in which we show that the sign of a permutation is a surjective group homomorphism, meet cosets and prove Lagrange’s Theorem.
At the end of last lecture, we defined the sign of a permutation to be , where is the number of transpositions we need to get . Remember that this only makes sense because we proved that “sign is well-defined”. All the hard work is in showing that it is really well defined! After that it is easy to show it is indeed a surjective group homomorphism . We call the kernel of the alternating group . As the last result of the second chapter, we stated that any subgroup of contains either no odd permutations or exactly half its permutations are odd. That is Question 3 on Sheet 2.
We then started on Chapter 3 and defined cosets: For a subgroup and an element , the left coset is , and the right coset is . We saw some examples in and in . To prepare for Lagrange’s Theorem, we defined a partition of a set to be a collection of subsets such that “the union gives all of ” (or “every element is in at least one ”) and the are pairwise disjoint (or “every element is in at most one ”). The most important result of this chapter is Lagrange’s Theorem: For a finite group , the size of any subgroup divides the size of . We prove this by showing that the left cosets of partition , and all have the same size (as ). Notice that we only need to be a finite group right at the end: the partitioning and the bijection between and work even for infinite groups. We called the number of cosets the index of in . So we get . I left it to you to show one direction of the “same coset check”: if and only if .
Right at the end we started to see some corollaries of Lagrange: today it was that the element order divides the group order, i.e. the order of any element in a finite group divides the size of . Next time we’ll have some more consequences and corollaries.
Understanding today’s lecture
It is probably good practice if you get used to seeing fairly quickly what the sign of a certain permutation is. Start by some single cycles. When you move to more general permutations in disjoint cycle notation, remember that sign is a homomorphism! So for permutations “even times even is even” because “plus times plus is plus” (or because “even plus even is even” for integers), and “odd times even is odd” because “minus times plus is minus” and so on.
To get used to cosets, there are a few things you could do. You have plenty of examples of subgroups: write down the cosets for them. Check if you can find a pattern when the right and left cosets might be the same or different. Remember that cosets are really sets and not subgroups (except the subgroup itself viewed as a coset).
Make sure you are happy with the proof of Lagrange, it is an important result. Don’t forget to do the exercises given in lectures, for example showing that if then . Another way to test your understanding of Lagrange could be to prove it using right cosets instead (maybe even with your notes closed). Practice will make you good at things like how to show two sets are exactly the same set, and such tools that we use a lot.
Preparing for Lecture 9
We saw one Lagrange corollary today. Can you think of other ways Lagrange might be useful? For example, does it tell you anything about the structure of a prime order group? We will also meet equivalence relations next time. You could look up what that is to give yourself a head-start. Also make sure you can remember what a partition is, as we’ll need it again next time. We may get as far as the Fermat-Euler Theorem which is a result from Number Theory. (In any case we’ll start on our way to a group-theoretic proof of it.) This has to do with modular arithmetic, so do remind yourself of that. You could also look up Fermat’s Little Theorem (which is a special case) and think whether you know any proofs of it.
Going a little deeper
The word “well-defined” is always a little tricky the first few times you meet it. What exactly does it mean? In fact, it can mean several things in different contexts. In the context of the sign it means the following: our definition of the sign homomorphism seems to depend on a choice, namely the choice of how we write as a product of transpositions. So if I made a different choice, would I get a different sign? That would not be a very good function! (In fact, it would not be a function.) So we have to check that the outcome, namely the , does not depend on this choice, but that we get the same outcome whichever choice we made. This is what “well-defined” means here: it really is a function. We met “well-defined”ness again later in the lecture and will meet it often again in the future. It can mean in different contexts: does the function really land where we say it lands? That is what it meant in the context of the bijection between and . Or: if we define something using a representative, do we get the same answer regardless of which representative we use (that is very close to the sense we used it today)? Look out for it in “Applications of Lagrange”, which we will start next time. You can also read an entry on well-definedness on Tim Gowers’ blog.
There are other ways to define the sign of a permutation. For example, you could define it via the number of inversions: how many pairs of numbers are there with , i.e. where the images are “the wrong way round”? If you count this number of inversions and call it say , you can define . See for example this Wikipedia article. So for this definition, it is very clear that the sign is well-defined, there is no ambiguity in the definition. But it is much harder to show that it is a group homomorphism! In fact the two definitions are equivalent (as you can see on the Wikipedia article). But you have to put the work in somewhere: either define it our way and put the work into “it is well-defined”, or define it via inversions and put the work into “it is a homomorphism”, or I suppose define it both ways and put the work into “the two definitions are equivalent”.