Post #9: Combination, the operation that works on permutations to create a group

In post #8, I listed the eight permutations that comprise the symmetries of the square, which I call D4. The day before, we saw the Cayley table D4 for in terms of 2x2 matrices. Now we will show how to compose permutations in a way that the Cayley tables for matrices and permutations match exactly.

 

Let me take two elements of D4 that do not commute, R90° and M45°.

R90°✕M45° = M90°

Here is the tricky part. The order of the permutations is opposite the order of the matrix multiplications, so we start by switching them. Instead of

(1234)(12),

 

we will work with

 

(24)(1234).

 

What we will do is follow the path of each element, taking the cycles left to right, and doing the same with the elements inside a cycles. Like matrix multiplication this takes some practice, so let me go step by step through a few examples.

 

In this version, the 2-cycle happens first. Let's find the path of 1 to start.  

(24)(1234)

 

1 goes to 2, so we begin with

 

(12...).

 

Where does 2 go?

(24)(1234)

 

2 goes to 4 in the 2-cycle, then 4 goes to 1 in the longer cycle. This closes the first part of the answer.

 

(12)...

 

Now we check 3.

 

(24)(1234)

 

3 goes to 4, and our answer becomes

 

(12)(34...)

 

We need to follow the path of 4.

(24)(1234)

 

4 goes to 2, then 2 goes to 3, and our work is complete.

Final answer (24)(1234) = (12)(34).

 

This is the reflection through the y-axis, switching Quadrants 1 and 2,and also switching 3 and 4.

 

M45°✕R90° = M0°

 

Again, the first step is to put the permutations in the opposite order of the matrices.

 

(1234)(24)

  

(1234)(24) is not a permutation in standard form because both 2 and 4 appear more than once.

 

Step 1: Where does the number 1 go?

 

We start with 

 

(1... 

 

and figure out all the elements in order before we close the cycle. 

 

(1234)(24) 

 

1 goes to 2 in the first cycle, and 2 goes to 4 in the second. Our incomplete cycle becomes

 

(14...) 

 

and now we follow the path of 4.

 

(1234)(24)

 

4 goes to 1 in the first cycle and 1 is not in the second cycle, so our first cycle in the final answer is 

(14)...

Let's follow the path of 2.

 

(1234)(24)

 

So 2 goes to 3, and 3 is not in the second cycle.

 

(14)(23...)

 

The lowest unresolved number now 3, so let's follow its path.

 

(1234)(24)

 

3 goes to 4 in the first cycle, then 4 goes to 2 in the second, and our composition of permutations is done.

 

Final answer (1234)(24) = (14)(23).

 

This makes sense geometrically. M0° is a reflection through the x-axis, so Quadrants 1 and 4 switch places, as do 2 and 3.  

Tomorrow: Subgroups of Z8 and D4, complete with Hasse diagrams.