Extending Permutation Groups: Transpositions And Applications Of Permutations To Solid Geometry

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:

P  _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:

P  _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 then write:

pp' =

[1 2 3]

[3 2 1]

Now, find the inverse (p^-1) of the above. (Note: the inverse permutation 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. 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 Problem:

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