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 [Reference Herfort and Ribes7, 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.

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. [Reference Neukirch13, Satz 3], for example). Note that $U_p$ are free pro-p (cf. [Reference Ribes and Zalesskii15, 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 [Reference Ribes and Zalesskii15, 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.

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 $C_{K/N}(CN/N) = 1$ .

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 ([Reference Ribes and Zalesskii15, Theorem 4.6.9 (d)]), and so $R_l(V)$ is nilpotent (see [Reference Ribes and Zalesskii15, 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 [Reference Ribes and Zalesskii15, Exercise 7.7.8 (a)], and therefore, G is soluble.

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 [Reference Ribes16, 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 [Reference Ribes16, 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 [Reference Ribes16, 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.

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 [Reference Burns1, 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.

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. [Reference Ribes and Zalesskii15, 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.

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.

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. [Reference Ribes and Zalesskii15, 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.

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 $F_s\neq 1$ .

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 [Reference Haran and Zalesskii6, 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.

Proof. This is the construction of T in [Reference Haran and Zalesskii6, Lemma 5.12].

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 [Reference Ribes16, Proposition 5.2.3], this is a profinite space.

By [Reference Haran and Zalesskii6, 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 [Reference Haran and Zalesskii6, Proposition 5.4], the embedding problem

(3)

has a solution which is clearly an embedding.

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. [Reference Ribes and Zalesskii15, 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- $\mathcal {C}$ .

Proof. Suppose (iii) does not hold. If G is a profinite Frobenius group, then $G=K\rtimes C$ , where C is finite (see [Reference Ribes and Zalesskii15, Theorem 4.6.9]) and K and C are coprime. By [Reference Ribes and Zalesskii15, 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 [Reference Ribes and Zalesskii15, Theorem 4.6.8] to $\gamma (G)$ ). By [Reference Ribes16, 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 [Reference Haran and Zalesskii6, 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 ([Reference Ribes and Zalesskii15, 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 [Reference Ribes and Zalesskii15, 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.

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 [Reference Ribes and Zalesskii15, Theorem 9.1.9]). So, if $n>1$ , P embeds in $F_s$ (see [Reference Ribes and Zalesskii15, 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 [Reference Ersoy and Herfort2, 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. [Reference Ribes16, Corollary 7.1.3]), we deduce the result.

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 [Reference Ribes and Zalesskii15, 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 [Reference Guralnick and Haran3, Corollary 5.2], H embeds into a free profinite product $C\amalg C$ . As G is not dihedral, by [Reference Ribes16, 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 [Reference Ribes16, 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 [Reference Ribes16, 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. 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.

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 $G=A\amalg P$ .

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 [Reference Ribes16, 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.