Chapter 11: A  Calculation

Let's look at the Sylow subgroups of (Subscript[S, 8]). First factor 8! .

FactorInteger[8!]

In other words

8! = (2^7)(3^2)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[P[{1,2,3,4,5,6,7,8}], P[{1,3,5,7}]]

Order[H]

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 (Subscript[S, 8]). 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[P[{1,2,3,4,5,6,7,8}], P[{1,3,5,7}], P[{1,5}]]

Order[H]

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 (Subscript[S, 8]) conjugate to our subgroup H. The function  NumberOfConjugates  will do this. An 8-cycle and a transposition generate (Subscript[S, 8]):

S8 = Group[ P[{1,2,3,4,5,6,7,8}], P[{1,2}] ]

Then

NumberOfConjugates[S8,  H]