 While we can compose two or more cycles by rewriting both in Cauchy 2-line notation, we can also find the product directly. And for that it's helpful to keep in mind, everything belongs somewhere. We'll see what we mean by that. Let's try to find the product 1, 2 by 1, 2, 3. Now we already know what this product is going to be, but let's see how we can get that directly from the cycle notation. To help distinguish, for now, between the elements and the cycles, we'll represent the elements in a cycle in our normal text and use bold color when we want to refer to an element. So 1, bold red, is an element being permuted, while 1, 2 is a cycle. So in this product, we want to write it in terms of the cycles, and so that means we need to find the orbit. So we'll find the orbit of 1. Wait, scratch that. We'll find the orbit of bold red 1. And so here's why we made that distinction. We want to see what happens when we apply the cycles 1, 2, 1, 2, 3 to the element 1. So remember that in this notation, the factor that is nearer is applied first. So we apply the cycle 1, 2, 3 to 1. And remember this cycle sends 1 to 2. And so 1, 2, 3 applied to 1 gives us 2. And we have yet to apply the cycle 1, 2. So our next step, the cycle 1, 2 is going to send this element 2 back to 1. And so when we apply 1, 2 to 2, we get 1. So our cycles also permute 2 and 3, and so we should find the orbit of 2. So again, we'll apply our cycles 1, 2, 1, 2, 3 to 2. 1, 2, 3 applied to 2. This cycle sends 2 to 3. And so 1, 2, 3 applied to 2 gives us 3. And then applying 1, 2 to 3, well, this cycle sends 1 to 2 and 2 to 1. So it doesn't do anything at all to 3. And so our cycle applied to 2 gives us 3. Well, that's the first step in the orbit. We'll apply our cycle to 3. So again, the cycle 1, 2, 3 applied to 3 sends 3 to 1. And then the cycle 1, 2 applied to 1 sends 1 to 2. And now we're back where we started. And so our cycle is going to be 2, 3. Now note that this 1 by itself, while it's a cycle, it doesn't actually do anything. 1 gets mapped to 1. And since the cycle 1 doesn't actually permute anything, we'll omit it and say that our product of the two cycles is just the cycle 2, 3. This leads to the following idea. Suppose we have a permutation sigma on n elements. We can take any element and find its orbit, producing a cycle kappa 1. And if there are any elements not in the cycle, we can find the orbit of one of those elements and produce another cycle kappa 2. Then lather, rinse, repeat until all elements are in a cycle. And this gives us the cycle decomposition of sigma. So for example, let's try to find a cycle decomposition of this permutation. So we'll find our orbits. So starting with 1, we find that 1 goes to 2. 2 goes to 4. And 4 goes back to 1. And so 1, 2, 4 is one of the constituent cycles. Now the element 3 is not in this cycle. And so we find that 3 goes to 3. And so 3 by itself is a cycle. And finally, the element 5 is not in any cycle we've found so far. And so we find that 5 goes to 6. And 6 goes to 5. So 5, 6 forms a third cycle. And so that means sigma could be written as 1, 2, 4 by 3 by 5, 6. Or does it? And by now you've learned enough abstract algebra to know that we can't always guarantee commutativity. And so the order may make a difference. So let's introduce a new term, cycles like 1, 2, 4, 3 and 5, 6 are said to be disjoint because no element appears in both. And we claim the following. Let sigma and tau be disjoint cycles. Then sigma tau equals tau sigma. In other words, if we have disjoint cycles, they will commute. Now we should prove this. And so to prove this, we should show that sigma tau applied to x is the same as tau sigma applied to x for all x. Because remember, things that do the same thing are the same thing. So with this in mind, let's try to rewrite as a product of disjoint cycles. And so here we have three cycles. And we see that the cycles commute the elements 1, 2, 3, 4 and 5. Or again, to make it easier, we'll put the actual elements in bold color. So we find the orbit of 1. So if I apply this product of cycles to the element 1, so the first cycle says that 1 goes to 3. Now the second cycle says that 3 doesn't have anything done to it. And so finally the third cycle says that 3 goes to 4. And so the first part in our orbit, 1 goes to 4. So let's see what our product of cycles does to 4. So our first cycle sends 1 to 3, 3 to 5, and 5 to 1. And it doesn't do anything to 4. Our next cycle sends 2 to 4, and 4 to 5. And our last cycle doesn't do anything to 5. And so 4 gets sent to 5. And that's the next part of our cycle. Now we're not back to our starting point, so we'll apply our composition to 5. So the first cycle sends 5 back to 1. The next cycle doesn't do anything to 1. And the third cycle sends 1 to 3. Still not home, so we'll apply our composition to 3. The first cycle sends 3 to 5. The next cycle sends 5 back to 2. And the third cycle doesn't change 2. Still not home, so we apply our composition to 2. The first cycle does nothing. The second cycle sends 2 to 4. The third cycle sends 4 back to 1. And finally we're back where we started. And since the cycle includes all the permuted elements, there is no other cycle to find. Now we can generalize our process for any permutation sigma. And so we can conclude any permutation on n elements can be expressed as a unique product of disjoint cycles.