 So by now you should be able to prove at the theorem that any cycle can be expressed as a product of transpositions and in fact, we found that we could express a cycle like two five three as a product of two transpositions and So a question we can ask is can we find a different product of transpositions that also give us two five three? Let's make things a little bit more challenging and see where that takes us So again our elements don't necessarily have an order, but suppose we have one or suppose we create one and limit ourselves to transpositions of adjacent elements So the question is can we express a permutation as a product of transpositions of adjacent elements? So our cycle is two five three So let's use two three as our first transposition where we'll just take the Expected ordering of the elements one two three four five and so on so can we write two five three as two three times something And of course the answer is yes because two three has an inverse and we can calculate the value of that other permutation and it works out to be two five and So two five three is two three two five So we've expressed our cycle as a product of transpositions, but two five isn't a transposition of adjacent elements So what can we do? We might think of it as follows in two five There's a gap between the elements two and five and we want to find transpositions that fill that gap So we could either start at two and build towards five using the transposition two three Or we can start at five and build towards two using the transposition four five There's actually another transposition. We could use it starts at five and another one that starts at two Find them yourself Notice that if we write two five as a product of two three and some other permutation Then our original cycle two five three is going to be equal to two three two three row and Again, every transposition is its own inverse. So the consecutive two threes cancel and we're right back where we started So we don't want to use the transposition two three Let's use four five instead and if we write two five as a product four five and row we find and We might object to this because our cycle length has increased. We've gone from a transposition to a three cycle But we might take a closer look while our cycle length has increased Notice that we've reduced the gap. So instead of a gap between two and five We now have a smaller gap between two and four So let's continue reducing that gap. So again the gap here is now between two and four so let's use three four and write two four five as a product of three four and some other permutation and again, we can solve for tau and While we now have an even longer cycle notice that the elements in the cycle are consecutive and so now our cycle consists of adjacent elements So now let's start breaking this cycle down. So let's let two three four five Well, let's take that as a product of two three and something and we find Three four five. Let's find this as a product of three four and something and we find and our last remaining transposition four five is a transposition of adjacent elements and so here we've written our cycle two five three as a product of transpositions where we are only transposing adjacent elements So notice that while we could write two five three as two different products the number of transpositions in both cases was even here There's two and here. There's six Now let's see if that's just a coincidence We might try to find a decomposition of another cycle. How about two four five six? both as any way that we can and that gives us this or maybe we'll restrict ourselves to transpositions of adjacent elements in which case one possible decomposition looks like this and Here there's three transpositions and here There's five and so we see the number of transpositions in both decompositions is odd And this leads to the following idea It appears that given any permutation sigma we can express sigma by a sequence of transpositions but the number of transpositions seems to preserve parity in other words if an odd number of transpositions can be used to produce sigma then any sequence of transpositions that produce sigma will also have an odd number of transpositions and if an even number can be used to produce sigma then any sequence will also have an even number of transpositions and the evidence for this is well kind of Underwhelming because we've only shown it for two different cycles and for each we only did two Decompositions so we don't have a lot of evidence here to go on so let's see if we can find a proof