scholarly journals On the asynchronous rational group

2019 ◽  
Vol 13 (4) ◽  
pp. 1271-1284 ◽  
Author(s):  
James Belk ◽  
James Hyde ◽  
Francesco Matucci
Keyword(s):  
2019 ◽  
Vol 22 (5) ◽  
pp. 953-974
Author(s):  
Ángel del Río ◽  
Mariano Serrano

Abstract H. J. Zassenhaus conjectured that any unit of finite order and augmentation 1 in the integral group ring {\mathbb{Z}G} of a finite group G is conjugate in the rational group algebra {\mathbb{Q}G} to an element of G. We prove the Zassenhaus conjecture for the groups {\mathrm{SL}(2,p)} and {\mathrm{SL}(2,p^{2})} with p a prime number. This is the first infinite family of non-solvable groups for which the Zassenhaus conjecture has been proved. We also prove that if {G=\mathrm{SL}(2,p^{f})} , with f arbitrary and u is a torsion unit of {\mathbb{Z}G} with augmentation 1 and order coprime with p, then u is conjugate in {\mathbb{Q}G} to an element of G. By known results, this reduces the proof of the Zassenhaus conjecture for these groups to proving that every unit of {\mathbb{Z}G} of order a multiple of p and augmentation 1 has order actually equal to p.


2005 ◽  
Vol 48 (3) ◽  
pp. 445-454 ◽  
Author(s):  
Frédéric Patras ◽  
Christophe Reutenauer ◽  
Manfred Schocker

AbstractThe orthogonal projection of the free associative algebra onto the free Lie algebra is afforded by an idempotent in the rational group algebra of the symmetric group Sn, in each homogenous degree n. We give various characterizations of this Lie idempotent and show that it is uniquely determined by a certain unit in the group algebra of Sn−1. The inverse of this unit, or, equivalently, the Gram matrix of the orthogonal projection, is described explicitly. We also show that the Garsia Lie idempotent is not constant on descent classes (in fact, not even on coplactic classes) in Sn.


2013 ◽  
Vol 63 (6) ◽  
Author(s):  
Temha Erkoç ◽  
Utku Yilmaztürk

AbstractA finite group whose irreducible complex characters are rational valued is called a rational group. Thus, G is a rational group if and only if N G(〈x〉)/C G(〈x〉) ≌ Aut(〈x〉) for every x ∈ G. For example, all symmetric groups and their Sylow 2-subgroups are rational groups. Structure of rational groups have been studied extensively, but the general classification of rational groups has not been able to be done up to now. In this paper, we show that a full symmetric group of prime degree does not have any rational transitive proper subgroup and that a rational doubly transitive permutation group containing a full cycle is the full symmetric group. We also obtain several results related to the study of rational groups.


1998 ◽  
Vol 08 (04) ◽  
pp. 467-477 ◽  
Author(s):  
A. Giambruno ◽  
E. Jespers

Let ℚAn be the group algebra of the alternating group over the rationals. By exploiting the theory of Young tableaux, we give an explicit description of the minimal central idempotents of ℚAn. As an application we construct finitely many generators for a subgroup of finite index in the centre of the group of units of ℚAn.


1991 ◽  
Vol 34 (2) ◽  
pp. 325-331 ◽  
Author(s):  
P. H. Kropholler ◽  
B. Moselle

We calculate K0 of the rational group algebra of a certain crystallographic group, showing that it contains an element of order 2. We show that this element is the Euler class, and use our calculation to produce a whole family of groups with Euler class of order 2.


2009 ◽  
Vol 08 (05) ◽  
pp. 723-732 ◽  
Author(s):  
CHRIS MEEHAN ◽  
LUTZ STRÜNGMANN

We study subgroups R of the rational numbers ℚ having the property that for every pair of integers m, n such that gcd(m, n) = 1 and gcd(m, p) = gcd(n, p) = 1 whenever p is in the spectrum of R there is a unit u of R and an element r ∈ R such that un + rm = 1. These rings are closely related to weakly transitive separable groups. We prove that the property is dependent on the spectrum of the rational group in question and that the spectrum may be very complicated.


1987 ◽  
Vol 48 (3) ◽  
pp. 213-216 ◽  
Author(s):  
Ashwani K. Bhandari ◽  
I. B. S. Passi

1970 ◽  
Vol 67 (2) ◽  
pp. 237-242 ◽  
Author(s):  
Michael Singer

The purpose of this paper is to obtain quantitative results on invertible powers of (fractional) ideals in commutative separable algebras over dedekind domains. This is connected with the work of Dade, Taussky and Zassenhaus(2) on ideals in noetherian domains. We do not, however, make use of their paper, but rather draw on the general theorems on ideals in commutative separable algebras established by Fröhlich (3), in particular his qualitative result that some power of any given ideal is invertible. Our basic result (Theorem 1) concerns the invertibility of powers of a particular type of ideal, the componentwise dedekind ideals defined below. From this we deduce a general result (Theorem 2), which includes as a special case Theorem C of (2) for the case of separable field extensions; specifically, the (n – 1)th power of any ideal is invertible, where n is the dimension of the algebra. Although, as we show, it is possible to deduce Theorem 2 from (2), we have here an independent proof of one of the main results of (2) based entirely on the results in (3). As a further application of Theorem 1 we obtain a new result on ideals over the group ring of an abelian group over the ring of rational integers; the (t – 1)th power of such an ideal is invertible, where t is the maximum number of simple components of the rational group algebra of any Sylow subgroup. We also show that this is the best possible result when some Sylow subgroup whose rational group algebra has t components is cyclic.


2011 ◽  
Vol 03 (03) ◽  
pp. 265-306 ◽  
Author(s):  
ANDREW PUTMAN

We calculate the first homology group of the mapping class group with coefficients in the first rational homology group of the universal abelian ℤ/L-cover of the surface. If the surface has one marked point, then the answer is ℚτ(L), where τ(L) is the number of positive divisors of L. If the surface instead has one boundary component, then the answer is ℚ. We also perform the same calculation for the level L subgroup of the mapping class group. Set HL = H1(Σg; ℤ/L). If the surface has one marked point, then the answer is ℚ[HL], the rational group ring of HL. If the surface instead has one boundary component, then the answer is ℚ.


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