in which we prove that disjoint cycle notation works and get a first glimpse of the sign of a permutation.
We carried on from last time by proving that every permutation can be written (essentially uniquely) as a product of disjoint cycles. This gives us the cycle type of a permutation, a very useful bit of information, as we started to see in “order by cycle type”. To learn a lot more about the symmetric group, we then wanted to introduce the sign of a permutation. For this we first need to know that is generated by transpositions (that is, -cycles), and then we had to prove that no matter how we write a permutation as a product of transpositions (this is almost not unique at all, to quote my own lecturer), we get, for a specific permutation, either always an even number or an odd number of transpositions. (So that is the only degree to which it is not unique 🙂 .) We will use that next time to finally define the sign of a permutation properly.
Understanding today’s lecture
To get used to the “almost not unique at all” part of products of transpositions, take some permutation, perhaps a cycle or so, and see how many ways you can find to write it as a product of transpositions. Perhaps you can have a competition in your college who finds one that nobody else thought of 🙂 But don’t overdo it, perhaps you should stop at about seven… Example Sheet 2 has some permutations for you to compose and calculate the order (and sign, see next lecture) of. You can also make up your own.
Preparing for Lecture 8
How are we going to define the sign of a permutation, given the proof of “sign is well-defined”? Do you have any ideas? Can you think of any nice properties it will have? You could compare it with the “even and odd” property of integers to get some ideas.
Going a little deeper
Can you think of any other situations where we have a choice of writing something, but there is some invariant of the choice? Invariant means something that is the same for each choice you make. Some easy examples to start you thinking: If I have a chessboard, I can cover it with domino tiles (which are of a size that exactly covers two chess squares). Now, if I take out one of the corners, then I clearly can’t completely cover it with domino tiles any more, because we’ve now got an odd number of chess squares. The invariant here would be: chessboards which can be covered have an even number of squares. However, perhaps you can find an even nicer invariant: What if we delete two opposite corners? Can it be covered now? It has an even number of squares, but is that enough? I’ll let you think about that on. (I suppose it is really more “Numbers and Sets”y than “Groups”y.)