He encontrado que alguna farmacia puede tener existencias limitadas de ciertos medicamentos, mientras que otras pueden tener casi cualquier formato que se le ocurra y el habitual de dosis habitualidad apareció. En resumen, siempre se contiene el almacén de corroborar. Al mismo tiempo que el producto que más que gustaba ha resultado no estaba disponible en stock otro distinto por las Buenas costumbres también debe buscarse jefe no asн parezca. Por eso es importante disponer de un Plan B para actuar cuandod ello no ocurra. Ventaja de tomar un genérico en lugar de Asix Un genérico es más barato que el nombre de marca Uno de los mayores incentivos para someterse al Dónde comprar Lasix genérico en lugar de pagar la marca es que usted puede obtener un ahorrando importantes Lasix genérico. Por lo tanto, un Lasix genérico es en general mucho más barato que el homólogo de marca, así que una denominación genérica se hace posible para las personas que usan este medicamento con frecuencia. Un ejemplo: La compra de lurosemida en lugar de Lasix es una considerable ahorro para el presupuesto mensual de medicamentos.

People.math.carleton.ca

Irreducible characters which are zero on only Institute of Advance Studies in Basic Sciences, Zanjan, Iran Suppose that G is a …nite solvable group which has an irreducible which vanishes on exactly one conjugacy class.
show that G has a homomorphic image which is a nontrivial 2-transitivepermutation group. The latter groups have been classi…ed by Huppert.
We can also say more about the structure of G depending on whetheris primitive or not.
Mathematics Subject Classi…cation 2000: 20C15 20D10 20B20 be an irreducible character of a …nite group G. A well-known theorem on at least one conjugacy class of G. Groups having an irreducible characterthat vanishes on exactly one class were studied by Zhmud’in [11] (see also [1]).
Chillag [2, Lemma 2.4] has proved that if the restriction of vanishes on exactly one class of G, then G is a Frobenius group with a complement of order 2 and an abelian odd-order kernel.
Our purpose in this paper is to show that, if an irreducible character …nite solvable group G vanishes on exactly one conjugacy class, then G has ahomomorphic image which is a nontrivial 2-transitive permutation group. Thelatter groups have been classi…ed by Huppert: they have degree pd where p isprime, and are subgroups of the extended a¢ ne group A L(1; pd) except for sixexceptional degrees (see Remark 8 below).
We shall initially assume that our character is faithful, and make the followingassumptions: (*) G is a …nite group with a faithful irreducible character only one class which we denote by C. Furthermore, G has a chief factorK=L which is an elementary abelian p-group of order pd such that therestriction must be nonlinear, the latter condition clearly holds whenever G is solvable, but for the present we shall not assume solvability.
Proposition 1 Suppose (*) holds. Then C = K n L, K = hCi and L is equalto L0 := fu 2 G j uC = Cg. In particular, C consists of p-elements (since L doesnot contain a Sylow p-subgroup of K). Moreover, either: pd is the sum of pd distinct G-conjugate irreducible Proof. Since K is irreducible, the theorem of Burnside quoted above shows Now since K=L is an abelian chief factor, and from [9, (6.18)] that either (i) d is even, pd is the sum of pd distinct G-conjugate irreducible characters We shall consider these two cases separately.
In case (i) we note that, since C \ L = ;, the irreducible character is assumed to be faithful, L is contained in the centre Z(G) of G. On the otherhand, for each z 2 Z(G), (z) is a scalar of the form 1: Thus for each x 2 C wehave K for all z 2 Z(G). This shows that Z(G) is a normal is a nonlinear irreducible character of K, we conclude that Z(G) = L. Finally,since K=Z(G) is abelian, [9, (2.30)] shows that K n L = C.
K is an irreducible constituent of ( 1)K and so comparison normal subgroup L, and so K n L = C in this case as well.
Finally since jC [ f1gj > 1 jKj, therefore K = hCi. Finally, it is easily seen that L0 is a normal subgroup of G, and that L0 C = K n L is a union of cosets of L, we see that L C * L0 since C is not a subgroup. Therefore L0 C G and L Corollary 2 Under the hypothesis (*) every normal subgroup N of G eithercontains K (when N is irreducible) or is contained in L (when N is reducible).
In particular, K=L is the unique chief factor such that is reducible and K=L is the socle of G=L. Since K has a nonlinear irreduciblecharacter, K is not abelian and so L 6= 1.
Remark 3 Both cases (i) and (ii) in Proposition 1 can actually occur. Thegroup SL(2; 3) has three primitive characters of degree 2 which satisfy (*) (case(i) with jKj = 8 and jLj = 2 for each character), and S4 has an imprimitivecharacter of degree 3 which satis…es (*) (case (ii) with jKj = 12 and jLj = 4).
Proposition 4 Suppose that the hypothesis (*) and case (i) of Proposition 1hold. Then L = Z(G) has order p, K is an extraspecial p-group and Proof. Let z 2 L. Then for any x 2 C we have zx 2 C and so zx = y 1xy for some y 2 G. Since K=L is an elementary abelian p-group, zpxp = (zx)p =y 1xpy = xp, and so zp = 1.
represented faithfully as a group of scalar matrices by a representation a¤ording , it follows that L is cyclic and hence jLj = p.
(K) = L = Z(K) and so K is an extraspecial p-group.
is primitive. Indeed, otherwise there is a maximal induced character shows that G is 0 on each conjugacy class disjoint fromH. As is well-known every proper subgroup of a …nite group is disjoint fromsome conjugacy class, and so we conclude that C is the unique class such thatC \ H = ;. By Proposition 1 this implies that H \ K (1) = pd=2, we obtain a contradiction. Thus Proposition 5 Suppose that the hypothesis (*) and case (ii) of Proposition 1hold (so is imprimitive). Then there exists a subgroup M of index pd in G = G for some 2 Irr(M), G = MK and M \K = L = coreG(M).
Proof. As noted in the proof of Proposition 1 irreducible constituents i. Because K is irreducible, these constituents are K-conjugates (as well as G-conjugates): Let M := IG( 1) be the inertial subgroup…xing the constituent Then jG : Mj = pd and G = MK because K acts jK : M \ Kj, we conclude that M \ K = L: On the other hand, since G is0 on any class which does not intersect M , the hypothesis on that ux does not lie in any y 1M y, and hence ux 2 C. Thus with the notationof Proposition 1, coreG(M ) L0 = L. Since L is a normal subgroup contained in M , the reverse inequality is also true and so coreG(M ) = L.
The proof of the next result requires a theorem of Isaacs [10, Theorem 2] Let H be a …nite group with centre Z and K be a normal subgroup of H with Z = Z(K). Suppose that H centralizes K=Z and jHom(K=Z; Z)jjK=Zj. Then H=Z = K=Z Proposition 6 Under the hypothesis (*) the centralizer CG(K=L) equals K.
is primitive, then Proposition 4 shows that the hypotheses of Isaacs’theorem are satis…ed for H := CG(K=L) (the condition jHom(K=Z; Z)jjK=Zj is trivial since the irreducibility of H implies that Z is cyclic). Also,since K is irreducible, CG(K ) = Z (G) = L, and so Isaacs’theorem shows that is imprimitive, then using the notation of Proposition 5 we can show that M \ H = L where H := CG(K=L): Indeed, it is clear from Proposition5 that L M \ H. To prove the reverse inequality suppose that u 2 M \ H.
Then for each x 2 K we have xu = yux for some y 2 L. Choose i such that in x 1M x. Since this is true for all x 2 K, it follows from Proposition 5 thatu 2 coreG(M) = L. Thus M \ H = L as claimed. Finally H = H \ MK =(H \ M)K = LK = K as required.
Corollary 7 Under the hypothesis (*) G acts transitively by conjugation on thenontrivial elements of the vector space K=L and the kernel of this action is K.
Thus G=K is isomorphic to a subgroup of GL(d; p) which is transitive on thenonzero elements of the underlying vector space.
Remark 8 Huppert [8, Chapter XII Theorem 7.3] has classi…ed all solvable sub-groups S of GL(d; p) which are transitive on the nonzero vectors of the underly-ing vector space. Apart from six exceptional cases (where pd = 32; 52; 72; 112; 232 or 34), the underlying vector space can be identi…ed with the Galois …eld GF (pd)in such a way that S is a subgroup of the group t is an automorphism of the …eld. The group 1)d. A classi…cation for nonsolvable groups has been carried out by Her- ing [5], [6]. It is considerably more complicated to state and prove, but amongother things it shows that such groups have only a single nonsolvable compositionfactor (a summary is given in [8, page 386]).
Since the latter half of hypothesis (*) is certainly satis…ed in a solvable group, we can specialize to solvable groups and drop the condition that Theorem 9 Let G be a …nite solvable group which has an irreducible character which takes the value 0 on only one conjugacy class C. Let K := hCi : Then: (b) There is a unique normal subgroup L of G such that K=L is a chief factor of G and K n L = C (we set jK : Lj = pd).
(c) G=K acts transitively on the set (K=L)# of nontrivial elements of the vector space K=L and so is one of the groups classi…ed by Huppert.
is imprimitive, then G=L is a 2-transitive Frobenius group of degree Remark 10 We also note that (c) and Huppert’s classi…cation show that theinteger k in (a) is bounded. Indeed, since except in the six exceptional cases. Computations using GAP [4] show that inthe remaining cases k Proof. (a) Let k be the largest integer such that K we know that the restriction G(k+1) is reducible, and so G(k+1) (b), (c) and (d) follow from Proposition 1, Corollary 7 and Proposition 4.
(e) Let M be the subgroup de…ned in Proposition 5. Since jMj : Since G = MK and G=K acts transitively on (K=L)# we 1 by Proposition 1, so equality must hold throughout.
1. Hence M=L acts regularly on (K=L)# and so G=L = (M=L)(K=L) is a 2-transitive Frobenius group.
Remark 11 Not all groups having an irreducible character which takes 0 ona single conjugacy class satisfy the second half of hypothesis (*).
ple, the Atlas [3] shows that A5 has three characters with this property and itscentral cover 2 A5 also has three.
the required property and each of the groups L2(2k) (k = 3; 4; :::) appears tohave one such character (of degree 2k). It would be interesting to know if thesewere the only simple groups with this property, or whether a group with such acharacter can have more than one nonabelian composition factor (see Remark8).
Another question which can be asked is what can be said about the kernel of such a character; evidently this kernel is contained in the normal subgroupL0 := fu 2 G j uC = Cg.
[1] Ya. G. Berkovich and E.M. Zhmud’, Characters of Finite Groups. Part (2), Vol. 181, Mathematical Monographs, Amer. Math. Soc., Rhode Island,1999.
[2] D. Chillag, On zeros of characters of …nite groups, Proc. Amer. Math. Soc.
[3] J.H. Conway et al, Atlas of Finite Simple Groups, Clarendon Press, Oxford, [4] The GAP Group, GAP–Groups, Algorithms and Programming, Version 4.4.4 (2005) (http://www.gap-system.org).
[5] Ch. Hering, Transitive linear groups and linear groups which contain irre- ducible subgroups of prime order, Geometriae Dedicata 2 (1974), 425–460.
[6] Ch. Hering, Transitive linear groups and linear groups which contain irre- ducible subgroups of prime order II, J. Algebra 93 (1985) 151–164.
[7] B. Huppert, Endliche Gruppen I, Springer-Verlag, Berlin, 1967.
[8] B. Huppert and N. Blackburn, Finite Groups III, Springer-Verlag, Berlin, [9] I.M. Isaacs, Character Theory of Finite Groups, Academic Press, New York, [10] I.M. Isaacs, Character degrees and derived length of a solvable group, [11] E.M. Zhmud’, On …nite groups having an irreducible complex character with one class of zeros, Soviet Math. Dokl. 20 (1979) 795-797.

Source: http://people.math.carleton.ca/~jdixon/OneZero.pdf