## Monday, May 24, 2021

### Permutation Groups Revisited

Definition:   The set of all permutations of a set S is denoted by Sym(S).

The set of all permutations of the set {1,2,...,n} is denoted by Sn

Proposition. If S is any nonempty set, then Sym(S) is a group under the operation of composition of functions.

Theorem. Every permutation in Sn can be written as a product of disjoint cycles. The cycles that appear in the product are unique.

Proposition. If a permutation in Sn is written as a product of disjoint cycles, then its order is the least common multiple of the lengths of its cycles.

Definition. Any subgroup of the symmetric group Sym(S) on a set S is called a permutation group or group of permutations.

Definition. A permutation is called even if it can be written as a product of an even number of transpositions, and odd if it can be written as a product of an odd number of transpositions.

Proposition. The set of all even permutations of Sn is a subgroup of Sn.

Definition. The set of all even permutations of Sn is called the alternating group on n elements, and will be denoted by An.

Among the most interesting entities in abstract algebra is the permutation group. To fix ideas, consider the following example:

Let S = {1, 2, 3} and define the group G =  {The set of all permutations on S}

a) Write out all the members of G

b) Show that G obeys group properties and determine whether or not it is Abelian

Solution:

We may first determine how many members G has using the fact the number of elements will be n = 3!

= 3 x 2 x 1 = 6.

The components can then be found:

(1,2, 3), (1, 3, 2), (3, 2, 1), (3, 1, 2), (2, 1, 3) and (2, 3, 1)

These are not the elements themselves, which must be written as (2 x 3) matrices. For example, the identity element will be: e

[1 2 3]

[1 2 3]

The other elements we can write:

s 1 =

[1 2 3]

[2 3 1]

s 2 =

[1 2 3]

[3 1 2]

t 1 =

[1 2 3}

[1 3 2]

t 2 =

[1 2 3]

[3 2 1]

t 3 =

[1 2 3]

[2 1 3]

Given the above relationships to assorted groups members, we can now prepare a group table to show the operation is binary and G meets the Group definition. We have for G = {The set of all permutations on S}:

e---- s1--- s2--- t1 ---  t2 ---  t3

--------------------------------------------------

e  --   e  ---s1--- s2 --- t1 --- t2---  t3

s1  --s1--- s2--- e ----t2 ----t3 ----t1

s2  --s2----e  ---s1---  t3---   t1----  t2

t1  ---  t1-  t3-  t2 ---  e ---  s2 -----s1

t2  ----t2--- t1-- t3- - s1--- e------- s2

t3 -----t3 ---t2 ---t1 ---- s2 --- s1---- e

------------------------------------------------

The shaded highlights shown are intended to emphasize the symmetries in the table (which can also be found for the s  elements).  Is G Abelian? If we check any two elements and find they don’t meet the commutative property (4) then the answer is no. By inspection:

t1 •  t  2 = s 1 =

[1 2 3]

[2 3 1]

But  t 2  • t 1 = s 2     =

[1 2 3]

[3 1 2]

So, the commutative property is not met, and G is not Abelian.

Other examples: Groups of permutations related to 1:1 mappings (functions).

Let A be a set and let s  and t be permutations of A so that s and t are both one-to-one functions mapping A onto A. Then the composite function is s t   and will be a permutation if it is one-to-one and onto A.

Definitional illustrations:

A )If  a1(s  t) = a2(s  t), then (a1 s ) t = (a2 st and (a1 s) = (a1 s ) (since t  is 1:1) and a1 = a2.

Hence  (s  t)  is 1:1.

B) Let a  Î  A , then since t is onto A, \$  a ‘ Î  A such that a’t = a. Since s is onto A, \$ a" Î A  such that a’ = a” s. Then, a = a’ t = (a” st  = a” (s  t).

Example: Let A = {1, 2, 3, 4, 5} and

s =

[1 2 3 4 5]

[4 2 5 3 1]

In this case: 1 s = 4, 2 s = 2, 3 s = 5, 4 s = 3, and 5 s = 1

Now, let:   t =

[1 2 3 4 5]

[3 5 4 2 1]

Then: 1t = 3, 2t = 5, 3t = 4, 4t = 2, and 5t = 1

Then we can find the result  by matrix multiplication:

[1 2 3 4 5]

[2 5 1 4 3]

(Which the reader should be able to verify!)

--------------------

Extending Permutation Groups: Transpositions:

A transposition is a permutation which interchanges two numbers and leaves the others fixed. The inverse of a transposition T is equal to the transposition T itself, so that:

T2 = I (identity permutation, e.g the permutation such that I(i) = i for all i = 1,...n)

A permutation p of the integers {1, . . . n} is denoted by

[1 .. . .. n]

[p(1).. p(n)]

So, for example:

[1 2 3]
[2 1 3]

denotes the permutation p such that p(1) = 2, p(2) = 1, and p(3) = 3.

Now, let's look at EVEN and ODD permutations:

Let  P  n  denote the polynomial of n variables x
1,  x 2 ……x n which is the product of all the factors  x i .. x j  with i < j. That is:

n ( x
1,  x 2 ……x n) = P(x i .. x j)

The symmetric group S(n) acts on the polynomial  P  n by permuting the variables. For p
Î  S(n) we have:

n ( x_p(1), x_p(2). . .x_p(n)) = (sgn p)  P  n ( x
1,  x 2 ……x n)

where sgn p = + 1. If the sign is positive then p is called an even permutation, if the sign is negative then p is called an odd permutation. Thus: the product of two even or two odd permutations is even. The product of an even and an odd permutation is odd.

Back to transpositions! We just saw:

[1 2 3]
[2 1 3]

The above permutation is actually a transposition 2 <-> 1 (leaving 3 fixed). Now, let p' be the permutation:

[1 2 3]
[3 1 2]

Then pp' is the permutation such that:

pp'(1) = p(p'(1)) = p(3) = 3

pp'(2) = p(p'(2)) = p(1) = 2

pp'(3) = p(p'(3)) = p(2) = 1

It isn’t difficult to ascertain that: sgn (ps) = (sgn p) (sgn s) so that we may write:

pp' =

[1 2 3]

[3 2 1]

Now, find the inverse p^-1 of the above. (Note: the inverse permutation, denoted by p - 1  is defined as the map: p - 1  : Zn -> Zn),  Since p'(1) = 3, then  p - 1 (3) = 1

Since p'(2) = 1 then p - 1 (1) = 2

Since p'(3) = 2 then  p - 1 (2) = 3

Therefore:  p - 1   =

[1 2 3]
[ 2 3 1]

Disjoint permutations:

Expressing a permutation as a product of disjoint cyclic permutations is not hard at all, once one gets the hang of it. The key is to “cycle through” the mapping in the original to yield the different disjoint cycles, taking care to stop when the end element leads to a number (on the top of the original) that repeats. For example:

Express as disjoint permutations:

[1 2 3 4  5  6   7]

[4 5 6 7  3  1  2]

Solution: 1 goes into 4, 4 goes into 7, 7 goes into 2 – STOP! (Since 2 commences new cycle in next top position). So first disjoint cycle is: (1, 4, 7, 2).

Now, 2 goes into 5, 5 goes into 3, STOP! (3 repeats) So cycle is: (2, 5, 3). Then 3 goes into 6, and 6 into 1. Stop.

Answer: (1, 4, 7, 2)(2, 5, 3)(6).

Permutations applied to solid geometry (tetrahedron):

Consider now the ordered tetrahedron (vertices ordered by number) shown below:

Call the ordering '1234'. In terms of signage (sign rules - e.g. for (+) or (-) being followed, it's important to note that a segment (1 2) induces orientation (+1) in the associated complex, but a segment (2 1) induces (-1). This is how differing segments acquire negative signage in the complex.

The boundary of the tetrahedron, in terms of its four faces can than be written:

- (1 2 3) - (1 3 4) + (1 2 4) + (1 3 4)

Suggested Problems:

1) Let A = {1, 2, 3, 4, 5} and s  be the result (st)

Find: 1 t, 2t , 3t , 4t, and 5t

Compare to:  st, 2st , 3st , 4st, and 5st

2)Express  p =

[1 2 3 4]
[2 3 1 4]

as the product of transpositions, and determine the sign (+1 or -1) of the resulting end permutation