Abstract Algebra Theory and Applications - Computer Science ...

78 CHAPTER 4 PERMUTATION GROUPSExample 7. Consider the permutation(16)(253) = (16)(23)(25) = (16)(45)(23)(45)(25).As we can see, there is no unique way to represent permutation as the productof transpositions. For instance, we can write the identity permutation as(12)(12), as (13)(24)(13)(24), and in many other ways. However, as it turnsout, no permutation can be written as the product of both an even numberof transpositions and an odd number of transpositions. For instance, wecould represent the permutation (16) by(23)(16)(23)or by(35)(16)(13)(16)(13)(35)(56),but (16) will always be the product of an odd number of transpositions.Lemma 4.5 If the identity is written as the product of r transpositions,then r is an even number.id = τ 1 τ 2 · · · τ r ,Proof. We will employ induction on r. A transposition cannot be theidentity; hence, r > 1. If r = 2, then we are done. Suppose that r > 2. Inthis case the product of the last two transpositions, τ r−1 τ r , must be one ofthe following cases:(ab)(ab) = id(bc)(ab) = (ab)(ac)(cd)(ab) = (ab)(cd)(bc)(ac) = (ab)(bc).The first equation simply says that a transposition is its own inverse. Ifthis case occurs, delete τ r−1 τ r from the product to obtainid = τ 1 τ 2 · · · τ r−3 τ r−2 .By induction r − 2 is even; hence, r must be even.In each of the other three cases, we can replace τ r−1 τ r with the right-handside of the corresponding equation to obtain a new product of r transpositionsfor the identity. In this new product the last occurrence of a will