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Derivation of the Cycle Index Formula of the Affine Square(Q) International Journal of Algebra, Vol. 13, 2019, no. 7, 307 - 314 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/ija.2019.9725 Derivation of the Cycle Index Formula of the Affine Square(q) Group Acting on GF (q) Geoffrey Ngovi Muthoka Pure and Applied Sciences Department Kirinyaga University, P. O. Box 143-10300, Kerugoya, Kenya Ireri Kamuti Department of Mathematics and Actuarial Science Kenyatta University, P. O. Box 43844-00100, Nairobi, Kenya Mutie Kavila Department of Mathematics and Actuarial Science Kenyatta University, P. O. Box 43844-00100, Nairobi, Kenya This article is distributed under the Creative Commons by-nc-nd Attribution License. Copyright c 2019 Hikari Ltd. Abstract The concept of the cycle index of a permutation group was discovered by [7] and he gave it its present name. Since then cycle index formulas of several groups have been studied by different scholars. In this study the cycle index formula of an affine square(q) group acting on the elements of GF (q) where q is a power of a prime p has been derived using a method devised by [4]. We further express its cycle index formula in terms of the cycle index formulas of the elementary abelian group Pq and the cyclic group C q−1 since the affine square(q) group is a semidirect 2 product group of the two groups. Keywords: Cycle index, Affine square(q) group, Semidirect product group 308 Geoffrey Ngovi Muthoka, Ireri Kamuti and Mutie Kavila 1 Introduction [1] The set Pq = fx + b; where b 2 GF (q)g forms a normal subgroup of the affine(q) group and the set C q−1 = fax; where a is a non zero square in GF (q)g 2 forms a cyclic subgroup of the affine(q) group under multiplication. The semidirect product Pq C q−1 form a group known as the affine square(q) o 2 group denoted by Aff(q). The elements of Aff(q) are of the form fax + b such that b 2 GF (q) and a is a non zero square in GF (q)g. In this study we derive the cycle index formula of the affine square(p) group acting on GF (q) and express it in terms of the cycle index formulas of the elementary abelian group Pq and the cyclic group C q−1 . 2 2 Preliminary Notes Definition 2.1. [2] A group G is said to be a semidirect product group of N by H if; N C G and H < G, N \ H = feg and NH = G. We symbolically express this as G = N o H. Definition 2.2. [4] The M¨obiusfunction of any n 2 N is given by, 8 −1 if n is a square free with an odd number of prime factors <> µ(n) = 0 if n has a squared prime factor :>1 if n is a square free with an even number of prime factors: Definition 2.3. [3] The cycle index of the action of G on X is the poly- nomial (say over the rational field Q) in t1; t2; : : :; tn given by; m 1 X Z(G) = Z (t ; t ; ··· ; t ) = fmon(g)g G;x 1 2 n jGj g2G Note that if G has conjugacy classes K1;K2; : : :; Km with gi 2 Ki then m 1 X Z(G) = jK j mon(g ): jGj i i i=1 Theorem 2.4. [6] Let x be a permutation with cycle type (α1; α2; α3; : : :; αn) then; l l P i. the number π(x ) of cycles of length one in x is ijl iαi. i i P l ii. αl = l ijl π x µ(i) where µ is the M¨obiusfunction. The cycle index of the affine square(q) group 309 3 Main Results 3.1 The cycle index of the affine square(q) group Lemma 3.1. Let g 2 Aff(q) be such that g fixes only one element in GF (q), then CG(g) = C q−1 . 2 Proof. Since the Affine square(q) group acts transitively on GF (q) [5], then the stabilizers of each of the elements in GF (q) are conjugate and only intersect at the identity so it is enough to find the centralizer of any one element. The elements of Aff(q) which fix 0 2 GF (q) are of the form αx + 0 where α 0 α 2 GF (q). This can be written as; M = . 0 1 a b A general element of Aff (q) is of the form such that a is an non 0 1 a b −1 1 − b zero square element of GF (q) and = a a . 0 1 0 1 a b α 0 1 − b α −αb + b Now a a = . 0 1 0 1 0 1 0 1 a b α 0 For the element to centralize then, −αb + b = 0, implying 0 1 0 1 b(1 − α) = 0 ) b = 0 or α = 1. α 0 If α = 1, then is the identity which is centralized by every element 0 1 α 0 a 0 of the group and so the centralizers of are of the form 0 1 0 1 where a is a non zero square element in GF (q) thus the centralizer has order (q−1) (q−1) 2 which is a cyclic multiplicative group C (q−1) of order 2 . 2 Lemma 3.2. Let g 2 Aff(q) be such that g does not fix any element in GF (q), then CG(g) = Pq. Proof. Since the Affine square(q) group acts transitively on GF (q) [5], then the stabilizers of each of the elements in GF (q) are conjugate and only intersect at the identity so it is enough to find the centralizer of any one element. The elements of Aff(q) which do not fix any element of GF (q) are of the form 1 α x + α where α 2 GF (q). This can be written as; M = . 0 1 a b A general element of Aff (q) is of the form such that a is a non 0 1 a b −1 1 − b zero square element in GF (q) and = a a . 0 1 0 1 310 Geoffrey Ngovi Muthoka, Ireri Kamuti and Mutie Kavila a b 1 α 1 − b 1 αa Now a a = . 0 1 0 1 0 1 0 1 a b 1 α For the element to centralize then, αa = α, implying 0 1 0 1 α(a − 1) = 0 ) α = 0 or a = 1. 1 α If α = 0, then is the identity which is centralized by every element of 0 1 1 α 1 b the group and so the centralizers of are of the form where 0 1 0 1 b 2 GF (q) thus the centralizer has order q which is the elementary abelian group Pq. Theorem 3.3. Let p > 2 be a prime and q = pr; the cycle index formula of the affine square(q) group G acting on the q elements of GF (q) is given by; 0 1 1 q X q−1 Z = tq + (q − 1) t p + q ;(d)t t d ; (G;X) jGj @ 1 p 1 d A q−1 16=dj 2 1 where jGj = 2 q(q − 1), ;(d) is the Euler's phi function and X the q elements of the GF (q) : Proof. The elements of the Aff (q) group are partitioned into I; τ (the set of 1 elements that fix one element on the field GF (q)) and τ0 (the set of elements that do not fix any element of GF (q)). To derive the cycle index formula we need to find the number of τ0 and τ1 elements and the respective cycle types. Let g 2 τ1. Then from Lemma 3.1 CG(g) = C q−1 . 2 1 g 2 q(q − 1) ) jC j = 1 = q (1) 2 (q − 1) where Cg is the conjugacy class in G containing g. Let g 2 τ0, then from Lemma 3.2 CG(g) = Pq 1 q(q − 1) q − 1 ) jCgj = 2 = (2) q 2 1 2 q(q−1) The subgroup NG C q−1 = C q−1 , implying there are 1 = q conjugate 2 2 2 (q−1) cyclic groups C q−1 in G. 2 These cyclic groups intersect only at the identity. Thus; q − 1 jτ j = − 1 q (3) 1 2 The cycle index of the affine square(q) group 311 We find the number of τ0 elements by subtracting the number of τ1 elements and the identity from the order of G. It follows that; q − 1 q − 1 jτ j = q − − 1 q − 1 = q − 1 (4) 0 2 2 q−1 Since the length of a conjugacy class of g 2 τ0 is 2 from Equation 2 and the number of τ0 elements are q − 1 from 4, then there are two conjugacy classes of elements in τ0 but each element in τ0 is of order q: Therefore; 1 q − 1 q − 1 Z = (tq + :mon(x ) + :mon(x ) (G;X) jGj 1 2 1 2 2 +q(summation of all monomials of the nontrivial elements in cyclic subgroups C q−1 )); 2 where x1 and x2 are representatives of elements in the first and the second conjugacy classes respectively but since all τ0 are of order p then they will have the same monomial and thus Z(G;X) can be written as; 1 Z = (tq + (q − 1):monomial of an element in τ (G;X) jGj 1 0 +q(summation of all monomials of the nontrivial elements in cyclic subgroups C q−1 )) 2 That is, 0 1 1 X Z = tq + (q − 1) :mon(x) + q mon(g) (5) (G;X) jGj @ 1 A g2Cq−1nfIg where x 2 τ0.
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