Proof

Suppose that for some k < n we have $$\begin{array}{c}{p}^{-k}q\in G\text{and}{p}^{n-k}q\in G.\end{array}$$ Hence, also pⁿ = (p^n−k q)(q⁻¹p^k) = (p^n−kq)(p^−kq)⁻¹ ∊ G which is impossible. □

Proof

Since G is a group, therefore relations i) and ii) are obvious. Let y ∊ G, g∊ 𝔓 \ G. Then also g⁻¹ ∊ 𝔓 \ G which implies that γ = (γg)g⁻¹ ∊(𝔓 \ G)∘(𝔓 \ G), i.e. relation iii) holds. If φ∊ 𝔓 \ G and additionally g² ∊ 𝔓 \ G, then by Lemma 3.1 either φg ∊ 𝔓 \ G or φg⁻¹ ∊ 𝔓 \ G which implies φ = (φg)g⁻¹ = (φg⁻¹)g ∊ (𝔓 \ G)∘(𝔓 \ G), i.e. 𝔓 \ G⊂(𝔓 \ G)∘(𝔓 \ G) and by iii) relation 𝔓 = (𝔓 \ G) ∘(𝔓 \ G) holds.

Suppose that gφ, ψg and ψφ all belong to G. Then also $$\begin{array}{c}{g}^2=g\phi (\psi \phi {)}^{-1}\psi g\in G,\end{array}$$ which gives the contradiction.  □

Proof

From Lemma 2.5 and the Stirling formula [14] we get that there exists an increasing sequence of natural numbers {n(u) : u ∈ ℕ} such that S_n(u) ∖ S_n(u)(u, u) ≠ ∅ for every u ∈ ℕ. Let ${\tilde{p}}_u\in (S_{n(u)}\mathrm{\setminus }{S}_{n(u)}(u,u))$ for every u ∈ ℕ.

Let us fix k ∈ ℕ, k > 1, and let {I_u} be the increasing sequence of intervals (which means that i < j for any i ∈ I_u, j ∈ I_{u + 1}, u ∈ ℕ) forming a partition of ℕ and satisfying condition card I_u = k ⋅ n(u), for every u ∈ ℕ. Let us put $$J_u=[(k-2)n(u)+min{I}_u,(k-1)n(u)+min{I}_u]$$ for u ∈ ℕ.

The expected permutation p is defined in the following way $$\begin{array}{}p(in(u)+j+min{I}_u)=(i+1)n(u)+j+min{I}_u,\\ p((k-2)n(u)+j+min{I}_u)=(k-1)n(u)+{\tilde{p}}_u(j+1)-1+min{I}_u,\\ p((k-1)n(u)+j+min{I}_u)={\tilde{p}}_u^{-1}(j+1)-1+min{I}_u,\end{array}$$ for i = 0, 1, …, k – 3 and j = 0, 1,…, n(u) − 1.

The verification of condition p^k = id_ℕ is trivial, whereas the fact that pⁱ ∈ (𝔓 ∖ G) for every i = 1, 2, …,k − 1 is a consequence of the fact that composition of the following three mappings: the increasing mapping of interval [1, n(u)] onto interval J_u, the restriction of p to interval J_u and the increasing mapping of interval [(k − 1)n(u) + min I_u, max I_u] onto interval [1, n(u)], is equal to ${\tilde{p}}_u$ for every u ∈ ℕ.

Now, let us suppose that there exist permutations p_i ∈ 𝔓, i = 1, 2, …, n, such that $$p=p_n{p}_{n-1}\dots p_1$$ and $$\text{either}{p}_i\in \mathfrak{C}\text{or}{p}_i^{-1}\in \mathfrak{C},$$ for every i = 1, 2, …,n.

Let J be an interval of natural numbers. We denote by p_J and p_i,J the restrictions of permutations p and p_i, respectively, to sets J and $$p_{i-1}{p}_{i-2}\dots p_1{p}_0(J),$$ respectively, for each i = 1, 2, …,n, where p₀ := id_ℕ. Then the following decomposition holds $$\begin{array}{}{q}_{n,J}{p}_j{q}_{1,J}^{-1}=(q_n,Jpn,J{q}_{n-1,J}^{-1})(q_n-1,Jpn-1,J{q}_{n-2,J}^{-1})\dots (q_{2,J}{p}_{2,J}{q}_{1,J}^{-1})(q_{1,J}{p}_{1,J}{q}_{0,J}^{-1}),\end{array}$$(3) where q_i,J, 1 ≤ i ≤ n, is the increasing mapping of set p_ip_{i − 1} … p₁(J) onto interval [1, card (J)] and q_0,J := id_ℕ.

Let us notice that for each i = 1, 2, …,n we have q_i,J p_i,J $\begin{array}{}{q}_{i-1,J}^{-1}\end{array}$ ∈ S_card(J) and this is the b–connected permutation for b = c(p_i) or b = c( $\begin{array}{}{p}_i^{-1}\end{array}$), depending on the fact whether p_i ∈ ℭ or $\begin{array}{}{p}_i^{-1}\end{array}$ ∈ ℭ. Justification is needed only for the last property.

So let us suppose that there exists 1 ≤ i ≤ n such that p_i ∈ ℭ and simultaneously ξ = q_i,Jp_i,J $\begin{array}{}{q}_{i-1,J}^{-1}\end{array}$ is not the b–connected permutation. Then there exists subinterval Λ of interval [1, card (J)] such that the set ξ(Λ) is a union of at least (b + 1) MSI. Let 𝔍 be the family of mutually separated intervals of natural numbers forming the decomposition of set ξ(Λ) . Auxiliarly we take $$\begin{array}{}{\displaystyle \alpha =min{q}_{i-1,J}^{-1}(\mathrm{\Lambda })\text{and}\beta =max{q}_{i-1,J}^{-1}(\mathrm{\Lambda }).}\end{array}$$

Then from the fact that $\begin{array}{}{q}_{i-1,J}^{-1}\end{array}$ is the increasing mapping we get the following relations $$\begin{array}{}{\displaystyle \alpha =q_{i-1,J}^{-1}(min\mathrm{\Lambda })\text{and}\beta =q_{i-1,J}^{-1}(max\mathrm{\Lambda })}\end{array}$$(4) and $$\begin{array}{}[\alpha ,\beta ]\cap q_{i-1,J}^{-1}([1,\text{card}(J)])=q_{i-1,J}^{-1}(\mathrm{\Lambda }).\end{array}$$(5)

Let us notice one more property. For any Θ, Ξ ∈ 𝔍, Θ ≠ Ξ, $$\begin{array}{}\text{if}{q}_{i,J}^{-1}(\mathrm{\Theta })<q_{i,J}^{-1}(\mathrm{\Xi })\text{\hspace{0.17em}then the following relation holds}\\ (max{q}_{i,J}^{-1}(\mathrm{\Theta }),min{q}_{i,J}^{-1}(\mathrm{\Xi }))\cap p_i{q}_{i,J}^{-1}([1,\text{card}(J)])\ne \mathrm{\varnothing }.\end{array}$$(6)

From the relations (4), (5) and (6) we are able to deduce that the set p_i([α, β]), as well as the set ξ(Λ), is the union of at least (b + 1)MSI, since the numbers from the set p_i([α, β] ∖ $\begin{array}{}{q}_{i-1,J}^{-1}\end{array}$(Λ)) “do not fill” completely all the“holes” between sets $\begin{array}{}{q}_{i,J}^{-1}\end{array}$(Θ) and $\begin{array}{}{q}_{i,J}^{-1}\end{array}$(Ξ) for any Θ, Ξ ∈ 𝔍, Θ ≠ Ξ. Thereby we get the contradiction with definition of number b and, in consequence, with the assumption that ξ is not the b–connected permutation.

Thus, if we assume that $$\begin{array}{}v>n+{\displaystyle \underset{1\le i\le n}{max}\{c(q_i):q_i=p_i\text{\hspace{0.17em}if\hspace{0.17em}}{p}_i\in \mathfrak{C},q_i=p_i^{-1}\text{otherwise}\}}\end{array}$$ then p_u = q_n,J pJ $\begin{array}{}{q}_{1,J}^{-1}\end{array}$ ∉ S_n(u)(u, u) for J = J_u and for every u ∈ ℕ, u > v, which contradicts the assumption.□

Proof of Theorem 3.6

By item i) of Lemma 3.2 we have p^{– 1} ∈ 𝔓 ∖ 𝒢. Thus, from Theorem 3.7 there exists a family 𝒜_{p ^{– 1}} ⊂ ℭ × ℭ possessing the respective properties i) –iii) given above. In the sequel, if (σ, δ) ∈ 𝒜_{p ^{– 1}} then σ p^{– 1} = p^{– 1}δ ∈ 𝔓 ∖ 𝒢, because of item ii) of Lemma 3.2 and $$\begin{array}{}p(p^{-1}\delta )=\delta \in \mathfrak{C}\text{and}(\sigma p^{-1})p=\sigma \in \mathfrak{C}.\end{array}$$

It is sufficient to set $$\begin{array}{}{B}_p:=\{\sigma p^{-1}:(\sigma ,\delta )\in {\mathcal{A}}_{p^{-1}}\}.\end{array}$$

We get card ℬ_p = 𝔠 because of the condition iii) of Theorem 3.7. □

Let us finish this section with one more result concerning the semigroups with one generator.

Proof

Let p, q ∈ S ∖ G. If pq ∈ G then p^{– 1} = q(pq)^{– 1} ∈ S which is impossible. Similarly, if p^{– n} ∈ S for some n ∈ ℕ then also p^{– 1} = p^{n − 1}(p^{− n}) ∈ S which again contradicts the assumptions. □

Proof of Theorem 4.1

1. The first equality follows from the fact that $$\begin{array}{}p([1,n_k))=[1,n_k),\end{array}$$ for every k ∈ ℕ.

2. This inclusion is obvious (see [6]).

3. First we note that id_ℕ ∈ G ∩ ℭℭ. Next let us choose the infinite set K ⊂ ℕ of indices such that $$\begin{array}{}\underset{K\ni k\to \mathrm{\infty }}{lim}(n_{k+1}-n_k)=\mathrm{\infty }\end{array}$$(7) and $$\begin{array}{}{n}_{k+1}-n_k\ge 3,\end{array}$$ for every k ∈ K.

   Let us set p|_ℐ = id_ℐ, where $\begin{array}{}\mathcal{I}:=\bigcup _{k\in \mathbb{N}\mathrm{\setminus }K}[n_k,n_{k+1})\end{array}$ and $$p(n_k-1+i)=n_k-1+2i,$$ for every i = 1, 2, …,⌊(n_{k + 1} − n_k + 1)/2⌋, and p is the increasing mapping of interval (⌊(n_{k + 1} − n_k + 1)/2⌋, n_{k + 1}) onto set $$(n_{k+1},n_k]\mathrm{\setminus }p([n_k,n_k-1+\lfloor (n_{k+1}-n_k+1)/2\rfloor ])$$ for every k ∈ K. From (7) it follows that p ∈ ℭ𝔇 ∩ G. Certainly we have p^{– 1} ∈ 𝔇ℭ ∩ G.

   Let K₁ and K₂ form a partition of the set K such that both of them are infinite. Now if we put $$\begin{array}{}q(i)=\left\{\begin{array}{l}p(i)\text{for}i\in [n_k,n_{k+1})\text{and}k\in K_1,\\ p^{-1}(i)\text{for}i\in [n_k,n_{k+1})\text{and}k\in K_2,\end{array}\right.\end{array}$$ where p is defined as above and q(j) = j for j ∈ ℐ, then we get q ∈ 𝔇𝔇 ∩ G.

4. Let us fix the increasing sequence $\begin{array}{}\{k_n{\}}_{n=1}^{\mathrm{\infty }}\end{array}$ of positive integers such that $$\begin{array}{}{S}_n\mathrm{\setminus }{S}_n(u,u)\ne \mathrm{\varnothing },\end{array}$$(8) for every u = 1, 2, …, where n := n_{ku + 1} − n_ku. Let us also fix $$Pu\in S_n\mathrm{\setminus }{S}_n(u,u),$$ for every u = 1, 2, …, and define $$p(i)=p_u(i)$$ for every i ∈ [n_ku, n_{ku + 1}) and u = 1, 2, …, Further, we set p = id_N where $\begin{array}{}N:=\mathbb{N}\mathrm{\setminus }\bigcup _{u\in \mathbb{N}}[n_{k_u},n_{k_u+1}).\end{array}$ Certainly p ∈ G. We intend to prove that p ∉ 𝒢. Condition (8) implies that p is not a composition of the form $\begin{array}{}{q}_1^{-1}{q}_2\dots q_u^{(-1{)}^u},\end{array}$ either $\begin{array}{}{q}_1{q}_2^{-1}\dots q_u^{(-1{)}^{u-1}},\end{array}$ of permutations q_i ∈ ℭ such that c(q_i) ≤ u for i = 1, 2, …, u for each u ∈ ℕ, which in view of definition of the group generated by the given set of generators means that p∉ 𝒢.

The following last result of this paper could be deduced from the proof of Theorem 4.1 on the basis of its proof.

Proof

First we fix a Sierpiński’s family S of increasing sequences of positive integers. We can also suppose that if $\begin{array}{}\{n_k{\}}_{k=1}^{\mathrm{\infty }}\in \mathbf{S}\text{\hspace{0.17em}then\hspace{0.17em}}\underset{k\to \mathrm{\infty }}{lim}(n_{k+1}-n_k)=\mathrm{\infty }.\end{array}$

Next, with each $\begin{array}{}\{n_k{\}}_{k=1}^{\mathrm{\infty }}\in \mathbf{S}\end{array}$ we connect a family $$\begin{array}{}{G}^{\prime }(\{n_k{\}}_{k=1}^{\mathrm{\infty }})\subset G(\{n_k{\}}_{k=1}^{\mathrm{\infty }})\end{array}$$ defined in the following way. Firstly, for every divergent permutation p on ℕ we set $$\begin{array}{}q(n_k)=n_{p(k)},k\in \mathbb{N},\end{array}$$ and $$\begin{array}{}q{|}_I={\text{id}}_I,\text{where}I=\mathbb{N}\mathrm{\setminus }\{n_k{\}}_{k=1}^{\mathrm{\infty }}.\end{array}$$

Secondly, let us define G′ to be the family of all permutations q defined in this way. Certainly we have $$\begin{array}{}q{|}_{\{n_k{\}}_{k=1}^{\mathrm{\infty }}}\in \mathfrak{D}(\{n_k{\}}_{k=1}^{\mathrm{\infty }})\end{array}$$ for every $\begin{array}{}\{n_k{\}}_{k=1}^{\mathrm{\infty }}\in \mathbf{S}\end{array}$ and permutation p ∈ 𝔇, and since any two different sequences $\begin{array}{}\{n_k{\}}_{k=1}^{\mathrm{\infty }}\text{\hspace{0.17em}and\hspace{0.17em}}\{m_l{\}}_{l=1}^{\mathrm{\infty }}\end{array}$ from S are almost disjoint we get the relations $$\begin{array}{}{\displaystyle \sum (\phi )\mathrm{\setminus }\sum (q)\ne \mathrm{\varnothing }\text{and}\sum (q)\mathrm{\setminus }\sum (\phi )\ne \mathrm{\varnothing }}\end{array}$$ which hold for any two $\begin{array}{}q\in G^{\prime }(\{n_k{\}}_{k=1}^{\mathrm{\infty }})\text{\hspace{0.17em}and\hspace{0.17em}}\phi \in G^{\prime }(\{m_l{\}}_{l=1}^{\mathrm{\infty }}).\end{array}$ Both these families G′ possess the power of continuum since card 𝔇 = 𝔠.

The above relations from the Theorem 2.1 can be deduced. □