 One special type of permutation is called a transposition, and these are also very important in our study of permutation groups. So a permutation in general is going to rearrange things, but a transposition is going to swap just two elements. I'm going to replace one thing with something, and I'm going to replace the something with the first thing. So in cycle notation, this transposition is also a two-cycle. X is going to be replaced with Y. Y is going to be replaced with X. So here's a very, very, very, very useful result. Every permutation can be written as a product of transpositions. Now it's very important to read the fine print on this, and here in this particular case there isn't any. And what's noteworthy here is what this theorem is not claiming. The theorem is not claiming that this product is unique. I may have different products of transpositions that give me a given permutation. The transpositions might not be disjoint. The transpositions may actually have elements in common. And so, for example, let's see if we can do that. So here is a permutation, and I want to find two different products of transpositions that are both equal to this same permutation. Now, if we want to do this, it may actually be easiest to write our permutation sigma in Coshino. So let's see what our product of cycles does to our elements A, B, C, D, and E. So again, the first cycle is applied, then the second cycle is applied. So A, first cycle says do nothing to it, and then second cycle A replaces with B. So B do nothing followed by replace it with C. C replace it with D, then do nothing. D replace it with C, but then C gets replaced with A. So D goes to A. Third cycle, third last term, E do nothing followed by do nothing. So this product of cycles gives us this particular permutation. So now let's apply transpositions to get the permutation, and we'll record those transpositions as we go. So remember, a transposition involved is switching to and only to elements, and our goal is to end up with B, C, D, A, E. So what can I do? Well, maybe I'll get a B into the first location. So I'll move B into the first location, which means I have to move A back to where B was. So my first transposition is going to be A, B. I'm going to swap A and B, that's recorded as this cycle A, B, and importantly, nothing else changes. So that is going to be C, D, E. So my first positions all set up. So I want to get a C into this second position, so maybe I'll swap these two, C and A. So that's going to give me the transposition C, A, A, C. And again, only these two change, all the rest, B, D, and E, they all stay the same. And let's say I have B, I have C, now I want to swap these two, A and D. So I'll swap these two, that's my transposition A, D, and all the other terms stay the same. And then now I have my final permutation as I want it to be. And it's the product of the transpositions A, B, followed by A, C, followed by A, D. Now I want to find two different products of transpositions that give me this particular permutation. So let's wipe out that sequence of transpositions and try to find a different one. So let's see. So the first time I worked by getting the B, C, D, A, and E. So maybe this time I'll work from the other end, I'll try to get an A in this last position here. So that means I need to move A over to here and D replaces A. So there's my first transposition A, D, and again all the others stay in the same place. Now I want to get a D in this third position, so I'll swap D and C. That puts those there. And then again all the other things stay in the same place. I want to get a C in this position, so I'll swap C and B. And there's my third transposition, all the others stay in place. And so here I have a transposition A, D, followed by C, D, followed by B, C. And again two very different products of transpositions that give you exactly the same permutation. So again part of the goal of abstract algebra is to build up certain habits of thinking mathematically. And so here's a very useful mathematical habit which is constantly asking yourself whether there is an underlying pattern or is it merely a coincidence. So in this particular case we wrote this permutation sigma as the product of transpositions, either A, D, A, C, A, B, again A, B first, then A, C, then A, D. The other way we could have written this permutation is A, D, C, D, B, C. And they're different, but there is something the same. They both have this transposition A, D. And so the question is, is that an important pattern? Maybe there's something special about these or is it a coincidence? And so could be a avoided it. If it's an actual pattern, there's no way we'd be able to avoid including A, D. But if it's a coincidence then we could have avoided it. So let's see. So again my permutation takes A, B, C, D, E and transforms it into B, C, D, A, E. And so the question is, I want to avoid swapping A and D. So is there any way that I could have done that? So let's see. Well here's another way to look at it. Is it a coincidence that neither of these included E? So let's see if I can avoid that. Let's see if I can start by including a transposition that involves E. So let's see this as a product of transpositions. Let's start off by including E and maybe I'll swap E and, how about A? So I'll swap E and A. So there's my first transposition. E swaps with A and again B, C and D are all the same. And let's see. Well okay so I want to get E and B. I'll swap those two so I get B in the first location. And again all the others stay the same. Let's see B is good. I'll swap E and C so I'll get C in my second location. Again all the others stay the same. I'll swap E and D so I get D in the right place. All the others stay the same. And then I'll finally swap E and A. So A is in the right location. E is in the right location and all the others stay the same. And so this permutation can be written as this product of transpositions. And if we look carefully we see two things happen. First of all we don't have the transposition A, D. It's nowhere in this list. In fact none of the transpositions we used. A, D, A, C, A, B, B, C, D, A, D. None of these transpositions are in here. So the transpositions that we get not only is the product not unique, but there seems to be very little that the set of transpositions have in common. And so this asking whether it's a pattern or a coincidence it certainly seems to be a coincidence that both of these had A, D in them because we could easily have avoided it. In fact it seems that there's no actual pattern. This has three transpositions. This has three transpositions. This has five transpi... Hold on, wait a minute. That's three. That's three. That's five. Wait a minute. There is something of a pattern starting to emerge. So again in the spirit of from one problem make many we might find other products of transpositions that give us that permutation. And if we do that one of the things we'll find is that all of those decompositions do share one feature. If a composition of transpositions make up that permutation sigma the number of transpositions that we require while the number is going to vary the nature of those is going to vary every set of decompositions is going to be odd. We're going to have an odd number of transpositions. And maybe again we'll make many problems out of one if I pick some other permutations and find transpositions that give us those permutations. One of the things we're going to find is that if I can write a permutation as a product of an odd number of transpositions then any other decompositions that produce that permutation will also require an odd number of transpositions. And likewise any permutation that can be written as a product of an even number of transpositions will always be expressible as a product of an even number of transpositions even though the actual number is going to vary. And after a while we'll accumulate enough evidence that these observations seem like they do form an underlying pattern and so it will be worth trying to prove our claim. And we'll take a look at that next.