# Random ubiquitous transformation semigroups

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## Abstract

A *smallest generating set* of a semigroup is a generating set of the smallest cardinality. Similarly, an *irredundant generating set**X* is a generating set such that no proper subset of *X* is also a generating set. A semigroup *S* is *ubiquitous* if every irredundant generating set of *S* is of the same cardinality. We are motivated by a naïve algorithm to find a small generating set for a semigroup, which in practice often outputs a smallest generating set. We give a sufficient condition for a transformation semigroup to be ubiquitous and show that a transformation semigroup generated by *k* randomly chosen transformations asymptotically satisfies the sufficient condition. Finally, we show that under this condition the output of the previously mentioned naïve algorithm is irredundant.

## Keywords

Computational semigroup theory Transformation monoids Asymptotic behavior## 1 Introduction

*X*of a semigroup

*S*is a

*smallest generating set*, also known as

*minimum generating set*, if every subset of

*S*with cardinality strictly smaller than |

*X*| does not generate

*S*. The size of a smallest generating set is known as the

*rank of S*. Similarly, an

*irredundant generating set for S*is a generating set

*X*such that no proper subset of

*X*is a generating set for

*S*. Of course, the notions of a smallest generating set, irredundant generating set, and the rank have a natural interpretation for groups and other algebraic objects. The question of finding a smallest generating set or a rank is a classical one, see for example [1, 13] in the case of quasigroups and [7, 8, 9] in the case of semigroups. However, from a computational perspective this is, in general, not an easy problem. In particular, there is no known efficient algorithm to find the rank of a given

*S*, besides examining most of its subsets. As such, fast naïve algorithms are sometimes used to obtain small, but not necessarily smallest, generating sets. The simplest of them is Algorithm 1.

The algorithm applies to both groups and semigroups. The advantages of the Greedy algorithm are its speed and that it requires no a priori knowledge about the object. The latter might be seen as a drawback if some structural information is known. For semigroups this algorithm can be improved by taking into account its \({\mathcal {J}}\)-class structure.

*S*be a semigroup and let 1 be a symbol which is not in

*S*. Define \(S^1 = S \cup \{1\}\) to be a semigroup such that for all \(x, y \in S\) the product \(x \cdot y\) in \(S^1\) is the same as the product in

*S*, and \(x \cdot 1 = 1 \cdot x = x\) for all \(x \in S^1\). It is routine to verify that the operation \([\cdot ]\) on \(S^1\) is associative. If \(A, B \subseteq S^1\) define \(A x = \{ a \cdot x : a \in A\}\) and similarly define

*xA*and

*AxB*. Define a relation on

*S*by

*preorder*on

*S*. Clearly, the relation

*S*and the preorder \(\le \) induces a partial order on the equivalence classes of \({\mathcal {J}}\), which we will also denote by \(\le \) if the distinction is clear from the context. We say that the list \(s_{1, 1}, \ldots , s_{1, n_1}, s_{2, 1}, \ldots , s_{2, n_2} \ldots s_{k, n_k}\) of all elements of

*S*is ordered according to the preorder \(\le \) if \(\{s_{i, 1}, \ldots , s_{i, n_i}\}\) is an equivalence class of \({\mathcal {J}}\) for all

*i*and if \(s_{i, k} \ge s_{j, m}\) then \(i \le j\). Using this idea we can state Algorithm 2.

If *S* is a group, then \(S^1 x S^1 = S\) for every \(x \in S\). Hence there is a single \({\mathcal {J}}\)-class in *S* and so any permutation of elements of *S* is ordered according to the preorder \(\le \), and so the SmallGeneratingSet algorithm does not perform any better than Greedy. For proper semigroups the algorithm is particularly useful if the \({\mathcal {J}}\)-class structure is known in advance, for example if the semigroup was enumerated using the *Froidure–Pin algorithm* [6, 11] or algorithms appearing in [5]. It is easy to come up with examples for which SmallGeneratingSet might return a generating set which is not a smallest generating set or even an irredundant generating set, for example any non-trivial finite group *G*. Even though the algorithm is naïve, it performs surprisingly well in practice. For instance, we ran the SmallGeneratingSet algorithm on all \(836\,021\) semigroups (up to (anti-)isomorphism) of size 7, available in *SmallSemi* [4]. In all cases the generating set found was a smallest generating set.

*transformation monoid*on

*n*points, that is the set of all functions from \(\{1, \ldots , n\}\) to itself. The set \({\mathcal {T}}_n\) forms a semigroup under the composition of functions. In the following table, we consider every subgroup of \({\mathcal {T}}_3\) up to conjugation. Observe that—for most of them—the size of the generating set output by SmallGeneratingSet is equal to the rank or is one greater (Table 1).

Subsemigroups of \({\mathcal {T}}_3\)

Rank | Size of the output | ||||||
---|---|---|---|---|---|---|---|

1 | 2 | 3 | 4 | 5 | 6 | 7 | |

1 | 7 | 3 | 1 | 0 | 0 | 0 | 0 |

2 | – | 32 | 25 | 11 | 3 | 1 | 0 |

3 | – | – | 38 | 50 | 23 | 9 | 2 |

4 | – | – | – | 23 | 28 | 6 | 6 |

5 | – | – | – | – | 5 | 7 | 2 |

The main motivation for this paper is to provide mathematical justification as to why SmallGeneratingSet algorithms often returns a smallest generating set. In order to do so, we consider properties of transformation semigroups picked at random, in a certain way. We say that a semigroup *S* is *ubiquitous* if every irredundant generating of *S* is also a smallest generating set. Alternatively, if *r* is the rank of *S*, then *S* is ubiquitous if every irredundant generating set is of size *r*.

First we will provide a sufficient condition for a transformation semigroup to be ubiquitous.

### Theorem 1.1

Let \(S \le {\mathcal {T}}_n\) and suppose that *X* is a generating set for *S* such that \({\text {rank}}(xyz) < {\text {rank}}(y)\) for all \(x, y, z \in X\). Then *S* is ubiquitous.

Even though we restrict our attention to transformation semigroups in this paper, Theorem 1.1 can be generalised to include semigroups of partial bijections as well. We follow the approach of Cameron [2] of choosing a random transformation semigroup. That is for some \(k \ge 1\) we choose *k* transformations of degree *n* with uniform probability and consider the semigroup generated by them. We show that most transformation semigroups are ubiquitous.

### Theorem 1.2

Let \(k \ge 1\), and let \({\mathbb {P}}_k(n)\) be the probability that for \(x_1, \ldots , x_k \in {\mathcal {T}}_n\), chosen with uniform probability, the semigroup \(\langle x_1, \ldots , x_k \rangle \) is ubiquitous. Then \({\mathbb {P}}_k(n) \rightarrow 1\) as \(n \rightarrow \infty \) exponentially fast.

Even though SmallGeneratingSet does not return an irredundant generating set in general, we show that under the assumptions of Theorem 1.1 the output is irredundant. Hence the final result of the paper is as follows.

### Theorem 1.3

Let \(k \ge 1\), and let \({\mathbb {W}}_k(n)\) be the probability that for \(x_1, \ldots , x_k \in {\mathcal {T}}_n\), chosen with uniform probability, SmallGeneratingSet returns a smallest generating set for a semigroup \(\langle x_1, \ldots , x_k \rangle \). Then \({\mathbb {W}}_k(n) \rightarrow 1\) as \(n \rightarrow \infty \) exponentially fast.

Here we only look at the asymptotic behaviour of transformation semigroups, however the same question can be investigated for any other infinite family of semigroups, for example symmetric inverse monoids on \(\{1,\ldots , n\}\), or binary relations on *n* points.

## 2 Preliminaries

In this section we give the definitions and notation needed in the remainder of the paper.

### Definition 2.1

*S*be a semigroup and let \(x, y \in S\). The

*Green’s relations*\({\mathcal {L}}\), \({\mathcal {R}}\), \({\mathcal {J}}\), and \({\mathcal {D}}\) are the following equivalence relations on

*S*:

Let \(x \in S\). Then \(L_x\), \(R_x\), and \(D_x\) denote the equivalences classes of \({\mathcal {L}}\), \({\mathcal {R}}\), and \({\mathcal {D}}\), respectively, containing *x*. If *S* is finite, then \({\mathcal {D}} = {\mathcal {J}}\), for a proof see [10]. Since we are only interested in finite semigroups we will not make any distinction between the \({\mathcal {D}}\) and \({\mathcal {J}}\) relations.

Throughout the paper, we write elements of \({\mathcal {T}}_n\) on the right of their argument and we write functions from a subset of \({\mathbb {R}}^n\) to \({\mathbb {R}}\) on the left. This is done in agreement with two different notations prevalent in algebra and analysis.

*image of f*is the set \({\text {im}}(f) = (\{1, \ldots , n\})f\). A

*transversal of f*is a set \({\mathfrak {T}} \subseteq \{1, \ldots , n\}\) such that

*f*is injective on \({\mathfrak {T}}\) and \(({\mathfrak {T}})f = {\text {im}}(f)\). The

*rank of f*is \({\text {rank}}(f) = |{\text {im}}(f)| = |{\mathfrak {T}}|\), where \({\mathfrak {T}}\) is a transversal of

*f*. The

*kernel of f*, denoted by \(\ker (f)\), is the equivalence relation defined by

*kernel class of f*containing \(x \in \{1, \ldots , n\}\) is the set

### Lemma 2.2

- (i)
if \(f {\mathcal {L}} g\) then \({\text {im}}(f) = {\text {im}}(g)\);

- (ii)
if \(f {\mathcal {R}} g\) then \(\ker (f) = \ker (g)\);

- (iii)
if \(f {\mathcal {D}} g\) then \({\text {rank}}(f) = {\text {rank}}(g)\).

## 3 Sufficient condition for ubiquitous semigroups

In this section we prove Theorem 1.1. We will do so in a series of lemmas. The first of which is the following easy observation about products in the \({\mathcal {D}}\)-classes. Recall that if \(x, y \in S\), then by \(D_x\) we denote the \({\mathcal {D}}\)-class containing *x* and \(x \le y\) if and only if \(S^1 x S^1 \subseteq S^1 y S^1\).

### Lemma 3.1

Let *S* be a semigroup, and let \(z_1 \cdots z_m \in D_x\) where \(x, z_1, \ldots , z_m \in S\). Then \(x \le z_i \cdots z_j\) under the preorder on *S* for all \(i,j \in \{1, \ldots , m\}\) with \(i \le j\).

### Proof

Next we give a condition for a semigroup *S* which restricts allowed products in a given \({\mathcal {D}}\)-class.

### Lemma 3.2

*X*be a generating set for

*S*, and let \(x \in X\) be such that \({\text {rank}}(y_1xy_2) < {\text {rank}}(x)\) for all \(y_1, y_2 \in X\) where \(y_1, y_2 \ge x\). Then only the following products

### Proof

First observe that if \(x^2 \in D_x\), then both *x* and \(x^2\) have the same rank by Lemma 2.2, in other words \(|{{\text {im}}(x)}| = |{{\text {im}}(x^2)}|\). However, since *x* is a finite degree transformation and \({\text {im}}(x^2) \subseteq {\text {im}}(x)\), it follows that \({\text {im}}(x) = {\text {im}}(x^2)\), and so *x* acts as a bijection on \({\text {im}}(x)\). Hence \({\text {rank}}(x^3) = {\text {rank}}(x)\), contradicting the hypothesis of the lemma. Therefore \(x^2 \notin D_x\), and since \(S^1 x^2 S^1 \subseteq S^1 x S^1\), it follows that \(x^2 < x\) under the preorder on *S*. Similarly, for every \(y_1, y_2 \in X\) such that \(y_1, y_2 \ge x\), it follows from Lemma 2.2 that \(y_1 x y_2 < x\), since \({\text {rank}}(y_1 x y_2) < {\text {rank}}(x)\) and \(S^1 y_1 x y_2 S^1 \subseteq S^1 x S^1\).

*i*by Lemma 3.1. Hence there are \(k \in {\mathbb {N}}\), \(n_1, \ldots , n_{k}, m_2, \ldots , m_{k} \ge 1\), and \(m_1, m_{k + 1} \ge 0\) such that

*i*and

*j*. Here we are assuming that

### Corollary 3.3

Let \(S \le {\mathcal {T}}_n\), let *X* be a generating set for *S*, and let \(x \in X\) be such that \({\text {rank}}(z_1xz_2) < {\text {rank}}(x)\) for all \(z_1, z_2 \in X\) where \(z_1, z_2 \ge x\). Then \(pxuys \notin D_x\) for all \(p, u, s \in S^1\) and any \(y \in X\) such that \(x {\mathcal {D}} y\).

### Proof

*x*occurs twice in the product, which is a contradiction according to Lemma 3.2. \(\square \)

Finally, we prove Theorem 1.1. Observe that if a transformation semigroup \(S \le {\mathcal {T}}_n\) is such that all irredundant generating sets have the same cardinality, then every irredundant generating set is a smallest generating set.

**Theorem** 1.1 *Let *\(S \le {\mathcal {T}}_n\)*and suppose that **X**is a generating set for **S**such that *\({\text {rank}}(xyz) < {\text {rank}}(y)\)*for all *\(x, y, z \in X\). *Then **S**is ubiquitous.*

### Proof

Let \(X' \subseteq X\) be irredundant. Then \({\text {rank}}(xyz) < {\text {rank}}(y)\) for all \(x, y, z \in X'\). It is sufficient to show that every irredundant generating set is of the same cardinality. Moreover, without loss of generality we may assume that *X* is irredundant and show that every irredundant generating set is of size \(|{X}|\).

*Y*be an irredundant generating set for

*S*. Let \(\le _d\) be a total order defined on \({\mathcal {D}}\)-classes of

*S*such that if

*D*and \(D'\) are \({\mathcal {D}}\)-classes of

*S*and \(D \le D'\) under the partial order of \({\mathcal {D}}\)-classes, then \(D \le _d D'\). Let \(\{D_1, \ldots , D_d\}\) be the set of all \({\mathcal {D}}\)-classes of

*S*, indexed so that \(D_d<_d \ldots <_d D_1\). For \(k \in \{1, \ldots , d\}\), define

*Y*, and so

By the definition of the total order \(\le _d\), the \({\mathcal {D}}\)-class \(D_{1}\) is maximal, and so both *X* and *Y* intersect \(D_1\) non-trivially. For any \(i \ge 2\) and \(x_1, \ldots , x_i \in X_1\), it follows from Corollary 3.3 that \( x_{1} \cdots x_{i} \notin D_{x_1} = D_{1}\), and so \(D_{1} = X_1\). The same argument shows that \(D_{1} = Y_1\), and so \(X_1 = Y_1 = D_1\).

For \(k \ge 1\), suppose that \(|{X_k}| = |{Y_k}|\) and \(\langle X_k \rangle = \langle Y_k \rangle \). If \(X \cap D_{k + 1} = \varnothing \), then \(D_{k + 1} \subseteq \langle X_k \rangle = \langle Y_k \rangle \). Hence \(X_{k + 1} = X_k\) and \(Y_{k + 1} = Y_k\), and thus \(|{X_{k + 1}}| = |{Y_{k + 1}}|\) and \(\langle X_{k + 1} \rangle = \langle Y_{k + 1} \rangle \).

*X*is irredundant and \(x \notin X_k \cup X' \subseteq X\). Hence if \(t = |{Y \cap D_{k + 1}}|\) then \(X' = X \cap D_{k + 1}\), or in other words \(X_{k + 1} = X_k \cup X'\) and \(Y_{k + 1} = Y_k \cup Y'\). Therefore, \(|{X_{k + 1}}| = |{X_k}| + t = |{Y_k}| + t = |{Y_{k + 1}}|\) and \(\langle X_{k + 1} \rangle = \langle Y_{k + 1} \rangle \). We will now show that \(t = |{Y \cap D_{k + 1}}|\).

*y*is equal to a product of elements of \(X_{k + 1}\) by (2). It follows from Corollary 3.3 that if \(x_1 \cdots x_m \in D_{k + 1}\) where \(x_1, \ldots , x_m \in X\) then there is at most one \(i \in \{1, \ldots , m\}\) such that \(x_i \in X \cap D_{k + 1}\), otherwise some subword of \(x_1 \cdots x_m\) would not be an element of \(D_{k + 1}\). Since \(y \notin Y_k \cup Y'\), the irredundancy of

*Y*implies that \(y \notin \langle Y_k, Y' \rangle = \langle X_k, X' \rangle \). It follows that \(y = p_1 \cdots p_m x s_1 \cdots s_l\) for some \(x \in X \cap D_{k + 1} {\setminus } X'\), \(m, l \ge 0\) and \(s_i, p_i \in X_k\). Hence \(y \in \langle x, X_k \rangle \). Since \(x, y \in D_{k + 1}\), it follows that there are \(a, b \in S^1\) such that

*t*. Therefore \(t = |{Y \cap D_{k + 1}}|\) and by the previous paragraph \(\langle X_{k + 1} \rangle = \langle Y_{k + 1} \rangle \) and \(|{X_{k + 1}}| = |{Y_{k + 1}}|\).

By induction it follows that \(\langle X_k \rangle = \langle Y_k \rangle \) and \(|{X_k}| = |{Y_k}|\) for all \(k \in \{1, \ldots , d\}\). In particular, \(X_d = X\) and \(Y_d = Y\), and thus \(|{X}| = |{Y}|\), as required. \(\square \)

## 4 SmallGeneratingSet

In this section we return to the motivating question about the algorithm SmallGeneratingSet. First, we note that SmallGeneratingSet might return a generating set which is not irredundant. For example, if the semigroup under investigation is a group of size at least 2, the algorithm can first pick an identity and so return a generating set which includes an identity. However, we show that under the assumptions of Theorem 1.1 the generating set returned by SmallGeneratingSet is irredundant.

### Lemma 4.1

Let \(S \le {\mathcal {T}}_n\) and suppose that *X* is a generating set for *S* such that \({\text {rank}}(xyz) < {\text {rank}}(y)\) for all \(x, y, z \in X\). Then SmallGeneratingSet returns an irredundant generating set.

### Proof

*i*is the largest integer such that \(x_i \in X {\setminus } I\) and \(x_i {\mathcal {D}} x_j\). Then there are \(a_1, \ldots , a_{k_a}, b_1,\ldots , b_{k_b} \in I\) such that

The following result is then immediate from Theorem 1.1.

### Corollary 4.2

Let \(S \le {\mathcal {T}}_n\) and suppose that *X* is a generating set for *S* such that \({\text {rank}}(xyz) < {\text {rank}}(y)\) for all \(x, y, z \in X\). Then SmallGeneratingSet returns a smallest generating set.

## 5 Asymptotics

The main aim of this section is to show that if for some fixed \(k \ge 1\) we choose \(x_1, \ldots , x_k \in {\mathcal {T}}_n\) with uniform probability, then the probability \({\mathbb {P}}_k(n)\) that \(\langle x_1, \ldots , x_k \rangle \) is ubiquitous and the probability \({\mathbb {W}}_k(n)\) that SmallGeneratingSet returns a smallest generating set for \(\langle x_1, \ldots , x_k \rangle \) both tend to 1 as *n* increases.

### Lemma 5.1

- (i)
there is \(x \in X\) such that \(\langle x \rangle \) is a group;

- (ii)
there are distinct \(x, y \in X\) such that \({\text {rank}}(xyx) = {\text {rank}}(y)\);

- (iii)
there are mutually distinct \(x, y, z \in X\) such that \({\text {rank}}(xyz) = {\text {rank}}(y)\).

### Proof

Suppose that \({\text {rank}}(xyz) = {\text {rank}}(y)\) for some \(x, y, z \in X\) and suppose that not all *x*, *y*, and *z* are distinct. If \(x = y = z\), then \({\text {rank}}(x^3) = {\text {rank}}(x)\), which is only possible if *x* acts bijectively on \({\text {im}}(x)\). However, in that case \(\langle x \rangle \) is a group. Hence we only need to consider the case that where exactly two of *x*, *y*, and *z* are equal.

### 5.1 Preliminary counting results

For \(n, r\in {\mathbb {N}}\) such that \(r \le n\), define \({\mathcal {A}}(n,r)\) to be the set of partitions of \(\{1,\ldots ,n\}\) into *r* non-empty components.

### Lemma 5.2

### Proof

A function \(f \in {\mathcal {T}}_n\) is called idempotent if \(f^2 = f\). We prove the lemma by finding the number of idempotent transformation of \({\mathcal {T}}_n\) of rank *r* in two ways. Denote this number by *N*. It can be shown that *f* is an idempotent if and only if \((x)f = x\) for all \(x \in {\text {im}}(f)\).

*r*, then there are \(\left( {\begin{array}{c}n\\ r\end{array}}\right) \) choices for the \({\text {im}}(f)\) and for every point in \(\{1,\ldots ,n\} {\setminus } {\text {im}}(f)\) there are

*r*choices in \({\text {im}}(f)\) to map to. Hence

*f*is an idempotent and \(A_1, \ldots , A_r\) are kernel classes of

*f*then \((A_i)f \in A_i\) for all \(i \in \{1, \ldots , r\}\), and so there are \(\prod _{i = 1}^r |A_i|\) choices for the \({\text {im}}(f)\). Hence

Since \(|{\mathcal {A}}(n, r)| = \left\{ \begin{array}{c} {n} \\ {r}\end{array}\right\} \), the following easy upper bound for the Stirling numbers is an immediate consequence of Lemma 5.2.

### Corollary 5.3

*W*(

*x*) is a well-defined function on \({\mathbb {R}}^+\). In the literature

*W*(

*x*) is known as

*Lambert W function*or

*product logarithm*, see e.g. [3]. The value \(\Omega = W(1)\) is known as the

*omega constant*and it satisfies \(\Omega e^\Omega = 1\), with the numerical value \(\Omega = 0.5671439\dots \) .

### 5.2 \({\pmb {\mathbb {G}}}_n\) tends to zero

We begin by obtain an expression for \({\mathbb {G}}_n\) in terms of *n*.

### Lemma 5.4

### Proof

*x*acts as a bijection on \({\text {im}}(x)\). There are

*x*such that

*x*acts bijectively on \({\text {im}}(x)\). That is, if \(|{\text {im}}(x)| = r\), then there are \(\left( {\begin{array}{c}n\\ r\end{array}}\right) \) choices for \({\text {im}}(x)\),

*r*! ways of bijectively mapping \({\text {im}}(x)\) to itself, and \(r^{n - r}\) ways to map every point from \(\{1,\ldots ,n\} {\setminus } {\text {im}}(x)\) to \({\text {im}}(x)\). Since \(|{\mathcal {T}}_n| = n^n\), the probability of randomly choosing \(x \in {\mathcal {T}}_n\) such that \(\langle x \rangle \) is a group is

In order to prove that \({\mathbb {G}}_n \rightarrow 0\) as \(n \rightarrow \infty \) we use an auxiliary function for which we prove some analytical properties. Also recall that \(\Omega \in {\mathbb {R}}\) is a unique constant which satisfies \(\Omega e^\Omega = 1\).

### Lemma 5.5

Let \(F:(0,1) \rightarrow {\mathbb {R}}\) be given by \(F(x) = x\log (x^{-1} -1)+x\). Then *F* has a unique maximum at \(\alpha = \frac{\Omega }{1 + \Omega } \in (0,1)\) and \(F(\alpha ) = \Omega < 1\).

### Proof

*F*(

*x*) is continuous on (0, 1), and \(F(x) \rightarrow 0\) as \(x \rightarrow 0\) and \(F(x) \rightarrow -\infty \) as \(x \rightarrow 1\). The first and second derivative are continuous and given by

*F*has a unique maximum at \(\alpha \) implicitly given by

Finally, we conclude this section by describing the asymptotic behaviour of \({\mathbb {G}}_n\).

### Proposition 5.6

The probability \({\mathbb {G}}_n\), that \(\langle x \rangle \) is a group where \(x \in {\mathcal {T}}_n\) is chosen with uniform distribution, tends to 0 exponentially at the rate less than \(1 - \Omega \).

### Proof

*F*is continuous on (0, 1) we conclude that \(\max _{x \in M_n} F(x)\rightarrow \max _{x\in (0,1)}F(x)\) as \(n\rightarrow \infty \). Therefore

### 5.3 \({\pmb {\mathbb {T}}}_n\) tends to zero

Recall that for \(n, r\in {\mathbb {N}}\) such that \(r \le n\), \({\mathcal {A}}(n,r)\) denotes the set of partitions of \(\{1,\ldots ,n\}\) into *r* non-empty components. Similarly, define \({\mathcal {B}}(n,r)\) to be the set of subsets of \(\{1, \ldots , n\}\) of cardinality *r*. Then \(|{\mathcal {B}}(n,r)|= \left( {\begin{array}{c}n\\ r\end{array}}\right) \).

### Lemma 5.7

### Proof

Let \(x, y \in {\mathcal {T}}_n\) be such that \({\text {rank}}(xyx) = {\text {rank}}(y)\). We first show that \({\text {im}}(xy)\) is contained in a transversal of *x*. Let \({\mathfrak {T}}\) be a transversal of *xyx*. Then *xyx* is injective on \({\mathfrak {T}}\) by definition, and so *x* is injective on \(({\mathfrak {T}})xy\). Hence \({\text {im}}(xy) = ({\mathfrak {T}})xy\) is contained in a transversal of *x*.

*x*. Then there are \(\left( {\begin{array}{c}n\\ r\end{array}}\right) r!\) choices for

*x*. Since

*x*is injective on \({\text {im}}(xy)\), there are

*x*. Since \(({\text {im}}(x))y = {\text {im}}(xy) = {\text {im}}(y)\), there are \(\left\{ \begin{array}{c} {r} \\ {k}\end{array}\right\} k!\) ways for

*y*to map \({\text {im}}(x)\) to \({\text {im}}(y)\). Finally, \((\{1, \ldots , n\}) {\setminus } {\text {im}}(x))y\subseteq {\text {im}}(y)\), and so there \(k^{n - r}\) for

*y*to map \((\{1, \ldots , n\}) {\setminus } {\text {im}}(x))\) to \({\text {im}}(y)\). Hence there are in total

*y*. Therefore

Next, we simplify the expression for \({\mathbb {T}}_n\).

### Lemma 5.8

### Proof

*s*. Note that every \(A_i\) is non-empty, so \(k \le s \le n - (r - k)\). Now suppose that only

*B*is fixed, then for every value of \(s \in \{k, \ldots , n + k - r\}\), there are \(\left( {\begin{array}{c}n\\ s\end{array}}\right) \) choices for \(\bigcup \{A_b : b \in B\}\), and there are \(\left\{ \begin{array}{c} {n-s} \\ {r-k}\end{array}\right\} \) many choices to choose \(\{A_b : b \notin B\}\). Hence we can write

Finally, we prove the main lemma of this section.

### Lemma 5.9

There exist \(r \in (0,1)\) and \(c > 0\) such that \({\mathbb {T}}_n \le c n^{7/2}r^n\).

### Proof

Note that by Stirling’s approximation there are constants \(a, b> 0\) such that \(a n^n e^{-n} \le n! \le b n^{n + \frac{1}{2}} e^{-n}\) for all \(n\in {\mathbb {N}}\).

*F*is continuous on a compact set \([0, 1]^2\), and so has a maximum. Hence we only need to consider the boundary of the domain and stationary points of

*F*, that is points in \([0,1]^2\) where \(\partial F/\partial x = 0 = \partial F/\partial y\). However, while it can be immediately be deduced from plots, using any mathematical software, that the maximum of

*F*is strictly less than 1, we show it here analytically. To this end, define the functions \(F_1, F_3: [0, 1] \rightarrow {\mathbb {R}}\) and \(F_2: [0,1]^2 \rightarrow {\mathbb {R}}\) by

*W*is the Lambert-W function. It follows that \(F_1\) is bounded from above by \(\max \{F_1(0), F_1(1), F(x_0)\}\). A simple algebraic manipulation gives

*F*(

*x*,

*y*) continuous on \([0,1]^2\) there is \(\varepsilon > 0\) and \(\beta < 1\) such that \(F(x, y) \le \beta \) for all \(x \in [1/2 - \varepsilon , 1]\) and all \(y \in [0, 1]\).

Finally, recall that \(x_0 = 1/(1 + W(e^{-1})) > 0.78\) and \(F_1(x)\) is increasing on \([0, x_0)\). We observe that if \(x \in [0, 1/2 - \varepsilon ] \subseteq [0, x_0)\) then \(0 = F_1(0) \le F_1(x) \le F_1(1/2 - \varepsilon ) < F_1(1/2) = 1\). Since \(F_2(x, y) \le 1\), it follows that \(F(x, y) \le F_1(1/2 - \varepsilon ) < 1\) for all \(x \in [0, 1/2 - \varepsilon ]\) and all \(y \in [0, 1]\). Therefore \(F(x, y) \le \max \{\beta , F_1(1/2 - \varepsilon )\} < 1\) for all \(x, y \in [0, 1]\), as required. \(\square \)

The following is an immediate corollary of Lemmas 5.7 and 5.9.

### Corollary 5.10

The probability \({\mathbb {T}}_n\), that \({\text {rank}}(xyx) = {\text {rank}}(y)\) where \(x, y \in {\mathcal {T}}_n\) are chosen with uniform distribution, tends to 0 as \(n \rightarrow \infty \) exponentially fast.

### 5.4 \({\pmb {\mathbb {V}}}_n\) tends to zero

We start by finding an expression for \({\mathbb {V}}_n\) in terms of *n*. The argument is similar to the proof of Lemma 5.7.

### Lemma 5.11

### Proof

If \(x, y, z \in {\mathcal {T}}_n\) are such that \({\text {rank}}(xyz) = {\text {rank}}(y)\). We first show that \({\text {im}}(xy)\) is contained in a transversal of *z*. Let \({\mathfrak {T}}\) be a transversal of *xyz*. Then *xyz* is injective on \({\mathfrak {T}}\) by definition, and so *z* is injective on \(({\mathfrak {T}})xy\). Hence \({\text {im}}(xy) = ({\mathfrak {T}})xy\) is contained in a transversal of *z*.

*z*. Note that \(t \le r\) and \(t \le k\). Then there are \(\left( {\begin{array}{c}n\\ r\end{array}}\right) \left\{ \begin{array}{c} {n} \\ {r}\end{array}\right\} r!\) choices for

*x*and \(\left( {\begin{array}{c}n\\ k\end{array}}\right) k!\) choices for

*z*. Since

*z*is injective on \({\text {im}}(xy)\), there are

*z*. Since \(({\text {im}}(x))y = {\text {im}}(xy) = {\text {im}}(y)\), there are \(\left\{ \begin{array}{c} {r} \\ {t}\end{array}\right\} t!\) ways for

*y*to map \({\text {im}}(x)\) to \({\text {im}}(y)\). Finally, \((\{1, \ldots , n\}) {\setminus } {\text {im}}(x))y\subseteq {\text {im}}(y)\), and so there \(t^{n - r}\) ways for

*y*to map \((\{1, \ldots , n\}) {\setminus } {\text {im}}(x))\) to \({\text {im}}(y)\). Hence there are in total

*y*. Therefore

Finally, we prove the main two lemmas of this section. This is an analogue of Lemma 5.9.

### Lemma 5.12

### Proof

*n*times their maximal value we obtain after some algebraic manipulation

*G*is continuous and bounded on

*G*to the closure \({{\overline{X}}}\). It remains to find the maximum of

*G*, which we do in the last lemma of this section.

### Lemma 5.13

There exists \(r \in (0, 1)\) such that \(G(x, y, z) \le r\) for all \(x, y, z \in [0, 1]\) such that \(z \le \min (x, y)\).

### Proof

*G*on the boundary \({\overline{X}}{\setminus } X\). Clearly for either \(x=0\), \(y = 0\) or \(z = 0\) we have \(G(x,y,z)=0\). If \(x=1\)

*G*are as follows

*G*(

*x*,

*y*,

*z*). Note that \(G(x, y, z) > 0\) if \(x, y, z \ne 0\), and so

*G*(

*x*,

*y*,

*z*) are of the form \((\alpha , \alpha , \gamma )\).

The following is an immediate corollary of Lemmas 5.12 and 5.13, which concludes the proof of Theorems 1.2 and 1.3.

### Corollary 5.14

The probability \({\mathbb {V}}_n\), that \({\text {rank}}(xyz) = {\text {rank}}(y)\) where \(x, y, z \in {\mathcal {T}}_n\) are chosen with uniform distribution, tends to 0 as \(n \rightarrow \infty \) exponentially fast.

## Notes

### Acknowledgements

Open access funding provided by TU Wien (TUW).

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