 So far, we've seen that every permutation can be written as a product of disjoint cycles, and every cycle can be written as a product of transpositions. And we have some evidence that the number of transpositions required reserves parity. But how many transpositions are required? To answer this question, let's consider the parity of the identity transformation. The advantage of looking at the identity is that we know it's going to be in every permutation group. So suppose the identity can be written as the product of n transpositions. We'll show that the number of transpositions must be... Well, let's figure it out. So suppose we could write the identity as a product of n transpositions. Let's consider the last two transpositions, which are actually at the front. Remember, we apply the transpositions from right to left. And so there are three cases. First, those last two transpositions might be the same, A, B, A, B. Second, they might be disjoint, A, B, C, D. And third, they might share one element, A, B, A, C. Let's consider those three cases separately. So since A, B is its own inverse, then this product A, B, A, B is the identity, and we can omit these last two from our product. And thus we can write the identity as a product of two fewer transpositions. The other possibility, if the last two transpositions are disjoint, because they're disjoint, we can switch them. Because the product of disjoint cycles is commutative, A, B, C, D is the same as C, D, A, B. So the last transposition does not move A. Or if the last transpositions share exactly one element, we can rewrite the product A, B, A, C as B, C, A, B. And again, the last transposition does not move A. And so this leads to a lemma, let sigma be a product of n greater than two transpositions with A, B as the last transposition. Then sigma could be rewritten as the product of n minus two transpositions, or n transpositions, where the last transposition does not move A. While we started with the last two transpositions, associativity means that this applies to any two consecutive transpositions. So remember, lather, rinse, repeat. So it appears we can move any element to the first transposition, giving us our identity, easy product, tau prime one, tau prime two, and so on, all the way up to A, x, where none of the towels move A, and x is some other element. But wait a minute, if none of the towels move A, then the identity applied to x is going to give us A, and none of the towels can restore it, so E can't be the identity. What went wrong? The only way to avoid this is to keep the transposition affecting A from reaching the front. And this can only happen if it encounters a copy of itself. And if it does, this will reduce the number of transpositions by two. And so this gives us a second lemma. Suppose the identity could be written as a product of n transpositions, where n is greater than two. Then it could be rewritten as a product of n minus two transpositions. And so now we're ready to conclude. Suppose we can write the identity as the product of an odd number of transpositions. We can reduce the number by two until we have one transposition left. But the identity can't be equal to a transposition. And consequently, we can't write the identity as a product of an odd number of transpositions. And so the identity can only be written as a product of an even number of transpositions. So now let's put it all together. Consider any other permutation sigma. And suppose we could write sigma using an odd number of transpositions, and also using an even number of transpositions. So sigma is some product of an even number of transpositions. And it's also some product of an odd number of transpositions. Now, if we find the inverse of sigma, we can invert either of these two products. And let's just invert the product of the odd number of transpositions. So the inverse will be, and consequently, when we multiply sigma inverse by sigma, we get. And then E, the identity, sigma inverse sigma, can be rewritten as the product of an odd number of transpositions. But that's impossible because the identity can only be written as a product of an even number of transpositions. And consequently, we can't write sigma as an odd number of transpositions, and also as an even number of transpositions. It's got to be one or the other. And so the parity of a permutation is invariant. Now, suppose rho and sigma are two even permutations. Their product will also be even. And so the even permutations are closed under composition. And consequently, the even permutations on n elements form a subgroup of Sn. This group is important enough that it has its own name. It's called the alternating group.