By Olexandr Ganyushkin

The target of this monograph is to offer a self-contained creation to the fashionable concept of finite transformation semigroups with a powerful emphasis on concrete examples and combinatorial purposes. It covers the subsequent themes at the examples of the 3 classical finite transformation semigroups: differences and semigroups, beliefs and Green's kin, subsemigroups, congruences, endomorphisms, nilpotent subsemigroups, displays, activities on units, linear representations, cross-sections and variations. The e-book comprises many routines and ancient reviews and is directed, firstly, to either graduate and postgraduate scholars searching for an advent to the idea of transformation semigroups, yet must also turn out necessary to tutors and researchers.

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Deﬁne α(i) = j if j is not the root and α(i) = ∅ otherwise. One veriﬁes that the correspondence Γ → α is a bijection from the set of all labeled trees on {1, 2, . . , n + 1} to the set of all nilpotent elements in PT n . 6 it follows that n(PT n ) (n + 1)n−1 1 = . 6 we get the following identity. 10 n S(n, k + 1)k! = (n + 1)n−1 . 1 Let S be a nonempty set such that 0 ∈ S. For all a, b ∈ S set a · b = 0. Show that (S, ·) is a semigroup (it is called a semigroup with zero multiplication or a null semigroup).

Let us count the number of such bijections of rank k. The set A = dom(α) can be chosen in nk diﬀerent ways. The set B = im(α) can be independently chosen in nk diﬀerent ways. If A and B are ﬁxed, there are exactly k! diﬀerent bijections from A to B. Hence we have exactly n n k · k · k! bijections of rank k. Since k can be an arbitrary integer between 0 and n, the statement of the theorem is obtained by applying the sum rule. Partial injections α : N → N are also called partial bijections or partial permutations on N.

This inverse element is usually denoted by a−1 (this can now be justiﬁed by the requirement that it is unique). 5 it follows that in the case of n > 1 the semigroups Tn and PT n are not inverse semigroups. 7 The semigroup IS n is an inverse semigroup. Proof. Let α ∈ IS n and β ∈ VPT n (α). Let us try to analyze in which case β is a partial permutation. 4 consist of one element each. 4(b). 4(c) imply that β must be undeﬁned on N\im(α). Thus VPT n (α) ∩ IS n contains a unique element, implying that IS n is an inverse semigroup.