Finite skew braces with isomorphic non-abelian characteristically simple additive and circle groups

Cindy (Sin Yi) Tsang

doi:10.1515/jgth-2021-0044

Article Publicly Available

Finite skew braces with isomorphic non-abelian characteristically simple additive and circle groups

Published/Copyright: September 28, 2021

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From the journal Journal of Group Theory Volume 25 Issue 2

Abstract

A skew brace is a triplet ( A , ⋅ , ∘ ) , where ( A , ⋅ ) and ( A , ∘ ) are groups such that the brace relation x ∘ ( y ⋅ z ) = ( x ∘ y ) ⋅ x - 1 ⋅ ( x ∘ z ) holds for all x , y , z ∈ A . In this paper, we study the number of finite skew braces ( A , ⋅ , ∘ ) , up to isomorphism, such that ( A , ⋅ ) and ( A , ∘ ) are both isomorphic to T n with 𝑇 non-abelian simple and n ∈ N . We prove that it is equal to the number of unlabeled directed graphs on n + 1 vertices, with one distinguished vertex, and whose underlying undirected graph is a tree. In particular, it depends only on 𝑛 and is independent of 𝑇.

1 Introduction

A skew brace is a triplet ( A , ⋅ , ∘ ) , where ( A , ⋅ ) and ( A , ∘ ) are groups such that the so-called brace relation

x ∘ ( y ⋅ z ) = ( x ∘ y ) ⋅ x - 1 ⋅ ( x ∘ z )

holds for all x , y , z ∈ A , and here x - 1 denotes the inverse of 𝑥 in ( A , ⋅ ) . We note that ( A , ⋅ ) and ( A , ∘ ) must have the same identity element. Often ( A , ⋅ ) is referred to as the additive group, and ( A , ∘ ) as the circle or multiplicative group, of the skew brace. Despite the terminology, the group ( A , ⋅ ) need not be abelian. A skew brace ( A , ⋅ , ∘ ) with abelian additive group ( A , ⋅ ) is called a brace or sometimes a classical brace.

Braces were introduced by W. Rump [12] in order to study non-degenerate set-theoretic solutions of the Yang–Baxter equation which are involutive. L. Guarnieri and L. Vendramin [10] later extended this notion to skew braces to study all such solutions which are not necessarily involutive. Enumerating all skew braces (up to isomorphism) has thus become a problem of interest. As the reader would guess, an isomorphism between two skew braces ( A 1 , ⋅ 1 , ∘ 1 ) and ( A 2 , ⋅ 2 , ∘ 2 ) is a bijective map ϕ : A 1 → A 2 such that

ϕ ⁢ ( x ⋅ 1 y ) = ϕ ⁢ ( x ) ⋅ 2 ϕ ⁢ ( y ) , ϕ ⁢ ( x ∘ 1 y ) = ϕ ⁢ ( x ) ∘ 2 ϕ ⁢ ( y )

for all x , y , ∈ A 1 . In [10], an algorithm to compute finite skew braces of small order is also given, which was later improved by [1]. In [10], they also showed that there is a connection between skew braces and regular subgroups in the holomorph, as we shall explain.

We shall only consider skew braces with a finite underlying set. Given any finite groups 𝐺 and 𝑁 of the same order, define

b ( G , N ) = # { isomorphism classes of skew braces ⁢ ( A , ⋅ , ∘ ) such that ( A , ⋅ ) ≃ N and ( A , ∘ ) ≃ G } .

Recall that a subgroup of Perm ⁢ ( N ) , the group of permutations of 𝑁, is said to be regular if its action on 𝑁 is both transitive and free. For example, both λ ⁢ ( N ) and ρ ⁢ ( N ) are regular, where

{ λ : N → Perm ⁢ ( N ) , λ ⁢ ( η ) = ( x ↦ η x ) , ρ : N → Perm ⁢ ( N ) , ρ ⁢ ( η ) = ( x ↦ x η - 1 ) ,

are the left and right regular representations of 𝑁. The holomorph of 𝑁 is

Hol ⁢ ( N ) = ρ ⁢ ( N ) ⋊ Aut ⁢ ( N ) = λ ⁢ ( N ) ⋊ Aut ⁢ ( N ) .

Let us further put

R ⁢ ( G , N ) = { regular subgroups of ⁢ Hol ⁢ ( N ) ⁢ which are isomorphic to ⁢ G } .

Note that Aut ⁢ ( N ) acts on this set via conjugation in Hol ⁢ ( N ) .

Proposition 1.1

For any finite groups 𝐺 and 𝑁 of the same order, we have

b ⁢ ( G , N ) = # ⁢ ( R ⁢ ( G , N ) / Aut ⁢ ( N ) ) ,

the number of orbits of R ⁢ ( G , N ) under conjugation by Aut ⁢ ( N ) .

Proof

See [10, Theorem 4.2 and Proposition 4.3] or [13, Appendix]. ∎

Now, regular subgroups in the holomorph are also known to be related to Hopf–Galois structures. More specifically, define

e ( G , N ) = # { Hopf–Galois structures of type ⁢ N on a Galois extension L / K with Gal ( L / K ) ≃ G } .

We refer the reader to [7, Chapter 2] for the definitions and background.

Proposition 1.2

For any finite groups 𝐺 and 𝑁 of the same order, we have

e ⁢ ( G , N ) = | Aut ⁢ ( G ) | | Aut ⁢ ( N ) | ⋅ # ⁢ R ⁢ ( G , N ) .

Proof

This follows from work of [9, 3], or see [7, Chapter 2]. ∎

From Propositions 1.1 and 1.2, we see that

b ⁢ ( G , N ) = 0 if and only if e ⁢ ( G , N ) = 0 .

However, the numbers b ⁢ ( G , N ) and e ⁢ ( G , N ) need not be equal, and the calculation of the former is more difficult in general because one also needs to take the conjugation action of Aut ⁢ ( N ) into account.

Assumption

In the rest of this paper, we shall take 𝐺 to be a finite non-abelian characteristically simple group. Equivalently, this means G = T n for some finite non-abelian simple group 𝑇 and n ∈ N .

By drawing tools from graph theory and some consequences of the classification of finite simple groups (CFSG), in [14], the author obtained the following closed formula for the number e ⁢ ( G , G ) .

Theorem 1.3

We have the formula

e ⁢ ( G , G ) = 2 n ⁢ ( n ⁢ | Aut ⁢ ( T ) | + 1 ) n - 1 .

For n = 1 , by Proposition 1.2 and Theorem 1.3, we have

e ⁢ ( G , G ) = # ⁢ R ⁢ ( G , G ) = 2 , and so R ⁢ ( G , G ) = { λ ⁢ ( G ) , ρ ⁢ ( G ) } .

This fact was first shown in [6]; also see [15, Theorem 1.3] for generalization to quasisimple groups. From this and Proposition 1.1, we get b ⁢ ( G , G ) = 2 .

For n ≥ 2 , however, the size of R ⁢ ( G , G ) becomes much larger, and it is no longer obvious what the value of b ⁢ ( G , G ) ought to be simply from Theorem 1.3. Nonetheless, we are still able to determine b ⁢ ( G , G ) . To state our result, we first need to introduce some definitions.

Recall that a tree is a connected graph which has no cycle, or equivalently, a graph in which any two vertices can be connected by a unique simple path. It is known that a graph on n + 1 vertices is a tree if and only if it has exactly 𝑛 edges and is connected.

Definition 1.4

In this paper, we shall use directed tree to mean a directed graph whose underlying undirected graph is a tree. For brevity, put

N n = { 1 , … , n } and N 0 , n = N n ⊔ { 0 } .

Let T ⁢ ( n ) be the set of all labeled directed trees on n + 1 vertices, labeled by N 0 , n . Two directed graphs Γ 1 , Γ 2 ∈ T ⁢ ( n ) are equivalent if there is a bijection

ξ ∈ Map ⁢ ( N 0 , n , N 0 , n ) with ⁢ ξ ⁢ ( 0 ) = 0

such that, for any vertices v , v ′ ∈ N 0 , n , there is an arrow v → v ′ in Γ 1 precisely when there is an arrow ξ ⁢ ( v ) → ξ ⁢ ( v ′ ) in Γ 2 . We shall write Γ 1 ∼ Γ 2 in this case.

While e ⁢ ( G , G ) depends upon both 𝑛 and 𝑇, the number b ⁢ ( G , G ) turns out to depend only on 𝑛. Our main theorem is the following.

Theorem 1.5

We have the equality

b ( G , G ) = # ( T ( n ) / ∼ ) ,

the number of equivalence classes of ∼ on the set T ⁢ ( n ) .

Observe that, unlike Theorem 1.3, here we do not have a closed formula for b ⁢ ( G , G ) . Let us briefly explain why this is expected to be difficult.

Remark 1.6

The equivalence relation ∼ in Definition 1.4 simply means that we can forget about the labeling of the vertices except the vertex 0. We then see that # ( T ( n ) / ∼ ) is equal to the size of the set

T ⋆ ( n ) = { unlabeled directed trees on ⁢ n + 1 ⁢ vertices with a distinguished vertex (root) } .

The underlying undirected graph of an element Γ ∈ T ⋆ ⁢ ( n ) may be regarded as an (unordered) rooted tree.^[1] Now, the famous Cayley’s formula tells us that the number of labeled trees on n + 1 vertices is ( n + 1 ) n - 1 . It was also shown in [8] that, for each 1 ≤ d ≤ n , we have

(1.1) # { labeled trees on ⁢ n + 1 ⁢ vertices in which a specified vertex has degree d } = ( n - 1 d - 1 ) n n - d .

The formula for e ⁢ ( G , G ) in Theorem 1.3 was obtained using (1.1). However, to compute b ⁢ ( G , G ) , we need to consider unlabeled trees instead, and so far, there is no known closed formula for the number of unlabeled (rooted) trees; see [16, A000055 and A000081]. We also need to take the orientation of the directed trees into account, which adds even more complexity into the problem. It is possible that there does not even exist a closed formula for b ⁢ ( G , G ) .

2 Description of endomorphisms

The endomorphisms of 𝐺 play an important role in the proof of both Theorems 1.3 and 1.5. Following the notation in [14], let us give a description of these endomorphisms.

For each i ∈ N n , we shall write

T ( i ) = 1 × ⋯ × 1 × T × 1 × ⋯ × 1 ( T ⁢ is in the ⁢ i ⁢ th position )

and use x ( i ) to denote an arbitrary element of T ( i ) . For convenience, let T ( 0 ) denote the trivial subgroup and x ( 0 ) the identity element.

Proposition 2.1

Elements of End ⁢ ( G ) are precisely the maps

( x ( 1 ) , … , x ( n ) ) ↦ ( φ 1 ⁢ ( x ( θ ⁢ ( 1 ) ) ) , … , φ n ⁢ ( x ( θ ⁢ ( n ) ) ) ) ,

where θ ∈ Map ⁢ ( N n , N 0 , n ) , and φ i ∈ Hom ⁢ ( T ( θ ⁢ ( i ) ) , T ( i ) ) for each i ∈ N n . Moreover, the above map lies in Aut ⁢ ( G ) precisely when 𝜃 is injective with image equal to N n and φ i is bijective for each i ∈ N n .

Proof

This basically follows from the proof of [4, Lemma 3.2], which gives a description of elements of Aut ⁢ ( G ) . ∎

In view of Proposition 2.1, we introduce the following notation.

Definition 2.2

Given any f ∈ End ⁢ ( G ) , we shall write

f ⁢ ( x ( 1 ) , … , x ( n ) ) = ( φ f , 1 ⁢ ( x ( θ f ⁢ ( 1 ) ) ) , … , φ f , n ⁢ ( x ( θ f ⁢ ( n ) ) ) ) ,

where θ f ∈ Map ⁢ ( N n , N 0 , n ) , and φ f , i ∈ Hom ⁢ ( T ( θ f ⁢ ( i ) ) , T ( i ) ) for each i ∈ N n . To ensure that θ f is uniquely determined by 𝑓, we shall assume that

(2.1) φ f , i ⁢ is non-trivial when ⁢ θ f ⁢ ( i ) ≠ 0 .

Note that, in this case, φ f , i is necessarily bijective because 𝑇 is simple. Let us define θ f ⁢ ( 0 ) = 0 so that we may regard θ f ∈ Map ⁢ ( N 0 , n , N 0 , n ) . Also, we shall write φ f , 0 ∈ Hom ⁢ ( T ( 0 ) , T ( 0 ) ) for the trivial map for convenience.

3 Characterization of regular subgroups

In [14], the present author computed e ⁢ ( G , G ) by first giving a characterization of R ⁢ ( G , G ) in terms of trees. Let us recall this characterization.

Using a combinatorial argument, it was shown in [14, Section 3] that

(3.1) every element of ⁢ R ⁢ ( G , G ) ⁢ lies in ⁢ InHol ⁢ ( G ) = ρ ⁢ ( G ) ⋊ Inn ⁢ ( G ) ,

where Inn ⁢ ( G ) denotes the inner automorphism group of 𝐺. We remark that the proof of (3.1) requires the fact that 𝑇 has solvable outer automorphism group, which is known as the Schreier conjecture and is a consequence of CFSG.

Recall that a pair ( f , g ) , with f , g ∈ End ⁢ ( G ) , is fixed point free (fpf) if

f ⁢ ( x ) = g ⁢ ( x ) ⁢ holds exactly when ⁢ x = 1 .

Given any f , g ∈ End ⁢ ( G ) , we can define a subgroup of InHol ⁢ ( G ) by setting

(3.2) G ( f , g ) = { ρ ⁢ ( g ⁢ ( x ) ) ⁢ λ ⁢ ( f ⁢ ( x ) ) : x ∈ G } .

It is easy to check that G ( f , g ) is regular, in which case G ( f , g ) ≃ G , if and only if ( f , g ) is fpf; see [5, Proposition 1]. Since 𝐺 has trivial center, these are all the regular subgroups isomorphic to 𝐺 in InHol ⁢ ( G ) by [5, Proposition 6], so

(3.3) R ⁢ ( G , G ) = { G ( f , g ) : f , g ∈ End ⁢ ( G ) ⁢ such that ⁢ ( f , g ) ⁢ is fpf } .

We are thus reduced to the problem of determining when a pair

( f , g ) with ⁢ f , g ∈ End ⁢ ( G )

is fpf. Plainly, ( f , g ) is fpf if and only if the equations

(3.4) φ f , i ⁢ ( x ( θ f ⁢ ( i ) ) ) = φ g , i ⁢ ( x ( θ g ⁢ ( i ) ) ) ,

for 𝑖 ranging over N n , have no common solution except

x ( 1 ) = ⋯ = x ( n ) = 1 .

Here the notation is as in Definition 2.2. To study this system of equations, in [14], the author developed a graph-theoretic method, as follows.

Definition 3.1

Given any f , g ∈ End ⁢ ( G ) , define Γ { f , g } to be the undirected multigraph with vertex set N 0 , n , and we draw one edge e i between θ f ⁢ ( i ) and θ g ⁢ ( i ) for each i ∈ N n . Note that Γ { f , g } has n + 1 vertices and 𝑛 edges, so it is a tree precisely when it is connected.

The n + 1 vertices of Γ { f , g } should be viewed as the identity element x ( 0 ) and the variables x ( 1 ) , … , x ( n ) . For each i ∈ N n , the edge e i represents (3.4), and

(3.5) ( 3.4 ) is equivalent to ⁢ { x ( θ f ⁢ ( i ) ) = ( φ f , i - 1 ∘ φ g , i ) ⁢ ( x ( θ g ⁢ ( i ) ) ) if ⁢ θ f ⁢ ( i ) ≠ 0 , x ( θ g ⁢ ( i ) ) = ( φ g , i - 1 ∘ φ f , i ) ⁢ ( x ( θ f ⁢ ( i ) ) ) if ⁢ θ g ⁢ ( i ) ≠ 0 ,

by condition (2.1). Thus, we can solve equations (3.4) by “following the edges” in the graph Γ { f , g } . Let us illustrate the idea via two examples.

Example 3.2

Take n = 3 , and let f , g ∈ End ⁢ ( G ) be such that

f ⁢ ( x ( 1 ) , x ( 2 ) , x ( 3 ) ) = ( φ f , 1 ⁢ ( x ( 2 ) ) , φ f , 2 ⁢ ( x ( 0 ) ) , φ f , 3 ⁢ ( x ( 1 ) ) ) , g ⁢ ( x ( 1 ) , x ( 2 ) , x ( 3 ) ) = ( φ g , 1 ⁢ ( x ( 1 ) ) , φ g , 2 ⁢ ( x ( 3 ) ) , φ g , 3 ⁢ ( x ( 0 ) ) ) .

The associated graph Γ { f , g } is given by

and is a tree. If f ⁢ ( x ( 1 ) , x ( 2 ) , x ( 3 ) ) = g ⁢ ( x ( 1 ) , x ( 2 ) , x ( 3 ) ) holds, then based on the observations in (3.5), we see that

the edge 0 - 1 tells us that x ( 1 ) = ( φ f , 3 - 1 ∘ φ g , 3 ) ⁢ ( x ( 0 ) ) = 1 ,
the edge 0 - 3 tells us that x ( 3 ) = ( φ g , 2 - 1 ∘ φ f , 2 ) ⁢ ( x ( 0 ) ) = 1 ,
the edge 1 - 2 tells us that x ( 2 ) is determined by x ( 1 ) with
x ( 2 ) = ( φ f , 1 - 1 ∘ φ g , 1 ) ⁢ ( x ( 1 ) ) .
Since we already know that x ( 1 ) = 1 , this implies that x ( 2 ) = 1 also.

We conclude that the pair ( f , g ) is fpf.

Example 3.3

Take n = 3 , and let f , g ∈ End ⁢ ( G ) be such that

f ⁢ ( x ( 1 ) , x ( 2 ) , x ( 3 ) ) = ( φ f , 1 ⁢ ( x ( 0 ) ) , φ f , 2 ⁢ ( x ( 2 ) ) , φ f , 3 ⁢ ( x ( 2 ) ) ) , g ⁢ ( x ( 1 ) , x ( 2 ) , x ( 3 ) ) = ( φ g , 1 ⁢ ( x ( 1 ) ) , φ g , 2 ⁢ ( x ( 3 ) ) , φ g , 3 ⁢ ( x ( 3 ) ) ) .

The associated graph Γ { f , g } is given by

and is not a tree. The cycle 2 - 3 - 2 may be thought of as representing

( φ f , 3 - 1 ∘ φ g , 3 ∘ φ g , 2 - 1 ∘ φ f , 2 ) ⁢ ( x ( 2 ) ) = x ( 2 ) .

It is a consequence of CFSG that 𝑇 has no fpf automorphism, whence there exists σ ( 2 ) ∈ T ( 2 ) with σ ( 2 ) ≠ 1 such that

( φ f , 3 - 1 ∘ φ g , 3 ∘ φ g , 2 - 1 ∘ φ f , 2 ) ⁢ ( σ ( 2 ) ) = σ ( 2 ) .

We then see that ( f , g ) has a fixed point other than the identity, namely

( x ( 1 ) , x ( 2 ) , x ( 3 ) ) = ( 1 , σ ( 2 ) , ( φ g , 2 - 1 ∘ φ f , 2 ) ⁢ ( σ ( 2 ) ) ) .

We conclude that the pair ( f , g ) is not fpf.

The above ideas may be formalized, and we have the following criterion.

Proposition 3.4

A pair ( f , g ) , with f , g ∈ End ⁢ ( G ) , is fpf if and only if the graph Γ { f , g } is a tree.

Proof

See [14, Propositions 2.5 and 2.9]. ∎

Combining (3.3) and Proposition 3.4, we obtain

(3.6) R ⁢ ( G , G ) = { G ( f , g ) : f , g ∈ End ⁢ ( G ) ⁢ such that ⁢ Γ { f , g } ⁢ is tree } .

We remark that the proof of Theorem 1.3 given in [14] uses this characterization of R ⁢ ( G , G ) and formula (1.1) with 0 as the specified vertex.

4 Counting orbits of regular subgroups

In this section, we shall prove Theorem 1.5. To do so, by Proposition 1.1, we need to consider the conjugation action of Aut ⁢ ( G ) on elements of R ⁢ ( G , G ) , and we shall again use graph theory. Let us first refine Definition 3.1 as follows.

Definition 4.1

Given any f , g ∈ End ⁢ ( G ) , define Γ ( f , g ) , → to be the directed multigraph obtained from Γ { f , g } by replacing the edge e i with an arrow from θ f ⁢ ( i ) to θ g ⁢ ( i ) for each i ∈ N n . The arrow → in the subscript indicates that the orientation is chosen to go from θ f ⁢ ( ⋅ ) to θ g ⁢ ( ⋅ ) .

Recall the definition in (3.2) and also the equivalence relation ∼ on the set T ⁢ ( n ) in Definition 1.4. The key is the next theorem.

Theorem 4.2

Let ( f , g ) and ( f ′ , g ′ ) , with f , g , f ′ , g ′ ∈ End ⁢ ( G ) , be such that the graphs Γ { f , g } and Γ { f ′ , g ′ } are trees. Then the subgroups G ( f , g ) and G ( f ′ , g ′ ) are Aut ⁢ ( G ) -conjugates exactly when Γ ( f , g ) , → ∼ Γ ( f ′ , g ′ ) , → .

From (3.6) and Theorem 4.2, we see that G ( f , g ) ↦ Γ ( f , g ) , → induces a well-defined injection

{ Aut ( G ) -orbits of R ( G , G ) } → { equivalence classes of ∼ on T ( n ) } .

This map is also surjective. Indeed, given any Γ in T ⁢ ( n ) , label the 𝑛 arrows in Γ as a 1 , … , a n , where the numbering may be randomly chosen. For each i ∈ N n , simply define θ f ⁢ ( i ) to be the tail and θ g ⁢ ( i ) to be the head of a i . Then, clearly, Γ = Γ ( f , g ) , → for f , g ∈ End ⁢ ( G ) given by

f ⁢ ( x ( 1 ) , … , x ( n ) ) = ( x ( θ f ⁢ ( 1 ) ) , … , x ( θ f ⁢ ( n ) ) ) , g ⁢ ( x ( 1 ) , … , x ( n ) ) = ( x ( θ g ⁢ ( 1 ) ) , … , x ( θ g ⁢ ( n ) ) ) .

Theorem 1.5 now follows from Proposition 1.1.

In the next two subsections, we shall prove Theorem 4.2, from which Theorem 1.5 would follow. Before we proceed, let us make two observations.

Lemma 4.3

Let ( f , g ) and ( f ′ , g ′ ) , with f , g ∈ End ⁢ ( G ) , be such that Γ { f , g } and Γ { f ′ , g ′ } are trees. Then G ( f , g ) and G ( f ′ , g ′ ) are Aut ⁢ ( G ) -conjugates precisely when there exist ψ , π ∈ Aut ⁢ ( G ) such that

(4.1) ( f ′ , g ′ ) = ( ψ ∘ f ∘ π , ψ ∘ g ∘ π ) .

Proof

Recall that ( f , g ) and ( f ′ , g ′ ) are fpf by Proposition 3.4, so then G ( f , g ) and G ( f ′ , g ′ ) are both regular subgroups. They are Aut ⁢ ( G ) -conjugates if and only if there exist ψ ∈ Aut ⁢ ( G ) and π ∈ Perm ⁢ ( G ) such that, for all x ∈ G , we have

(4.2) ρ ⁢ ( g ′ ⁢ ( x ) ) ⋅ λ ⁢ ( f ′ ⁢ ( x ) ) = ψ ⁢ ρ ⁢ ( g ⁢ ( π ⁢ ( x ) ) ) ⁢ ψ - 1 ⋅ ψ ⁢ λ ⁢ ( f ⁢ ( π ⁢ ( x ) ) ) ⁢ ψ - 1 = ρ ⁢ ( ( ψ ∘ g ∘ π ) ⁢ ( x ) ) ⋅ λ ⁢ ( ( ψ ∘ f ∘ π ) ⁢ ( x ) ) .

Note that λ ⁢ ( G ) and ρ ⁢ ( G ) intersect trivially because 𝐺 has trivial center. It then follows that equation (4.2) holds for all x ∈ G exactly when (4.1) is satisfied. Thus, it remains to prove that 𝜋 must be a homomorphism in this case. Since ψ ∈ Aut ⁢ ( G ) , from (4.1), we deduce that f ∘ π and g ∘ π are both homomorphisms, so for all x , y ∈ G , we have

π ⁢ ( x ⁢ y ) ≡ π ⁢ ( x ) ⁢ π ⁢ ( y ) ( mod ker ⁡ ( f ) ∩ ker ⁡ ( g ) ) .

But ( f , g ) is fpf, whence ker ⁡ ( f ) and ker ⁡ ( g ) intersect trivially. We see that 𝜋 is indeed a homomorphism. ∎

Lemma 4.4

Let f ∈ End ⁢ ( G ) and ψ , π ∈ Aut ⁢ ( G ) . Then we have

θ ψ ∘ f ∘ π = θ π ∘ θ f ∘ θ ψ in ⁢ Map ⁢ ( N 0 , n , N 0 , n ) ,

and for each i ∈ N n , we also have

φ ψ ∘ f ∘ π , i = φ ψ , i ∘ φ f , θ ψ ⁢ ( i ) ∘ φ π , ( θ f ∘ θ ψ ) ⁢ ( i ) .

Here the notation is as in Definition 2.2.

Proof

We compute that

( x ( 1 ) , … , x ( n ) ) ↦ π ( … , φ π , i ⁢ ( x ( θ π ⁢ ( i ) ) ) , … ) ↦ f ( … , ( φ f , i ∘ φ π , θ f ⁢ ( i ) ) ⁢ ( x ( ( θ π ∘ θ f ) ⁢ ( i ) ) ) , … ) ↦ ψ ( … , ( φ ψ , i ∘ φ f , θ ψ ⁢ ( i ) ∘ φ π , ( θ f ∘ θ ψ ) ⁢ ( i ) ) ⁢ ( x ( ( θ π ∘ θ f ∘ θ ψ ) ⁢ ( i ) ) ) , … ) ,

and from here, the claims are clear once we verify condition (2.1). More precisely, we need to check that, for each i ∈ N n , we have

(4.3) φ ψ , i ∘ φ f , θ ψ ⁢ ( i ) ∘ φ π , ( θ f ∘ θ ψ ) ⁢ ( i ) ⁢ is non-trivial when ⁢ ( θ π ∘ θ f ∘ θ ψ ) ⁢ ( i ) ≠ 0 .

Since ψ , π ∈ Aut ⁢ ( G ) , the maps θ π and φ ψ , i are bijections, and φ π , ( θ f ∘ θ ψ ) ⁢ ( i ) is a bijection when ( θ f ∘ θ ψ ) ⁢ ( i ) ≠ 0 . It follows that (4.3) is equivalent to

φ f , θ ψ ⁢ ( i ) ⁢ is non-trivial when ⁢ θ f ⁢ ( θ ψ ⁢ ( i ) ) ≠ 0 ,

and this indeed holds by (2.1). ∎

Assumption

In the rest of this paper, we shall fix pairs ( f , g ) and ( f ′ , g ′ ) , with f , g , f ′ , g ′ ∈ End ⁢ ( G ) , to be such that Γ { f , g } and Γ { f ′ , g ′ } are trees.

4.1 Proof of Theorem 4.2: forward implication

This direction is very easy to prove. Suppose that G ( f , g ) and G ( f ′ , g ′ ) are Aut ⁢ ( G ) -conjugates. Then

( θ f ′ , θ g ′ ) = ( θ π ∘ θ f ∘ θ ψ , θ π ∘ θ g ∘ θ ψ ) for some ⁢ ψ , π ∈ Aut ⁢ ( G )

by Lemmas 4.3 and 4.4. Since ψ , π ∈ Aut ⁢ ( G ) , the maps

θ ψ , θ π : N 0 , n → N 0 , n ⁢ are bijections with ⁢ θ ψ ⁢ ( 0 ) = 0 = θ π ⁢ ( 0 ) .

For any vertices v , v ′ ∈ N 0 , n , we then see from Definition 4.1 that

there is an arrow ⁢ v → v ′ ⁢ in ⁢ Γ ( f , g ) , → ⇔ ( v , v ′ ) = ( θ f ⁢ ( i ) , θ g ⁢ ( i ) ) for some ⁢ i ∈ N n ⇔ ( v , v ′ ) = ( θ f ⁢ ( θ ψ ⁢ ( j ) ) , θ g ⁢ ( θ ψ ⁢ ( j ) ) ) for some ⁢ j ∈ N n ⇔ ( θ π ⁢ ( v ) , θ π ⁢ ( v ′ ) ) = ( ( θ π ∘ θ f ∘ θ ψ ) ⁢ ( j ) , ( θ π ∘ θ g ∘ θ ψ ) ⁢ ( j ) ) for some ⁢ j ∈ N n ⇔ ( θ π ⁢ ( v ) , θ π ⁢ ( v ′ ) ) = ( θ f ′ ⁢ ( j ) , θ g ′ ⁢ ( j ) ) for some ⁢ j ∈ N n ⇔ there is an arrow ⁢ θ π ⁢ ( v ) → θ π ⁢ ( v ′ ) ⁢ in ⁢ Γ ( f ′ , g ′ ) , → .

It thus follows that Γ ( f , g ) , → and Γ ( f ′ , g ′ ) , → are equivalent via θ π . This proves the forward implication of Theorem 4.2.

4.2 Proof of Theorem 4.2: backward implication

This direction is a lot harder to prove. Suppose that Γ ( f , g ) , → and Γ ( f ′ , g ′ ) , → are equivalent, which means that there exists a bijection

ξ ∈ Map ⁢ ( N 0 , n , N 0 , n ) with ⁢ ξ ⁢ ( 0 ) = 0

such that, for any vertices v , v ′ ∈ N 0 , n , there is an arrow v → v ′ in Γ ( f , g ) , → precisely when there is an arrow ξ ⁢ ( v ) → ξ ⁢ ( v ′ ) in Γ ( f ′ , g ′ ) , → . By Lemmas 4.3 and 4.4, we need to show that there exist ψ , π ∈ Aut ⁢ ( G ) such that

(4.4) ( θ f ′ , θ g ′ ) = ( θ π ∘ θ f ∘ θ ψ , θ π ∘ θ g ∘ θ ψ ) ,

and for each i ∈ N n , we have the equalities

(4.5) φ f ′ , i = φ ψ , i ∘ φ f , θ ψ ⁢ ( i ) ∘ φ π , ( θ f ∘ θ ψ ) ⁢ ( i ) , φ g ′ , i = φ ψ , i ∘ φ g , θ ψ ⁢ ( i ) ∘ φ π , ( θ g ∘ θ ψ ) ⁢ ( i ) .

In other words, we need to find bijections

θ ψ , θ π ∈ Map ⁢ ( N 0 , n , N 0 , n ) with ⁢ θ ψ ⁢ ( 0 ) = 0 = θ π ⁢ ( 0 )

satisfying (4.4), as well as isomorphisms

(4.6) φ ψ , i ∈ Iso ⁢ ( T ( θ ψ ⁢ ( i ) ) , T ( i ) ) , φ π , i ∈ Iso ⁢ ( T ( θ π ⁢ ( i ) ) , T ( i ) ) for ⁢ i ∈ N n

satisfying both equations in (4.5).

First, we consider condition (4.4). Let us take θ π = ξ , and we shall define θ ψ via the next proposition. It will be helpful to recall Definition 4.1.

Proposition 4.5

For each i ∈ N n , there is a unique θ ψ ⁢ ( i ) ∈ N n such that

( θ π - 1 ⁢ ( θ f ′ ⁢ ( i ) ) , θ π - 1 ⁢ ( θ g ′ ⁢ ( i ) ) ) = ( θ f ⁢ ( θ ψ ⁢ ( i ) ) , θ g ⁢ ( θ ψ ⁢ ( i ) ) ) .

Moreover, the map

θ ψ ∈ Map ⁢ ( N 0 , n , N 0 , n ) , where we define ⁢ θ ψ ⁢ ( 0 ) = 0 ,

is a bijection.

Proof

For i ∈ N n , there is an arrow θ f ′ ⁢ ( i ) → θ g ′ ⁢ ( i ) in Γ ( f ′ , g ′ ) , → by definition, so there is an arrow θ π - 1 ⁢ ( θ f ′ ⁢ ( i ) ) → θ π - 1 ⁢ ( θ g ′ ⁢ ( i ) ) in Γ ( f , g ) , → . This means that

( θ π - 1 ⁢ ( θ f ′ ⁢ ( i ) ) , θ π - 1 ⁢ ( θ g ′ ⁢ ( i ) ) ) = ( θ f ⁢ ( j i ) , θ g ⁢ ( j i ) )

for some j i ∈ N n , which is unique, for otherwise, there would be two distinct edges joining θ π - 1 ⁢ ( θ f ′ ⁢ ( i ) ) and θ π - 1 ⁢ ( θ g ′ ⁢ ( i ) ) in Γ { f , g } , and Γ { f , g } would not be a tree. Similarly, distinct 𝑖 give rise to distinct j i because Γ { f ′ , g ′ } is a tree. The claims are then clear by setting θ ψ ⁢ ( i ) = j i . ∎

We have thus chosen bijections θ ψ and θ π such that (4.4) is satisfied. Now, we also need to show that equations (4.5), for 𝑖 ranging over N n , have a common solution (4.6). Let us put

N n , f = { i ∈ N n : θ f ′ ⁢ ( i ) = 0 } = { i ∈ N n : θ f ⁢ ( θ ψ ⁢ ( i ) ) = 0 } , N n , g = { i ∈ N n : θ g ′ ⁢ ( i ) = 0 } = { i ∈ N n : θ g ⁢ ( θ ψ ⁢ ( i ) ) = 0 } ,

where the latter equalities follow from (4.4) and the fact that θ π is bijection with θ π ⁢ ( 0 ) = 0 . By Definition 2.2, we have

{ φ f ′ , i ∈ Hom ⁢ ( T ( θ f ′ ⁢ ( i ) ) , T ( i ) ) , φ f , θ ψ ⁢ ( i ) ∈ Hom ⁢ ( T ( θ f ⁢ ( θ ψ ⁢ ( i ) ) ) , T ( θ ψ ⁢ ( i ) ) ) , φ g ′ , i ∈ Hom ⁢ ( T ( θ g ′ ⁢ ( i ) ) , T ( i ) ) , φ g , θ ψ ⁢ ( i ) ∈ Hom ⁢ ( T ( θ g ⁢ ( θ ψ ⁢ ( i ) ) ) , T ( θ ψ ⁢ ( i ) ) ) .

Thus, the first and second equations in (4.5), respectively, trivially hold for i ∈ N n , f and i ∈ N n , g . So we only need to find (4.6) which satisfies

(4.7) φ f ′ , i = φ ψ , i ∘ φ f , θ ψ ⁢ ( i ) ∘ φ π , ( θ f ∘ θ ψ ) ⁢ ( i ) ⁢ ⁢ for ⁢ i ∈ N n ∖ N n , f ,

(4.8) φ g ′ , i = φ ψ , i ∘ φ g , θ ψ ⁢ ( i ) ∘ φ π , ( θ g ∘ θ ψ ) ⁢ ( i ) ⁢ ⁢ for ⁢ i ∈ N n ∖ N n , g .

We are in a situation similar to (3.4). Again, we shall use graph theory and represent these equations as edges on a graph, as follows.

Definition 4.6

Let us put

Ψ = { 1 ψ , … , n ψ } and Π = { 1 π , … , n π } .

Define Γ Ψ , Π to be the undirected simple graph with vertex set Ψ ⊔ Π , and we draw one edge e i , f between i ψ and ( θ f ∘ θ ψ ) ⁢ ( i ) π for each i ∈ N n ∖ N n , f , as well as one edge e i , g between i ψ and ( θ g ∘ θ ψ ) ⁢ ( i ) π for each i ∈ N n ∖ N n , g . For all i ∈ N n , we have θ ψ ⁢ ( i ) ∈ N n , and so ( θ f ∘ θ ψ ) ⁢ ( i ) ≠ ( θ g ∘ θ ψ ) ⁢ ( i ) because Γ { f , g } has no loop. This means that Γ Ψ , Π is indeed a simple graph. Also, Γ Ψ , Π is a bipartite graph, with parts Ψ and Π.

Definition 4.7

For each i ∈ N n , define

{ Φ i , f : Iso ⁢ ( T ( θ ψ ⁢ ( i ) ) , T ( i ) ) → Iso ⁢ ( T ( θ f ′ ⁢ ( i ) ) , T ( ( θ f ∘ θ ψ ) ⁢ ( i ) ) ) when ⁢ i ∉ N n , f , Φ i , g : Iso ⁢ ( T ( θ ψ ⁢ ( i ) ) , T ( i ) ) → Iso ⁢ ( T ( θ g ′ ⁢ ( i ) ) , T ( ( θ g ∘ θ ψ ) ⁢ ( i ) ) ) when ⁢ i ∉ N n , g ,

respectively, by setting

{ Φ i , f ⁢ ( φ ) = φ f , θ ψ ⁢ ( i ) - 1 ∘ φ - 1 ∘ φ f ′ , i when ⁢ i ∉ N n , f , Φ i , g ⁢ ( φ ) = φ g , θ ψ ⁢ ( i ) - 1 ∘ φ - 1 ∘ φ g ′ , i when ⁢ i ∉ N n , g .

Note that these Φ i , f and Φ i , g are well-defined bijections by condition (2.1).

The vertices in Ψ and Π, respectively, should be viewed as the variables

φ ψ , 1 , … , φ ψ , n and φ π , 1 , … , φ π , n

which we wish to solve. For each i ∈ N n , the edge e i , f represents (4.7) when i ∉ N i , f , and e i , g represents (4.8) when i ∉ N i , g . Observe that

(4.9) { ( 4.7 ) is equivalent to ⁢ φ π , ( θ f ∘ θ ψ ) ⁢ ( i ) = Φ i , f ⁢ ( φ ψ , i ) for ⁢ i ∉ N n , f , ( 4.8 ) is equivalent to ⁢ φ π , ( θ g ∘ θ ψ ) ⁢ ( i ) = Φ i , g ⁢ ( φ ψ , i ) for ⁢ i ∉ N n , g .

Therefore, analogous to the case of (3.4), we may solve equations (4.7) and (4.8) by “following the edges” in the graph Γ Ψ , Π . Let us first consider an explicit example to illustrate the idea.

Example 4.8

Take n = 3 , and suppose that

f ⁢ ( x ( 1 ) , x ( 2 ) , x ( 3 ) ) = ( φ f , 1 ⁢ ( x ( 1 ) ) , φ f , 2 ⁢ ( x ( 3 ) ) , φ f , 3 ⁢ ( x ( 2 ) ) ) , g ⁢ ( x ( 1 ) , x ( 2 ) , x ( 3 ) ) = ( φ g , 1 ⁢ ( x ( 0 ) ) , φ g , 2 ⁢ ( x ( 1 ) ) , φ g , 3 ⁢ ( x ( 1 ) ) ) , f ′ ⁢ ( x ( 1 ) , x ( 2 ) , x ( 3 ) ) = ( φ f ′ , 1 ⁢ ( x ( 1 ) ) , φ f ′ , 2 ⁢ ( x ( 2 ) ) , φ f ′ , 3 ⁢ ( x ( 3 ) ) ) , g ′ ⁢ ( x ( 1 ) , x ( 2 ) , x ( 3 ) ) = ( φ g ′ , 1 ⁢ ( x ( 2 ) ) , φ g ′ , 2 ⁢ ( x ( 0 ) ) , φ g ′ , 3 ⁢ ( x ( 2 ) ) ) .

The graphs Γ ( f , g ) , → and Γ ( f ′ , g ′ ) , → , respectively, are equal to

according to Definition 4.1. We may take ξ = θ π to be the bijection

θ π ⁢ ( 0 ) = 0 , θ π ⁢ ( 1 ) = 2 , θ π ⁢ ( 2 ) = 1 , θ π ⁢ ( 3 ) = 3 ,

and following Proposition 4.5, we see that θ ψ in turn is given by

θ ψ ⁢ ( 0 ) = 0 , θ ψ ⁢ ( 1 ) = 3 , θ ψ ⁢ ( 2 ) = 1 , θ ψ ⁢ ( 3 ) = 2 .

Then the graph Γ Ψ , Π is equal to

where we have used solid and dotted lines, respectively, to denote the edges e i , f and e i , g . To get a solution to (4.7) and (4.8), we can first pick

φ ψ , 1 ∈ Iso ⁢ ( T ( θ ψ ⁢ ( 1 ) ) , T ( 1 ) )

to be any isomorphism. Then we “follow the edges” in Γ Ψ , Π and define

{ φ π , 1 = Φ 1 , g ⁢ ( φ ψ , 1 ) , φ π , 2 = Φ 1 , f ⁢ ( φ ψ , 1 ) , φ ψ , 2 = Φ 2 , f - 1 ⁢ ( φ π , 1 ) , φ ψ , 3 = Φ 3 , g - 1 ⁢ ( φ π , 1 ) , φ π , 3 = Φ 3 , f ⁢ ( φ ψ , 3 ) ,

recursively. Equations (4.7) and (4.8) are then all satisfied by (4.9). We have thus obtained a common solution (4.6) to equations (4.5).

The idea of Example 4.8 is somewhat similar to that of Example 3.2. But unlike Proposition 3.4, in this case, we do not need Γ Ψ , Π to be a tree for the idea to work in general. Indeed, in the special case that

f ′ = f = Id G , g ′ = g = the trivial map , θ π = Id N 0 , n = θ ψ ,

the graph Γ Ψ , Π has exactly 𝑛 edges, namely

which is not a tree, but certainly (4.5) has a solution (4.6) corresponding to

ψ = Id G = π .

Nevertheless, we do need Γ Ψ , Π to have no cycle.

Proposition 4.9

There is no cycle in Γ Ψ , Π .

Proof

Suppose for contradiction that Γ Ψ , Π has a cycle 𝔠, whose length must be even, say 2 ⁢ r with r ∈ N , because Γ Ψ , Π is bipartite. Then 𝔠 has the shape

where i 1 , … , i r , j 1 , … , j r ∈ N n . For each j ∈ N n , by Definition 4.6, there are at most two edges in Γ Ψ , Π with j ψ as an endpoint, namely

{ e j , f : j ψ - ( θ f ∘ θ ψ ) ⁢ ( j ) π defined when ⁢ ( θ f ∘ θ ψ ) ⁢ ( j ) ≠ 0 , e j , g : j ψ - ( θ g ∘ θ ψ ) ⁢ ( j ) π defined when ⁢ ( θ g ∘ θ ψ ) ⁢ ( j ) ≠ 0 .

Putting i r + 1 = i 1 , we then deduce that

{ θ f ⁢ ( θ ψ ⁢ ( j s ) ) , θ g ⁢ ( θ ψ ⁢ ( j s ) ) } = { i s , i s + 1 } for all ⁢ 1 ≤ s ≤ r .

But then, by Definition 3.1, this implies that there is an edge joining i s and i s + 1 in Γ { f , g } , so we have a cycle

in Γ { f , g } , which is impossible because Γ { f , g } is a tree. ∎

For convenience, let us refine Definition 4.6 as follows.

Definition 4.10

Define Γ Ψ , Π , ↔ to be the symmetric directed graph that is obtained from Γ Ψ , Π by replacing every edge by a pair of arrows going in the opposite directions. In other words, for each i ∈ N n , we have the arrows

in Γ Ψ , Π , ↔ when i ∉ N n , f and i ∉ N n , g , respectively.

The next lemma is not necessary, but it makes the proof slightly cleaner.

Lemma 4.11

There is no vertex in Γ Ψ , Π which has degree 0.

Proof

For any i ∈ N n , we cannot have θ f ⁢ ( θ ψ ⁢ ( i ) ) and θ g ⁢ ( θ ψ ⁢ ( i ) ) both equal to 0 because Γ { f , g } has no loop, so plainly vertices i ψ in Ψ have degree at least one. To show that vertices j π in Π have degree at least one, it is enough to show that, for any j ∈ N n , one of the equations

θ f ⁢ ( θ ψ ⁢ ( i ) ) = j and θ g ⁢ ( θ ψ ⁢ ( i ) ) = j

has a solution i ∈ N n . By (4.4), they may be respectively rewritten as

θ f ′ ⁢ ( i ) = θ π ⁢ ( j ) and θ g ′ ⁢ ( i ) = θ π ⁢ ( j ) .

Since Γ { f ′ , g ′ } is connected, there must be an edge containing θ π ⁢ ( j ) in Γ { f ′ , g ′ } . This means that indeed one of these two equations is solvable in i ∈ N n . ∎

We shall now complete the proof of the backward implication of Theorem 4.2. Recall that we have already picked bijections θ π and θ ψ for which (4.4) is satisfied. It remains to exhibit a solution (4.6) to equations (4.7) and (4.8). As illustrated in Example 4.8, we shall do so by “following the edges” in the graph Γ Ψ , Π . More precisely, we shall carry out the following steps for each connected component Γ of Γ Ψ , Π . Notice that Γ cannot be a singleton by Lemma 4.11, and so it contains a vertex in Ψ.

(1) Fix a vertex i 0 , ψ in Ψ lying in Γ, and pick φ ψ , i 0 ∈ Iso ⁢ ( T ( θ ψ ⁢ ( i 0 ) ) , T ( i 0 ) ) to be any isomorphism.

(2) Let 𝑣 be any other vertex in Γ. Since Γ is a tree by Proposition 4.9, there exists a unique simple path in Γ joining i 0 , ψ and 𝑣, which corresponds to a simple directed path 𝔭 in Γ Ψ , Π , ↔ . Let us write p = a ⋅ q as a concatenation, where 𝔮 is a simple directed path starting at i 0 , ψ , and 𝔞 is an arrow ending at 𝑣. Let d ∈ N be the length of 𝔭, so d - 1 would be the length of 𝔮.

There are two cases to consider.

In the case that 𝑑 is odd, we have
v = j π with ⁢ j = ( θ h ∘ θ ψ ) ⁢ ( i ) and a = a i , h
for some i , j ∈ N n and ( h , h ) ∈ { ( f , f ) , ( g , g ) } , as illustrated below.

Assuming that φ ψ , i ∈ Iso ⁢ ( T ( θ ψ ⁢ ( i ) ) , T ( i ) ) has been chosen, we define
φ π , j = Φ i , h ⁢ ( φ ψ , i ) , so then φ h ′ , i = φ ψ , i ∘ φ h , θ ψ ⁢ ( i ) ∘ φ π , ( θ h ∘ θ ψ ) ⁢ ( i )
is satisfied, as observed in (4.9).
In the case that 𝑑 is even, we have
v = i ψ with ⁢ j = ( θ h ∘ θ ψ ) ⁢ ( i ) and a = a i , h - 1
for some i , j ∈ N n and ( h , h ) ∈ { ( f , f ) , ( g , g ) } , as illustrated below.

Assuming that φ π , j ∈ Iso ⁢ ( T ( θ π ⁢ ( j ) ) , T ( j ) ) has been chosen, we define
φ ψ , i = Φ i , h - 1 ⁢ ( φ π , j ) , so then φ h ′ , i = φ ψ , i ∘ φ h , θ ψ ⁢ ( i ) ∘ φ π , ( θ h ∘ θ ψ ) ⁢ ( i )
is satisfied, as observed in (4.9).

In either case, equation (4.7) or (4.8) represented by 𝔞 is satisfied.

Therefore, by induction on 𝑑, the above allows us to define φ ψ , i and φ π , j for all vertices i ψ ∈ Ψ ∩ Γ and j π ∈ Π ∩ Γ such that equations (4.7) and (4.8) represented by the edges in Γ are all satisfied. Applying this to all connected components of Γ Ψ , Π then gives us a solution (4.6) to equations (4.7) and (4.8). This completes the proof.

5 Counting unlabeled directed trees with a root

Let T ⋆ ⁢ ( n ) be the set defined as in Remark 1.6, and write t n ⋆ for its size. We shall end this paper by briefly discussing the problem of computing t n ⋆ .

Clearly, t 0 ⋆ = 1 and t 1 ⋆ = 2 . It is also easy to see that t 2 ⋆ = 7 , with

as the elements of T ⋆ ⁢ ( 2 ) , where the root is being placed at the top of graph. For larger values of 𝑛, we can somewhat use induction to compute t n ⋆ .

Definition 5.1

Let Γ be any unlabeled directed tree with a root, and write Γ un for its underlying undirected graph. Let 0 denote the root, and let 𝑣 be any other vertex of Γ.

Let Γ v be the unlabeled directed tree with root 𝑣 which is obtained from Γ by removing all the vertices v ′ such that the unique simple path in Γ un joining 0 and v ′ does not contain 𝑣. We call Γ v the subtree of Γ at 𝑣.
The vertex 𝑣 is a child of Γ if the unique simple path in Γ un joining 0 and 𝑣 has length one. In this case, this path corresponds to an arrow 𝔞 in Γ. We call 𝑣 an in-child if 𝔞 ends at 0, and an out-child if 𝔞 starts at 0.

Proposition 5.2

Suppose that we are given tuples

( n 1 , n 2 ) , ( k 1 , 1 ⁢ … , k 1 , n 1 ) , ( k 2 , 1 , … , k 2 , n 2 )

of non-negative integers such that

n = n 1 + n 2 , n 1 = ∑ s = 1 n 1 s ⁢ k 1 , s , n 2 = ∑ s = 1 n 2 s ⁢ k 2 , s ,

where an empty sum represents 0. Then the number of Γ ∈ T ⋆ ⁢ ( n ) such that,

for each 1 ≤ s ≤ n 1 , the number of in-children 𝑣 whose subtrees Γ v have exactly 𝑠 vertices is equal to k 1 , s ,
for each 1 ≤ s ≤ n 2 , the number of out-children 𝑣 whose subtrees Γ v have exactly 𝑠 vertices is equal to k 2 , s ,

is given by

∏ s = 1 n 1 ( t s - 1 ⋆ + k 1 , s - 1 k 1 , s ) ⋅ ∏ s = 1 n 2 ( t s - 1 ⋆ + k 2 , s - 1 k 2 , s ) ,

where an empty product represents 1.

Proof

A similar statement is true for unlabeled rooted trees; see [11, p. 386] for example. The only difference is that our trees are directed, which means that the in-children and out-children have to be treated differently.

To count the number of Γ ∈ T ⋆ ⁢ ( n ) satisfying (a) and (b), note that Γ may be viewed as a collection of subtrees Γ v at the children 𝑣. Also, the subtrees Γ v at the in-children 𝑣 are independent from those at the out-children 𝑣. For each ℓ = 1 , 2 and 1 ≤ s ≤ n ℓ , at the k ℓ , s vertices 𝑣 which are

{ in-children if ⁢ ℓ = 1 , out-children if ⁢ ℓ = 2 ,

and whose subtrees Γ v are to have exactly 𝑠 vertices, we can choose any k ℓ , s elements of T ⋆ ⁢ ( s - 1 ) to be these Γ v with repetition allowed. There are

( ( t s - 1 ⋆ k ℓ , s ) ) = ( t s - 1 ⋆ + k ℓ , s - 1 k ℓ , s )

such choices. Multiplying over 𝑠 and ℓ then yields the claim. ∎

Given any m ∈ N , let A ⁢ ( m ) be the set of all 𝑚-tuples k = ( k 1 , … , k m ) of non-negative integers such that m = k 1 + 2 ⁢ k 2 + ⋯ + m ⁢ k m , and put

P ⁢ ( m ; k ) = ∏ s = 1 m ( t s - 1 ⋆ + k s - 1 k s )

in this case. From Proposition 5.2, we then deduce that

(5.1) t n ⋆ = ∑ m ∈ N 0 , n ∑ k 1 ∈ A ⁢ ( m ) ∑ k 2 ∈ A ⁢ ( n - m ) P ⁢ ( m ; k 1 ) ⁢ P ⁢ ( n - m ; k 2 ) .

This gives us a way to compute t n ⋆ inductively, but the calculation gets very complicated. The better approach is perhaps to consider the associated generating function given by

z ↦ ∑ n = 0 ∞ t n ⋆ ⁢ z n ,

and then use (5.1) to study its properties. But this is a completely different problem. We shall therefore content ourselves with computing t n ⋆ for n ≤ 10 by applying formula (5.1) directly.

(a)

𝑛	t n ⋆
1	1
2	7
3	26
4	107
5	458

(b)

𝑛	t n ⋆
6	2058
7	9498
8	44947
9	216598
10	1059952

The calculation was done in Magma [2]. For 1 ≤ n ≤ 7 , the computation only took a few seconds in total. However, the set A ⁢ ( n ) gets very difficult to compute very quickly. For n ≥ 8 , the code gets very slow, and we were only able to run it up to n = 10 . We note that the skew braces (for which the additive and circle groups are isomorphic to the finite same characteristically simple group) enumerated by this calculation are not covered by [10, 1] mentioned in the introduction.

Finally, the code used is a straightforward implementation of (5.1), so we shall not include it here. The interested reader can find the code in the arXiv version of this paper (arXiv:2009.00266).

Acknowledgements

The author thanks Dr. Chao Qin (Chris King) for helping her calculate t n ⋆ for n ≤ 10 by running the Magma code on his computer. She would also like to thank the referee for helpful comments.

Communicated by: Andrea Lucchini

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Received: 2021-03-12

Revised: 2021-09-03

Published Online: 2021-09-28

Published in Print: 2022-03-01

Articles in the same Issue

https://doi.org/10.1515/jgth-2021-0044