Generating subgroups

I have just finished going through book 2 of M336 Groups and Geometry and I am making reasonable progress. As usual I seem to be working at roughly half the speed that you are supposed to but I expect I will get through enough material to make life easier when the course actually starts in October. So now I have read four books in total from M336 and M381 and I am alternating between the two courses; i.e. when I finish a book from one course I start on a book from the other course. I will be looking at book 3 from M381 next which is on congruence.

Before I leave book 2 of M336, there are a few things I want to say about it. Firstly, I have noticed that the exercises in these level 3 courses require much more from the student. At level 2 it was pretty much spoon feeding - i.e. the books would give examples of how to do something and then there would be exercises where you repeat what you just learnt. Here at level 3 the questions are more demanding as the way to do an exercise is often left up to the student. I welcome this as I like thinking things through for myself, but at times it can be very time consuming. What has surprised me is that a certain amount of 'hand-waving' has appeared both in the text and in the answers to exercises; i.e. rather than thrashing out an answer exhaustively, a bit of 'well, the answer is like this because of such and such.' It can be a bit frustrating if the details are skipped over and a bit of shock after the intense rigour of M208.

I want to put down here some bits of this book which I thought ventured slightly into the hand-waving arena. This part of the book was the most difficult and the most time consuming and so it is worth going through so that later I have something to look back on for revision.

How can you generate subgroups? In M208 we have already met cyclic subgroups which are generated from integer powers of an element in a group. For example, if we consider the group of symmetries of the regular hexagon and label a rotation about its centre through π/3 as r then we can generate a cyclic subgroup from r as

$$<r>=\left\{r^{0}, r, r^{2}, r^{3}, r^{4}, r^{5}\right\}$$

Now the new idea is to expand this notion of generating subgroups to include two or more distinct elements of a group. In the text it says that "Informally, we define the subgroup generated by x,y,... to be the set obtained by forming all possible combinations of copies of x,y,.. and their inverses." In the margins it does say that a formal definition is given later in unit IB4 but I found it difficult to understand this without some formal definition. Here is my version of a formal definition for two distinct elements of a group x and y:-

$$<x,y>=\left\{uv:u,v \in \left\{ (x)^{i},(y)^{i}:i,j \in Z \right\} \right\}$$

Now it is clear that integer powers of x and integer powers of y are just the cyclic subgroups <x> and <y> and so we can rewrite this as

$$<x,y>=\left\{uv:u,v \in <x>\cup<y> \right\}$$

In other words we are taking combinations of elements taken from a union of the subgroups generated by x and y. This does not mean that the resulting subset is a subgroup. This still has to be verified by seeing if the subset satisfies the subgroup properties. One thing that becomes clear straight away is that <x,y> contains the elements of <x> and the elements of <y>. This is because if u and v are drawn from <x>, say, then because <x> is a cyclic subgroup, then the combinations of uv are also in <x>. What is new are products of elements which are not common to both sets.

Let's take an example. If we consider the regular hexagon again and call the symmetry that is a reflection in the horizontal axis s, then s is self-inverse and

$$<s>=\left\{e,s\right\}$$

If we also consider the half-turn about the centre r³ then this is also self-inverse and

$$<r^{3}>=\left\{e,r^{3}\right\}$$

But what about <r³,s>? We see from the above discussion that this is going to contain e, r³, and s but it also must contain combinations of elements which are not common to both sets, namely sr³ and r³s. These turn out to be the same element since sr³ is a reflection of some sort which is self-inverse so

$$sr^{3}=(sr^{3})^{-1}=r^{-3}s^{-1}=r^{3}s$$

Hence we have the subset

$$<r^{3},s>=\left\{e, r^{3}, s, r^{3}s\right\}$$

which does indeed turn out to be a subgroup of the symmetries of the regular hexagon and is isomorphic to the Klein group.