in which we meet the quaternions and start exploring matrix groups.
We started last lecture with the result on groups of order and finished the proof today: Any group of order is either abelian and iso to one of , or non-abelian and iso to one of (dihedral) or (quaternion). We proved this by looking at elements of certain orders: if we have an element of order or only elements of order ( and) , it is easy. Given an element of order and an element which is not in , we used the fact that is normal to show that (commutes with and) is either or . Adding the restriction or gives us four cases, which we investigated.
One of those four cases gave us some new relations we didn’t know about yet, which might lead to a new group (or to some contradictions). We saw by looking at a certain matrix group that such a group, called quaternions, does indeed exist, and we can write it out in several different ways.
Having mentioned them a few times in the beginning of the course, we finally properly defined the general linear groups (for or ) as those matrices over which are invertible. The determinant gives us a group homomorphism which has kernel , the special linear group. I left you to work out the quotient using the isomorphism theorem.
Understanding today’s lecture
The “groups of order 8” proof is quite long and convoluted with lots of cases, so it needs some time to get your head round it. Have a look at what techniques we used, and whether you’ve seen them/used them in other contexts. We had the Direct Product Theorem several times, we used normality, and we used conjugation quite a lot to find out what possible relations between and there can be. Perhaps you can write out a short summary, or bullet points of the important steps, to help you get an overview of the proof.
Preparing for Lecture 19
Next lecture we will explore actions of the general linear group. We can act on (or ) viewing matrices as linear maps. We can also act on the set of matrices by conjugation. We will look at orthogonal groups as well and prove that orthogonal matrices are isometries.
Going a little deeper
In the proof of the “groups of order 8” result, we boldly claimed that if we have of order and of order with , then our group is (isomorphic to) . How can we be so sure? What if there are some other relations lurking that we just haven’t found out about? In general, given a group in terms of generators and relations, deciding whether two specific products of a sequence of generators (also called “words”) are the same group element or not can be undecidable. (Look up “word problem” for groups to find out more, but it is quite advanced.) But we are in a case which is manageable. The relation, equivalently written as , tells us how to “commute s past s”, so we can put all the s at the front, where they will cancel each other out to leave either no or one . The s at the end also reduce to for . So we can easily check that has exactly elements, and our group also has elements, so we must be ok. More formally, what we are doing is defining a group homomorphism from to our group of order , sending to and to . The point of all those relations is to check that this really does give a group homomorphism (cf properties of group homs and the related example sheet questions). Then we see it is surjective onto our group of size , so as we have finite groups of the same size, it must be a bijection, i.e. also injective.