 Continuing our discussion of transpositions, earlier we showed that if I can write the identity as a product of transpositions ending in UV, what I can do is I can sweep this element up to the first transposition, and because this is the only time this element U is moved, this product can't be the identity. And the problem we ran into is while that worked if we had an odd number of transpositions, it also worked if we had an even number of transpositions, and it suggests we can't actually write the identity as a product of transpositions. So we know that the identity can be written as a product of an even number of transpositions, so something keeps this from happening. Somehow the transposition involving U vanishes, and it seems that we need the following lemma. Let E be expressed as a product of K transpositions ending in something. The process of moving the transposition involving U to the front will invariably eliminate a transposition involving U. That is the only way we can prevent this situation where U gets all the way to the front, because if U gets all the way to the front, we cannot write the identity as a product of transpositions. So somewhere along the way a transposition involving U has to be eliminated so that you can't get to the front. Well actually the proof of this is pretty easy. If it doesn't happen, we can't write E as a product of transpositions. So if E can be expressed as a product of transpositions, then somehow U has to be eliminated because otherwise we can't do it. Now we should actually look a little bit beyond the proof. The proof is a very very very very simple proof, and we might actually ask ourselves, how does it happen? We know it has to happen, but how can a transposition involving U be eliminated? How do you eliminate such a transposition? Well we can take a note from our earlier observation. Back when we assumed that the identity could be written as a product of K plus 1 transpositions, we considered the last two transpositions, but we did assume that they were different because if they are the same, their product is going to be the identity, and we don't have to include them. And so that means that the only way we can really eliminate a transposition of the form UX is somewhere along the way. We have to encounter two identical transpositions, UX, and the one before it is also UX. And if we do that, if we do encounter those, their product is the identity, and we can eliminate both of them at the same time. And that means that this product of K transpositions, in order for it to be able to express the identity, this product has to include somewhere along the way this product of a transposition with itself, and that means I can eliminate two transpositions. And so while we did take the easy proof, if we look a little bit deeper, if we go farther into the proof, a better proof gives us a more useful result, not only do we eliminate a transposition involving U, but in fact we eliminate two transpositions. And so we reduce the number of transpositions down to K minus 2. And so let's go back to our original claim. Suppose we could write E as a product of K plus 1 transpositions. Well, it follows that we should be able to eliminate a transposition involving an element. We should be able to eliminate two transpositions, which knocks us down to E as a product of K minus 1 transpositions. But, lather, rinse, repeat. And what I can do is I can take that last transposition, U, V, and I can move U to the front. I can start to move it to the front, but somewhere along the way I'll eliminate two transpositions. And now I can write E as a product of K minus 3 transpositions. And I'll repeat the process. And remember, our assumption was that K plus 1 was actually an odd number, which means that if I continuously eliminate two transpositions, eventually I'll be able to reduce this down to the identity as a product of 1 transposition, which is impossible because a transposition is not going to give us the identity. And there's our proof. Now, it's worth noting the following. We began by thinking, hey, this might be a good proof by induction because we thought, well, it seems to be that the key step here is if we could write the identity as a product of K transpositions. If we know something about K, we know something about K plus 1, or we want to prove something about K plus 1. So this looks like it's a good proof by induction. And so, because we wanted to prove something was impossible for K plus 1, we tried out a proof by contradiction. We assumed the identity could be written as a product of K plus 1 transpositions. Except we found a gaping hole in our proof. Our process implies that we could not even write the identity as a product of any number of transpositions. Now, we dug ourselves out of the hole. We figured out that somewhere in this reduction process, we'd actually lose two transpositions. Now, here's the important thing. We could actually go back and clean up our proof. And it's a good exercise to do that because if we go back and clean up our proof, the really important thing that we found is that if I can write a product of K transpositions that gives us the identity, I can rearrange that to be a product of K minus 2 transpositions that will also give us the identity. And, lather, rinse, repeat. We write this as a product of K minus 4 transpositions that give us the identity, and so on. And that means that K has to be an even number because I'll keep knocking down the number of transpositions by 2 until I end up with zero transpositions that give us the identity, or if K is odd, one transposition which is impossible. And so, if I think about that, this is the only important thing I have to prove, and then all of the scaffolding is actually unnecessary, and I can just get rid of it. But let's keep it in because here's an important moral here. The proof is more about the process than the result. I didn't need all of these steps here, but I had to go through these steps to identify what was actually important. And while I might omit these steps in the write-up of the proof, it's important that we go through these steps because they give us insight into how we get to this last important step. And so, let's tie that together. We've proven that the identity permutation has even parity. And so, here's another useful habit to get into as a mathematician. You want to look for the corollaries. What's a corollary? Well, roughly speaking, a corollary is a result that you can prove from a theorem, lemma, proposition, or whatever. But the proof is one or two lines. It's a really, really short proof. So, for example, here's a corollary. Once you know the identity permutation has even parity, then you know that any permutation and its inverse have to have the same parity. So, if I can write this as a product of some number of transpositions, if I could write this as a product of some number of transpositions, then the product of these two has to be the identity which has to have even parity. And these are either both odd or both even. Also, another useful result here is that if I have a permutation with some order, if that order is odd, then I know that the permutation itself must have even parity. And again, that's because sigma to the k has to give us the identity, and we know that's an even parity permutation. How about the actual permutation theorem? So, let's go ahead and take a look at that. So, as far as I have a permutation and I could write it as a product of transpositions, and maybe I could write it as a different product of transpositions. Well, since the permutations are equal, then the product of transpositions here is the same as the product of transpositions there. Now, I do know, because we're dealing with a group, that sigma has an inverse, so its inverse can be written as a product of permutations, and so it's the inverse. I'll apply that product of permutations to both sides. I'll multiply by the inverse here, multiply by the same thing over there. Over on the left-hand side, this has to be the identity, and over on the right-hand side, what I have is this product of permutation. So, let's think about that here. Now, I know that M and M' have to have the same parity. Their product is the identity. There has to be an even number of transpositions over on the left-hand side. But, again, because this product is also the identity, there have to be an even number of transpositions over here on the right-hand side. So, that says that M' and N have to have the same parity. So, M and M' have the same parity, M' and N have the same parity, so that M and N have to have the same parity. And that proves our parity theorem. The parity of a permutation is the same regardless of the transpositions used to produce it. If a permutation could be expressed using an odd number of transpositions, then any expression will use an odd number. If it can be expressed using an even number, then any expression will use an even number.