_x)$ is conjugate to $G_x$ for every $x\in X$ .

Unless said otherwise, all groups in this paper are profinite, subgroups are closed, homomorphisms are continuous and groups are generated in the topological sense.

2 Case (ii) of Pop’s theorem

We begin this section with an easy generalization of [6, Theorem 3.2].

Lemma 2.1. Let F be a non-abelian free pro-p group and G a finite soluble subgroup of $Aut(F)$ of order coprime to p. If the subgroup of fixed elements $Fix_G(F)\neq F$ , then $Fix_G(F)$ is infinitely generated.

Proof. We use induction on $|G|$ .

Suppose G is of prime order and suppose on the contrary $Fix_G(F)$ is finitely generated. Let  $f\in F\setminus Fix_G(F)$ be an element of F and $H=\langle Fix_G(F), g(f)\mid g\in G\rangle $ . Then H is a finitely generated G-invariant subgroup of F, and so by [7, Theorem 3.2], $Fix_G(H)$ is infinitely generated. As $Fix_G(F)=Fix_G(H)$ , we arrive at a contradiction. This gives us the base of induction.

Suppose now that $|G|$ is not of prime order, and let A be a proper normal subgroup of G. Then either $Fix_A(F ) = F$ or, by the induction hypothesis, $Fix_A(F)$ is infinitely generated. In particular, $Fix_A(F)$ is non-abelian. Note that if $Fix_G(F) = Fix_A(F )$ , then $Fix_A(F )\neq F$ , so $Fix_G(F) = Fix_A(F )$ is infinitely generated. Otherwise, by the induction hypothesis again, $Fix_{G/A}(Fix_A(F))=Fix_G(F)$ is infinitely generated as required.

Lemma 2.2. Let G be a projective group and S a Sylow p-subgroup of G. If $rank(S)=r$ , then there exists an open subgroup U of G containing S such that for any open subgroup V of U, the maximal pro-p quotient $V_p$ of V is free pro-p of rank $\geq r$ . Moreover, if $r=1$ , then $V=R_p(V)\rtimes \mathbb {Z}_p$ with $R_p(V)$ coprime to p.

Proof. Since S is the intersection of all open subgroups U containing S, we can view this as the inverse limit $S=\varprojlim\limits _{S\leq U\leq _o G} U$ . It follows that $S=S_p=\varprojlim\limits _{S\leq U\leq _o G} U_p$ is the inverse limit of the maximal pro-p quotients $U_p$ of U (cf. [13, Satz 3], for example). Note that $U_p$ are free pro-p (cf. [15, Proposition 7.7.7]), and S is also a p-Sylow subgroup of U. Since an epimorphism maps p-Sylow subgroups onto p-Sylow subgroups, we deduce that $rank(U_p)\leq rank(S)$ . Since $rank(S)=r$ , $U_p$ must be free pro-p of rank r for some U.

Let V be an open subgroup of U. Then the image of V in $U_p$ is free pro-p of rank $\geq rank(U_p)=r$ (see [15, Theorem 3.6.2 (b)]). But this image is a quotient of $V_p$ . So $V_p$ is free pro-p of rank $\geq r$ .

For the last part of the statement, assume S is infinite cyclic. Then $V_p\cong \mathbb {Z}_p$ (by the first paragraph), and the natural epimorphism $V\longrightarrow V_p$ to its maximal pro-p quotient splits; so its kernel $R_p(V)$ intersects $S\cap V$ trivially. Hence, $R_p(V)$ is coprime to p.

The proof of the following lemma was obtained in communication with Pavel Shumyatsky.

Lemma 2.3. Let $G=K\rtimes C$ be a semidirect product of a profinite group K and a finite group C such that $C_{K}(C)=1$ . Let $N\leq K$ be normal in G and L be the preimage of $C_{K/N}(CN/N)$ in K. Suppose that every L-conjugate of C is N-conjugate to C. Then $C_{K/N}(CN/N)=1$ .

Proof. By hypothesis, for any $l\in L$ , there exists an element of $n\in N$ such that $C^l=C^n$ . This means that $N_L(C)N=L$ . As $N_L(C)\leq N_K(C)=C_K(C)=1$ , we deduce that $N=L$ . This means precisely that $C_{K/N}(CN/N) = 1$ .

Lemma 2.4. Let $p\neq l$ be primes, and let $C=A\rtimes C_l$ be a semidirect product of elementary abelian p-group A and a cyclic group of order l. Let $G=K\rtimes C$ be a semidirect product of a projective profinite group K and C such that $C_K(c)=1$ for every $1\neq c\in C$ . Suppose that for any C-invariant subgroup L of K, every maximal finite subgroup of $L\rtimes C$ is L-conjugate to C. Then G is soluble.

Proof. We shall first show that a q-Sylow subgroup of K is cyclic for any $q\neq p$ . Suppose not. Then by Lemma 2.2, there exists an open subgroup U of K such that the maximal pro-q quotient of the core of U in G is non-cyclic free pro-q, and so replacing U by its core, we may assume that U is normal. Let $R_q(U)$ be the kernel of the natural epimorphism of U onto $U_q$ . Since $R_q(U)$ is characteristic, it is normal in G and $UC/R_q(U)=U_q\rtimes C$ . As $U_q$ is non-abelian free pro-q and $C_{U_q}(A)\neq U_q$ by hypothesis, we deduce from Lemma 2.1 that $C_{U_q}(A)$ is infinitely generated; in particular, $C_{U_q}(A)$ is not trivial.

Let L be the preimage of $C_{U_q}(A)$ in G. By hypothesis, all maximal finite subgroups of $L\rtimes C$ are conjugate to C, and all maximal finite subgroups of $R_q(U)\rtimes C$ are conjugate to C. Then all maximal finite p-subgroups of $L\rtimes A$ are conjugate to A, and all maximal finite p-subgroups of $R_q(U)\rtimes A$ are conjugate to A. As $R_q(U)A$ is normal in $LA$ , it follows that any L-conjugate of A is in $R_q(U)A$ , and so is $R_q(U)$ -conjugate to A. Thus, $L\rtimes A$ and $R_q(U)\rtimes A$ satisfy the premises of Lemma 2.3, and applying it, one deduces that $C_{U_q}(A)$ is trivial, a contradiction. Thus, q-Sylow subgroups of K are cyclic.

Now by Lemma 2.2, there exists an open normal subgroup V of G contained in K such that $V\cong R_q(V)\rtimes \mathbb {Z}_q$ and $R_q(V)$ is coprime to q – in particular, putting $q=l$ , we have that $R_l(V)$ is coprime to l. Hence, $R_l(V)\rtimes C_l$ is a profinite Frobenius group ([15, Theorem 4.6.9 (d)]), and so $R_l(V)$ is nilpotent (see [15, Corollary 4.6.10]). Since $R_l(V)$ is projective, it is cyclic – in particular, a p-Sylow subgroup of $R_l(V)$ is cyclic. As a torsion free virtually cyclic pro-p group is cyclic, we deduce that all Sylow subgroups of K are cyclic. Then K is metacyclic by [15, Exercise 7.7.8 (a)], and therefore, G is soluble.

We finish this section proving Theorem 1.1 for finitely many factors only and postpone the proof for the general free profinite product until the next section.

Theorem 2.5. Let $G=\coprod _{i=1}^n G_i$ be a free profinite product of profinite groups and H a prosoluble subgroup of G that is not a subgroup of a free factor up to conjugation. If the intersections $H \cap G_i^g$ , $g \in G$ , $i=1, \ldots , n$ are not all pro-p for some prime p, then $H\cong \widehat {\mathbb {Z}}_\pi \rtimes C$ is a profinite Frobenius group (i.e., C is cyclic, $|C|$ divides $p-1$ for every $p\in \pi $ and $[c,z]\neq 1$ for all $1\neq c\in C, 0\neq z\in \widehat {\mathbb {Z}}_\pi $ ).

Proof. By Theorem 1, if $H\cap G_i^g$ for $g\in G$ and $ i=1, \ldots , n$ are not all pro-p and none of them is H, then they are finite and conjugate in H. Since for $i\neq j$ no nontrivial subgroup of $G_i$ can be conjugate in G to a subgroup of $G_j$ (as can be easily seen looking at the quotient $\prod _{i=1}^n G_i$ of G), it follows that $H\cap G_i^g$ is nontrivial for one i only, say for $i=1$ . Then the intersection of the normal closure of $G_2, \ldots , G_n$ in G with H is normal in H and torsion free (as every finite subgroups is conjugate to a subgroup of a free factor [16, Corollary 7.1.3]). Thus, $H=K\rtimes C$ with $C= H\cap G_{1}^g$ for some $g\in G$ .

If H is soluble, then by [16, Corollary 7.1.8], $H\cong \widehat {\mathbb {Z}}_\pi \rtimes C_n$ is a Frobenius profinite group, where $C_n$ is cyclic of order n dividing $p-1$ for every $p\in \pi $ .

Thus, to finish the proof, we just need to prove that H is soluble. Choose a minimal normal elementary abelian p-subgroup A in C and a cyclic subgroup $C_l$ of prime order l coprime to p. Replacing C by $A\rtimes C_l$ , if necessary, we may assume w.l.o.g. that $C=A\rtimes C_l$ . By [16, Corollary 7.1.5 (a)], $C_K(c)=1$ for every $1\neq c\in C$ . By Theorem 1, all maximal finite subgroups of H and of any subgroup containing C are conjugate to C. Then by Lemma 2.4, H is soluble.

3 Subgroups of free pro- $\mathcal {C}$ products

In this section, $\mathcal {C}$ will denote a class of finite groups closed for subgroups, quotients and extensions. We shall use it, however, mainly for the class of all finite groups or the class $\mathcal {S}$ of all finite soluble groups.

Definition 3.1. A system E of subgroups of a pro- $\mathcal {C}$ group G is a map $\Sigma : T\longrightarrow \textit{Subgrps}(G)$ , where T is a profinite space and $\textit{Subgrps}(G)$ denotes the set of all closed subgroups of G. We shall find it convenient to write $G_t$ for $\Sigma (t)$ . We shall say that a system E is continuous if, for any open subgroup U of G, the set $T(U) = \{t\in T\mid G_t\leq U\}$ is open in T. A continuous system $\{H_s\mid s\in S\}$ over a profinite space S such that for every $s\in S$ , there exists a unique $t\in T$ with $H_s\leq G_t$ will be called a continuous subsystem of the continuous system $\{G_t\mid t\in T\}$ .

Lemma 3.2 [16, Lemma 5.2.1].

Let $\{G_t\mid t\in T\}$ be a system of subgroups of a pro- $\mathcal {C}$ group G. The following statements are equivalent:

1. (i) $\{G_t\mid t\in T\}$ is continuous.

2. (ii) $\{(g,t)\in G\times T\mid g\in G_t\}$ is closed in $G \times T$ .

3. (iii) $E = \bigcup _{t\in T} G_t$ is closed in G.

Definition 3.3. A pro- $\mathcal {C}$ group G is said to be a free (internal) pro- $\mathcal {C}$ product of a continuous system $\{G_t\mid t\in T\}$ of subgroups if

(i) $G_t \cap G_s = 1$ for all $t\neq s$ in T;

(ii) for every pro- $\mathcal {C}$ group H, every continuous map $\beta : \bigcup _{t\in T} G_t\longrightarrow H$ , whose restriction $\beta _t:=\beta _{|G_t}$ is a homomorphism for every $t\in T$ , extends uniquely to a homomorphism $\psi : G \longrightarrow H$ .

We shall use $\coprod _{t\in T} G_t$ to denote the free profinite product, $\coprod _{t\in T}{}^{\mathcal {C}} G_t$ to denote the free pro- $\mathcal {C}$ product and $\coprod _{t\in T}{}^{s} G_t$ to denote the free prosoluble product. The reader can find in [16] the proof of the existence and uniqueness of a free pro- $\mathcal {C}$ product as well as basic properties of it.

We are ready to prove Theorem 1.1.

Proof of Theorem 1.1. Suppose H is not Frobenius and for no prime p the intersections ${H \cap G_t^g, t\in T, g\in G}$ are all pro-p. By [12, (1.7)] (or by [5, Lemma 4.5.4]), for any clopen partition $T=V_1\cup V_2\cup \cdots \cup V_n$ , we have $G=G(V_1)\amalg G(V_2) \amalg \cdots \amalg G(V_n)$ , where $G(V_i)=\langle G_t\mid t\in V_i\rangle $ , and denoting by $\Lambda $ all such clopen partitions of T, we have $G_t=\bigcap _{\lambda \in \Lambda } G(V_\lambda )$ , where $V_\lambda $ is an element of the partition $\lambda $ containing t. By Theorem 2.5, H is contained in a conjugate $G(V_\lambda )^{g_\lambda }$ for each $\lambda \in \Lambda $ , and this conjugate is unique since $G(V_i)\cap G(V_j)^k=1$ whenever $i\neq j$ or $k\not \in G_j$ by [15, Theorem 9.1.12]. Then H must be contained in the intersection $\bigcap _{\lambda \in \Lambda }G(V_\lambda )^{g_\lambda }$ of such conjugates, and so $H\leq G_t^{g}$ for some $t\in T, g\in G$ .

Remark 3.4. One can use also [16, Theorem 5.3.4] instead of [12] and a projective limit argument from [5, Remark 4.5.9] to perform the proof of Theorem 1.1.

In the rest of the section, we shall use only finite sets of indices of a free pro- $\mathcal {C}$ product, thereby avoiding general notions, as discussed in Definition 3.1 and Lemma 3.2.

Lemma 3.5. Let $G= G_1 * G _2$ be a free (abstract) product of nontrivial finite groups such that $|G_1|+|G_2|>4$ . Then for some $g\in G$ , there exists a finite index subgroup U of G that decomposes as a free product $G_1 * G^g_2 * M * F$ with F a nontrivial free group.

Proof. Choose $1\neq g_i\in G_i$ , and put $g=g_1g_2$ . Then the subgroup H generated by $G_1$ and $G_2^g$ is a free product $H=G_1* G_2^g$ , and a proper subgroup of G as no element from $G_2\setminus \{1\}$ is a product of elements of $G_1$ and $G_2^g$ . Denote by $C_G$ the Cartesian subgroup of G: the kernel of the epimorphism onto the direct product of factors. Note that $C_G$ is free of finite index in G, and since G is not dihedral, (by hypothesis) $C_G$ is free non-abelian. As $C_H$ is free of the same rank as $C_G$ , $C_H$ is of infinite index in $C_G$ by the Schreier formula, and therefore, $[G:H]=\infty $ . Hence, by the analog of M. Hall theorem for free products (see [1, Theorem 1.1]), H is a free factor of a subgroup of finite index U of G (i.e., $U=H* L$ ), and using the Kurosh subgroup theorem for L, we have a decomposition $L=*_{i=1}^2 *_{g_i} (L\cap G_i^{g_{i}}) * F_0$ , where $F_0$ is a free group. Here, $F_0$ can be trivial. If this is the case, apply the above argument to the decomposition $U=H*L$ to get a subgroup of finite index $U_1=H*L^u* L_1$ with $L_1=*_{i=1}^2 *_{g_i} (L_1\cap G_i^{g_{i}}) * F_1$ , where $F_1$ is free (possibly trivial). Putting $V_1=L^u*L_1$ , we get $U_1=H*V_1$ with $V_1$ having a nontrivial free normal subgroup $F_2$ of finite index. Setting $U_2$ to be the kernel of the epimorphism $U_1\longrightarrow V_1/F_2$ that sends H to the trivial group and $V_1$ naturally onto $V_1/F_2$ , we get the Kurosh decomposition $U_2=H*F_2* M$ with $F_2$ nontrivial. Thus, for certain subgroup of finite index U, we achieve a nontrivial decomposition $U=H*M* F$ , where F is nontrivial. This finishes the proof.

Lemma 3.6. Let $G= G_1 \amalg G _2$ be a free profinite product of nontrivial profinite groups such that $|G_1|+|G_2|>4$ . Then for some $g\in G$ , there exists an open subgroup U of G that decomposes as a free profinite product $G_1 \amalg G^g_2\amalg U\amalg F$ , where F is a nontrivial free profinite group.

Proof. Suppose first that $G_1,G_2$ are finite and so G is the profinite completion of the abstract free product $G^{abs}=G_1 *G_2$ . By Lemma 3.5, there exists $g\in G^{abs}$ and a finite index subgroup $U^{abs}$ of $G^{abs}$ such that $G_1 * G^g_2 * M^{abs} * F^{abs}$ with $F^{abs}$ a nontrivial free group (of finite rank). It follows that $U=\widehat U^{abs}=G_1\amalg G_2^g\amalg M \amalg \widehat F^{abs}$ is a free profinite product with $\widehat F^{abs}$ nontrivial free profinite, as required.

Suppose now $G_1$ , $G_2$ are arbitrary. Let $Q_i$ be a nontrivial finite quotient of $G_i$ and

$$ \begin{align*}f:G\longrightarrow Q=Q_1 \amalg Q_2\end{align*} $$

the epimorphism induced by the natural epimorphisms $f_i:G_i\longrightarrow Q_i$ . By the previous paragraph, there exists an open subgroup $U_Q$ such that $U_Q=Q_1\amalg Q_2^q\amalg M\amalg F_Q$ for some $q\in Q$ and a nontrivial free profinite group $F_Q$ . Put $U=f^{-1}(U_Q)$ . Then by a profinite version of the Kurosh subgroup theorem (cf. [15, Theorem 9.1.9]), $U=(G_1\amalg G_2^g)\amalg \left (\coprod _{i=1}^2{}\coprod (U\cap G_i^{g_{i}})\right ) \amalg F$ , where F is a free profinite group. As $ker(f)$ is generated by $ker(f_1),\ker (f_2)$ as a normal subgroup, the kernel of the natural projection $U\longrightarrow F$ contains $ker(f)$ , and so one has the following commutative diagram of epimorphisms:

Since $F_Q$ is nontrivial, so is F. The proof is finished.

To prove a prosoluble version of Lemma 3.6, we shall need the following simple lemma about finite soluble groups.

Lemma 3.7. Let $p, q$ be prime numbers. Let $H=C_p\times C_q$ be a cyclic group of order $pq$ if $p\neq q$ and $H=C_p$ if $p=q$ . Then there exists a prime r such that $pq| r-1$ and a semidirect product $R= \mathbb {F}_{r}\rtimes H$ of a field $\mathbb {F}_{r}$ of r elements with H such that R is generated by an element of order p and an element of order q.

Proof. Since an arithmetic progression $\{1+pqk | k=1, 2 \ldots \}$ contains infinitely many primes by Dirichlet’s theorem, there exists a prime r such that $pq|r-1$ . Consider a semidirect product $R= \mathbb {F}_{r}\rtimes H$ with the faithful action of H by multiplication by units of $\mathbb {F}_r$ . Then for any nonzero element $m \in F_{r}$ we get $R=\langle C_p , C_q^m\rangle $ . Indeed, the generators $c_1$ of $C_p$ and $c_2$ of $C_q^m$ have order p and q, respectively, and their commutator $[c_1,c_2]$ has order r. Therefore, $\langle C_p , C_q^m\rangle $ contains elements of order $p, q$ and r and so must coincide with G.

Lemma 3.8. Let $G= G_1 \amalg ^s G _2$ be a free prosoluble product of nontrivial prosoluble groups such that $|G_1|+|G_2|>4$ . Then for some $g\in G$ , there exists an open subgroup U of G that decomposes as a free prosoluble product $G_1 \amalg ^s G^g_2\amalg ^s U_1\amalg ^s U_2 \amalg ^s F_s$ , where $F_s$ is a non-abelian free prosoluble group and $U_i=\coprod _{ g_i}{}^s (U\cap G_i^{g_{i}})$ . Moreover, if $G_i$ is infinite, then $U_i$ is nontrivial.

Proof. Let $C_p$ and $C_q$ be quotient groups of $G_1$ and $G_2$ of prime orders p and q, respectively.

Let $R=\mathbb {F}_{r}\rtimes H$ be a finite group from Lemma 3.7 generated by elements $a,b$ of order p and q, and assume w.l.o.g that $a\in H$ . Let $f_1:G_1 \longrightarrow \langle a\rangle $ and $f_2:G_2\longrightarrow \langle b \rangle $ be the natural epimorphisms with kernels $N_1$ and $N_2$ . Consider an epimorphism $f : G \longrightarrow R$ extending $f_1$ and $f_2$ . Then $ U= f^{-1}(H)$ is a proper open subgroup of G, and so by a prosoluble version of the Kurosh subgroup theorem (cf. [15, Theorem 9.1.9]), $U=\coprod _{i=1}^2{}^s\coprod _{ D_i}{}^s (U\cap G_i^{g_{i}}) \amalg F_s$ , where $D_i=U\backslash G/G_i$ , $g_i$ runs through representatives of the double cosets $D_i$ and $F_s$ is a free prosoluble group of rank $1+[G:U]-\sum _{i=1}^2|U\backslash G/G_i|$ .

Observe that $G_1, G_2^g\leq U$ for some $g\in f^{-1}(c)$ , where c is an element conjugating b into H. Therefore, we can rewrite $U=G_1 \amalg ^s G_2^{g} \amalg ^s U_1\amalg ^s U_2 \amalg ^s F_s$ . Note that $|U\backslash G/G_1|=|H\backslash R/\langle a\rangle |$ , and taking into account that $Hra=Har^a$ , we can see that the action of $\langle a\rangle $ on $H\backslash R$ is isomorphic to the action of $\langle a\rangle $ on $\mathbb {F}_r$ , so $|H\backslash R/\langle a\rangle |=\frac {r-1}{p}+1\geq 2$ . Similarly, $|U\backslash G/G_2|=\frac {r-1}{q}+1\geq 2$ , and so if $G_i$ is infinite, the decomposition of $U_i$ into a free profinite product contains at least one nontrivial factor $U\cap G_i^{g_{i}}=N_i^{g_i}$ distinct from $G_1$ , $G_2^g$ .

We show that $F_s$ is nontrivial if $|H|>2$ . Indeed,

$$ \begin{align*}1+[G:U]-\sum_{i=1}^2|U\backslash G/G_i|=1+r-\frac{r-1}{p}-\frac{r-1}{q}-2)=(r-1)(1-\frac{1}{p}-\frac{1}{q}),\end{align*} $$

which is not $0$ unless H is of order $2$ .

If H in Lemma 3.7 has order $2$ , then $p=q=2$ . In this case, $U_1\amalg ^s U_2$ is nontrivial, since at least one $G_i$ has order $>2$ , and as $|U\backslash G/G_i|\geq 2$ , we have a factor $U\cap G_i^{g_i}\neq 1$ in $U_i$ .

To get $F_s$ nontrivial in this case, we put $G^{\prime }_1=G_1\amalg ^s G_2^g$ , $G^{\prime }_2=U_1\amalg ^s U_2$ and write U as ${U=G^{\prime }_1\amalg ^s G^{\prime }_2}$ . Then applying the argument above to this decomposition, we get an open subgroup $V\leq U$ such that $V=G^{\prime }_1\amalg ^s (G^{\prime }_2)^{g'}\amalg ^s V_1=G_1\amalg ^s G_2^g\amalg ^s (G^{\prime }_2)^{g'}\amalg ^s V_1$ . If $(G^{\prime }_2)^{g'}\amalg ^s V_1$ has a quotient of order $p\neq 2$ , then the above argument applied to V produces a needed open subgroup with $F_s$ nontrivial. Otherwise, the kernel K of the epimorphism $V\longrightarrow C_2$ to a group of order 2 that sends $G_1\amalg ^s G_2^g$ to the trivial group and $(G^{\prime }_2)^{g'}, V_1$ epimorphically to $C_2$ has the Kurosh decomposition $K=G_1\amalg ^s G_2^g\amalg ^s (K\cap (G^{\prime }_2)^{g'})\amalg ^s (K\cap V_1) \amalg ^s F_s$ with $F_s\cong \mathbb {Z}_2$ .

Finally, to get a non-abelian factor which is free, we can put $G^{\prime }_1=G_1\amalg ^s G_2^g$ and apply the argument of the preceding paragraph to $G^{\prime }_1 \amalg ^s F_s$ getting an open subgroup $W=G^{\prime }_1\amalg ^s (F_s)^{w}\amalg ^s F^{\prime }_s=G_1\amalg ^s G_2^{g}\amalg ^s F_s^{w}\amalg ^s F^{\prime }_s$ for some $w\in W$ with nontrivial free prosoluble group $F^{\prime }_s$ , so that one has a non-abelian free factor $F_s^w\amalg ^s F^{\prime }_s$ . This finishes the proof.

Lemma 3.9. Let $G=G_1\amalg ^s G_2$ be a free prosoluble product of nontrivial prosoluble groups such that $|G_1|+|G_2|>4$ . Then for some $g\in G$ , the group G contains $G_1\amalg ^s G_2^g \amalg ^s U$ such that $|U|=|G|$ .

Proof. By Lemma 3.8, G contains a free prosoluble product

$$ \begin{align*}G_1 \amalg^s G^g_2\amalg^s U_1\amalg^s U_2\amalg^s F_s,\end{align*} $$

where $U_i$ is isomorphic to a free product of open subgroups of $G_i$ , $i=1,2$ and a non-abelian free prosoluble group $F_s$ . Moreover, $U_1$ and $U_2$ are nontrivial if $G_1,G_2$ are infinite. Hence, if one of $G_i$ is infinite, say $G_1$ and w.l.o.g. $|G_1|\geq |G_2|$ , then $|U_1|=|G|=|U_1 \amalg U_2\amalg F_s|$ , and so the result is proved.

If $G_1,G_2$ are finite, the result is obvious since $F_s\neq 1$ .

4 Relatively projective groups

In this section, we shall follow the terminology [6] that concerns relatively projective groups. In particular, the actions of profinite groups on profinite spaces will be on the right. We also continue to use $\mathcal {S}$ for the class of all finite soluble groups and $\mathcal {C}$ for an arbitrary class of finite groups closed for subgroups, quotients and extensions.

Definition 4.1. A group pile (or just pile) is a pair $\mathbf {G} = (G,T)$ consisting of a profinite group G, a profinite space T and a continuous action of G on T. The stabilizer of $t\in T$ will be denoted by $G_t$ .

Definition 4.2. A morphism of group piles $\alpha : (G,T) \longrightarrow (G',T')$ consists of a group homomorphism $\alpha : G \longrightarrow G'$ and a continuous map $\alpha : T \longrightarrow T'$ such that $\alpha (tg) = \alpha (t){\alpha (g)}$ for all $t \in T$ and $g \in G$ . The above morphism is an epimorphism if $\alpha (G) = G'$ , $\alpha (T) = T'$ and for every $t' \in T'$ , there is $t \in T$ such that $\alpha (t) = t'$ and $\alpha (G_t) = {G'}_{t'}$ . It is rigid if $\alpha $ maps $G_t$ isomorphically onto ${G'}_{\alpha (t)}$ , for every $t \in T$ , and the induced map of the orbit spaces $T/G \longrightarrow T'/G'$ is a homeomorphism.

Definition 4.3. An embedding problem for a pile $\mathbf {G}$ is a pair $(\varphi ,\alpha )$ of morphisms of group piles, where $\varphi : \mathbf {G} \to \mathbf {A}$ and $\alpha : \mathbf {B} \to \mathbf {A}$ , such that $\alpha $ is an epimorphism.

(1)

It is a $\mathcal {C}$ -embedding problem if $G,B,A$ are pro- $\mathcal {C}$ . It is rigid if $\alpha $ is rigid. A solution to the embedding problem is a morphism $\gamma : \mathbf {G} \to \mathbf {B}$ such that $\alpha \circ \gamma = \varphi $ .

Definition 4.4. A pile $\mathbf {G}=(G,T)$ is $\mathcal {C}$ -projective if every finite rigid $\mathcal {C}$ -embedding problem for $\mathbf {G}$ has a solution. In this case, we say that G is $\mathcal {C}$ -projective relative to the family of point stabilizers $\{G_t\mid t\in T\}$ .

Remark 4.5. Definition 4.4 is slightly more restrictive than the definition of relative projectivity from [6, Section 3]; it becomes equivalent if one adds the condition $G_t\neq G_{t'}$ whenever $t\neq t'\in T_0=\{t\mid G_t\neq 1\}$ (see [6, Remark 5.8]). In Definition 4.4, we assume that G acts on a profinite space and the family is the family of point stabilizers. However, the notion does not depend on the choice of the space (as long as the family of point stabilizers is the same). Indeed, if $(G,T)$ is $\mathcal {C}$ -projective, then $G_{t_1}\cap G_{t_2}=1$ for $t_1\neq t_2\in T$ and any pile $(G,T')$ with the same property such that $\{G_t\mid t\in T\}\cup \{1\}=\{G_t\mid t\in T'\}\cup \{1\}$ is $\mathcal {C}$ -projective (see [6, Proposition 5.7]).

Remark 4.6. If $\mathcal {C}_2\subset \mathcal {C}_1$ are two classes of finite groups closed for subgroups, quotients and extensions, then a $\mathcal {C}_1$ -projective pro- $\mathcal {C}_2$ pile $(G,T)$ is $\mathcal {C}_2$ -projective, since every finite $\mathcal {C}_2$ -embedding problem is a $\mathcal {C}_1$ -embedding problem. In particular, if G is pro- $\mathcal {C}$ and $(G,T)$ is projective, then $(G,T)$ is $\mathcal {C}$ -projective for any class of finite groups closed for subgroups, quotients and extensions.

Proposition 4.7. Let $\mathbf {G}=(G,T)$ be a $\mathcal {C}$ -projective pile and $T_0=\{t\in T\mid G_t\neq 1\}$ . If G is finitely generated and $G_t, t\in T$ are all pro-p, then $T_0$ is closed in T and $|T_0/G|\leq d$ , the minimal number of generators of G. Moreover, $\sum _t d(G_t)\leq d(G)$ , where t runs through representatives of G-orbits in $T_0$ .

Proof. Suppose, on the contrary, $\sum _t d(G_t)> d(G)$ . Then there exists an epimorphism $\varphi :G\longrightarrow A$ to a finite group A such that there are nontrivial non-conjugate subgroups $G_{t_1}, \ldots , G_{t_{n}}$ with the images $A_1=f(G_{t_1}), \ldots , A_{n}=f(G_{t_{n}})$ – none of them are contained in a conjugate of the other and with $\varphi (\bigcup _{t\in T} G_t)=\bigcup _{i=1}^n A_i$ such that $\sum _i d(A_i)> d(G)$ . Define $X=\bigcup _{i=1}^n (A_{i}\backslash A)$ so that $\mathbf {A}=(A,X)$ is a finite pile. Let $f:F_{\mathcal {C}}\longrightarrow A$ be an epimorphism from a free pro- $\mathcal {C}$ group $F_{\mathcal {C}}$ . Form a free pro- $\mathcal {C}$ product $B=\coprod _{i=1}^n{}^{\mathcal {C}} A_{i}\amalg ^{\mathcal {C}} F_{\mathcal {C}}$ . Put $Y=\bigcup _{i=1}^n (A_{i}\backslash B)$ . Then $ B$ acts on Y on the right, and $\mathbf {B}=(B,Y)$ is a pile. Let $\alpha :\mathbf {B}\longrightarrow \mathbf {A}$ be a rigid morphism of piles defined by sending $A_i$ to their copies in A. Since $(G,T)$ is $\mathcal {C}$ -projective, the embedding problem

(2)

admits a solution $\gamma :(G,T)\longrightarrow (B,Y)$ (see [6, Proposition 5.4]). Note that $\gamma (G_{t_i})$ is conjugate to $A_i$ . So the natural epimorphism $B\longrightarrow \prod _{i=1}^n A_i$ restricts surjectively on $\gamma (G)$ . As $A_i$ are nontrivial finite p-groups, the minimal number of generators of $\prod _{i=1}^n A_i$ is greater than d, a contradiction. Thus, $T_0/G$ is finite and so $T_0$ is closed in T. This finishes the proof.

If $G_t$ are not pro-p, then $|T/G|$ can be greater than the minimal number of generators of G (see [15, Theorems 9.5.3 and 9.5.4]), but we do not know whether it is always finite.

We shall frequently use the following lemma on relative projectivity of subgroups of free pro- $\mathcal {C}$ products from [6]. The statement is slightly different, but the proof is the same.

Lemma 4.8 [6, Lemma 5.12].

Let $G=(\coprod _{x\in X}{}^{\mathcal {C}} G_x) \coprod {}^{\mathcal {C}} P$ be a free pro- $\mathcal {C}$ product of a continuous family of its subgroups $G_x$ and a projective pro- $\mathcal {C}$ group P. Let H be a closed subgroup of G. Put

$$ \begin{align*}\mathcal{H} = \{H \cap G_x^g \mid x \in X,\ g \in G\}.\end{align*} $$

Then there is a profinite H-space T such that

1. (a) $\mathcal {H} = \{H_t \mid t \in T\}$ ;

2. (b) $H_t\cap H_{t'}=1$ for $t\neq t'$ ;

3. (c) $\mathbf {H} = (H,T)$ is a $\mathcal {C}$ -projective pile.

Moreover, T is the disjoint union $T=\bigcup _{x\in X} G_x\backslash G$ on which H acts by multiplication on the right viewed as a quotient space of $X\times G$ via the map $X\times G\longrightarrow T$ given by $(x,g)\rightarrow (x, G_xg)$ .

Proof. This is the construction of T in [6, Lemma 5.12].

Remark 4.9. Suppose H in Lemma 4.8 is second countable. Put $T_0=\{t\in T\mid G_t\neq 1\}$ . Then by [16, Proposition 5.4.2], there exists an epimorphism $\mu : \overline T_0\longrightarrow \widetilde T$ of H-spaces such that

(a) $\widetilde T$ is second countable;

(b) $\mu _{|T_0}$ is an injection;

(c) $H_t=H_{\mu (t)}$ for every $t\in \overline T_0$ ;

(d) $\mu $ is a homeomorphism on the H-orbits of $\overline T_0$ .

It follows then that $(H,\widetilde T)$ is a pile and $\mathcal {H}\cup \{1\} = \{H_t \mid t \in \widetilde T\}\cup \{1\}$ . Then $(H,\widetilde T)$ is $\mathcal {C}$ -projective by Remark 4.5.

In the next proposition and in the rest of the section, we shall use the external definition of a free pro- $\mathcal {C}$ product over a profinite space as in [16, Section 5.1], which is equivalent to Definition 3.3 in the sense that given a continuous system of pro- $\mathcal {C}$ groups $\{G_x\mid x\in X\}$ over a profinite space X, there exists a pro- $\mathcal {C}$ group G containing $G_x$ as subgroups such that $G=\coprod _{x\in X}{}^{\mathcal {C}}\ G_x$ is a free pro- $\mathcal {C}$ product in the sense of Definition 3.3. The external definition is quite involved, based on the notion of profinite sheaves that are not used in the paper. Hence, we shall use the notation of a free pro- $\mathcal {C}$ product as before even if we define a group as an external free pro- $\mathcal {C}$ product.

Proposition 4.10. Let $(H,T)$ be a $\mathcal {C}$ -projective pile such that there exists a continuous section $\sigma :T/H\longrightarrow T$ . Then H embeds into a free pro- $\mathcal {C}$ -product $G=\coprod _{s\in \Sigma }{}^{\mathcal {C}}\ G_s\coprod {}^{\mathcal {C}} P$ , where $\Sigma =Im(\sigma )$ , $G_s$ is an isomorphic copy of $H_s$ and P is a projective pro- $\mathcal {C}$ group admitting a homomorphism $f:P\longrightarrow H$ such that $H=\langle f(P), H_s\mid s\in \Sigma \rangle $ . Moreover, $\{H_t\mid t\in T\}=\{G_s^g\cap H\mid g\in G, s\in \Sigma \}$ .

Proof. Let $\alpha :G\longrightarrow H$ be an epimorphism that extends f and sends $G_s$ onto their isomorphic copies in H. Put $S=\bigcup _{s\in \Sigma } G_s\backslash G$ . We view S as the quotient space of the profinite space $\Sigma \times G$ via the map $\pi \colon \Sigma \times G \to S$ given by $(s,g) \mapsto (s, G_s g)$ (cf. Lemma 4.8). By [16, Proposition 5.2.3], this is a profinite space.

By [6, Construction 4.3], $(G,S)$ is a pile, and so extending $\alpha $ to $S\rightarrow T$ by $\alpha (G_sh)=s\alpha (h)$ , one gets a rigid epimorphism $\alpha :(G,S)\longrightarrow (H,T)$ . Since $(H,T)$ is $\mathcal {C}$ -projective, by [6, Proposition 5.4], the embedding problem

(3)

has a solution which is clearly an embedding.

Corollary 4.11. Let $G = \coprod _{x\in X} G_x$ be a free profinite product of pro- $\mathcal {C}$ groups over a profinite space X. Then every second countable pro- $\mathcal {C}$ subgroup H of G is isomorphic to a closed subgroup of the free pro- $\mathcal {C}$ product

$$ \begin{align*}\coprod_{\sigma\in \Sigma}{}^{\mathcal{C}}\ H_\sigma \amalg{}^{\mathcal{C}} F_{\mathcal{C}} ,\end{align*} $$

where each nontrivial $ H_\sigma $ equals $H\cap G_x^{g}$ for some $g\in G$ and $F_{\mathcal {C}}$ is a second countable free pro- $\mathcal {C}$ group.

Proof. Let H be a second countable pro- $\mathcal {C}$ subgroup of G. By Remark 4.9, there exists a second countable profinite space T on which H acts continuously such that $(H,T)$ is a projective pile and

$$ \begin{align*}\{H_t\mid t\in T\}\cup \{1\}= \{H \cap G_x^g \mid x \in X,\ g \in G\}\cup \{1\}.\end{align*} $$

So there exists a continuous section $\sigma : T/H\longrightarrow T$ (cf. [15, Lemma 5.6.7]). By Remark 4.6, $(H,T)$ is $\mathcal {C}$ -projective. Then from Proposition 4.10, we get an embedding of H into a free pro- $\mathcal {C}$ product

$$ \begin{align*}\coprod_{\sigma\in \Sigma}{}^{\mathcal{C}} H_\sigma\amalg^{\mathcal{C}} F_{\mathcal{C}},\end{align*} $$

where $\Sigma =Im(\sigma )$ and $F_{\mathcal {C}}$ is second countable free pro- $\mathcal {C}$ .

Theorem 4.12. Let $\mathbf {G}=(G,T)$ be a projective pile with G prosoluble. Then one of the following holds:

1. (i) There is a prime p such that all $G_t, t\in T$ are pro-p and $(G,T)$ is $\mathcal {S}$ -projective.

2. (ii) $G= \widehat {\mathbb {Z}}_\pi \rtimes C$ is a profinite Frobenius groups (i.e., C is cyclic, $|C|$ divides $p-1$ for any $p\in \pi $ and $[c,z]\neq 1$ for all $1\neq c\in C, 0\neq z\in \widehat {\mathbb {Z}}_\pi $ ). Moreover, $T_0=\{ t\in T\mid G_t\neq 1\}$ is closed in T, $(G,T_0)$ is projective and $|T_0/G|=1$ . Besides, $\{ G_t\mid t\in T_0\}=\{C^z\mid z\in \widehat {\mathbb {Z}}_\pi \}$ .

3. (iii) $|T_0|=1$ and $G_t=G$ for the unique $t\in T_0$ .

Proof. Suppose (iii) does not hold. If G is a profinite Frobenius group, then $G=K\rtimes C$ , where C is finite (see [15, Theorem 4.6.9]) and K and C are coprime. By [15, Lemma 4.6.5], $G=\varprojlim G/V $ , where $G/V=(K/V)\rtimes (HV/V)$ and V runs through the collection of all open normal subgroups of G contained in K. Put $G_V=G/V$ and let $\varphi _V:G\longrightarrow G/V$ be the natural epimorphism. Let $A_1, \ldots A_n$ be the set of representatives of the conjugacy classes of the set $\{\varphi (G_t), t\in T\}$ of images of $G_t$ in $G_V$ . Set $X_V=\bigcup _{i=1}^n A_i\backslash G_V$ . Then ${\mathbf G_V}=(G_V,X_V)$ is a finite pile; moreover, as (iii) does not hold, we can assume w.l.o.g. that $|X_V|>1$ . Let $f:F\longrightarrow G_V$ be an epimorphism of a free profinite group of finite rank to $G_V$ . Form a free profinite product $B=\coprod _{i=1}^n A_{i}\amalg F$ . Set $Y=\bigcup _{i=1}^n (A_{i}\backslash B)$ . Then B acts on Y on the right, and $\mathbf {B}=(B,Y)$ is a pile. Thus, one has a rigid embedding problem

(4)

where $\alpha $ extends f and the natural isomorphisms of $A_i$ to their copies in $G/V$ . Since $\mathbf {G}$ is projective, there exists a solution $\gamma $ of (4). Then $\gamma (G)$ is a Frobenius group (as $ker(\gamma )\leq K$ and so one can apply [15, Theorem 4.6.8] to $\gamma (G)$ ). By [16, Corollary 7.1.8 (c2)] (a profinite Frobenius group can only appear in (a) and (c2) of this corollary, but (a) cannot hold by assumption on $G_V$ ), the group $\gamma (G)$ is a semidirect product of a cyclic projective group and cyclic finite group (i.e., $\gamma (G)=\gamma (K)\rtimes \gamma (C)$ with $\gamma (K)$ projective cyclic and $\gamma (C)$ finite cyclic). As it happens for every V, we deduce that $K\cong \widehat {\mathbb {Z}}_\pi $ and C is finite cyclic. As $G_t\cap G_{t'}=1$ for $t\neq t'$ (see [6, Proposition 5.7]), the normalizer of any subgroup $H_t\leq G_t$ is contained in $G_t$ . Hence, $K\cap G_t=1$ for every $t\in T$ , and since we assumed that $G\neq G_t$ , we deduce that $G_t$ are finite cyclic.

By the profinite version of Schur-Zassenhaus theorem ([15, Theorem 2.3.15]), all complements of $\widehat {\mathbb {Z}}_\pi $ in G are conjugate so that $G_t\cong C$ for each $t\in T_0$ and $|T_0/G|=1$ . It follows that $\{ G_t\mid t\in T_0\}=\{C^z\mid z\in \widehat {\mathbb {Z}}_\pi \}$ . Thus, (ii) holds in this case.

Suppose now that G is not Frobenius. We show that for some prime p, $G_t$ is pro-p for all $t\in T$ . Suppose, on the contrary, $G_t$ are not all pro-p for any prime p. Then there exists an epimorphism $\varphi :G\longrightarrow A$ to a finite group A such that A is non-cyclic and not Frobenius (see [15, Theorem 4.6.9 (a), (c)]), and the images of $\varphi (G_{t}), t\in T$ in A are not all finite p-groups for any prime p. As before, let $A_1, \ldots A_n$ be the set of representatives of the conjugacy classes of the set $\{\varphi (G_t), t\in T\}$ of images of $G_t$ in A. Set $X=\bigcup _{i=1}^n A_i\backslash A$ . Then $\mathbf {A}=(A,X)$ is a finite pile; moreover, as (iii) does not hold, we can assume w.l.o.g. that $|X|>1$ . Let $f:F\longrightarrow A$ be an epimorphism of a free profinite group of finite rank to A. Then $\mathbf {A}=(A,X)$ is a pile. Form a free profinite product $B=\coprod _{i=1}^n A_{i}\amalg F$ . Set $Y=\bigcup _{i=1}^n (A_{i}\backslash B)$ . Then B acts on Y on the right, and $\mathbf {B}=(B,Y)$ is a pile. Thus, one has a rigid embedding problem

(5)

where $\alpha $ extends f and the natural isomorphisms of $A_i$ to their copies in A. Since $\mathbf {G}$ is projective, there exists a solution $\gamma $ of (5). But $\gamma (G)$ is Frobenius by Theorem 2.5, contradicting that $\alpha (\gamma (G))=\varphi (G)$ is not.

Thus, all $G_t$ s are pro-p. It remains to show that $(G,T)$ is $\mathcal {S}$ -projective. But this follows from Remark 4.6. The proof is finished.

Lemma 4.13. Let $G_s=\coprod _{i=1}^n{}^s K_i \amalg ^s P$ be a free prosoluble product of finite p-groups $K_i$ and a finitely generated prosoluble projective group P. Then $G_s$ is projective relative to $\{K_i^g\mid g\in G_s, i=1, \ldots , n\}$ .

Proof. We observe first that $G_s$ is a subgroup of $L_s=\coprod _{i=1}^n{}^s (K_i\times C_i)$ , where $C_i$ is a cyclic group of order $p^2$ . Indeed, the kernel U of the homomorphism $L_s\longrightarrow C_{p^2}$ that sends all $C_i$ isomorphically to $C_{p^2}$ and all $K_i$ to 1 has the Kurosh decomposition $U=\coprod _{i=1}^n K_i \amalg ^s F_s$ , where $F_s$ is a free prosoluble group of rank $1+(n-1)p^2-n=(p^2-1)n-p^2+1\geq 2$ unless $n=1$ (see [15, Theorem 9.1.9]). So, if $n>1$ , P embeds in $F_s$ (see [15, Lemma 7.6.3]), and we have an embedding of $G_s$ into $L_s$ .

If $n=1$ , then $K_1\amalg ^s P$ embeds in $K_1\amalg ^s K_1\amalg ^s P$ so that we can use the previous embedding.

Hence, by Lemma 4.8, we may assume w.l.o.g. that $G_s=\coprod _{i=1}^n{}^s K_i$ . By [2, Theorem 1.4], $G_s$ retracts to a free profinite product $\coprod _{i=1}^n K_i$ such that the images of $K_i$ are conjugate to the corresponding free factors, and so applying Lemma 4.8 again together with the fact that any maximal finite subgroup of $G_s$ is conjugate to one of $K_i$ s (cf. [16, Corollary 7.1.3]), we deduce the result.

Proof of Theorem 1.5 .

Suppose G is projective relative to $\{G_t, t\in T\}$ (i.e., there is a projective pile $(G,T)$ such that $\{G_t\mid t\in T\}$ is the family of all point stabilizers of T (see Remark 4.5)). Then, by Theorem 4.12, items (i) or (ii) or (iii) hold.

Converse. Suppose (i) or (ii) or (iii) hold. If (iii) holds, then G is clearly projective relative to $\{G_t\mid t\in T\}$ . If (ii) holds, then by [3, Corollary 5.2], G embeds into a free profinite product $G_0=C\amalg C$ . Put $T_C=C\backslash G$ . By Lemma 4.8, $(G,C\backslash G_0)$ is projective, and therefore, so is $(G,T_C)$ by Remark 4.9. Since $\{G_t\mid t\in T_C\}=\{C^z\mid z\in \widehat {\mathbb {Z}}_\pi \}=\{G_t\mid t\in T_0\}$ , $(G, T)$ is projective by Remark 4.5.

Suppose now (i) holds. To prove that $(G,T)$ is projective, we need to solve a finite rigid embedding problem

(6)

Choose representatives $x_1,x_2,\ldots x_n$ of the orbits of the action of A on X, put $A_i=A_{x_i}$ and choose $B_i=B_{y_i}$ for $y_i\in Y$ such that $\alpha (B_i)=A_i$ . Let $f:F\longrightarrow B$ be an epimorphism of a free profinite group F of finite rank to B. As A is soluble, $\alpha f$ factors through a free prosoluble group $F_s$ of the same finite rank as F (i.e., there are epimorphisms $\eta :F\longrightarrow F_s$ and $f_s:F_s\longrightarrow A$ such that $\alpha f=f_s\eta $ ). Form a free profinite product $B'=\coprod _{i=1}^n B_{i}\amalg F$ and a free prosoluble product $B_s=\coprod _{i=1}^n{}^s A_{i}\amalg ^s F_s$ . Put $Y'=\bigcup _{i=1}^n (B_{i}\backslash B')$ and $Y_s=\bigcup _{i=1}^n (B_{i}\backslash B_s)$ . Then $B', B_s$ act on $Y',Y_s$ on the right, and $\mathbf {B}'=(B',Y')$ , $\mathbf {B}_s=(B_s,Y_s)$ are piles. Thus, we have the following commutative diagram:

(7)

where $\pi :B'\longrightarrow B$ and $\alpha _s:B_s\longrightarrow B$ are the epimorphisms that send $B_i$ to their copies in B and $\pi _{|F}=\eta $ . Since $(G,T)$ is $\mathcal {S}$ -projective, the embedding problem $(\varphi , \alpha _s)$ admits a solution $\gamma _s:(G,T)\longrightarrow (B_s,Y_s)$ (see [6, Proposition 5.4]) so that $\alpha _s\gamma _s=\varphi $ . By Lemma 4.13, $(B_s,Y_s)$ is projective and so by [6, Proposition 5.4], again the following embedding problem

(8)

has a solution r, so that $\rho r=id$ . Then $\pi r\gamma _s$ is a solution of (6) because $\alpha \pi r\gamma _s=\alpha _s\rho r\gamma _s=\alpha _s\gamma _s=\varphi $ . This finishes the proof.

The converse implication from (i) is, in fact, a generalization of Lemma 4.13 that we shall state as follows:

Corollary 4.14. Let $(G,T)$ be a pile such that $G_t$ are pro-p and G is prosoluble. If $(G,T)$ is $\mathcal {S}$ -projective, then $(G,T)$ is projective.

Proof of Corollary 1.6 follows from Proposition 4.7 and Theorem 1.5.

Proof of Corollary 1.7 . By [6, Example 5.2], $(G,S)$ and $(G_s,T)$ , where $S=\bigcup _{x\in X} G_x\backslash G$ and ${T=\bigcup _{x\in X} G_x\backslash G_s}$ , are projective and $\mathcal {S}$ -projective piles, respectively. Therefore, by Corollary 4.14, $(G_s, T)$ is projective. Note that $\alpha $ extends to a rigid epimorphism of piles $\alpha :\mathbf {G}=(G,S)\longrightarrow \mathbf {G}_s=(G_s,T)$ . Hence, by [6, Proposition 5.4]. the embedding problem

(9)

has a solution which is the needed retraction.

Theorem 4.15. Let $G=\coprod _{x \in X} G_x$ be a free profinite product over a profinite space X. A second countable prosoluble group H is isomorphic to a subgroup of G if and only if one of the following holds:

1. (i) H is isomorphic to a subgroup of a second countable free prosoluble product $H_s=\coprod _{t\in T}{}^s H_t$ of a continuous subsystem of pro-p groups $\{H_t\mid t\in T\}$ of the continuous system $\{G_x^g\mid x\in X, g\in G\}$ over $Y=\bigcup _{x\in X} G_x\backslash G$ , where T is a second countable profinite space;

2. (ii) H is isomorphic to a Frobenius group $\widehat {\mathbb {Z}}_\pi \rtimes C$ , with C a finite cyclic subgroup of $G_x$ for some $x\in X$ ;

3. (iii) H isomorphic to a subgroup of a free factor $G_x$ .

Proof. Suppose H is a subgroup of G. By Remark 4.9, there is a second countable profinite space $\widetilde T$ on which H acts continuously such that $(H,\widetilde T)$ is a projective pile and

$$ \begin{align*}\{H_t\mid t\in \widetilde T\}= \{H \cap G_x^g \mid x \in X,\ g \in G\}.\end{align*} $$

Then by Theorem 1.5, either (ii) or (iii) hold, or all $H_t$ are pro-p and H is $\mathcal {S}$ -projective relative to

$$ \begin{align*}\{H \cap G_x^g \mid x \in X,\ g \in G\}.\end{align*} $$

As $\widetilde T$ is second countable, $\widetilde T\longrightarrow \widetilde T/H$ admits a continuous section $\sigma :\widetilde T/H\longrightarrow \widetilde T$ (see [15, Lemma 5.6.7]), and so by Proposition 4.10, H embeds into a free prosoluble product $\coprod _{t\in T}{}^s H_{t}\amalg ^s F_s$ and a free prosoluble group $F_s$ , where $T=im(\sigma )$ . Now by Lemma 3.9, $\coprod _{t\in T}{}^s H_t\amalg ^s F_s$ embeds in $\coprod _{t\in T}{}^s H_t$ if $|T|>3$ . Otherwise, it embeds into $\coprod _{t\in T}{}^s H_t\amalg ^s \coprod _{t\in T}{}^s H_t$ by Lemma 3.9 again; that finishes the proof of this implication.

Converse. If (iii) holds, there is nothing to prove. Suppose (ii) holds. Then by [3, Corollary 5.2], H embeds into a free profinite product $C\amalg C$ . As G is not dihedral, by [16, Corollary 5.3.3], the normal closure is a free profinite product of copies of $G_x$ , and so $C\amalg C$ in turn embeds in G as needed.

Suppose now (i) holds. As any second countable profinite space embeds into second countable homogeneous, we may assume that T is homogeneous and it suffices to embed $H_s$ into G. By Corollary 1.7, $H_s$ retracts into $\coprod _{t\in T} H_t$ , so it suffices to embed $\coprod _{t\in T} H_t$ into G. Consider $G\times T$ as a continuous family $\{G\times \{t\}, \mid t\in T\}$ over T, and form a free profinite product $\coprod _{t\in T}(G\times \{t\})$ . Then by [16, Theorem 5.5.6], $\coprod _{t\in T} H_t$ embeds into $\coprod _{t\in T}G\times \{t\}$ , and so it suffices to embed $\coprod _{t\in T}(G\times \{t\})$ into G. By Lemma 3.6, G contains $G\amalg F$ , where F is a nontrivial free profinite group and so is homeomorphic to T. By [16, Corollary 5.3.3], the normal closure N of G in $G\amalg F$ is $\coprod _{f\in F} \coprod _{f\in F} G^{f}\cong \coprod _{t\in T}(G\times \{t\}) $ , as required.

Proof of Theorem 1.4. ‘Only if’ is just a particular case of Theorem 4.15 as well as the converse for the cases (ii) and (iii).

Moreover, suppose (i) of Theorem 1.4 holds. By Remark 4.9, there is a second countable profinite H-space T such that

$$ \begin{align*}\{H \cap H_i^h \mid \ h \in H_s, i=1,\ldots, n\} = \{H_t \mid t \in T\}\end{align*} $$

and $\mathbf {H} = (H,T)$ is a $\mathcal {S}$ -projective pile. Then by Theorem 1.5, $(H,T)$ is projective. By Corollary 1.6, $T_0=\{t\in T\mid G_t\neq 1\}$ is closed and $T_0/H$ is finite and $H_t$ are finitely generated.

It follows that $\{H_t\mid t\in T_0\}$ is a continuous subfamily of $\{G_x^g\mid x\in X, g\in G\}$ , so we can apply Theorem 4.15 again to finish the proof.

5 Questions 1.2 and 1.3

Proposition 5.1. Let $G_s=A \amalg ^s P$ be a prosoluble free product of a pro-p group A and a prosoluble projective group P. Then $(G_s, T)$ is projective, where $T=A\backslash G_s$ .

Proof. By Lemma 4.8, $(G_s , T)$ is an $\mathcal {S}$ -projective pile, where $T=A\backslash G_s$ . So by Corollary 4.14, $(G_s, T)$ is projective.

The next Corollary answers Question 1.3.

Corollary 5.2. Let A be a pro-p group and P a prosoluble projective group. Then $G=A \amalg P$ admits a prosoluble retract $G_s=A \amalg ^s P$ .

Proof. By Proposition 5.1, $(G_s, T)$ is a projective pile, where $T=A\backslash G_s$ . But $|T/G_s|=1$ , and so by Proposition 4.10, $G_s$ embeds into $G=A\amalg P$ .

We finish the paper answering Question 1.2 in the second countable case, but assuming that the free factors are prosoluble rather than pro-p.

Theorem 5.3. Let $G = G_1\amalg G_2$ be a free profinite product of prosoluble groups $G_1$ and $G_2$ . Then every second countable prosoluble subgroup of G is isomorphic to a closed subgroup of $L=G_1 \amalg ^s G_2$ .

Proof. If $G_1, G_2$ are of order $\leq 2$ , then $G=L$ , and there is nothing to prove. Suppose both $G_1$ and $G_2$ are nontrivial and $|G_1|+|G_2|>4$ .

Let H be a prosoluble subgroup of G. Then by Corollary 4.11, H embeds into a free prosoluble product $\coprod _{i=1}^2{}^s\coprod _{g\in D_i}{}^s (H\cap G_i^g)$ , where g runs through the closed set $D_i$ of representatives of the double cosets $G_i\backslash G/H$ , $i=1,2$ . This embeds in turn into $\coprod _{i=1}^2{}^s\coprod _{g\in D_i}{}^s G_i^g$ . Thus, it suffices to embed $\coprod _{i=1}^2{}^s\coprod _{g\in D_i}{}^s G_i^g$ into L.

Note that $|G|=|L|$ . By Lemma 3.9, L contains a free prosoluble product $G_1 \amalg ^s G_2^l \amalg ^s U$ for some $l\in L$ , where $|U|=|L|$ . Thus, $|G|=|L|=|U|=2|U|$ . Therefore, there exists a homeomorphism $\sigma :G\longrightarrow U_1\cup U_2$ , where $U_1, U_2$ are two copies of U. By [16, Corollary 5.3.3], the normal closure of $G_1 \amalg ^s G_2^l$ in $G_1 \amalg ^s G_2^l \amalg ^s U$ is

$$ \begin{align*}\coprod_{u\in U}{}^s (G_1 \amalg^s G_2^l)^u=\coprod_{i=1}^2{}^s\coprod_{u\in U_i}{}^s G_i^{l_iu},\end{align*} $$

where $l_1=1,l_2=l$ . Thus, there is an isomorphism $\coprod _{i=1}^2{}^s\coprod _{g\in G}{}^s G_i^g\longrightarrow \coprod _{i=1}^2{}^s\coprod _{u\in U_i}{}^s G_i^{l_iu}$ induced by a homeomorphism $G\longrightarrow U_1 \cup U_2$ and isomorphisms $id:G_1\longrightarrow G_1$ , $G_2\longrightarrow G_2^l$ . This finishes the proof.