Software and Calculations

(Some of the calculations in this notebook look different from those in the text because of
last minute changes in the software).

For computing cosets there are functions  LeftCosets  and RightCosets  . They take as arguments a group G and a subgroup K and produce a list of the elements of G partitioned into cosets with K itself as the first coset. To illustrate, let's repeat example 10.1. We have

A4  =  Group[P[{1, 2, 3}],  P[{2, 3, 4}]]

and

V  =  Group[P[{1, 2}, {3, 4}] ,  P[{1, 3}, {2, 4}]]

Then

LeftCosets[A4, V]

If you only want a representative from each coset you can use   LeftCosetReps  or  
RightCosetReps:

LeftCosetReps[A4, V]

These are just the first elements from each coset. We can check that V satisfies 10.3(i) by
computing its right cosets and comparing them with the left cosets:

RightCosets[A4, V]

So you can multiply two cosets together to get a third:

{P[{1,  2,  3}], P[{1,  3,  4}],  P[{1,  4,  2}],  
        P[{2,  4,  3}]}  .  {P[{1,  2,  4}],  P[{1,  3,  2}],  
        P[{1,  4,  3}],  P[{2,  3,  4}]}

We can also verify that V is normal in (Subscript[A, 4]) by calculating its conjugates. First we have
the function   Conjugate[a,b] which conjugates b by a . For example, if

a  =  P[{1, 2, 3}]

b  =  P[{1, 4}]

then

Conjugate[a, b]

You can conjugate all the elements of a set by a at once:

Conjugate[ a, Elements[V] ]

which confirms that V is normal in (Subscript[A, 4]).

Example 10.5

In (Subscript[S, 5]) we have the subgroup (Subscript[F, 20])(see exercise 8.19):

F20  =  Group[P[{1, 2, 3, 4, 5}] ,  P[{1, 2, 4, 3}]]

A pair of generators of (Subscript[S, 5]) is

a  =  P[{1, 2, 3, 4, 5}]
b  =  P[{1, 2}]

To check whether (Subscript[F, 20]) is normal in (Subscript[S, 5]) you conjugate the generators of (Subscript[F, 20]) first by a and then by b and look whether the resulting sets lie in (Subscript[F, 20]). In fact since a belongs to (Subscript[F, 20]) you need only check

Conjugate[b, Generators[F20]]

which does not lie in (Subscript[F, 20]). So  (Subscript[F, 20]) is not normal.

Example 10.6

Let (Subscript[G, 72]) (see exercise 8.16) be the permutation group

G72  =  
      Group[ P[{1, 2, 3}] ,  P[{1, 4}, {2, 5}, {3, 6}] ,  
        P[{1, 5, 2, 4}, {3, 6}] ]

(Subscript[G, 72]) has order 72:

Order[G72]

Let K be the subgroup

K  =  Group[ P[{1, 2, 3}] ,  P[{4,  5, 6}] ]

Since these two 3-cycles commute with one another, 

					K ≅  ℤ/3ℤ × ℤ/3ℤ. 

The first element of the list of generators of (Subscript[G, 72]) is also a generator of K. So to check that 
K ⊲ (Subscript[G, 72])  we need only conjugate the generators of K by the remaining two generators of (Subscript[G, 72]):

Conjugate[ P[{1, 4}, {2, 5}, {3, 6}] ,  Generators[K]]
Conjugate[ P[{1, 5, 2, 4}, {3, 6}] ,  Generators[K]]

Therefore K is a normal subgroup of (Subscript[G, 72]). The quotient group  L := (Subscript[G, 72])/ K  has order  72/9=8 . We want to determine which group of order 8 it is. Since each coset has 9 elements, we will tell Mathematica not to print out the entire list of cosets when it computes L:

L  =  LeftCosets[G72, K];

First we look at representatives from each of the cosets.

LeftCosetReps[G72, K]

The first coset, L[[1]], is K = (Subscript[1, L]). The second,  L[[2]] = (1 2)K . Now

					( (1 ((2) K ))^2) =  (1 ((2))^2)K  =  K .

In other words, it has order 2. Since the representatives of the cosets L[[3]] and L[[6]] have order 2 as well, so do the cosets themselves. Now

L[[4]] . L[[4]]

which is just L[[1]], and

L[[5]] . L[[5]]

as well. So L[[4]] and L[[5]] also have order 2. This leaves L[[7]] and L[[8]]. We can see that

				( (1 4)(2 5 3 6))^(-1) =  (1 4)(2 6 3 5) .

So these two are inverse to each other. Do they in fact have order 4 ?

L[[7]] . L[[7]]

Comparing this with our list of coset representatives, we see that this must be  L[[6]] ,
which has order 2. So  L[[6]] = L(([[7]])^2) and  L[[8]] = ((L[[7]])^3). It begins to look as if L might be isomorphic to (Subscript[D, 4])(see equation (7.1)). To check this, we need
generators σ and τ satisfying

(σ^4) = 1    (τ^2) = 1    τσ = (σ^(-1)) τ  .

Let's try   σ =  L[[7]] and  τ = L[[2]]. To see that they generate L we compute

L[[2]] . L[[7]]
L[[2]] . L[[6]]
L[[2]] . L[[8]]

Thus

L[[2]] . L[[7]]  =  L[[4]]
L[[2]] . ((L [[7]])^2)  =  L[[2]] . L[[6]]  =  L[[3]]
L[[2]] . ((L [[7]])^3)  =  L[[2]] . L[[8]]  =  L[[5]] .

So  L[[7]] and   L[[2]] do generate L. Now let's see whether they satisfy the right
relation. We must check whether

L(([[7]])^(-1)).L[[2]] = L[[8]].L[[2]] = L[[4]]  :

\(\(\(
\)\(L[\([8]\)] . L[\([2]\)]\)\)\)

which is L[[4]]. So indeed

L  ≅ (Subscript[D, 4])