 This lecture is part of an online math course on group theory and will be about free groups So we'll start by quickly reviewing free abelian groups and Then see how much of this theory extends to free possibly none abelian groups. So free abelian group on various elements a b c and so on is a sort of universal Abelian group generated by These elements a b c and so on So universal means it should be as big as possible So we we don't put on any unnecessary relations like consisting a should have all the three and so on and it's pretty obvious What the free abelian group on this is if we've got elements a b c in an abelian group Then then the group subgroup generated them consists of all elements of the form n1 a plus n2 b and so on where the n i are in z and That the most universal way of doing that is just to take a sum of copies of the group z So this z is going to be generated by a this one by b and so on So the free abelian group on a finite set of generators is just a sum of copies of z It has a universal property so if if f means the free abelian group on Elements a b c or whatever then if we've got any group a any abelian group a and We've got certain elements a Let's call them a b c in this group a then there's a unique homomorphism From the free abelian group which takes the element a here to that to the element a and there and so on so the number of In particular the rank of a free abelian group, which is the number of generators is well defined For instance the free abelian group on two generators is not Isomorphic to the free abelian group on three three generators and the easy way to see this is the number of homomorphisms from the free group on n generators to Say z modulo 2z It's just 2 to the n because each of the n generators can be mapped to one of the two elements here so so If you know if you're given a free abelian group you can determine its number of generators just by counting homomorphisms to z over 2z. That's if it's fine actually generated if it's infinitely generated you And Subgroups of free abelian groups are easy to describe So if we've got a subgroup of a free abelian group, which we will take to be z to the n these subgroups are all free abelian of rank Less than or equal to n. Moreover, we can choose a basis A1, A2, An, so the subgroup is of the form generated by n1, A1, n2, A2 and so on where the n i's are The negative integers and the proof of this just see the classification of finitely generated free finitely generated abelian groups Because the way we classified finitely generated abelian groups was that we showed Implicitly showed they were the quotient of a free abelian group by a subgroup and then showed that we could choose the subgroup in this form So that our finitely generated group is a sum of cyclic groups of orders n1, n2 and so on or At least if the n i are finite And Finally, we'll just draw a quick picture of a free abelian group So a free abelian group on let's do two generators might look something like this so here we've got two generators a red generator and a green generator and the points of the Free abelian group just form that the elements of a free abelian group just form the points of a lattice so these black spots Correspond to the points of the group and the red and the green arrows Just tell you what happens if you multiply some element by one of the generators and you can see you're just getting a square Lattice or the obvious analog in higher dimensions So that's free abelian groups and we're now going to try and find analogs of all these for None abelian groups So first of all We'd better construct a free abelian group Well before constructing free abelian groups. Let's first construct free monoids So a monoid is like a group without Inverses so we just it has an identity and a multiplication and it's associative and the identity is Behaves in the obvious way, but we don't assume that there are inverses And then there's an obvious way to construct a free monoid on a B. C This is going to be all words In a B C and so on And the word has the fairly obvious meaning. It's just formed by putting letters together So the words are as follows first of all, there's the empty word Where you put no elements at all and we denote that by one because if I denoted the empty word by by writing nothing at all You wouldn't be able to see it and then we can have words Let's do two generators weird of a B a a a b b a b b then a a a and so on and It's pretty obvious that this gives you a sort of universal monoid It has the universal property if you've got the universal monoid F and any other monoid M and you choose say two elements in M. There would be a Unique homomorphism from the universal monoid on a and B to M that takes a to this element and B to that element and so on So free monoids are easy to construct Also, if you've got a monoid we want to find quotient monoids and this is a little bit tricky You can't start using cosets of monoids because that's that sort of required the use of cosets However, we can fit we can quotient out by an equivalence relation. So if you've got a monoid Supposedly want to identify some elements of the monoid. So we want to identify the element X with Y say then we form the smallest Equivalence relation on M such that It contains whatever Equivalence as we want we might have X we might want to make X 1 equal to Y 1 and X 2 equals to Y 2 So we insist that X I eat is equivalent to Y I and then we want to say if A is equivalent to B and C A is equivalent to C B and AC is equivalent to BC and If we form this equivalence relation, you can check that the equivalence classes also form a monoid and a multiplication And this is a sort of quotient of M It but by forcing various elements of M to be equal And now you can form a free group We just take the free monoid on elements a a to the minus 1 B B to the minus 1 and so on and quotient by My forcing a a to the minus 1 equals a to the minus 1 a equals 1 and same for B C and so on so This gives us a free group and You can check it has the universal property if G is a group with elements a B C and so on we can map The free group on elements a B and C to G by mapping little a to big a Little B to big B and so on so it has the correct universal property In particular, we can see the rank is well-defined because the rank of a free group on n generators is Given as follows the number of homomorphisms from the free group on n generators To Z over 2 Z is again just 2 to the power of the rank just as it was for the AB in case So the free group on 2 generators is not the same as the free group on free generators for example So the problem is how do we write out the? elements of a Free group, so let's try writing out the elements of the free monoid. We have one a B a to the minus 1 B to the minus 1 then we get a squared a B a And then we get all sorts of things like a B to the minus 1 a a to the minus 1 and so on and We notice that although in the free monoid, this would be an element element not equal to one in the free group. This is equal to one and Similarly, if you get more complicated elements say a a B B to minus 1 a to the minus 1 B a to the minus 1 We notice that we can simplify this element Because B B to minus 1 is equal to one But then we can simplify it a bit further because we can cancel out the a and the a to the minus 1 And if you think about it a bit, you'll see that every word is equivalent to a reduced word Where reduced means it does not contain X the minus 1 or X the minus 1 X for any of the generating elements and Now we have the following question if two reduced words are distinct Are they different in the free group? And it sounds very plausible And in fact, it's true, but we would like to prove it and If you multiply one word by the inverse of the other you see this is equivalent to the following question if a reduced word is None empty is it not equal to one in the free group Well in order to do in order to show that this is correct. We end up with the following problem given a reduced word in a a to minus 1 B B to minus 1 and so on find a group with elements a a to minus 1 B B to minus 1 and so on So that this word is Not equal to one here. I'm being a little bit careless I'm using these elements a and b for for elements of the group g I'm also using them for the basis elements of the free group But since you're mapping the elements of the free group to these elements of g It won't cause too much harm if we just use the same letter for both And if you can do this that will show that any um None empty reduced word is Not the identity in the free group because it maps to a non trivial element in this group g And we can do this as follows in fact if the word Has length n we can take G to be the symmetric group on n plus one letters. So this is the symmetric group And the best way to prove this is just to do an example So suppose I've got a word a squared b a to minus 1 B a b a squared b to the minus 3 and I want to ask you can you find a group With elements a and b such that this element is not equal to 1 So this is a reduced word and you know, it's if you're given this It's not at all obvious how you write down a group such that this an explicit group such that this Word is obviously not equal to one unless you're kind of cheap by using the free group. Well, you can do this as follows So this is a word of length 1 2 3 4 5 6 7 8 9 10 11 12 So I'm going to write down 13 points and then I'm going to let a Act like this because I've got an a squared and then B is going to act like this and Then a is going to act like this. So this is going to be a to minus 1 and Then b is going to act here. So the green is going to be b and the red is going to be a So here's a we've got another b. I should have chosen a shorter word. This is getting a bit boring Then we've got an a squared and then we've got a b to the minus 3 which means So So if the elements a and b act like this then then I've indeed found an Element of the symmetric group on 13 points such that this element is none trivial because it maps that point to that point Well, the trouble is these red and green things are not actually permutations yet However, I can turn them into permutations just by completing a cycle. So I'm going to I'm going to let a be the following permutation. I can let it do that and then I could maybe let it do that and That and I can complete it like that and then I could just let it fix these points here And I can do exactly the same thing for b So I can let b fix these points and map that point back to there and map that point to there and map that point there and map these points like this So here I've written down two permutations a permutation a and a permutation b such that this permutation is not the identity and And let's just think about why it's possible to do that. So what we're doing is we're taking a few points and We've got a partial permutation And we want to know can we complete it to a Permutation well, there are two cases when you couldn't do that. So first of all if if I had A partial permutation that did this then I couldn't complete it because The action on this point wouldn't be defined. There would be two point I mean, there'd be two things that I was trying to map this point to Similarly, if I had it this way around that would be impossible because there'd be two points trying to map to that So that couldn't be a bijection. Well, if you look at this, this would correspond to a permutation x x to minus one and and this would correspond to x x to minus one so so These would only arise if you start with the with a non reduced word You see if I started with a a non reduced word here say if I did a b b to minus one a and try to do this construction Then I would run into a problem because a would be trying to do this Which would be fine, but b would be trying to do this and I couldn't complete b to a permutation. So the the proof would break down for this So that's where we use the fact that words are reduced. We want to be able to extend permutations to the whole set So I think that construction makes it obvious that any non reduced word in the free group is indeed not the identity Well, we've actually proved a bit more than that we've proved free groups are Residually finite So what does this mean? Well a group G is residually finite means that for any element G not equal to one we can find a homomorphism from the group to a finite group G so G has image Not equal to one and that's exactly what we did for the symmetric group where we took this finite group to be a to be a symmetric group on on One more Generic one more elements than that than the length of G In case you're wondering if there are any groups that aren't residually finite, you notice the rational numbers under addition is not Residually finite Similarly the real numbers under addition is not residually finite and any infinite Simple group is also not residually finite which is kind of obvious because if it's an infinite simple group It doesn't have any non-trivial homomorphisms onto finite groups. So Infinite groups are most infinite groups you come across are in fact not residually finite So this is actually quite a quite an unusual property And by the way, it's tempting to define the free group and so why not Define the free group To be the set of reduced words Well, you can the problem is Defining the product is a bit tricky Because you then have to say you take the union of two reduced words and cancel out Anything that matches and proving associativity of this Product is kind of tricky Because the problem is you may get things like this you might have a word a B And then you might have a word a to minus one B to the minus one and then you might have a another word a a B and If you multiply these two then the a's cancel out and the B's cancel out And if you multiply these two then nothing cancels out and You've then it's it's it's really quite a tricky combinatorial problem to show that The product defined by taking reduced words and multiplying by canceling things out is associative It's you have to look at a large number of slightly tricky cases So it's possible to do this, but it's it's the sort of argument where it's really easy to make mistakes So I think it's better to define the free group in a way that makes it obviously exist and then Prove that it consists of the reduced words I Just finished by trying to draw a picture of the free group on two generators So we had a picture of the free a billion group on two generators And the free group on two generators is kind of looks a bit like this. So first of all, I'm going to take the point here and I Can map the first generator So I'm going to take this point to be the identity and then we have the first generator can act on it and The second generator can act on it Like this. So I get some more points Where the second generator Does something and Now I want to apply the first generator to things that the second generator Moved and I can do it sort of like this and I've got to leave a lot of room oops Because I need to leave room for the second generator act on all these points. So the second generator is going to act like this And there are going to be lots more points here and so on and now I Need to act on with the first generator again. So the second generator on these points here and I have to These going through all these points and through all these points and what you notice is that I'm really running out of room and If I did one or two more iterations, I just wouldn't have room to draw these things anymore because the trouble is free groups on more than one generator sort of exponentially increasing you can see the number of Words of length n on the generators increases as an exponential function of n So as soon as I get to words of length more than about three or four I certainly don't have room to draw them on a Euclidean piece of paper so The free a billion group was okay because the number of words just increases like a polynomial So you can draw them in Euclidean space, which is plenty of room for polynomially growing functions But you can't really draw free groups in Euclidean space. What we should really do is draw them in hyperbolic space because hyperbolic space has an exponentially increasing amount of room if you Put some sort of as you go further away from the origin. So you can draw Free non-Abelian groups very nicely in hyperbolic space in some sense free a billion group groups live in Euclidean space and Free non-Abelian groups live in hyperbolic space Well, there's one final question So for free a billion groups I said that any subgroup of a free a billion was a free a billion group of Rank most equal to the group you started with and you ask if there's an analog of this for none a billion free groups and That's almost true. First of all Any subgroup of a free a billion group is free a billion. However, the rank might increase So you may think the free a billion group on three generators is bigger than the free a billion group on two generators But you're wrong because the free a billion group on three generators is actually a subgroup of the free a billion group on two Generators and next lecture we will explain this rather bizarre sounding result