We started by going over parity of cycles.

In a net diagram with only two elements and one small rung connecting them, it would be a 2-cycle.  Thus 2-cycles are odd, since they require one (an odd number) rung.  While in a net diagram with 3 elements, a small rung connecting the first two and a large rung connecting the first and third, it would be a 3-cycle.  This shows that 3-cycles are even, since they require 2 (an even number) rungs.

Theorem:
Odd-cycles have even parity, and even-cycles have odd parity.

Proof:
Rashid's Proof - Build a net diagram and count the rungs.

Permutations have cycle structure, also.  This is a description of how the permutation moves the elements around.  From a net diagram, we can describe this motion as follows:  After creating a net diagram, read from the bottom straight to the  top to find where the elements "go".

1 2 3 4 5 6 7 8
| | | | | | | |
|-+-| | | | | |
| |-+-| | | | |
|-+-+-+-+-+-| |
| | | | |-+-+-|
| | | | |-| | |
| | | | | | | |
7 4 1 2 6 8 3 5

In this figure, 1 "goes to" 3, 2 "goes to" 4, etc...  Build the cycle structure as follows:

Starting with the lowest unused number, see where it goes.  Then follow that number, etc..., until we get back to the first number.  In this example, we see 1 -> 3 -> 7 -> 1, and the cycle is done.  The notation for this cycle is (1,3,7).  Now, 2 is the least number we haven't investigated, so let's see what cycle 2 is in.  2 -> 4 -> 2, so we have the cycle (2,4).  The last cycle is (6,5,8).  We write this cycle as:  (1, 3, 7)(2, 4)(6, 5, 8).

*As a note, we do not have to write the cycles exactly as shown above. They can be written in many other ways as long as each cycle contains the same numbers, moving in the same way.  For example, the notation (4, 2)(6, 5, 8)(3, 7, 1) describes the same permutation.

Using the net diagram above we set up some rungs (as shown) on it to give us a parity (odd), then we added an extra rung to transpose 1 and 7, it changed the cycle structure to (1)(3, 7)(2, 4)(6, 5, 8)  And when we added another rung to transpose 8 and 3, the cycle structure became (1)(2, 4)(3, 6, 5, 8, 7).

Theorem: CASE 1- If the rung connects elements in a different cycle, it joins the cycles.
         CASE 2- If the rung connects elements in the same cycle, it splits the cycle into 2 cycles.

Proof: *proved with net diagrams*

Corollary: Each rung in the net diagram changes the number of cycles by 1, either up or down.

Fact: Home position with n elements has n cycles.

Consider a 12-cycle:
What is the least number of rungs needed for this permutation?

If there are no rungs there would be twelve cycles, so we need 11 rungs in order to make 1-cycle, since the Theorem above says that each rung changes the number of cycles by exactly 1, either up or down.

Theorem: To build an n-cycle, we need at least (n-1) rungs. In fact, you can do it with (n-1) but not fewer.

Question:
What is the least number of transpositions that will generate this permutation?

Home:   12345678
        43186275
Cycles:(1,3,2,6,5,8,4)(7)
         7-cycles     1-cycle
    6 transpositions  0 transpositions
            6+0= 6 transpositions

Theorem:  In general, a permutation on n elements, consisting of c cycles, can be built in a net diagram with n - c cycles, but no fewer.