SOME CONGRUENCES ASSOCIATED WITH THE EQUATION X α = X β IN CERTAIN FINITE SEMIGROUPS

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1 SOME CONGRUENCES ASSOCIATED WITH THE EQUATION X α = X β IN CERTAIN FINITE SEMIGROUPS THOMAS W. MÜLLER Abstract. Let H be a finite group, T n the symmetric semigroup of degree n, and let α, β be integers with 0 < α < β. We establish congruences modulo an arbitrary prime for the number s n (α, β, H) of solutions of the equation X α = X β in the wreath product H T n, where n = p, p+1, p+, p+. Our results generalise well-known congruences for the number of idempotent elements in T n, to which they reduce for α = 1, β =, and H the trivial group. Our approach is based on the exponential generating function for the sequence (s n (α, β, H)) n 0, which is computed, among other things, in J. Combin. Theory (A) 105 (004), Bell s formalism of exponential polynomials is used to reduce the complexity of the necessary mod p calculations. Key-words: equations in symmetric semigroups, wreath products, congruences modulo primes, idempotents, exponential polynomials. MSC-Index: 05A15, 11A07, 0M0 1. Introduction In [], Harris and Schoenfeld study the number u n of idempotent elements in the symmetric semigroup T n, where n 0. Among other things, they note the curious congruences u p+1 mod p, (1) u p+ 7 mod p, () u p+ 8 mod p, () valid for each prime number p. These congruences are easily proved by analysing the explicit formula ([, Theorem 1]) n ( ) n u n = ν n ν. ν ν=0 One may also observe that, for an arbitrary prime p, u p 1 mod p. (4) For a finite semigroup H with unit element 1 and integers α, β, n with 1 α < β and n 0, let s n (α, β, H) be the number of solutions of the equation X α = X β (i.e., the most general semigroup equation in one variable) in the wreath product H T n. 1 In [4], among other things, the asymptotics of the sequence s n (α, β, H) is found in the case 1 See the beginning of [4, Sec. ] for the precise definition. 1

2 T. W. MÜLLER when H is a finite group; cf. [4, Theorem ]. The present note arose from the author s curiosity to learn how (1) (4) would generalise to congruences for the latter sequences.. Exponential polynomials and the sequences s(α, β, H) We shall find it convenient to use (some of) Bell s formalism of exponential polynomials as developed in [1]. Let (ψ n ) n 0 be a sequence of independent variables (constants in our case), and define a sequence (φ n ) n 0 of polynomials φ n = φ n (ψ 1,..., ψ n ) by the equation n 0 ( φ n x n /n! = exp n 1 ) ψ n x n /n!, (5) in accordance with [1, Eq. (4.41)]. Among other things, Bell obtains the congruences (see Equations (6.4) and (6.6) in [1]) φ p ψ p 1 + ψ p mod p, (6) φ p+1 ψ p ψ 1 ψ p + ψ p+1 mod p, (7) φ p+ (ψ 1 + ψ )ψ p + ψ 1 ψ p+1 + ψ p+ + ψ p+ 1 + ψ ψ p 1 mod p, (8) φ p+ (ψ 1 + ψ )ψ p+1 + ψ 1 ψ p+ + ψ p+ + (ψ 1 + ψ 1 ψ + ψ )ψ p + ψ p+ 1 + ψ ψ p ψ ψ p 1 mod p, p >, (9) which we shall use. Here, p denotes an arbitrary prime, except for (9), where the case p = is excluded. The connection with the problem raised in the introduction is given by the fact that the exponential generating function of the sequence (s n (α, β, H)) n 0 with H a finite group is an exponential function. To be more precise, define a sequence of functions ( α (x)) α 0 by n 0 0 (x) = x, Then we have (see Theorem 1 in [4]) ( 1 s n (α, β, H)x n /n! = exp H α (x) = x exp( α 1 (x)), α 1. (10) µ β α 1 µ F (β α)( H α ( H x) ) ) µ, (11) µ where, for a positive integer m, F H (m) denotes the m-th Frobenius number of H; that is, the number of solutions in H of the equation x m = 1 (for a generalisation in a different direction of the special case where H = 1 see [, Prop. ]). It is easy to deduce from (10) that, for α 1, α (x) = n n 1 n n n nα α 1 x n 1+ +n α+1 = nc n (α)x n /n!, (1) n 1! n α! n 1 where n 1,...,n α 0 c n (α) := n 1,...,n α 0 n 1 + +n α=n 1 ( n 1 n 1,..., n α ) α 1 λ=1 n n λ+1 λ, (1)

3 CONGRUENCES ASSOCIATED WITH THE EQUATION X α = X β with the convention that an empty product equals 1. Hence, given integers α, β with 0 < α < β, and a finite group H, and letting, in the setting of Bell s formalism, φ n = s n (α, β, H) for n 0, we have ψ n = H n 1 µ β α ν 1,...,ν µ 1 ν 1 + +ν µ=n We also have the explicit formula s n (α, β, H) = n m=1 m κ=1 which follows from (11) and (1). (β α) n! c ν1 (α) c νµ (α) F H, n 1. (14) µ µ (ν 1 1)! (ν µ 1)! µ 1,...,µ κ β α ν 1,...,ν m 1 µ 1 + +µ κ=m ν 1 + +ν m=n n! H n κ κ! (ν 1 1)! (ν m 1)! c ν 1 (α) c νm (α)f H ( β α µ 1 ) F H ( β α µ κ ) µ 1 µ κ, n 1, (15). The coefficients c n (α) The following auxiliary result collects together the information concerning the coefficients c n (α), which we shall need. Lemma 1. Let α be a positive integer, and let p be a prime number. Then we have the following. (i) c 1 (α) = c (α) = 1. { 1, α = 1 (ii) c (α) =, α. (iii) c p+1 (α) 1 mod p. { } 1, α = 1 (iv) c p+ (α), α 1, α = 1 (v) c p+ (α) = 7, α = 9, α mod p. mod p. Proof. (i) The fact that c 1 (α) = c n (1) = 1 for α 1 and n 1, respectively, is clear. For n = and α, the sum on the right-hand side of (1) has α summands of the form n i = 1 for some i [α] and n λ = 0 for λ i. If i = 1, the corresponding summand is 1 0 0} 0 {{ 0} 0 = 1; for i < α, the corresponding summand is α factors i factors } 0 {{ 0} 0 = 0; α i 1 factors

4 4 T. W. MÜLLER while for i = α the contribution is } 0 0 {{ 0} = 0. Thus, c (α) = 1 for α 1, as α factors claimed. (ii) For α, we get contributions from tuples (n 1,..., n α ) of the form n i = for some given i [α] and n λ = 0 for λ i; or from tuples of the form n i = n j = 1 for given indices i, j [α], i j, and n λ = 0 for λ i, j. The same analysis as in the proof of Part (i) shows that the first group of summands contributes a total of 1 to the sum. Concerning the second group of summands one observes that only the case where n 1 = n = 1, n λ = 0 (λ ) gives a non-zero contribution, which is ; more precisely, for α =, this contribution is 1 1 =, while, for α, this contribution is ( } 0 {{ 0} 0 ) =. Hence, c (α) = for α, as claimed. α factors (iii) We may assume that α. For n = p + 1, only those α summands on the righthand side of (1) contribute modulo p, where n i = p for some i [α], and n λ = 0 for all λ i. For i = 1, the contribution is p 0 0} 0 {{ 0} 0 = 1; for i < α, the contribution α factors is i factors 0 p p 0 0} 0 {{ 0} 0 = 0; α i 1 factors while, for i = α, the contribution is } 0 0 {{ 0} 0 0 p = 0. Thus, c p+1 (α) 1 mod p, as α factors claimed. (iv) For n = p + and α, the right-hand sum in (1) has contributions modulo p for two kinds of α-tuples (n 1,..., n α ): those with n i = p + 1 for some given i [α], and n λ = 0 for λ [α]\{i}; and those with n i = p, n j = 1 for given indices i, j [α], and n λ = 0 for λ [α]\{i, j}. By an analysis similar to that in the proof of (i), the first group of summands contributes 1 in total. Summands from the second group will be zero if min{i, j}, or if i j, so we only need to consider the cases i = 1, j = and j = 1, i =. In the first case the product λ nn λ+1 λ contains a factor p 1, so is congruent to 0 modulo p whereas, in the second case, the product in question is 1 p = 1 for α =, and 1 p p 0 0} 0 {{ 0} 0 = 1, α factors if α. Thus, for α, the second group of summands also contributes 1 modulo p, whence our claim. (v) For α, contributions to the right-hand sum in (1) come from tuples (n 1,..., n α ) with (a) n i = p+ for some given i [α], and n λ = 0 for λ i; (b) n i = p+1, n j = 1 for indices i, j [α], and n λ = 0 for λ i, j; (c) n i = p, n j = for distinct indices i, j [α], and n λ = 0 for λ i, j; (d) n i = p, n j = n k = 1 for distinct indices i, j, k [α], and n λ = 0 for λ i, j, k. An analysis as in the proof of Part (i) shows that the total contribution coming from the group of summands in (a) is 1. From summands of type

5 (b) we get the contribution CONGRUENCES ASSOCIATED WITH THE EQUATION X α = X β 5 (p + ) [ (p + 1) α factors ] + 1 p+1 (p + 1) 0 0} 0 {{ 0} 0 4 mod p α factors coming from the tuples (p + 1, 1, 0,..., 0) and (1, p + 1, 0,..., 0), respectively, with all other contributions of this type vanishing. Similarly, for the group of summands of type (c), the total contribution computes as (p + 1)(p + ) [ p α factors ] + p p 0 0} 0 {{ 0} 0 p mod p α factors by Fermat s little theorem. Finally, for the summands of type (d), the total contribution is found to be (p + 1)(p + ) [ ] p } {{ 0} p p 1 1} {{ 0} p p mod p. α factors α factors α factors Hence, c p+ (α) 9 mod p for α ; while, for α = the contributions of types (a) (c) occur, but none of type (d), so that c p+ () 7 mod p, as claimed. 4. Some congruences for the sequences s(α, β, H) Given the information collected so far, we are in a positition to answer the question raised in the introduction. Theorem 1. Let α, β be integers with 0 < α < β, let H be a finite group, and let p be a prime not dividing β α. Then we have the following. (i) s p (α, β, H) F H (β α) mod p. (ii) s p+1 (α, β, H) F H (β α) ( F H (β α) + H ) mod p. (iii) For p, we have s p+ (α, β, H) F H (β 1) ( 4 H F H (β 1) + F H (β 1) + H ), α = 1, β even 4 H F H (β 1) + F H (β 1) + H F H (β 1)F H ( β 1 ) + H ( F H (β 1) + F H ( β 1 )), α = 1, β odd F H (β α) ( 4 H F H (β α) + F H (β α) + 4 H ), α, β α 4 H F H (β α) + F H (β α) + H F H (β α)f H ( β α ) + H ( F H (β α) + F H ( β α )), α, β α while, for p = and β α, we have mod p s 4 (α, β, H) F H (β α) mod. (17) (16)

6 6 T. W. MÜLLER (iv) For p 5, we have s p+ (α, β, H) H F H (β 1) + 15 H F H (β 1) + 9 H F H (β 1) +F H (β 1) 4, α = 1;, β 1 H F H (β 1) + 15 H F H (β 1) + 9 H F H (β 1) +F H (β 1) H F H ( β 1 ) +15 H F H (β 1)F H ( β 1 ) + H F H(β 1) F H ( β 1 ), α = 1, β odd, β 1 () H F H (β 1) + 15 H F H (β 1) + 9 H F H (β 1) +F H (β 1) H F H ( β 1 ) + H F H (β 1)F H ( β 1 ), α = 1, β even, β 1 () H F H (β 1) + 15 H F H (β 1) + 9 H F H (β 1) +F H (β 1) H F H ( β 1 ) + 6 H F H ( β 1 ) +15 H F H (β 1)F H ( β 1 ) + H F H(β 1) F H ( β 1 ) + H F H (β 1)F H ( β 1 ), α = 1, β 1 (6) 1 H F H (β ) + 7 H F H (β ) + 9 H F H (β ) +F H (β ) 4, α = ;, β 1 H F H (β ) + 7 H F H (β ) + 9 H F H (β ) +F H (β ) H F H ( β ) + 15 H F H (β )F H ( β ) + H F H (β ) F H ( β ), α =, β even, β () 1 H F H (β ) + 7 H F H (β ) + 9 H F H (β ) +F H (β ) H F H ( β ) + H F H (β )F H ( β ), α =, β odd, β () 1 H F H (β ) + 7 H F H (β ) + 9 H F H (β ) +F H (β ) H F H ( β ) + 6 H F H ( β ) +15 H F H (β )F H ( β ) + H F H (β )F H ( β ) + H F H (β ) F H ( β ), α =, β (6) 7 H F H (β α) + 7 H F H (β α) + 9 H F H (β α) +F H (β α) 4, α ;, β α mod p. 7 H F H (β α) + 7 H F H (β α) + 9 H F H (β α) +F H (β α) H F H ( β α ) + 15 H F H (β α)f H ( β α ) + H F H (β α) F H ( β α ), α, β α, β α 7 H F H (β α) + 7 H F H (β α) + 9 H F H (β α) +F H (β α) H F H ( β α ) + H F H (β α)f H ( β α ), α, β α, β α 7 H F H (β α) + 7 H F H (β α) + 9 H F H (β α) +F H (β α) H F H ( β α ) + 6 H F H ( β α ) +15 H F H (β α)f H ( β α ) + H F H(β α) F H ( β α ) + H F H (β α)f H ( β α ), α, 6 β α (18) Moreover, for p = and β α, we have s 5 (α, β, H) F H (β α)( H + 1) mod, (19)

7 CONGRUENCES ASSOCIATED WITH THE EQUATION X α = X β 7 while for p = and β α, we have s 6 (α, β, H) F H (β α) mod. (0) Proof. Combining (14) with Lemma 1, we obtain the following equations and congruences: ψ 1 = F H (β α), (1) { 0, β α ψ = H F H (β α) + H F H ( β α ), β α, () H F H (β 1), α = 1;, β 1, H (F H (β 1) + F H ( β 1 )), α = 1, β odd, β 1 (), H (F H (β 1) + F H ( β 1 ), α = 1, β even, β 1 (), H (F H (β 1) + 6F H ( β 1 ψ = ) + F H( β 1 )), α = 1, β 1 (6), 9 H F H (β α), α ;, β α, H (F H (β α) + F H ( β α )), α, β α, β α, H (9F H (β α) + F H ( β α )), α, β α, β α, H (9F H (β α) + 6F H ( β α ) + F H( β α )), α, 6 β α. () ψ p 0 mod p, (4) ψ p+1 H F H (β α) mod p, (5) H F H (β 1), α = 1, β even H (F H (β 1) + F H ( β 1 )), α = 1, β odd ψ p+ 4 H F H (β α), α, β α H (F H (β α) + F H ( β α )), α, β α mod p, (6)

8 8 T. W. MÜLLER and ψ p+ H F H (β 1), α = 1, β even, β 1 () H (F H (β 1) + 4F H ( β 1 )), α = 1, β odd, β 1 () H (F H (β 1) + F H ( β 1 )), α = 1, β even, β 1 () H (F H (β 1) + 4F H ( β 1 ) + F H( β 1 )), α = 1, β odd, β 1 () 1 H F H (β ), α =, β odd, β () H (7F H (β ) + 6F H ( β )), α =, β even, β () H (7F H (β ) + F H ( β )), α =, β odd, β () H (7F H (β ) + 6F H ( β ) + F H( β )), α =, β even, β () 7 H F H (β α), α ;, β α H (9F H (β α) + 6F H ( β α )), α, β α, β α H (9F H (β α) + F H ( β α )), α, β α, β α H (9F H (β α) + 6F H ( β α ) + F H( β α )), α, 6 β α (7) Assertions (i), (ii) of Theorem 1, as well as Parts (iii), (iv) for p and p 5, respectively, follow immediately from (1) (7) plus the congruences (6) (9). On the other hand, given Lemma 1, proving (1) (7) is, of course, purely routine. As an example, take a closer look at (6), which is fairly representative. By assumption, we have p β α, thus a divisor µ of β α is invertible modulo p, and cancellation of the factor p in the numerator on the right-hand side of (14) can only come from factors (ν i 1)! in the denominator. Hence, modulo p, we only need to consider summands with µ and ν i p+1 for some i [µ]. Rewriting H p+1 H mod p by Fermat s little theorem, we thus find that { } 0, β α ψ p+ H F H (β α)c p+ (α) + H F H ( β α )c mod p. 1(α)c p+1 (α), β α Using Parts (i), (iii), and (iv) of Lemma 1 to evaluate c 1 (α), c p+1 (α), and c p+ (α), Congruences (6) follow immediately. The special cases (17), (19), (0) in Parts (iii) and (iv) are easily dealt with via the explicit formula (15). mod p. References [1] E. T. Bell, Exponential polynomials, Annals of Math. () 5 (194), [] A. Dress and T. Müller, Decomposable functors and the exponential principle, Adv. in Math. 19 (1997), [] B. Harris and L. Schoenfeld, The number of idempotent elements in symmetric semigroups, J. Combin. Theory (1967), [4] C. Krattenthaler and T. W. Müller, Equations in finite semigroups: explicit enumeration and asymptotics of solution numbers, J. Combin. Theory (A) 105 (004), School of Mathematical Sciences, Queen Mary & Westfield College, University of London, Mile End Road, London E1 4NS, United Kingdom