 Now, a cycle is a special type of permutation, we'll now consider a special type of cycle known as a transposition. So a transposition is a permutation that acts on just two elements. In cycle notation, a transposition tau will be a two-cycle AB. And note that we don't actually require that the two elements be consecutive because the elements of our permutation might not have an order. Now we've already determined that we could take a permutation and write it as a product of cycles. Can we go further and write a permutation as a product of transpositions? Well let's find out. Now here we really want to rely on the idea that the whole point of abstract algebra is to be able to solve equations. So let's think about finding the equation. So a cycle like 1, 4, 3, 2 permutes the elements 1, 2, 3, and 4. And so we might consider transpositions of these elements. So let's pick a random transposition, say 1, 2, and ask, can we write our cycle as a product of 1, 2, and some other cycles? Well let's find out. So if possible, we want to write 1, 4, 3, 2 as the product of the transposition 1, 2, and some other cycles. And this means we want to solve the equation 1, 4, 3, 2 equals 1, 2 times sigma, where sigma is some other permutation. So let's think about this. Since 1, 2 is a transposition, then its inverse is also 1, 2. And so if we want to solve for sigma, we'll left multiply by the inverse, and so we get, and we find, and so we can write 1, 4, 3, 2 as the product 1, 2, applied to 1, 4, 3. Now remember Lather-Rinz-Repeat, well we found 1, 4, 3, 2 equals 1, 2 times 1, 4, 3. We might have continued and tried to write 1, 4, 3 as a product of a transposition and another cycle. But which one? Now let's think about that. Since 1, 2 is its own inverse, if we were able to write 1, 4, 3 as 1, 2, applied to some other permutation row, then 1, 4, 3, 2, well that was 1, 2 times 1, 4, 3, and 1, 4, 3 is 1, 2 times row. But since 1, 2 is its own inverse, these two factors would cancel and we'd be left with just row. And that doesn't take us anywhere, it takes us right back to the cycle we started with. You might try another cycle, well how about something like 5, 6. But then if we try to write 1, 4, 3 as the product of 5, 6 and some other permutation, then again 5, 6 is its own inverse, so we'll left multiply and this product doesn't simplify any further. But if we take say 1, 3 where both elements of the transposition are elements permuted by 1, 4, 3, then we get 1, 4, 3 is 1, 3 times some unknown permutation row and we'll solve for row and find. And so 1, 4, 3 is 1, 3 times 1, 4 and so our cycle 1, 4, 3, 2 can be written as the product of 3 transpositions. And in fact the preceding suggests that any cycle can be expressed as a product of transpositions. And since any permutation can be expressed as a product of cycles, this means that any permutation can be expressed as a not necessarily unique product of not necessarily disjoint transpositions. Well this is a theorem and we need to prove it, but you should do your own homework. Still if you actually find these cycle decomposition you'll get some idea of what you have to do in order to prove that theorem and some of the things you have to watch out for. So let's try to express 2, 5, 3 as a product of transpositions. So again we'll start with a random transposition but let's make sure it includes two of the elements in our cycle. So we'll start with the transposition, we'll have a 2, 5 and solve 2, 5, 3 is 2, 5 times some of the permutation. And we solve and again this cycle consisting of just one element is really an identity so we only need the transposition 3, 5. And so we find 2, 5, 3 is 2, 5 composed with 3, 5.