# Chapter 11: A Calculation # # # # Let's look at the Sylow subgroups of S8. First factor 8! . # > ifactor(8!); 7 2 (2) (3) (5) (7) # # The Sylow 5-subgroups and Sylow 7-subgroups are cyclic, generated by # 5-cycles and 7-cycles respectively. A Sylow 3-subgroup is generated by # two disjoint 3-cycles. A Sylow 2-subgroup has order 2^7 = 128 and is # harder to find. The order of any element in it is a power of 2. So # let's begin our list of generators with an 8-cycle, say # # (1 2 3 4 5 6 7 8) # # If we add an arbitrary 4-cycle we will get a group which is too big. # Now the square of this 8-cycle is # # (1 3 5 7)(2 4 6 8) # # So let's take # # (1 3 5 7) # # as the second generator. # > H := Group( [[1,2,3,4,5,6,7,8]], [[1,3,5,7]] ): > Ord(H); 32 # # So we have to add another element to our list of generators. If we # take an arbitrary 2-cycle we will get the whole group S8. But the # square of our 4-cycle is # # (1 5)(3 7) . # # So let's add # # (1 5) # # to our set of generators. # > H := Group( [[1,2,3,4,5,6,7,8]], [[1,3,5,7]], [[1, 5]] ): > Ord(H); 128 # # We have found a 2-Sylow subgroup! # # Theorem 11.2 and Corollary 11.1 tell us that the number of Sylow # 2-subgroups is odd and divides # 8! /128 = 315. It would be helpful to know how many there really are. # By the second Sylow theorem (Theorem 11.3) all Sylow 2-subgroups are # conjugate to one another. So we must determine the number of subgroups # of S8 conjugate to our subgroup H. The function NumberOfConjugates # will do this. An 8-cycle and a transposition generate S8: # > S8 := Group( [[1,2,3,4,5,6,7,8]], [[1,2]] ): # # Then # > NumberOfConjugates( S8, H ); 315