www.amepc.org/tau © Translational Andrology and Urology. All rights reserved. While Virginia Johnson, was born as Virginia Eshleman on February 11, 1925 in Springfield, Missouri, she came to be known as the research assistant, wife, and former wife of William H. Masters, M.D. (1,2). She was the “soft” side of the “Masters and Johnson sexual research team”. Dr. Masters was a gynecologist on facu...

Extensive discussions and debates about the advantages of using technology to create a shared space among learning participants have been presented in studies in the field of e-learning. One of the approaches in using or adopting technology for learning is through the use of online discussion forums, which as reported has beneficial impact on the teaching and learning process. Online discussion...

1 Centre Interdisciplinaire de Recherche en Réadaptation et en Intégration Sociale de l’Université Laval, Centre de Recherche de l’Institut Universitaire en Santé Mentale de Québec, Faculté de Médecine, Université Laval, Quebec, QC, Canada 2 Centre Hospitalier le Vinatier, Université de Lyon, F-69003, Université Claude Bernard Lyon I, EA 4615, Bron, Lyon, France 3 Berenson-Allen Center for Noni...

The order-disorder transition is studied in a system of a scalar nonconserved order parameter. We use this well studied system to show that the application of the methods of topology and geometry reveals that our knowledge of the kinetic pathways by which the order-disorder transition proceeds is far from being complete. We show that in two-dimensional (2D) and 3D systems there are three dynami...

The Masters Swimming Program is over 40 years old and has achieved international status, with thousands of participants competing in five-year age categories between 18 and 99. Early studies of Masters swimmers by age groups found an increase in times for most events of about 1% per year, and later studies showed a significant correlation with the age-associated decline in maximal oxygen uptake...

Solution. The 11-Sylow subgroup is Z/11Z; the 5-Sylow subgroup is Z/5Z. By Sylow’s theorem, the 11-Sylow subgroup is normal. Hence, the group is a semi-direct product of its 5 and 11-Sylow subgroups. Since Aut(Z/11Z) = Z/10Z has a unique subgroup of order 5, there are up to isomorphism exactly two groups of order 55: the abelian group Z/55Z and the group with presentation ⟨x, y|x = y = 1, xyx−1...

Solution: For the action mentioned in the hint, the stabilizer of the coset eH (e is the identity) is H since gH = H implies that g ∈ H. The kernel K of the homomorphism of G into the permutation group of the cosets is a normal subgroup of G which is the intersection of stabilizers of all cosets. In particular the kernel K of the above homomorphism is a subgroup of H. Its index in G is equal to...

Let G have order 2013. For p = 3, 11, 61 denote by np the number of Sylow p-groups in G. By Sylow’s theorems we have 61 | (n61 − 1) and n61|33, which is possible only for n61 = 1. Hence the 61-Sylow subgroup B is unique and therefore normal in G. Similarly, n11 | 3× 61 and 11 | (n11 − 1) yields n11 = 1; and the unique 11-Sylow subgroup A is normal in G. Note that A ∩ B is the trivial subgroup {...