By American Mathematical Society

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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.