BACKGROUND The number of new world records has decreased substantially in most athletic events in recent years. There has been enormous growth in participation at Masters events, and older athletes have been competing at the highest levels with much younger athletes. However, the progression of athletic performance over time has not been well investigated in Masters athletes. OBJECTIVE AND ME...

UNLABELLED The aim was to determine the incidence of symptoms of decompression sickness (DCS) in dive masters and instructors in relation to number of dives and possible risk factors. STUDY DESIGN Retrospective cohort study of dive masters and instructors in Sweden. STUDY BASE: All dive masters and instructors listed with PADI, NAUI and CMAS in Sweden as of January 1st 1999 (2380 divers). M...

High-intensity interval training (HIIT) improves peak power output (PPO) in sedentary aging men but has not been examined in masters endurance athletes. Therefore, we investigated whether a six-week program of low-volume HIIT would (i) improve PPO in masters athletes and (ii) whether any change in PPO would be associated with steroid hormone perturbations. Seventeen male masters athletes (60 ± ...

Solution. Since (ab)(abc) = (bc) and (abc)(ab) = (ac), it is easy to see that the center of S3 is the trivial subgroup. Therefore, the group of inner automorphisms of S3 is isomorphic to S3 and has size 6. On the other hand, S3 has 3 transpositions. These must be permuted by an automorphism of S3, and a nontrivial automorphism induces a nontrivial permutation. This gives an injective homomorphi...

Solution. Let np denote the number of p-Sylow subgroups. Since n11 is 1 mod 11 and divides 12, we know n11 is either 1 or 12. If it is 1, then it is normal and we are done. So instead, assume n11 is 12. Then there must be 120 elements of order 11 in G, leaving only 12 more elements available. Since n3 is 1 mod 3 and divides 44, we know that n3 is 1, 4 or 22. If it is 1, then again it is normal ...

Solution. The prime factorization of 185 is 5*37. Given a group of order 185, Let n5 be the number of subgroups of order 5 and n37 the number of subgroups of order 37. Since subgroups of order 5 and 37 will be Sylow subgroups, by the Sylow Theorems n5 and n37 have to be 1 (mod 5) and 1 (mod 37), respectively and divide 37 and 5, respectively. Hence, n5 = n37 = 1 and the 5 and 37 Sylow subgroups...