Minimum degrees of finite rectangular bands, null semigroups, and variants of full transformation semigroups

For a positive integer $n$, the full transformation semigroup $T_n$ consists of all self maps of the set $\{1,\ldots,n\}$ under composition. Any finite semigroup $S$ embeds in some $T_n$, and the least such $n$ is called the (minimum transformation) degree of $S$ and denoted $\mu(S)$. We find degrees for various classes of finite semigroups, including rectangular bands, rectangular groups and null semigroups. The formulae we give involve natural parameters associated to integer compositions. Our results on rectangular bands answer a question of Easdown from 1992, and our approach utilises some results of independent interest concerning partitions/colourings of hypergraphs. As an application, we prove some results on the degree of a variant $T_n^a$. (The variant $S^a=(S,\star)$ of a semigroup $S$, with respect to a fixed element $a\in S$, has underlying set $S$ and operation $x\star y=xay$.) It has been previously shown that $n\leq \mu(T_n^a)\leq 2n-r$ if the sandwich element $a$ has rank $r$, and the upper bound of $2n-r$ is known to be sharp if $r\geq n-1$. Here we show that $\mu(T_n^a)=2n-r$ for $r\geq n-6$. In stark contrast to this, when $r=1$, and the above inequality says $n\leq\mu(T_n^a)\leq 2n-1$, we show that $\mu(T_n^a)/n\to1$ and $\mu(T_n^a)-n\to\infty$ as $n\to\infty$. Among other results, we also classify the $3$-nilpotent subsemigroups of $T_n$, and calculate the maximum size of such a subsemigroup.


Introduction
Denote by T X the full transformation semigroup over the set X, which consists of all selfmaps of X under composition.When X = {1, . . ., n} for some integer n we denote T X by T n .Cayley's Theorem states that every semigroup S embeds in some T X with |X| ≤ |S| + 1; see [37,Theorem 1.1.2].The (minimum transformation) degree of a semigroup S is defined to be the cardinal µ(S) = min |X| : X ̸ = ∅, S embeds in T X .
(The requirement that X ̸ = ∅ is exclusively to establish the convention that a semigroup of size 1 has degree 1.) Several authors have calculated µ(S) for various classes of finite (semi)groups; for semigroups see especially the works of Easdown [15][16][17] and Schein [51,52], but note that in some papers degrees are defined in terms of representations by partial transformations.(Writing µ ′ (S) for the degree with respect to partial transformations, the reader might like to verify that µ(S) − 1 ≤ µ ′ (S) ≤ µ(S).)See also [4,21,35] for computational studies, [34,53,58] for connections to radical theory, and [3,18,23,36,38,39,49,50,59] for degrees of finite groups.The topic is also closely related to enumeration of (classes of) semigroups by size [9-11, 25, 41].We also mention the very recent study of Margolis and Steinberg [43], which was written after the current article but published before it, and concerns the class of Rhodes semisimple semigroups.
Apart from trivially small exceptions, none of the semigroups we study here belong to this class, so our techniques are necessarily different from theirs.
The initial source of motivation for the current paper came from semigroup variants.Recall that the variant of a semigroup S with respect to a fixed element a ∈ S is the semigroup S a = (S, ⋆), where the sandwich operation ⋆ is defined by x ⋆ y = xay for x, y ∈ S. Variants were introduced by Hickey [32,33], building on earlier ideas of Lyapin [40] and Brown [5], and have been studied by many others since.The papers [2,12,19,44,56,57] study variants of full transformation semigroups; see also [13,14] for a categorical approach.Sandwich operations also play an important role in computational semigroup theory [22].
Despite the simple definition, the structure of a variant T a n is vastly more complicated than that of T n itself.See for example Figure 1, which gives egg-box diagrams for T 4 and a variant T a 4 , both produced with GAP [27,46].(Egg-box diagrams display the structure of a semigroup as determined by Green's relations [29], which are themselves defined below; see [8,37] for more details.)Nevertheless, it was shown in [19] that variants T a n can be embedded in (ordinary) transformation semigroups of relatively small degree.For the next statement, which is [19,Theorem 1.4], the rank of a transformation is the size of its image.
Theorem 1.1.If a ∈ T n and r = rank(a), then T a n embeds in T 2n−r .
In fact, it was shown in [19] that T It was shown in [19] that (for very simple reasons) µ(T a n ) in fact attains the upper bound of 2n − r for r ≥ n − 1.One of our main results, Theorem 5.15 below, improves this by showing that µ(T a n ) = 2n − r for r ≥ n − 6.The situation is very different, however, when the sandwich element a ∈ T n has minimum possible rank 1, where (1.2) becomes n ≤ µ(T a n ) ≤ 2n − 1.For rank(a) = 1, we show that: • µ(T a n ) achieves the upper bound of 2n − 1 if and only if n ≤ 15 (Proposition 5.23), • the ratio µ(T a n )/n tends to 1 as n → ∞ (Theorem 5.27), • the difference µ(T a n ) − n tends to ∞ as n → ∞ (Theorem 5.28).
En route to proving the above results on variants, we conduct an analysis of various other classes of semigroups.Section 3 concerns rectangular bands.The main result of this section is Theorem 3.9, which gives the degree for such a band, extending previously-known results on left and right zero semigroups [17,28].Table 2 gives several calculated values.The proof of Theorem 3.9 utilises new results on hypergraphs, which we believe are of independent interest; see especially Theorem 3.6.As a further application we also treat rectangular groups in Theorem 3.12.
Section 4 concerns null semigroups, and semigroups we call right null semigroups (these include null semigroups and right zero semigroups as special cases).Among other things, we classify all null subsemigroups of full transformation semigroups and calculate the maximum size of such subsemigroups; see Theorems 4.4 and 4.7.Curiously, it transpires that null and left zero semigroups of the same size have the same degree.We also consider (3-)nilpotent semigroups in Section 4, and Theorem 4. 15 gives the maximum size of a 3-nilpotent subsemigroup of T n ; Table 4 gives some calculated values.This result naturally complements the article [11], which enumerates abstract 3-nilpotent semigroups by size.The degree of a uniform right null semigroup is given in Theorem 4.27; calculated values are given in Table 6.We also apply the results of Section 4 to give an example of a semigroup S such that µ(S a ) < µ(S) for all a ∈ S, answering a question from [19].
The above-mentioned results on variants of T n are then given in Section 5. 1 2 3 3 ) in standard two-line notation.
Proof.This is clear for k = 0, and for k ≥ 1 we use the Binomial Theorem: Corollary 2.2.For integers n ≥ k ≥ 0 and t ≥ 0, we have Proof.This is clear for t = 0 and for t > k.For 1 ≤ t ≤ k, we repeatedly apply Lemma 2.1: Two closely-related functions ξ, α : N → N will play an important role throughout, as well as an associated real function L : R + → R + .Here R + is the set of positive reals.For n ∈ N, we define (with the n notation introduced at the start of the subsection): 3) The numbers ξ(n) and α(n) appear as Sequences A003320 and A056155 on the OEIS [1].Some calculated values are given in Table 1.The ξ and α functions played a role in [28] in relation to left zero subsemigroups of T n ; they will appear in several other contexts in the current paper.
To define the function L : R + → R + , first consider f : R + → R defined by f (x) = x + x ln x.Since f ′ (x) = 2 + ln x, we see that f (x) is strictly increasing for x ≥ e −2 .Since f (e −1 ) = 0, it follows that f restricts to a bijection {x ∈ R : x > e −1 } → R + .We then define L to be the inverse of this bijection.Thus, for every t ∈ R + , L(t) is the unique real x > e −1 satisfying t = x + x ln x.Note that L is also an increasing function, and that t ≥ L(t) ⇔ t ≥ 1 ⇔ L(t) ≥ 1. (iii) For any n ∈ N, α(n) is one of ⌊x⌋ or ⌈x⌉, where x = L(n), and ξ(n) = max ⌊x⌋ n−⌊x⌋ , ⌈x⌉ n−⌈x⌉ .
Remark 2.5.It follows that the 'max' in the definition of α(n) in (2.3) is only needed for n = 2. Indeed, for n ≥ 3, and writing u 1 = ⌊u⌋ and u 2 = ⌈u⌉, where u + u ln u = n, we have is even and the other is odd.
The functions ξ and α will also come up in another, somewhat indirect, way.We explore this in the next subsection.

Compositions
Recall that a composition of a natural number n is a tuple σ = (s 1 , . . ., s r ) of positive integers satisfying s 1 + • • • + s r = n, in which case we write σ ⊨ n and |σ| = r; we call r the length of σ.If also s 1 ≥ • • • ≥ s r , then σ is a partition of n, and we write σ ⊢ n.Certain parameters associated to compositions will play an important role in all that follows, as well as certain numerical functions/sequences defined in terms of them.
Perhaps the simplest parameter associated to a composition σ = (s 1 , . . ., s r ) is its product: For integers 1 ≤ r ≤ n, we define Clearly π(n) = max r π r (n).The numbers π(n) and π r (n) are well understood, and a proof of the following can be found for example in [28].The first part follows from the fact that the maximum value of σ, for σ ⊨ n with |σ| = r, occurs when the entries of σ are 'almost equal', in the sense that they are all within of each other (as The second part then involves showing that the product of almost-equal numbers with a given sum occurs when almost all of the numbers are 3.
For a semigroup S, we denote by S 1 the monoid obtained by adjoining an identity element to S if necessary; so S = S 1 if S is a monoid.Green's L , R and J relations are defined for x, y ∈ S by Green's H and D relations are defined by H = L ∩ R and D = L ∨ R, where the latter is the join in the lattice of equivalence relations on S. We have An element x of a semigroup S is regular if x = xyx for some y ∈ S (equivalently, if x = xyx and y = yxy for some y ∈ S), while x is an idempotent if x = x 2 .We write Reg(S) and E(S) for the sets of all regular elements and all idempotents of S, respectively, neither of which is a subsemigroup in general; both are non-empty when S is finite.Obviously E(S) ⊆ Reg(S).
An element x of a semigroup S is a left zero of S if xa = x for all a ∈ S. A left zero semigroup is a semigroup in which every element is a left zero: i.e., S satisfies the law xy = x.Every left zero semigroup is L -simple, meaning that all elements are L -related.By a left zero subsemigroup of a semigroup S we mean a subsemigroup that happens to be a left zero semigroup (but note that its elements might not be left zeros of S itself).
Right zeros and right zero (sub)semigroups are defined analogously, and they have analogous properties.
If a semigroup S has a left zero and a right zero, then it has a unique left zero and a unique right zero, which are equal, and hence the unique (two-sided) zero of S. A null semigroup is a semigroup S for which S 2 = {xy : x, y ∈ S} = {z} for some z ∈ S, which is then necessarily the zero of S.
A rectangular band is (isomorphic to) a semigroup of the form P × Q, where P is a left zero semigroup and Q a right zero semigroup; multiplication obeys the rule (p 1 , q 1 )(p 2 , q 2 ) = (p 1 , q 2 ).If |P | = p and |Q| = q we say the rectangular band is p × q.Green's relations on a rectangular band P × Q are particularly easy to describe: Further, H is the equality relation, and D = J is the universal relation.As extreme cases, a left zero semigroup of size p is a p × 1 rectangular band, with a similar statement for right zero semigroups.

Full transformation semigroups
The full transformation semigroup T X consists of all self-maps of the set X, under composition.When X is the set n = {1, . . ., n} for a positive integer n, we write T X = T n .For more on transformation semigroups, see [8,26,37].
For f ∈ T X we write im(f ) = {xf : x ∈ X} for the image of f , rank(f ) = |im(f )| for the rank of f , and ker(f ) = (x, y) ∈ X × X : xf = yf for the kernel of f .Green's relations on T X are given by Fix some subset Ω of T X .For any non-empty subset A ⊆ X, any equivalence relation ε on X (denoted ε ⪯ X), and any cardinal 1 ≤ r ≤ |X|, we write When there is no chance of confusion, we generally write L A = L A (T X ), and so on.So the sets L A (∅ ̸ = A ⊆ X), R ε (ε ⪯ X) and D r (1 ≤ r ≤ |X|) are the L -, R-and J = D-classes of T X , respectively.For an equivalence ε ⪯ X, we write ∥ε∥ = |X/ε| for the number of ε-classes.
At times it will be convenient to work with the kernel partition of a transformation f ∈ T X , defined by Ker(f ) = X/ker(f ).This is a set partition of X, and |Ker(f )| = ∥ker(f )∥ = rank(f ).
A transformation f ∈ T X will be represented in tabular form as f = A i a i i∈I to indicate that im(f ) = {a i : i ∈ I} and a i f −1 = A i for all i ∈ I; we additionally assume that the indexing set I is faithful in the sense that a i ̸ = a j whenever i ̸ = j, so that rank(f ) = |I|.Note that It is well known and easy to see that f = A i a i i∈I is idempotent if and only if a i ∈ A i for all i.
It will also be convenient to describe here an operation ⊕.Suppose we have a set of pairwisedisjoint non-empty sets X i (i ∈ I), and a collection of transformations f i ∈ T X i (i ∈ I).Writing X = i∈I X i , we define i∈I f i to be the unique transformation of X whose restriction to each X i is f i .For non-empty subsets Ω i ⊆ T X i (i ∈ I), we write i∈I Ω i for the set of all i∈I f i with each f i ∈ Ω i .In particular, i∈I T X i is a subsemigroup of T X , and is isomorphic to the (external) direct product i∈I T X i .
For a fixed non-empty subset A of the set X, we have a subsemigroup These subsemigroups have been studied extensively in the literature [20,24,45,47,48,54,55], and will play an important role in the current work.Note that for any a ∈ T X with im(a) = A, the subsemigroup T X (A) is precisely the principal left ideal While the semigroup T X is regular, the same is not true of T X (A) in general.It was shown in [48] that an element f of T X (A) is regular (in the semigroup T X (A)) if and only if Xf = Af ; an equivalent formulation [20] is that A saturates ker(f ), meaning that every ker(f )-class contains an element of A.

Left and right zero transformation semigroups
For a subset Ω of a finite semigroup S, we write λ(Ω) and ρ(Ω) for the maximum size of a left or right zero subsemigroup of S contained in Ω.In particular, λ(S) and ρ(S) are the maximum sizes of a left or right zero subsemigroup of S. Now consider a finite semigroup S, and a left zero subsemigroup T of S. Since the elements of T are all L -related in T , it follows that T is contained in a single L -class of S, and hence in a single (regular) D-class; let these classes be L and D, respectively.Noting that all left zeros are idempotents, we have T ⊆ E(L).It is easy to see that E(L) is a left zero semigroup for any regular L -class L. It follows from all this, and the analogous considerations of right zero subsemigroups, that The paper [28] classified the left and right zero subsemigroups of a finite full transformation semigroup T n , and calculated the maximum size of such subsemigroups; see also [17].It will be convenient to interpret these results in our current set-up, using the ξ and π parameters from Subsections 2.1 and 2.2.For 1 ≤ r ≤ n, we will write As discussed in Subsection 2.4, the L -and R-classes contained in the D-class D r (T n ) are, respectively, the sets of the form The size of E(L A ) is r n−r , as any idempotent from L A maps A identically, and maps n \ A arbitrarily into A. The size of E(R ε ) is equal to the product of the sizes of the ε-classes, as any idempotent from R ε maps each ε-class onto a single element of that class.By maximising these values, one obtains the following, which is [28, Propositions 3.1 and 3.2].
Theorem 2.9.For 1 ≤ r ≤ n, the maximum size of a left or right zero semigroup contained in Remark 2.10.It is easy to check that for fixed r, the sequences λ r (n) and ρ r (n) are nondecreasing in n ≥ r.
Remark 2.11.Keeping in mind the formula for ρ r (n) = π r (n) from Lemma 2.7(i), we of course have ρ n (n) = 1 for all n.Less trivially, we have The values of λ(n) and ρ(n) are found by maximising λ r (n) and ρ r (n) over all 1 ≤ r ≤ n.The following is [28,Theorems 3.3 and 3.4].
Theorem 2.12.For n ≥ 1, the maximum size of a left or right zero subsemigroup of T n is equal to We will also need to extend Theorem 2.9 to the semigroups for the largest size of a left or right zero semigroup contained in D r (T n (k)), respectively.Note that D r (T n (k)) is generally not a D-class of T n (k); however, it contains a single regular D-class, as follows from results of [20,Section 5].
Proof.For any A ⊆ k of size r, it is clear that L A (T n ) ⊆ D r (T n (k)).Since the L -classes contained in D r (T n ) have a common number of idempotents, it quickly follows that λ r (n, k) = λ r (n).
For the statement concerning ρ r (n, k), consider a regular R-class R contained in D r (T n (k)).Let ε be the common kernel of the elements of R, and let the ε-classes be A 1 , . . ., A r .Since R is regular, k saturates ε, so it follows that B i = A i ∩ k is non-empty for each i ∈ r; let η ⪯ k be the equivalence with classes B 1 , . . ., B r .Every idempotent from R is uniquely determined by (and uniquely determines) its restriction to Moreover, the above correspondence is reversible; given any η ⪯ k with ∥η∥ = r we can find a suitable ε ⪯ n by arbitrarily assigning the remaining elements of n \ k to η-classes.The result now follows.
One could then obtain results concerning the largest left and right zero subsemigroups of T n (k) by maximising the values of λ r (n, k) = λ r (n) and ρ r (n, k) = ρ r (k) over r ∈ k.Note that the latter maximum will be ρ(k) = π(k), as in Theorem 2.12, but the former is max{λ r (n) : 1 ≤ r ≤ k}, and is not necessarily equal to λ(n) = ξ(n).

Rectangular bands and hypergraphs
In this section we calculate the (minimum transformation) degree µ(B) of an arbitrary finite rectangular band B, thereby answering a question of Easdown [17].Since all p × q rectangular bands are isomorphic, the degree of such a band depends only on the parameters p and q.Accordingly, for p, q ∈ N we define β(p, q) = min{n : T n contains a p × q rectangular band}.
Since rectangular bands are D-simple, any such band contained in T n is in fact contained in a single D-class D r (T n ).Accordingly, for p, q, r ∈ N we define β r (p, q) = min{n : D r (T n ) contains a p × q rectangular band}, and of course β(p, q) = min r β r (p, q).In the above definition of β r (p, q), we interpret min ∅ = ∞.
As we noted in Subsection 2.3, the special cases of q = 1 and p = 1 correspond to left and right zero semigroups, respectively.In these cases, values of β(p, q) and β r (p, q) can be quickly deduced from the results of [17,28] stated in Theorems 2.9 and 2.12 above.The following statement uses the ξ and π parameters from Subsections 2.1 and 2.2.(ii) For any q, r ≥ 1, we have When p, q ≥ 2, we cannot calculate the numbers β(p, q) and β r (p, q) by finding the largest rectangular bands contained in T n and D r (T n ).The reason for this is that even if T n contained a rectangular band of size pq, it may not contain one with the correct dimentions, p × q.In fact, it follows from results of [28,Section 4] that the largest rectangular band contained in T n (or in D r (T n )) is always a left or right zero semigroup, so such an approach is bound to fail.Also, since a rectangular band is the direct product of a left and right zero semigroup of appropriate sizes, and since µ(S × T ) ≤ µ(S) + µ(T ) for any semigroups S and T , we have an upper bound of β(p, q) ≤ β(p, 1) + β (1, q).This upper bound turns out to be an over-estimate, however, as can be seen by inspecting Table 2.
Our approach for p, q ≥ 2 will be via hypergraphs.After proving some results on hypergraphs in Subsection 3.1, which we believe are of independent interest (see especially Theorem 3.6), we return to rectangular bands in Subsection 3.2 (see Theorem 3.9).We give further applications of our results to rectangular groups in Subsection 3.3 (see Theorem 3.12).

Hypergraphs
A hypergraph is a pair H = (V, E), where V is a set of vertices, and E is a set of non-empty subsets of V called (hyper)edges.If each edge has size r, we say that H is r-uniform and call it an r-hypergraph; obviously we must have r ≤ |V | if H is non-empty (i.e., has at least one edge).By an r-partition of an r-hypergraph H we mean a partition V = {V 1 , . . ., V r } of the vertex set such that each edge is a transversal of V (i.e., contains a unique point from each block V i of V).We call H r-partite if such an r-partition exists.
Given a partition V = {V 1 , . . ., V r } of a set V , the complete r-partite r-hypergraph H V has vertex set V , and its edges are all the transversals of V.The number of edges of and hence has at most σ V edges.
From the above discussion we have the following well-known result, expressed in terms of the π parameters from Subsection 2.2.Lemma 3.2.Let n, r ≥ 1 be integers.Then there exists a non-empty r-partite r-hypergraph with n vertices if and only if r ≤ n, in which case the maximum number of edges in such a hypergraph is π r (n).
Our main goal in this subsection is to prove Theorem 3.6 below, which gives the corresponding result for hypergraphs with some specified number of partitions.The key technical step in the proof is Proposition 3.5, which shows that r-hypergraphs with many r-partitions give rise to r-partite r-hypergraphs on smaller vertex sets.The proof of the proposition is inductive, and relies on the next two lemmas.Lemma 3.3.Suppose H is an r-hypergraph with at least one edge, and at least rt + 1 distinct r-partitions for some integer t ≥ 1.Then there exists a pair of vertices that belong to the same block in at least t + 1 of the partitions, but not in all of them.
Then w and v m belong to the same block in at least t + 1 of the partitions (all the V i for i ∈ I m ).
Aiming for a contradiction, suppose w and v m belong to the same block in all of the partitions.
Consider a hypergraph H with vertex set V , and let u, v ∈ V be distinct vertices.Write H/{u, v} for the hypergraph obtained by identifying u and v. Formally, the vertex set of H/{u, v} is obtained from V by removing the two vertices u, v and replacing them by a single vertex denoted uv; all edges of H involving neither of u, v are still edges of H/{u, v}; all other edges of H/{u, v} are of the form {uv, w 1 , . . ., w k } where at least one of {u, w 1 , . . ., w k }, {v, w 1 , . . ., w k } or {u, v, w 1 , . . ., w k } is an edge of H.If H is r-uniform, then H/{u, v} might not be.Also note that H/{u, v} could have fewer edges than H, but it cannot have more.
Lemma 3.4.Suppose H is an r-hypergraph with n vertices and q edges, with at least t + 1 distinct r-partitions, for some integer t ≥ 1. Suppose vertices u and v belong to the same block in t of the r-partitions, but not in all of them.Then H/{u, v} is an r-hypergraph with n − 1 vertices and q edges, and has at least t distinct r-partitions.
Proof.For simplicity, we write For each edge A of H, let A ′ be the corresponding edge of H/{u, v}, obtained by replacing u and/or v by uv if necessary, as explained before the lemma.To show that H ′ has q edges, we need to show that the map Then A and B must be one of {u} ∪ C, {v} ∪ C or {u, v} ∪ C.But {u, v} ∪ C cannot be an edge of H, since u, v belong to the same block in some r-partition of H. On the other hand, {u} ∪ C and {v} ∪ C cannot both be edges of H, since u, v belong to distinct blocks in some r-partition of H.This all shows that A = B.

The previous paragraph also shows that
Finally, suppose V 1 , . . ., V t are distinct r-partitions of H in each of which u, v belong to the same block.For each 1 by replacing the pair of vertices u, v (from H) by the single vertex uv (from H ′ ).Then by the above characterisation of the edges of H, it is clear that each V ′ i is an r-partition of H ′ , and that these are still distinct.
Proposition 3.5.If H is an r-hypergraph with n vertices and q ≥ 1 edges, with at least r l−1 + 1 distinct r-partitions for some integer l ≥ 1, then there is an r-partite r-hypergraph with n − l vertices and q edges.
Proof.The proof is by induction on l.Suppose first that l = 1.So H has at least two distinct r-partitions, say V 1 and V 2 .Since these are distinct, there exist vertices u, v belonging to the same block in V 1 but not in V 2 .By Lemma 3.4 (with t = 1), H/{u, v} is an r-partite r-hypergraph with n − 1 = n − l vertices and q edges.Now suppose l ≥ 2. Since r l−1 + 1 = r • r l−2 + 1, it follows from Lemma 3.3 (with t = r l−2 ) that there are vertices u, v of H that belong to the same block in at least r l−2 +1 of the partitions, but not in all of them.By Lemma 3.4 (with t = r l−2 + 1), H/{u, v} is an r-hypergraph with n − 1 vertices and q edges, and with at least r l−2 + 1 = r (l−1)−1 + 1 distinct r-partitions.By induction there exists an r-partite r-hypergraph with (n − 1) − (l − 1) = n − l vertices and q edges.
Here is the main result of this subsection.Theorem 3.6.Let n, p ≥ 1 and r ≥ 2 be integers, and let l = ⌈log r p⌉.Then there exists a non-empty r-hypergraph with n vertices and at least p distinct r-partitions if and only if r +l ≤ n, in which case the maximum number of edges in such a hypergraph is π r (n − l).
Proof.If p = 1 then l = 0, and the result reduces to Lemma 3.2.For the rest of the proof we assume that p ≥ 2. It follows from the definition of l that l ≥ 1, and Suppose there exists a non-empty r-hypergraph with n vertices and q(≥ 1) edges, and with p distinct r-partitions.Since p ≥ r l−1 + 1, Proposition 3.5 applies, and it tells us there is an r-partite r-hypergraph with n − l vertices and q edges.It then follows from Lemma 3.2 that r ≤ n − l (i.e., r + l ≤ n) and that q ≤ π r (n − l).
It remains to show that if r + l ≤ n there exists an r-hypergraph with n vertices and π r (n − l) edges, with (at least) p distinct r-partitions.To do so, suppose r + l ≤ n.By Lemma 3.2 we may fix an r-hypergraph H on n − l vertices with π r (n − l) edges.Let V = {V 1 , . . ., V r } be an r-partition of H. Now let H ′ be the r-hypergraph obtained from H by adding l isolated vertices.Then H ′ has n vertices and π r (n−l) edges, and (at least) r l distinct r-partitions (each of the new vertices, independently, can go in any of the blocks V 1 , . . ., V r ).Since r l ≥ p, we are done.

Rectangular bands
We now use the results of the previous subsection to calculate the degree of an arbitrary finite rectangular band, answering a question of Easdown [17].The first result shows how to construct certain 'large' such bands.Its proof shares some ideas with that of Theorem 3.6.
For each function f : l → r, and each transversal It is easy to check that these compose according to the rule e(f, A) • e(g, B) = e(f, B).It follows that the set of all such e(f, A) is a rectangular band contained in D r (T n ).Since there are r l functions l → r, and π r (n − l) transversals of V, the band has the stated dimensions.
Lemma 3.8.If D r (T n ) contains a p × q rectangular band, where p, q ≥ 2, then with l = ⌈log r p⌉ we have π r (n − l) ≥ q.
Proof.Suppose B is a p × q rectangular band contained in D r (T n ).Since D 1 (T n ) is a right zero semigroup, and since p ≥ 2, we have r ≥ 2. Let be the sets of all images and kernel partitions of the elements of B.
Since each element of B has rank r, each member of I is a subset of n of size r, which means that H = (n, I) is an r-hypergraph.Moreover, given A ∈ I and K ∈ K we have A = im(f ) and K = Ker(g) for some f, g ∈ B, and then since f g ∈ B ⊆ D r it follows that A is a transversal of K.This means that each member of K is an r-partition of H.
Since B has p R-classes and q L -classes, it follows that |I| = q and |K| = p.This all means that H is an r-hypergraph with n vertices and q edges, and with p distinct r-partitions.It then follows from Theorem 3.6 that π r (n − l) ≥ q.
We can now give an expression for β(p, q), which we recall denotes the degree of a p × q rectangular band.The cases where one or both of p, q is 1 were treated in Theorem 3.1.Theorem 3.9.If p, q ≥ 2, then the degree of a p × q rectangular band is equal to Proof.At the beginning of Section 3 we explained that β(p, q) = min r≥2 β r (p, q), where β r (p, q) = min{n : D r (T n ) contains a p × q rectangular band}.
It therefore remains to show that for n, r ≥ 2, D r (T n ) contains a p × q rectangular band if and only if π r (n − l) ≥ q, where l = ⌈log r p⌉.The forwards implication is Lemma 3.8.
For the converse, suppose π r (n − l) ≥ q.Then of course we have n − l ≥ r, and so l ≤ n − r.Proposition 3.7 then tells us that D r (T n ) contains an r l × π r (n − l) rectangular band, and hence also a p × q rectangular band, since p ≤ r l and q ≤ π r (n − l).Remark 3.10.In principle, the expression for β(p, q) in Theorem 3.9 involves finding the minimum of an infinite sequence of numbers: i.e., the β r (p, q) for all r ≥ 2. In practice however, only a finite subsequence needs to be considered.For example, to find β(p, q), one could first find (say) m = β 2 (p, q).We then know that β(p, q) ≤ m, and since β r (p, q) ≥ r for all r it quickly follows that β(p, q) = min 2≤r≤m β r (p, q).Table 2 gives several values of β(p, q), calculated in this way.The table also gives the ranks r such that β(p, q) = β r (p, q).For some values of p, q there are multiple such r, in which case the table gives the minimum r.Remark 3.11.A band is a semigroup consisting entirely of idempotents.It is well known that every band is a semilattice of rectangular bands; see [37,Section 4.4] for a proof of this fact, and an explanation of the terminology.We leave it as an open problem to determine whether this structure (and Theorem 3.9) leads to a formula for the degree of an arbitrary finite band in terms of its underlying semilattice and constituant rectangular bands.The degree of a semilattice of groups was determined by Easdown in [16].p, q).If the (p, q) entry is n, r, then n = β(p, q) = β r (p, q), with r minimal.This means that n is minimal with the property that T n contains a p × q rectangular band, and ≤ r ≤ n is minimal with the property that D r (T n ) contains such a band.

Rectangular groups
Recall that a rectangular group is (isomorphic to) a direct product S = B × G, where B is a rectangular band and G is a group.If the band B is p × q, we say that S is a p × q rectangular group over G.An alternative characterisation of rectangular groups can be found in [37, Exercise 10, page 139]: they are precisely the D-simple regular semigroups whose idempotents form a subsemigroup.
The results of previous subsections can be applied to calculate the degree of an arbitrary finite rectangular group: Theorem 3.12.If S is a finite p × q rectangular group over G, then The numbers β r (p, q) are given in Theorems 3.1 and 3.9.
Proof.Consider a finite p × q rectangular group S = B × G, and for simplicity write and We must show that n = N .
By definition of n, there exists an embedding ϕ : S → T n .Since S is D-simple, im(S) is contained in a single D-class of T n , say D r (T n ).Since G embeds in S, it also embeds into a group H -class in D r (T n ); since such an H -class is isomorphic to the symmetric group S r , it follows that r ≥ µ(G).Similarly, D r (T n ) contains an isomorphic copy of B, and so n ≥ β r (p, q) by definition.This all shows that n ≥ N .
To show that n ≤ N , we must show that S embeds in T N .To do so, suppose r ≥ µ(G) is such that N = β r (p, q).By definition, D r (T N ) contains a p × q rectangular band C. Let T = c∈C H c be the union of all the H -classes of T N containing idempotents from C, so that T is a p × q rectangular group over S r (this follows quickly from the above-mentioned characterisation of rectangular groups from [37]).Since G embeds in S r (as r ≥ µ(G)), it follows that S = B × G embeds in B × S r ∼ = T , and hence in T N .Remark 3.13.As in Remark 3.10, the expression for µ(S) in Theorem 3.12 involves finding the minimum of an infinite sequence of numbers.However, we again have µ(S) = min µ(G)≤r≤m β r (p, q) for any a priori known upper bound m of µ(S).For example, we can take m = β(p, q) + µ(G), since S is the direct product of a p × q rectangular band with G.

(Right) null and nilpotent semigroups
In this section we study null subsemigroups (Subsections 4.1 and 4.2) of a full transformation semigroup T X , as well as some related kinds of subsemigroups: nilpotent subsemigroups (Subsections 4.3 and 4.4), and so-called right null subsemigroups (Subsection 4.5).As an application of our results on null semigroups, we give an example of a semigroup with a larger degree than one of its variants (Subsection 4.6).Our results on right null semigroups will be used in Section 5.

Null semigroups
Let X be an arbitrary set, possibly infinite.Obviously the zero of a null subsemigroup of T X is an idempotent.So we fix an idempotent ζ ∈ E(T X ), with the goal of classifying the null subsemigroups containing ζ.We write For such a ζ-system W, we define where To see that S ⊆ N (W, ζ), let f ∈ S and i ∈ I be arbitrary.We have Z i f ⊆ W i by definition.Moreover, for any g ∈ S we have Let f, g ∈ S be arbitrary.For any i ∈ I, and using f, g ∈ N (W, ζ), we have Now that we have characterised the null subsemigroups of T X , we wish to find the maximum size of such a subsemigroup.We begin by calculating the size of the semigroups N (W, ζ).It follows from the next lemma that this size depends only on the partitions Z = {Z i : i ∈ I} and W = {W i : i ∈ I}, indeed only on the cardinalities We now consider the finite case.For a positive integer n, we write ν(n) for the maximum size of a null subsemigroup of T n .We also write ν r (n) for the maximum size of a null subsemigroup of T n whose zero element has rank r.
Theorem 4.4.Let n be a positive integer.
(i) For any 1 ≤ r ≤ n, the maximum size of a null subsemigroup of T n whose zero element has rank r is equal to ν r (n) = ξ(n − r + 1).
(iii) The maximum size of a null subsemigroup of T n is equal to ν(n) = ν 1 (n) = ξ(n).Moreover, writing t = α(n), the set is a null subsemigroup of T n of size ν(n).Remark 4.6.Null semigroups played an important role in the investigation of subsemigroup chains in T n undertaken in [6].The null subsemigroups in [6] were contained in principal factors of D-classes, so in a sense all their elements had the same rank (apart from the adjoined zero element).By contrast, the elements of the null subsemigroups of T n considered here can have a range of ranks.
Turning Theorem 4.4(iii) around, we obtain the degree of a null semigroup:

Maximal null subsemigroups of T n
It is also possible to calculate the number of maximal null subsemigroups of T n .(By a maximal null subsemigroup we mean a null subsemigroup that is not properly contained in another null subsemigroup.)To this end, we write F (n) for this number.By Lemma 4.2, any maximal null subsemigroup of T n is of the form N (W, ζ) for some idempotent ζ, and some ζ-system W. So a first attempt to calculate F (n) would be to count the number of pairs (ζ, W) where ζ is an idempotent and W a ζ-system.However, due to a technicality explained below, this results in some over-counting.To get around this, it will be convenient to also write F 1 (n) for the number of maximal null subsemigroups of T n whose zero has rank 1.

Consider an idempotent
∈ T Z i be the constant map with image z i .Then each W i = {W i } is trivially a ζ i -system, and it is clear that with each factor N (W i , ζ i ) a null subsemigroup of T Z i whose zero has rank 1.Thus, we can uniquely specify a maximal null subsemigroup S of T n as follows: (i) Choose an integer partition σ = (s 1 , . . ., s r ) ⊢ n (which, recall, means that σ is a composition of n and s 1 ≥ • • • ≥ s r ).
(ii) Choose a set partition {Z 1 , . . ., Z r } of n with |Z i | = s i for each i.
(iii) Choose maximal null subsemigroups S i of T Z i , i ∈ r, each of which has a zero of rank 1.
The semigroup so-determined is then Given a partition σ as in (i), write f (σ) for the number of ways to choose the set partition in (ii), and g(σ) for the number of ways to choose the semigroups S i in (iii).Note that the number of choices available in (iii) does not depend on the particular choice made in (ii).It is therefore clear that and also that A formula for f (σ) is also well known, and easily calculated.For each i ∈ n, let σ i be the number of entries of σ equal to i. Then Thus, given (4.9)-(4.11), to compute F (n) it remains to understand the numbers F 1 (n), and we now show that The n ≤ 2 cases are easily dealt with, so we now assume that n ≥ 3. Clearly F 1 (n) is equal to n times the number of maximal null subsemigroups of T n whose zero is ζ = ( n 1 ), the constant map with image {1}, so it is enough to show that there are 2 n−1 − 2 such subsemigroups.As above, such a subsemigroup has the form N (W, ζ), where W = {W } for some subset {1} ⊆ W ⊆ n.For simplicity we denote this subsemigroup by As extreme cases, we observe that N (n) = N ({1}) = {ζ} is not maximal, as it is obviously contained in every other N (W ); recall that we are assuming n ≥ 3. On the other hand, given distinct subsets {1} ⊂ W 1 , W 2 ⊂ n, the subsemigroups N (W 1 ) and N (W 2 ) are incomparable in the inclusion order.Indeed, if x ∈ W 1 \ W 2 , then for any y ∈ n \ W 1 and z ∈ W 2 \ {1} we have It follows from all of this that the maximal null subsemigroups of T n with zero ζ = ( n 1 ) are precisely those of the form N (W ) with {1} ⊂ W ⊂ n.Since there are of course 2 n−1 − 2 of these, this completes the proof of (4.12).
Calculating the numbers F (n) using (4.9)-(4.12)requires the enumeration of all integer partitions of n, which is computationally challenging.However, F (n) may be computed more efficiently using a simple recurrence, as we now explain.Observe that we can also specify a maximal null subsemigroup S of T n in the following way: • Choose the kernel-class containing 1 of the zero of S. Denote this class by Z 1 .
• Choose a maximal null subsemigroup S 1 of T Z 1 whose zero has rank 1.
• Choose a maximal null subsemigroup S 2 of T n\Z 1 (with zero of any rank).
The semigroup so-determined is then S = S 1 ⊕ S 2 .If |Z 1 | = k, then we can perform the above steps in n−1 k−1 , F 1 (k) and F (n − k) ways, respectively.This then leads to the recurrence: with the numbers F 1 (k) given in (4.12).Table 3 gives some values of F (n) and F 1 (n), computed using (4.12) and (4.14).The numbers F (n) grow rapidly; for example, F (1000) has 2287 digits.At the time of writing, the numbers F (n) did not appear on the OEIS [1].

Nilpotent semigroups
The construction of the null semigroups N (W, ζ) in Subsection 4.1 can be generalised to nilpotent semigroups.Recall that a semigroup S is nilpotent if for some integer k ≥ 1 we have S k = {z} for some fixed element z ∈ S, which is necessarily a zero element of S; if k is minimal with respect to this property, we say that S is k-nilpotent.So 2-nilpotent semigroups are null.Here we give the details for 3-nilpotent subsemigroups of T X ; the construction for k ≥ 4 is easily adapted, though the enumeration becomes more complicated.
To describe 3-nilpotent subsemigroups of T X with zero ζ = Z i z i i∈I , we require two systems V = {V i : i ∈ I} and W = {W i : i ∈ I} such that z i ∈ V i ⊆ W i ⊆ Z i for all i.For such a pair of systems, it is easy to show that Strictly speaking, to ensure that N (V, W, ζ) is indeed 3-nilpotent, and not degenerately 2-nilpotent (i.e., null), we must have strict inclusions {z i } ⊂ V i ⊂ W i ⊂ Z i for at least one i ∈ I.In particular, this forces |X| ≥ 4. Conversely, any 3-nilpotent subsemigroup S with zero ζ is contained in some such N (V, W, ζ).Indeed, given S we define The size of N (V, W, ζ), as defined above, is given by In the finite case, say we have three compositions υ = (u 1 , . . ., u r ), τ = (t 1 , . . ., t r ) and σ = (s 1 , . . ., s r ), satisfying υ ⪯ τ ⪯ σ ⊨ n.We then calculate and note that n = s 1 + • • • + s r .For integers 1 ≤ i ≤ p and 1 ≤ j ≤ q we have i p−i j q−j ≤ (i + j) p−i (i + j) q−j = (i + j) (p+q)−(i+j) .It quickly follows that Note that when r = 1 we have size t n−t • u t−u , so the maximum is achieved at rank 1: i.e., a 3-nilpotent subsemigroup of T n of maximum size has a zero of rank 1.Thus, if we write κ(n) for the maximum size of a 3-nilpotent subsemigroup of T n , then we have For fixed 1 ≤ t ≤ n, note that t n−t • u t−u is maximised (for 1 ≤ u ≤ t) when u = α(t), and then u t−u = ξ(t); the functions α and ξ were defined in (2.3).We have therefore proved the following: Theorem 4.15.For an integer n ≥ 4, the maximum size of a 3-nilpotent subsemigroup of T n is equal to Table 4 gives calculated values of κ(n), along with the values of u and t for which κ(n) = t n−t • u t−u .These numbers do not appear on the OEIS [1]. Figure 2 shows the egg-box diagram of a 3nilpotent subsemigroup of T 5 of size κ(5) = 18, again produced with GAP [27,46].

Maximal 3-nilpotent subsemigroups of T n
As with null semigroups in Subsection 4.2, it is possible to calculate the number of maximal 3-nilpotent subsemigroups of T n , though the calculation is a little more delicate.Here, by a maximal 3-nilpotent subsemigroup we mean a 3-nilpotent subsemigroup that is not properly contained in another 3-nilpotent subsemigroup.To this end, we write • G(n) for the number of maximal 3-nilpotent subsemigroups of T n , and • G 1 (n) for the number of such subsemigroups whose zero has rank 1.
For reasons that will become clear shortly, we also need to define some further sequences.Specifically, we also write • H(n) for the number of null subsemigroups of T n that are not properly contained in any null or 3-nilpotent subsemigroup, and • H 1 (n) for the number of such subsemigroups whose zero has rank 1.
Now consider some maximal 3-nilpotent subsemigroup S of T n .So this must have the form S = N (V, W, ζ), as described in the previous subsection, and we write where Again, the inclusions in (4.16) do not all have to be strict, but to ensure that S is 3-nilpotent (and not null) there must be at least one i such that where each Moreover, since S is maximal 3-nilpotent, each • a null subsemigroup of T Z i that is not properly contained in any null or 3-nilpotent subsemigroup, depending on the strictness of the inclusions in (4.16) for that particular i ∈ r.In either case, the zero ζ i of each factor S i has rank 1.One can then follow either of the two methods in Subsection 4.2 to calculate the number G(n) of maximal 3-nilpotent subsemigroups of T n .
We just give the details for the second approach, and obtain a recurrence for G(n).The basis of the recursion is the observation that a maximal 3-nilpotent subsemigroup S can be specified in the following way: (ii) Choose a subsemigroup S 1 of T Z 1 whose zero has rank 1, and is either • maximal 3-nilpotent, or else • null, but not properly contained in any null or 3-nilpotent subsemigroup of T Z 1 .
(iii) Choose a subsemigroup S 2 of T n\Z 1 (with zero of any rank) that is either • maximal 3-nilpotent, or else • null, but not properly contained in any null or 3-nilpotent subsemigroup of T n\Z 1 , ensuring that S 1 and S 2 are not both null.
The semigroup so-determined is then S = S 1 ⊕ S 2 .If |Z 1 | = k, then we can perform step (i) in n−1 k−1 ways.If S 1 is null, then there are H 1 (k) and G(n−k) ways to perform steps (ii) and (iii), respectively.Otherwise, S 1 is 3-nilpotent, and there are G 1 (k) and G(n − k) + H(n − k) ways.This then leads to the recurrence: In order to implement this, it remains to understand the numbers G 1 (n), H 1 (n) and H(n).Formulae/recursions for these are given below in (4.19)-(4.21).
We begin with the H-numbers, and we first claim that Indeed, for n ≤ 3 we have H 1 (n) = F 1 (n), as there are no (properly) 3-nilpotent subsemigroups of T n for n ≤ 3. We now assume that n ≥ 4. Aiming for a contradiction, suppose S is a null subsemigroup of T n , with zero of rank 1, that is not properly contained in any null or 3-nilpotent subsemigroup.By the former, S is a maximal null subsemigroup.By symmetry, we may assume the zero of S is ζ = ( n 1 ), so that S = N (W, ζ), where W = {W } for some {1} ⊂ W ⊂ n.Since n ≥ 4, we either have |W | ≥ 3 or else |W | ≤ n − 2. But in these cases, respectively, we have: Since N (W ′ , W, ζ) and N (W, W ′′ , ζ) are both 3-nilpotent, we have reached the desired contradiction, and this completes the proof of (4.19).
As in Subsection 4.2, it is then easy to see that Keeping (4.19) in mind, we note that only the k ≤ 3 terms in the above sum are non-zero.
As for the G-numbers, we claim that As usual, this is clear for n ≤ 3, so we assume that n ≥ 4.And again, G 1 (n) is equal to n times the number of maximal 3-nilpotent subsemigroups of T n whose zero is ζ = ( n 1 ).So consider some such subsemigroup S, which has the form Here, S = N (V, W, ζ), for W = {W } and V = {V }, and we abbreviate this to S = N (V, W ).This time, then as in (4.13), it is easy to construct transformations belonging to S 1 \ S 2 and S 2 \ S 1 .)Thus, G 1 (n)/n is equal to the number of pairs (V, W ), as in (4.22).Specifying such a pair is equivalent to specifying the ordered partition 3), where S(k, l) is a Stirling number of the second kind.It is well known (and easy to verify using the standard Stirling recurrence) that S(k, 3) = (1 + 3 k−1 − 2 k )/2 for k ≥ 3, so (4.21) quickly follows.Table 5 gives some values of G(n) and G 1 (n), computed using (4.17)-(4.21).The numbers G 1 (n) appear as Sequence A052761 on the OEIS [1], while G(n) was not on the OEIS at the time of writing.Although F (n) > G(n) for n ≤ 6, the sequence G(n) quickly starts to out-grow F (n).For example, G(1000) has 2405 digits, and we mentioned earlier than F (1000) has 2287.

Right null semigroups
When studying variants in Section 5, an important role will be played by certain special kinds of semigroups we call right null semigroups.
Recall that a band is a semigroup consisting entirely of idempotents.Following Clifford [7], a semigroup S is a band of semigroups if we have S = b∈B S b , for some band B, where the S b are pairwise-disjoint subsemigroups of S, and S b S c ⊆ S bc for all b, c ∈ B. Of particular interest for us is the case in which: • B is a right zero semigroup (bc = c for all b, c ∈ B), • each S b is a null semigroup, say with zero z b ∈ S b , and We will call such a semigroup S = b∈B S b a right null semigroup.When |B| = 1, S is a null semigroup; when each |S b | = 1, S is a right zero semigroup; so right null semigroups simultaneously generalise both classes.
For a right null semigroup S as above, the idempotents E(S) = {z b : b ∈ B} form a right zero subsemigroup of S isomorphic to B; so we will typically identify E(S) with B. We call S uniform if the S b have a common size.If |B| = p and |S b | = q for all b ∈ B, we call S a p × q (uniform) right null semigroup, and then of course |S| = pq.In fact, it is easy to show that a p × q right null semigroup is simply a direct product of a p-element right zero semigroup and a q-element null semigroup.
In order to understand the right null subsemigroups of a full transformation semigroup T X , we make use of the following natural extension of the ζ-systems, and associated null semigroups, from Definition 4.1.
(ii) For integers p, q ≥ 2 we have where the parameters A, B, C, D in the above sets are all integers.
Proof.(i).This follows from Theorems 3.1(ii) and 4.7, given that p × 1 and 1 × q right null semigroups are right zero or null semigroups, respectively.
(ii).We now assume that p, q ≥ 2. By Lemma 4.25, the number ϱ(p, q) is the solution to the following optimisation problem: for all r, s 1 , . . ., s r , t 1 , . . ., t r ∈ N, subject to: In the remainder of the proof, we reformulate Problem 1 until we arrive at the desired result.Writing d i = s i − t i in the above, Problem 1 is equivalent to the following: In the above, some of the d i can be 0, but not all (as q ≥ 2).For a given sum for all r, t 1 , . . ., t r , D ∈ N, subject to: Adding an ordering constraint t 1 ≤ • • • ≤ t r of course does not change the solution to Problem 3. Additionally writing C = t r = max(t 1 , . . ., t r ), we can simplify the statement a little: Any solution to Problem 4 has t 1 , . . ., t r−1 , C ≥ 2. Indeed, C ≥ 2 follows from C D ≥ q ≥ 2, while if r ≥ 2 and t 1 = 1, then t 1 could be removed from the sum Also, for a given sum t 1 + • • • + t r−1 , the maximum value of t 1 • • • t r−1 occurs when at most two of the t i are 2, and the rest are 3. Problem 4 is then equivalent to: This gives the first claimed expression for ϱ(p, q), and we now work towards establishing the second.To do so, suppose we have some solution (A, B, C, D) to Problem 5, so that 2A + 3B + C + D = µ(S), where S is some p × q right null semigroup.Since C D ≥ q we must have D ≥ log C q, and so minimality forces D = ⌈log C q⌉.We similarly deduce from It remains to show that C is bounded above by ϱ(p, 1) + ϱ(1, q).To see this, note first that C ≤ 2A + 3B + C + D = µ(S).But S ∼ = R × N for some p-element right zero semigroup R and some q-element null semigroup N , so C ≤ µ(S) ≤ µ(R) + µ(N ) = ϱ(p, 1) + ϱ(1, q), as required.
Remark 4.28.When p, q ≥ 2, the second expression in Theorem 4.27(ii) reduces the calculation of ϱ(p, q) to three separate (finite) one-variable minimisations, one for each value of A = 0, 1, 2. It is then easy to compute the numbers ϱ(p, q); some values are given in Table 6.We also observe (from the proof) that the upper bound on C of ϱ(p, 1) + ϱ(1, q) could be replaced by any other a priori known upper bound on ϱ(p, q).
It is interesting to compare Table 6 with Table 2, which gives the degree β(p, q) of a p × q rectangular band.Indeed, consider a p × q rectangular band B and a q × p right null semigroup S (note the swapping of the parameters for S).Then we have for some left zero semigroup L, right zero semigroup R and null semigroup N , with |L| = |N | = p and |R| = q.In fact, not only do we have |L| = |N |, but we have already observed in Remark 4.8 that µ(L) = µ(N ).However, we do not necessarily have µ(B) = µ(S); for example, β(2, 2) = 4 while ϱ(2, 2) = 3.Thus, we have a natural example where Other examples of this phenomenon are not hard to find, and some can be constructed using other results from this paper.For example, let B be 2 × 2 rectangular band, and let L be a left zero semigroup and G a group with Consulting Table 2, we have µ(B) = 4 = β 2 (2, 2).It then follows from Theorem 3.12 that µ(B × G) = 4 as well.On the other hand, B × L is a 4 × 2 rectangular band, and so µ(B × L) = 5, again by Table 2.
Remark 4.29.The above results also lead to a formula for the degree of an arbitrary finite right null semigroup.Specifically, if S = b∈B S b is some such semigroup, then Indeed, it follows from Lemma 4.24 that if S embeds in T n then S is contained in some uniform right null subsemigroup of T n , which of course contains a p × q right null semigroup.
Remark 4.30.Of course one could also consider left null semigroups.These are defined as expected, and the uniform ones are of course direct products of left zero and null semigroups.A description of left null transformation semigroups is more complicated than for right null semigroups, and is omitted since it is not important for our central purposes (cf.Subsection 5.3).As an example of the difference in behaviour between left and right null transformation semigroups, the following can all be easily verified using GAP [27,46]: • The transformations ( 1 2 3 4 5 1 1 x 5 5 ), x = 1, 2, 4, 5, form a 2 × 2 (uniform) left null subsemigroup of T 5 .
The first two points show that the degree of a 2 × 2 left null semigroup is 5. Combined with the third point, it follows that the degree of an arbitrary left null semigroup can not be reduced to the uniform case, contrasting with the right null situation (cf.Remark 4.29).

Decreasing the degree by forming a variant
Before we move on, we pause to address another question asked in [19].
Consider a finite semigroup S, and an arbitrary element a ∈ S. It is natural to wonder how the degrees of S and the variant S a are related.Theorem 1.1 leads to the upper bound µ(S a ) < 2µ(S).Indeed, if n = µ(S), then any embedding ϕ : S → T n yields an embedding S a → T aϕ n , which we can follow with an embedding T aϕ n → T 2n−r , where r = rank(aϕ).Question 4.4 of [19] asks if there exists a finite semigroup S such that µ(S a ) < µ(S) for some a ∈ S.This can be quickly answered in the affirmative, and we do so with two examples.The first involves cyclic groups, so we begin with the following basic result.It is most likely well known, but proofs are provided for convenience.(ii) For every i ∈ n there exists g ∈ G such that i(gϕ) ̸ = i.
Proof.(i).Clearly im(ϕ) is contained in some group H -class H of T n , and H is contained in some D-class D r with 1 ≤ r ≤ n.But H ∼ = S r , so G embeds in S r , and hence in T r .Thus, by minimality of n = µ(G) we have n ≤ r, and so r = n.Thus, H = S n .
(ii).If not, then we may assume by symmetry that n(gϕ) = n for all g ∈ G. Since im(ϕ) ⊆ S n by (i), it follows that G embeds in S n−1 , contradicting n = µ(G).
To show that µ(S) ≥ n + 1, suppose to the contrary that there is an embedding θ : S → T n .Let ζ = 0θ, and let i ∈ im(ζ) be arbitrary, so i = iζ.By (ii) there exists g ∈ G such that i(gθ) ̸ = i, so it follows that i = iζ = i(0θ) = i(0θ)(gθ) = i(gθ) ̸ = i, a contradiction.Example 4.32.Let G be a cyclic group of prime order p, and let S = G ∪ {0} be the semigroup obtained by adjoining a zero element to G. It is well known [38] that µ(G) = p, so it follows from Lemma 4.31 that µ(S) = p + 1.On the other hand, the variant S 0 (with sandwich element 0) is a null semigroup of size p + 1.Thus, for p ≥ 5 we have µ(S 0 ) < p + 1 = µ(S), as follows quickly from Theorem 4.7; cf.Table 1.
In fact, the rapid growth of the ξ function means that the ratio µ(S)/µ(S 0 ) can be made arbitrarily large by choosing p large enough.For example, again consulting Table 1, if p has 20 digits, then so too does µ(S) = p + 1, while µ(S 0 ) = 30.By contrast, we have µ(S a )/µ(S) < 2 for any semigroup S and any a ∈ S, as noted earlier.
We can also use null and nilpotent semigroups to provide an example of a semigroup whose degree is strictly greater than the degrees of all its variants.Example 4.33.Consider the 3-nilpotent semigroup S = {x, y, z, 0} for which the only non-zero product is x 2 = y: • 0 x y z 0 0 0 0 0 x 0 y 0 0 y 0 0 0 0 z 0 0 0 0 We claim that: (i) µ(S) = 5, but (ii) µ(S a ) = 4 for all a ∈ S.
We first note that (ii) is clear, given the fact that any variant of a 3-nilpotent semigroup is null, and that ξ(3) = 2 and ξ(4) = 4; cf.Theorem 4.7 and Table 1.
(If we denote these transformations by x ′ , y ′ , z ′ and 0 ′ , respectively, then the only product not equal to To show that µ(S) ≥ 5, we must show that T 4 contains no subsemigroup isomorphic to S. To do so, and aiming for a contradiction, suppose T 4 does contain such a subsemigroup T ∼ = S. Since {y, z, 0} is a null subsemigroup of S, it follows that T contains a null subsemigroup of size 3.For 2 ≤ r ≤ 4, the biggest null subsemigroup of T 4 with a zero of rank r has size ν r (4) ≤ ν 2 (4) = ξ(3) = 2; cf.Theorem 4.4.It follows that the zero ζ of T has rank 1.By symmetry, we may assume that ζ = ( 41 ).As in Subsection 4.4, it follows that T is contained in the subsemigroup Again, by symmetry we can assume that V = 2 and W = 3. Denoting the elements of N by ) , e = ( 1 2 3 4 1 1 2 3 ) , the multiplication table for N is as follows: • We now turn to variants of finite full transformation semigroups.Recall that for a fixed transformation a ∈ T n , the variant T a n is the semigroup with underlying set T n and operation ⋆ defined by g ⋆ h = gah for all g, h ∈ T n .Recall from (1.2) that when rank(a) = r, we have The main guiding theme of the current section is Question 4.2 of [19], which asks whether the upper bound of 2n − r is in fact the exact value of µ(T a n ).We begin in Subsection 5.1 with some general results showing that any embedding of T a n in T 2n−r−1 (which would 'break' the upper bound of 2n − r) is rather restricted.We then apply these results in Subsection 5.2 to show that indeed µ(T a n ) = 2n − r when r is suitably large (r ≥ n − 6); see Theorem 5.15.Subsection 5.3 then considers the other extreme case of r = 1, where a is a constant map.Since µ(T a n ) does not depend on the actual choice of the constant map a, we denote it by µ(n), and the above bounds become n ≤ µ(n) ≤ 2n−1.By realising T a n as an n×n n−1 right null semigroup (for rank(a) = 1), we apply Theorem 4.27 to give a formula for µ(n) in Proposition 5.19.We then show in Proposition 5.23 that it is indeed possible to break the upper bound of 2n − 1, and in Theorem 5.24 that the sequence µ(n) is strictly increasing in n.Theorems 5.27 and 5.28, concern the asymptotic behaviour of µ(n), showing in particular that the ratio µ(n)/n tends to 1 as n → ∞, while the difference µ(n) − n grows without bound.We state some open problems in Subsection 5.4.

General results
For the duration of this subsection, we fix an integer n ≥ 2, and a transformation a ∈ T n with rank(a) = r.Since we wish to study the variant T a n , and since T a n ∼ = T gah n for any permutations g, h ∈ S n (see [26,Proposition 13.1.3]),we may assume without loss of generality that a is an idempotent with im(a) = r.
As noted above, we have µ(T a n ) ≤ 2n − r, and here we are interested in the question of whether we could in fact have strict inequality: µ(T a n ) < 2n − r.If we do, then T a n can be embedded in T 2n−r−1 .For the duration of this subsection, we write m = 2n − r − 1 for simplicity, and we hypothesise the existence of an embedding ϕ : T a n → T m .
Since ϕ obviously cannot exist for r = n (since then m = n − 1), we assume that r < n in all that follows.We will write f We will occasionally need to refer to Green's relations on the semigroups T a n and T m ; those on the former will be denoted L a , R a , and so on, to distinguish them from those on the latter, which will be denoted in the usual way.

Consider the set
which is easily seen to be a subsemigroup of T a n of size n n−r .Consequently, S ′ = Sϕ is a subsemigroup of T m of the same size.Also define Proof.Fix some f ∈ S and x ∈ F .By definition, we have x = yg ′ for some y ∈ m and g ∈ T n .Since f ∈ S we have a = af , so g ⋆ f = gaf = ga = gaa = g ⋆ a, which gives g Now that we have proved the first assertion, it follows that the elements of S In what follows, a crucial role will be played by the constant maps from T n , and their images in T m under ϕ.For i ∈ n, write e i = n i ∈ T n for the constant map with image {i}.We will frequently make use of the fact that e i f = e if for all i ∈ n and f ∈ T n .
Since the e i are all R a -related in T a n , the e ′ i = e i ϕ are all R-related in T m , so they have a common kernel, the classes of which we will denote by B 1 , . . ., B q .Thus, we may write Since each e ′ i is an idempotent, we have x ij ∈ B j for all i ∈ n and j ∈ q.
Proof.If we had q = 1 then e ′ 1 , . . ., e ′ n would be distinct transformations of rank 1, and so The next statement uses the ⊕ operation defined in Subsection 2.4.
Lemma 5.5.For any f ∈ T n and any i ∈ q, we have Proof.In T a n we have f ⋆ e 1 = e 1 , so it follows that f ′ e ′ 1 = e ′ 1 .The claim follows.
For each i ∈ q let F i = F ∩B i .The next result follows immediately from Lemmas 5.1 and 5.5.
Corollary 5.6.The image im(ϕ) is contained in Next we wish to find a lower bound on the rank of a ′ .For this, we require the following result concerning Green's ≤ J -preorder, defined on a semigroup S by x ≤ J y ⇔ x ∈ S 1 yS 1 .In a full transformation semigroup T X , we have f ≤ J g ⇔ rank(f ) ≤ rank(g).The proof uses the natural partial order on the idempotents E(S) defined by e ≤ f ⇔ e = ef = f e. Lemma 5.7.Suppose we have an embedding ψ : S → T , where S and T are finite regular semigroups, and where all elements of S are comparable in the ≤ J -preorder.Then for all x, y ∈ S, we have x ≤ J y in S if and only if xψ ≤ J yψ in T .
Proof.Clearly x ≤ J y ⇒ xψ ≤ J yψ.For the converse, suppose xψ ≤ J yψ.By assumption we have either x ≤ J y or x ≥ J y.In the former case we are done, so suppose instead that x ≥ J y.This implies that xψ ≥ J yψ, so in fact xψ J yψ.Since x ≥ J y in the regular semigroup S, it follows from [30, Theorem 1] that there exist idempotents e, f ∈ E(S) such that x J e ≥ f J y.This implies that the idempotents eψ, f ψ ∈ E(T ) satisfy eψ ≥ f ψ, and also eψ J xψ J yψ J f ψ.As T is finite, the natural order is trivial on J -classes (see for example [31,Result 6]), so that eψ = f ψ, and thus e = f as ψ is injective.We thus have x J e = f J y.In particular, x ≤ J y, as required.
Proof.Consider the set R = Reg(T a n ) of all regular elements of T a n .As shown in [12], R is a subsemigroup of T a n , and is a chain of J a = D a -classes: D a 1 < • • • < D a r .Fix some f i ∈ D a i for each i ∈ r, and assume that f 1 = e 1 and f r = a.Then using Lemma 5.7 and the above-mentioned characterisation of the ≤ J -ordering in T m , The result is then immediate.
In the case r = 1, a is a constant map, so that rank(a ′ ) attains the lower bound of q+r−1 = q just established.We show below that the lower bound is never attained when r ≥ 2; see Lemma 5.12.
First, however, we take a closer look at the transformation a ′ = aϕ.For each i ∈ q, denote the restriction a ′ | B i by a ′ i , and write r i = rank(a ′ i ).By Lemma 5.5 we have a Lemma 5.9.We have r i ≥ r for some i ∈ q.
Proof.Recall that ϕ is an embedding of T a n into T B 1 ⊕ • • • ⊕ T Bq .For i ∈ q, let ϕ i : T a n → T B i be the result of composing ϕ with the projection onto the ith coordinate.As in [14, Section 4.3], the set T = aT n a = a ⋆ T n ⋆ a is a subsemigroup of T a n isomorphic to T r .Consider the restrictions ψ = ϕ| T and ψ i = ϕ i | T for each i ∈ q, noting that ψ is injective.If all the ψ i were non-injective, then each ker(ψ i ) would contain the minimum non-trivial congruence on T (the congruences on T ∼ = T r form a chain [42]), but then so too would ker(ψ) = ker(ψ 1 ) ∩ • • • ∩ ker(ψ q ), meaning that ψ is non-injective, a contradiction.It follows that some ψ i : T → T B i is injective.If we write p for the minimum rank of an element in im(ψ i ), then as in the proof of Lemma 5.8, we have Thus, without loss of generality we may assume that r 1 ≥ r.It follows as well that |F 1 | ≥ r (as a ′ 1 ∈ T B 1 (F 1 ) and rank(a ′ 1 ) ≥ r).
Suppose first that 1 ≤ i ≤ r.Since a is an idempotent with image r we have i = ia, and so e i = e i a = e i aa = e i ⋆ a.Thus, e ′ i = e ′ i a ′ , and so x ij ∈ im(e ′ i ) = im(e ′ i a ′ ) ⊆ im(a ′ ).But x ij ∈ B j , and since im(a ′ ) ∩ B j = {z j } it follows that x ij = z j .Now suppose r < i ≤ n.Put k = ia, and let l ∈ r \ {k} be arbitrary (recall that r ≥ 2).Choose any f ∈ T n such that kf = i and lf = l.Since k, l ∈ im(a), and since a is an idempotent, we have k = ka and l = la.It follows that e k ⋆ f = e i and e l ⋆ f = e l , so that e ′ k f ′ = e ′ i and e ′ l f ′ = e ′ l .It follows from these, respectively, that x kj f ′ = x ij and x lj f ′ = x lj .But k, l ≤ r, so by the previous paragraph we have x kj = z j = x lj , and it follows that x ij = x kj f ′ = x lj f ′ = x lj = z j .This completes the proof of (5.11).

It now follows that e
for all i ∈ n.
Since the e ′ i are pairwise distinct, it follows that the set {x 11 , . . ., x n1 } has size n.But this set is contained in F , and this contradicts Lemma 5.3.Lemma 5.12.If r ≥ 2, then rank(a ′ ) ≥ q + r.
The following is a simple consequence of Lemmas 5.4 and 5.12: Corollary 5.13.If r ≥ 2, then rank(a ′ ) ≥ r + 2. Now that we have gathered all the general results we need, we proceed to consider separate cases for 'large' and 'small' r in the next two subsections.

The case r ≥ n − 6
We now consider the case in which the rank r of the sandwich element a is suitably large, meaning specifically that r ≥ n − 6.We will shortly prove that for such large r, the degree µ(T a n ) reaches its upper bound of 2n − r.For this we need the following simple lemma.Proof.Again we write m = 2n − r − 1, and aiming for a contradiction, we assume there is an embedding ϕ : T a n → T m .As usual, we may assume that a is an idempotent, and we keep the notation of Subsection 5.1, writing f ′ = f ϕ, and so on.Keeping Lemmas 5.1 and 5.3 in mind, we may assume by symmetry that ϕ maps T a n into T m (n − 2).Let t = rank(a ′ ), so that r + 2 ≤ t ≤ n − 2, by Corollary 5.13.We immediately obtain a contradiction if r + 2 > n − 2: i.e., if r ≥ n − 3.So for the rest of the proof we assume that n − 6 ≤ r ≤ n − 4. Incidentally, this implies n ≥ 5.
Let D be the D a -class of a in T a n .Since all elements of D ′ = Dϕ are D-related (in T m ) to a ′ , and since rank(a ′ ) = t, it follows that (Note that D t (T m (n − 2)) is not itself a D-class of T m (n − 2).)By [12,Theorem 5.7(v)], D is an r n−r × Λ rectangular group over S r , where Λ is the product of the sizes of the ker(a)-classes.By Lemma 5.14 we have Λ ≥ n − r + 1, so it follows that D contains an r n−r × (n − r + 1) rectangular band.So too therefore does D ′ , and hence also D t (T m (n − 2)).It follows, using the λ and ρ parameters from Subsection 2.5, that λ t (m, n − 2) ≥ r n−r and ρ t (m, n − 2) ≥ n − r + 1.
We will obtain the desired contradiction by showing that ρ t (n − 2) < k + 1 for all n − k + 2 = r + 2 ≤ t ≤ n − 2. (5.16) To do so, we consider the allowable values of k separately.In the following we make use of Remark 2.11, which tells us that ρ n−2−l (n − 2) = 2 l if 0 ≤ l ≤ n−2 2 .We only wish to apply this for l ∈ {0, 1, 2}, so since n ≥ 5, this could only be invalid when n = 5 and l = 2; however, in this case we have ρ n−2−l (n − 2) = ρ 1 (3) = 3 < 4 = 2 l , which is sufficient for our purposes.
• When k = 6, (5.16) holds since (The ≤ sign in the first of these is because of the (n, l) = (5, 2) case mentioned above.) Remark 5.17 We do not currently know if Theorem 5.15 holds for r = n − 7, or more generally.

The case r = 1
The previous subsection considered the case in which r ≥ n − 6 was suitably large.We now consider the other extreme case in which r = 1, meaning that the sandwich element a is a constant map.Since the value of µ(T a n ) does not depend on the particular choice of the constant map a, we denote it by µ(n) = µ(T a n ).Here (1.2) gives the bounds n ≤ µ(n) ≤ 2n − 1.The variant T a n has a rather simple structure when rank(a) = 1.To describe it, we may assume by symmetry that a = e 1 = n 1 .For each i ∈ n, define Then T n = S 1 ∪ • • • ∪ S n , with |S i | = n n−1 , and we note that only S 1 is a subsemigroup of T n .However, we have S i ⋆ S j = {e j } for all i, j ∈ n, (5.18) and this determines the entire multiplication table of T a n .In particular, each S i is a null subsemigroup of T a n with zero e i .Moreover, (5.18) exhibits T a n as an n × n n−1 (uniform) right null semigroup, in the language of Subsection 4.5.It therefore follows that µ(n) = µ(T a n ) = ϱ(n, n n−1 ), so we have the following: Let 0 < y < x − 1 be arbitrary, and set z = x − y, noting that z > 1.Also let k ∈ N be such that ky ≥ 1.Note that y, z and k depend only on x.Since k, y > 0 and z > 1 are constants (not depending on n), the function (yn) zn−k−2 dominates n n−1 , so there exists N ′ ∈ N such that (yn) zn−k−2 > n n−1 for all n > N ′ .We then take N = max(k + 2, N ′ ).
Theorem 5.27 tells us that the ratio µ(n)/n tends to 1 as n increases.In contrast to this, the difference µ(n) − n grows without bound: Theorem 5.28.For any integer k ≥ 2 we have µ(n) ≥ n + k for all n ≥ ⌈k ln k⌉ + 1.
For the proof we require the following lemma, concerning the L function defined in Subsection 2.1: Now write x = L(m), so that ξ(m) = max ⌊x⌋ m−⌊x⌋ , ⌈x⌉ m−⌈x⌉ by Lemma 2.4(iii).Since m is an integer, x is not an integer, so it follows that ⌊x⌋ < ⌈x⌉.Since m ≥ µ(n) > n ≥ 2, we also have m ≥ x, so that m ≥ ⌈x⌉ > ⌊x⌋.All of this implies that ⌊x⌋ m−⌊x⌋ and ⌈x⌉ m−⌈x⌉ are both strictly less than ⌈x⌉ m−⌊x⌋ .
We first claim that ⌈x⌉ ≤ n.Indeed, this is clear if x < e since then n ≥ k + 1 ≥ 3 = ⌈e⌉.Now suppose x ≥ e, so that 1 + ln x ≥ 2. Then 2x ≤ x(1 + ln x) = n + k − 1 ≤ 2n − 1, as n ≥ k, and so x < n.The claim follows as n is an integer.
Since x, k ≥ e −1 , it follows that x ≥ k, and again the claim follows as k is an integer.
a n is isomorphic to a local subsemigroup of T 2n−r of the form bT 2n−r b, where b ∈ T 2n−r satisfies rank(b) = n and rank(b 2 ) = r.Theorem 1.1 (and the fact that |T a n | = |T n |) leads to bounds on the degree of a variant: n ≤ µ(T a n ) ≤ 2n − r for a ∈ T n with r = rank(a).(1.2)

Lemma 4 . 2 .
Given an idempotent ζ of T X , a subset S of T X is a null semigroup with zero ζ if and only if ζ ∈ S ⊆ N (W, ζ) for some ζ-system W. Proof.(⇒).Suppose S is a null semigroup with zero ζ.Obviously ζ ∈ S. If f ∈ S, then from ζ = f ζ we obtain

Theorem 4 . 7 .
If S is a null semigroup of size p, then µ(S) = min{n : ξ(n) ≥ p}.Remark 4.8.Comparing Theorems 4.7 and 3.1(i), we see that null and left zero semigroups of the same size have the same degree.
(i) Choose the kernel-class containing 1 of the zero of S. Denote this class by Z 1 .

Lemma 4 . 31 .
Let G be a finite group, and let ϕ : G → T n be an embedding, where n = µ(G).

( i )
The image im(ϕ) is contained in the symmetric group S n .

ζ a b c d e ζ ζ ζ ζ ζ ζ ζ a ζ ζ ζ ζ ζ ζ b ζ ζ ζ a a a c ζ ζ ζ ζ ζ ζ d ζ ζ ζ ζ ζ ζ e ζ ζ ζ a a a
Since S contains an element with non-zero square, it follows that e ∈ T .Since all other products in T must equal ζ, it follows from the last row of the table that T = {ζ, a, b, e}.But this T has precisely two non-zero products, b • e = e • e = a ̸ = ζ.This is the desired contradiction5 The degree of T a n
′ are only distinguished by their restriction to m \ F .Every such restriction is a map m \ F → G, of which there are |G| m−|F | .Since |S ′ | = n n−r , the second assertion follows.Lemma 5.3.We have |F | ≤ n − 2. Proof.Suppose to the contrary that |F | = n − 1 + k where k ≥ 0. Then by Lemma 5.2 and Corollary 2.2, and since G ⊆ F , we have

.
The argument in the above proof breaks down when r = n − 7.Here k = 7, and (5.16) does not hold for the minimum value of t = n − 5, as we have ρ n−5 (n − 2) = 8 = k + 1.

Table 3 .
Values of F (n) and F 1 (n): F (n) is the number of maximal null subsemigroups of T n , while F 1 (n) is the number of such subsemigroups whose zero has rank 1.

Table 4 .
Values of κ(n), which is the maximum size of a 3-nilpotent subsemigroup of T n .Also shown are the values of u and t for which κ

Table 5 .
Values of G(n) and G 1 (n): G(n) is the number of maximal 3-nilpotent subsemigroups of T n , while G 1 (n) is the number of such subsemigroups whose zero has rank 1.