 This lecture is part of an online math course on group theory and will be about subgroups of free groups. More precisely, we'll be showing that subgroups or free groups are free. So for free Abelian groups, this is fairly straightforward and was just essentially part of the classification of finitely generated Abelian groups. So the first question is, what does a subgroup of a free group look like? Suppose you've got an index n subgroup. We're going to talk about finite index subgroups. The same thing applies to infinite index subgroups, but just slightly messier because we need infinite pieces of paper to draw them. So an index n subgroup of any group G is just the same as an action of G on n points. Well, it's not quite true because it should be a transitive action. And one of these points should be marked to show it's the base point. Now a transitive action of a free group on n points is really easy to draw. For example, here is a transitive action of a free group on two Gs. On several points, all I do is I fix my base point, which I put a circle around to indicate that it's special. And then I have, say I want an index, one, two, three, four, index seven subgroup. All I do is I choose some sort of action of the generators on these points. So one generator might do that, and then it might do this, and then it might do this. And then the orange one has to do something. All I've got to do is to make sure the graph is connected. So it might go like this. And then maybe it goes to there and maybe it fixes that one. So here I've got two different permutations acting on a set of seven points and they're acting transitively. So this generates an action of the free group on two generators on seven points. So the subgroup fixing this point is going to be an index seven subgroup. So this point counts as index seven subgroup. So that shows that subgroups of free groups are easy to write down. All you do is write down some points and write down a few permutations acting on them. And that gives you, and then you have to pick a base point and that gives you a subgroup. Well, what is this subgroup? Well, it's a set of all things which map this point to itself. And you can, so in order to find out what a word does you just have to follow it along here. So for instance, here's an example of something that fixes this point. So let's have the green one is going to be called A and the orange one is going to be called B. And then for example, if I fix this point, if I do A, then A, that gets me to here, then B, then B, and then B. So this element here is an element of the subgroup fixing this point. Or similarly, I could do A, sorry, no, B would get me to here and then I can do A, which gets me back to here and then I can do B inverse. So this is also in H. So what you see is the subgroup H is all paths from the base point to the base point. Well, it's not quite all paths because if I backtrack, then that counts as the same element. So what I should say is all paths up to homotopy. What does this mean? Well, homotopy means that I can insert something and it's inverse. So if I've got a path along like that, then I can change it to a path A, A to minus one and that still counts as the same path. What it means is instead of going like that, that and that, I may be going like that and then backtracking and then going like that and that. This notion of homotopy is the same as the notion of homotopic paths where you get an algebraic topology more or less except an algebraic topology, you describe paths as being continuous maps from the unit interval into something whereas here I'm describing paths combinatorially as just being a list of which edges you go along but it's more or less the same thing as homotopy. Anyway, what you see is that this index seven subgroup is just the fundamental group of this graph with respect to this base point where the fundamental group just means all paths opt to homotopy. And the fundamental group is usually noted by pi one, there are higher homotopy groups done by pi N but we aren't going to worry about those. So it's pi N of the graph with respect to this base point. So that tells us what a subgroup of a free group is, it's the fundamental group of some graph. So let's think about what fundamental groups of graphs are. Well, suppose I've got a graph that looks like this and I don't know what sort of edges all over the place with a base point and I want to know what its fundamental group is. Well, notice I can simplify this graph without changing the fundamental group. I can pick an edge with two different vertices and just sort of contract it to a point. So I'm going to identify all this to one point and this turns it into a different graph. So now it's really just this graph because I've contracted this to a single point. And you notice that any loop in this graph corresponds to a loop in this graph and vice versa. So these have the same fundamental group with respect to this base point. So I don't change the fundamental group of a graph if I keep contracting vertices as long as they've got different edge points. If they've got the same end point then I can't contract it because that gets rid of a loop. So I'm going to keep contracting edges whenever there are two different vertices. So if I take a connected graph and contract edges, I'm eventually going to end up with just one vertex. So the graph is going to look like this. And what's the fundamental group of this? Well, the fundamental group of this is just a free group on five generators because it's got five loops. So I've got one generator there and one there. And so it's kind of obvious that the paths without backtracking are the same as the reduced words in these loops, which is the same as the free group. So pi one is just a free group on the loops. So this shows that any subgroup of a free group is free because a subgroup is the fundamental group of a graph and the fundamental group of any graph is a free group. Well, we now have the question, how do you actually write down a set of generators for this free group? And that's quite easy and it's probably best just to illustrate by this by an example. So let's do the simplest non-trivial example. I'm going to look at a free group on two generators. I'm going to look at an action of it on two points and you have to think how many actions are there on two points? Well, I have to fix one point. And then each generator can either fix the two points or it can swap them and they can't both fix the two points because the graph has to be connected. So altogether there are three possible ways you can make this into a connected graph of two generators. So the free group on two generators has three possible subgroups. And let's write down one of them. Let's call this generator A and I'll call the green generator B. So we want the fundamental group of this graph. And so what I'm going to do is I'm going to contract it. So there's only one point. So I'm going to contract it by killing off this line here. So there are really now three, I'm pretending that this is just one point and there are now three loops. So the subgroup is a free group on three generators. And let's see what these generators are. Well, they're independent loops of this graph. So one of these generators is going to be A. And another one is going to correspond to this loop and you might think that's going to be A as well. Well, that wouldn't make a lot of sense because we've already got A there. But what you would have to remember is we're really going along here because we can track this to point, then going around here, then back again. So this corresponds to doing B to the minus one, then A and then B. And similarly, this loop here may look like as if it's just a generator B, but we have to get back to here. So we do B and then we have to do B again. So this gives us B squared. So the subgroup of index two is a free group on three generators, A, B squared and B, A, B minus one. Now you've seen in particular, we've got a free group on three elements is now a subgroup of the free group on two elements. This is a bit strange if you're used to a billion free groups because for a billion groups, a free a billion group on some number of generators can't be a subgroup of a free a billion group on a smaller number of generators, but free none a billion groups are a bit strange that a free group on three generators isn't really any bigger than a free group on two generators because it's in fact a subgroup of it. In fact, we can figure out how many generators index and subgroup has. So we now have the following question. Suppose G is of index K in FN. So this is free on N generators. How many generators do we need for G? Well, let's think about this. We draw the graph for G. So it has K points, K vertices because it is index K. And then how many edges does it have? Well, it must have K times N edges because for each of the N generators, we need to form a permutation of K points which needs N, so that's K edges for each generator. So now we work at the Euler characteristic of the graph. Well, the Euler characteristic is the number of vertices minus the number of edges which is going to be K minus KN. And now you notice the Euler characteristic does not change when we contract an edge because if we take an edge and change it to a single point, we've removed one edge and one vertex so we haven't changed the Euler characteristic. So if we contract this down to a point with a lot of loops, this still has Euler characteristic K minus KN which is equal to one minus the number of loops. So the number of loops is one plus KN minus K and this is the number of generators. So an index K subgroup of the free group on N generators is this number of generators. For example, when we had K equals two and N equals two that would be one plus four minus two which is three generators. So we found an index two subgroup of the free group on two generators did indeed was indeed a free group on three generators. So let's do one more example. Let's do a slightly more complicated one. We have a map from the free group on two generators onto the symmetric group on three points. So you remember the symmetric group on three points is generated by three elements, two elements A and B such that A squared equals one, B squared equals one and AB cubed equals one. So we get S three by taking the free group on two generators and adding these relations. And now I'm going to ask find generators for the kernel of this map from F two onto S three. Well, that sounds like a completely daft question. I just told you that the relations that give you S three of these three elements. So surely the kernel is just the free group on these three elements A squared, B squared and AB cubed. Well, no, that's wrong. The problem is these three elements don't generate the kernel. The kernel is generated by these three elements together with all their conjugates because the kernel has to be a normal subgroup and A squared, B squared and AB cubed won't, but they don't generate a normal subgroup. Moreover, the subgroup is index six. So we see by the formula we had before that the number of generators should be one plus six times two minus six, which is seven. So this is really a free group on seven generators. Well, how do we find these generators? Well, what we do is we write down an action of the free group on two generators on six points corresponding to this. So here are our six points and we're going to fix one of them as our base point. And now what does the generator A do? Well, A takes it to that and we want A squared to be one, so it does that. And then we want B squared to be one. So B might do something like this and A then does this and B might do this and A might do this. And notice that AB cubed is equal to one, which is why we end up with a hexagon here because with AB, AB, AB goes back to where we started. Now you notice that this particular diagram looks very symmetric. In particular, there's a symmetry of this graph taking any vertex to any other vertex. And that corresponds to the fact that this subgroup here is a normal subgroup because if we take the fundamental group of this point of the graph, it's conjugate by some element of the symmetric, some element of the free group, will be the fundamental group where we take some other element as base point, as very easy to check. So if this is a normal subgroup, all these fundamental groups must actually be the same. So there must be a symmetry of this graph taking each point to each other point. So this graph is really symmetric because this is a normal subgroup. Now we have to contract this down so that we've only got one point. So we can track some of the edges and let's pick some of the edges to contract. So what we're really doing is we're picking a maximal tree in this graph. So let's pick a tree that looks like this. So I'm going to use this maximal tree to sort of identify all these points. And now you see I've got one, two, three, four, five, six, seven loops left. So these seven loops are going to give me the seven generators. And in order to turn each loop into a generator, I've got to go along the tree to the beginning of that path, then go along the path and go along the tree back to the end. So this one corresponds to I do A to get to the beginning and then go A there. So I get an element A squared as an element of K. For this one, I do A, then B, then B, then go back along A. It's rather annoying that the composition of elements goes from right to left. So this element gives me that element, gives me this element of K. This loop here, I do A, B, A, then A, and then B to the minus one, A to the minus one. This one, similarly, I do A, B, A, B, B, and then A to the minus one, B to the minus one, A to the minus one. And this one, I do A, B, B to the minus one, A to the minus one, B to the minus one, A to the minus one. This one, I do B, and then I have to go back along here. So I do A to the minus one, B to the minus one, A to the minus one, B to the minus one, A to the minus one. And finally, this one, I have to go to the start of it. So I go A, B, A, B, A, and then I finish off with B. So here we have a set of seven generators of this free group on seven elements. Okay, so next lecture, I'll give some more examples of finding presentations of groups from free groups.