ON NECESSARY AND SUFFICIENT CONDITIONS OF SIMPLY REDUCIBILITY OF WREATH PRODUCT OF FINITE GROUPS

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Introduction. Simply reducible groups introduced by Wigner in [1] in connection with the questions of the quantum theory have found applications in other branches of physics as well (for example, [2-6]). The questions formulated above were noted in [7-9]. A positive solution of the problem of a finite simply reducible group solvability was published in a series of papers [10-12]. Questions of the portability of various properties of groups to wreath products were considered in [13-15]. In this paper we determine necessary and sufficient conditions for the simply reducibility of the wreath product of finite groups. In the first section we find necessary and sufficient conditions for the reality of the Let us prove that K is an elementary Abelian 2-group. We assume the contrary, let | k j |= s > 2 for some j. Without restricting the generality, we may assume that k = k , k = k 2,…, k = k s-1. Let us consider the element 2 j 3 j s j 2 g = k × hk1 of the group G, where 1 ¹ h Î H . Again according to the definition of the reality there will be found such an element i 1 n 1 n y = k × hk1 …hkn , h ,…, h Î H , of the group G, that g y = g -1. We have 2 ) 1 2 wreath product of two finite groups. In the second section we give necessary and sufficient conditions for the simply k -1(h -1 k k -1 = g -1 = g y = reducibility of the wreath product of a simply reducible group and a cyclic group of order 2, and we present an infinite series of simply reducible groups built using the construction of a wreath product. The standard group-theoretical notation is used in this paper (for example, [16]). We also use the following equivalent definition of a simply reducible group that is more convenient for computations. A finite real group is = (h-1 )kn …(h-1 )k1 k -1k hk1 k hk1 …hkn = i 2 2 1 i 1 n 2 n 1 1 n n 1 i 2 i 1 n = kk (h-1 )knkki …(h-1 )k1kki hk k hk …hk , 2 2 i whence it follows that kki = k -1 (in particular, k does not lie in a cyclic subgroup ák2 ñ ) and, therefore, called simply reducible if -1 k k -1 -1 k k -1 -1 k k -1 k k k k G G åq3 (g ) = å| C (g) |2, (h ) 1 2 = (hn ) n 2 …(h1 ) 1 2 h 1 i h1 1 …hnn . (1) gÎG gÎG Remembering, that k k -1 = k , k k -1 = k , where q (g) =|{h Î G | h2 = g}| and C (g ) is a 1 2 s 2 2 1 G G k k -1 = k ,…, k k -1 = k , we obtain (after comparing the centralizer of the element g . 3 2 2 s 2 s-1 Necessary conditions for simply reducibility right-hand and left-hand sides of relation (1)) the system of equations of the wreath product of finite groups 1 = h-1h , 1 = h-1h ,1 = h-1h ,…,1 = h-1 h , h = h-1h , 1 s 2 1 3 2 s-1 s-2 s s -1 s-1 s -2 3 2 2 1 1 s Theorem 1. Let H and K be finite groups, and their wreath product HsK is a simply reducible group. Then H is a real group, K is an elementary Abelian 2-group. Proof. According to its definition, a simply reducible from which s s-1 h = (h-1h )(h-1 h )…(h-1h )(h-1h )(h-1h ) = 1, group is real. We show that the reality G = HsK implies that H is real, and the group K must be an elementary Abelian 2-group. So, let the group G = HsK be real, k1 = 1, k2 ,…, kn are all the different elements of the group K. Let us take an arbitrary element h Î H and consider the element x = hk1 of the group G. According to the definition of the contradiction with the choice of h. Hence, the group K is an elementary Abelian 2-group. The theorem is proved. We show that the conditions formulated in Theorem 1 are sufficient for the realness of the group G = HsK. Let G = HsK, where the group H is real, and K is an elementary Abelian 2-group. We choose an arbitrary element g Î G and establish its realness. reality, there will be found such an element Situation 1. Let g = hk1 …hkn , where h ,…, h Î H . 1 n 1 n y = k × hk1 …hkn , h ,…, h ÎH , From realness H it follows that there exist such elements 1 n i i i 1 n 1 n of the group G, that x y = x-1. We have r ,…, r Î H , that hri = h-1, i = 1,…, n. The element (h-1)k1 = x-1 = xy = (h-1)kn …(h-1)k1 × k -1hk1 k × hk1 …hkn = y = rk1 …rkn , obviously, inverts g. n 1 i i 1 n 1 n = (h-1h )k1 …(h-1hh )ki …(h-1h )kn = (h-1hh )ki , Situation 2. Let g = k × hk1 …hkn , where k Î K и 1 1 i i n n i i 1 n -1 -1 i i whence it follows that ki = 1 and h = h hh . Thus, h1, …, hn Î H . Without restricting the generality, we may every element of the group H is conjugate in it with its inverse and, hence, H is a real group. assume, that the elements of the group K are arranged in such a way that kik = ki+m , i = 1,…, m = n / 2. According to this the element g (and any other element from K ) can be expressed by the following product m Situation 2. Let z-1 = s( y-1, x-1), z = s(x, y), x, y Î H . Then i i+m g = k ×Õhki hkik . i=1 We sort out for each product hihi+m , i = 1,…, m, such z-1gz = s( y-1, x-1)(h, f )s(x, y) = = (x-1, y-1)( f , h)(x, y) = ( f x , hy ) = (h, f ), an element f Î H , that (h h ) fi = (h h )-1 and assume from which f x = h and hy = f . From the equalities i i i+m i i+m f x = h and f v = h it follows, that x = uv, u Î C ( f ) . H m k -1 k k y v-1 y = k ×Õ( fihi ) i (hi fi ) i . Similarly, from the equalities h = f and h = f it i=1 follows, that y = tv-1 , where t Î C (h) . Thus, H Then m z = s(uv,tv-1) , where t Î C (h) and u Î C H H ( f ). i i i i g y = Õ(h-1 f -1)ki ( f -1h )kik × k k ´ i=1 The second assertion of the lemma follows from the isomorphism CH (h) @ CH ( f ). The lemma is proved. ´ m Õhki hkik k ×Õ( f h )ki (h-1 f )kik = Lemma 2. If g = (h, f ) and the elements h and f m i i+m i i i i i=1 i=1 m are not conjugate in H , then CG (g) = {(x, y) | x Î CH (h), y Î CH ( f )}, = k ×Õ(h-1 f -1 )kik ( f -1h )ki hkik hki ( f h )ki (h-1 f )kik = i i i i i i+m i i i i i=1 m in particular, i i i+m i i i i i i i = k ×Õ( f -1h h f h )ki (h-1 f -1h h-1 f )kik = CG (g) = CH (h) × CH ( f ) . i=1 m Proof. Let z Î CG (g ) . If z = (x, y) , then i+m i = k ×Õ(h-1 )ki (h-1)kik = g -1. g z = (x-1, y-1)(h, f )(x, y) = (hx , f y ) = (h, f ). i=1 A sufficient condition of simply reducibility From which x Î CH (h) and y Î CH ( f ) . If we assume of the wreath product of a simply reducible group and a cyclic group of order 2 Theorem 2. In order that the wreath product G=HsZ2 that z = s(x, y) , then g z = s( y-1, x-1)(h, f )s(x, y) = ( f x , hy ) = (h, f ). of a simply reducible group H and a cyclic group of order 2 be a simply reducible group it is necessary and sufficient that the equality which contradicts the disconjugacy of h and lemma is proved. Lemma 3. If g = s(h, f ) , then f . The å (3q4 (u) | C (u) | +3q2 (u) | C (u) |2 + | C (u) |3) = C (g ) = H H H H H (u,v)ÎH´H ,uÎvH = {( yh , y) | y Î C G (h × f )}∪{s( f -1t, th-1) | t Î C ( f × h)}, = 4 H å C (h) 2 + 3 å | C (h) |4. H H (2) hÎH be satisfied. H H ( h, f )ÎH ´H ,hÎ f H in particular, CG (g) = 2 × CH (h × f ) . In the following three lemmas, the structure of the centralizers of the group G elements is determined. Proof. Let z Î CG (g ) . Again we consider two Lemma 1. If g = (h, f ), f , h Î H , some v Î H , then and f v = h for situations. Situation 1. Let z = (x, y), x, y Î H . We show that C (g) = {(t, u), s(uv, tv-1) | t Î C (h), u Î C ( f )}, z Î CG (g ) when and only when y Î CH (hf ) and x = yh . G H H in particular, G H C (g) = 2 × C (h) 2 . We have s(h, f ) = z-1gz = (x-1, y-1)s(h, f )(x, y) = = s( y-1, x-1)(h, f )(x, y) = s( y-1hx, x-1 fy), Proof. Let z Î CG (g) ) . Let us cosider two situations. therefore y-1hx = h, x-1 fy = f . (3) Situation 1. Let z = (x, y), x, y Î H . Then Multiplying these equalities, we obtain y-1hfy = hf , z-1gz = (x-1, y-1)(h, f )(x, y) = (hx , f y ) = (h, f ), that is, y Î CH (hf ) . The first equality (3) in this case from which x Î CH (h) and y Î CH ( f ). gives x = yh . Inversely, let y Î CH (hf ) and x = yh . Then and v = y2. If z = s(x, y) , then the equality z2 = g is y-1hx = h and, consequently, x-1 fy = h-1 y-1hfy = satisfied when and only when = h-1hf = f , that is, equalities (3) are satisfied, hence, u = yx, v = xy. (6) z-1gz = g . Situation 2. Let z = s(x, y), x, y Î H . Then We show that the conditions (6) are equivalent to the fact that both u and v are conjugate in some element h Î H and g = z-1gz = s( y-1, x-1)s(h, f )s(x, y) = x = h-1t, y = ut-1h, t Î CH (u). (7) = s( y-1, x-1)( f , h)(x, y) = s( y-1 fx, x-1hy) Indeed, ( yx) y = xy , therefore, u and v are conjugate and the equality two equalities: z-1gz = g is equivalent to the following in H. Let and v = uh . Then from (6) it follows that h -1 x -1 u = v = xy = xux = u . y = ux-1 y-1 fx = h, x-1hy = f . (4) From which x = h-1t, y = ut-1h for some t Î CH (h). Multiplying the first equality by the second from the right, we obtain -1 The lemma is proved. Using Lemma 4, we find the sum of the cubes of the values of the function q. We have y-1 fhy = hf = ( fh)h , åq3 (g) = å q3 (s(u, v)) + å q3 ((u, v)) = from which y = h-1t for some t Î C ( fh) and G G G gÎG u,vÎH u,vÎH H = å q3 ((u, v)) = å q3 (u)q3 (v) + x = f -1 yh = f -1th-1h = f -1t. G H H u,vÎH (u,v)ÎH ´H ,uÏvH + å (q (u)q (v)+ | C (u) |)3 = å q3 (u)q3 (v) + Inversely, let x = f -1t and y = th-1 for some (u,v)ÎH ´H ,uÎvH H H H H H u,vÎH t Î C ( fh) . Then æ 3q4 (u) | C (u) | + ö H + å ç H H ÷. (8) ç 2 2 3 ÷ -1 -1 -1 (u ,v)ÎH ´H ,uÎvH è + 3qH (u) | CH (u) | + | CH (u) | ø and y fx = ht ff t = h x-1hy = t -1 fhth-1 = fhh-1 = f , Comparing (5) with (8) and taking into account that, by virtue of the simply reducibility of the group H , the following equality is true that is, equalities (4) are satisfied. The lemma is proved. Using lemmas 1-3, we find the sum of the squares of the centralizers of the elements of the group G. We have å q3 (u)q3 (v) = å C (h) 2 C ( f ) 2 , H H H H u ,vÎH h, f ÎH we obtain that condition (2) is necessary and sufficient for G G G å C (g) 2 = å | C (s(h, f )) |2 + å | C gÎG h, f ÎH h, f ÎH ((h, f )) |2 = simply reducibility of the wreath product of a simply reducible group with a cyclic group of order 2. The theorem is proved. = å | 2C h, f ÎH (hf ) |2 + å (2 | C H H ( h, f )ÎH ´H ,hÎf H (h) |2 )2 + The proof of Theorem 2 allows us to construct the following infinite series of simply reducible 2-groups. Corollary 1. Let G = HsZ2, where H is an elementary + å (|C (h) || C ( f ) |)2 = 4 H å C (h) 2 + Abelian 2-group of order 2n. Then G is simply reducible. ( h, f )ÎH ´ H ,h Ï f H H H H hÎH Proof. Let Z2 = {1, -1} , H = Z 2 ´…´ Z2 and + å | C (h) |2| C ( f ) |2 +3 å | C (h) |4. (5) h = (h1, …, hk ) . We find q(h). Obviously, if i h, f ÎH H H H ( h, f )ÎH ´H ,hÎ f H h = (1,…,1) , then q(h) = H = 2n . If h = -1 for at least Lemma 4. Let g Î G . Then one i , then q(h) = 0 , as the equation x2 = -1 is not 1) qG (g) = 0 , if g = s(u, v); solvable in group Z2 . As H is Abelian, then 2) qG (g) = qH (u)qH (v) , if g = (u, v) and, u and v C (h) = H for any h Î H . From which, are not conjugate in H ; En æ 3q4 (u) | C (u) | + ö 3) qG (g) = qH (u)qH (v) + CH (u) , if u and v are å ç H H ÷ = ç 2 2 3 ÷ conjugate in H. Proof. The assertion 1) is obvious, since the square of any element from G lies in H ´ H . (u,v)ÎH ´H ,uÎvH è + 3qH (u) | CH (u) | = 24n (4 + 3× 2n ) = = 4 H å C (h) 2 + 3 å + | CH (u) | ø | C (h) |4. Let g = (u, v) and z Î G . If z = (x, y) , then the H hÎH H ( h, f )ÎH ´H ,hÎ f H i equality z2 = g is satisfied when and only when u = x2 The corollary is proved. Conclusion. It is proved that the reality of H is the necessary condition of simply reducibility of the wreath product of the finite group H with the finite group K, and the group K must be an elementary Abelian 2-group. We also indicate sufficient conditions for simply reducibility of a wreath product of a simply reducible group with a cyclic group of order 2.

Sobre autores

S. Kolesnikov

Reshetnev Siberian State University of Science and Technology

Email: sgkolesnikov@sibsau.ru
31, Krasnoyarsky Rabochy Av., Krasnoyarsk, 660037, Russian Federation

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