On Groups of Even Order 论文

摘要

odd order are soluble. We shall use the term involution for a group element of order 2. If m is the total number of involutions of 65 and if we set n = g/m, the same method shows that 65 contains a normal subgroup V distinct from 5 such that the index of V is either 2 or is less than [n(n + 2)/2]! (where [x] denotes the largest integer not exceeding the real number x). If J is an involution in 65 and if n(J) is the order of its normalizer 91(J) in 5, then n < n(J). It then follows that there exist only a finite number of simple groups in which the normalizer of an involution is isomorphic to a given group.