 So this lecture is part of an online mathematics course on group theory and it will be about normal subgroups of groups. We will start by looking at groups of order six. So you remember we have found two examples of groups of order six. We've found the group S3 of all symmetries of a triangle and we found the group Z, modulo 6Z, the cyclic group of order six and we remember we saw this is actually isomorphic to Z over 3Z times Z over 2Z. And we'll start by looking at the subgroups of these groups. So for cyclic groups it's really rather easy to work out the subgroups. The group is generated by some element G and it's easy to check that all subgroups must be generated by some power of G and this power may as well you may as well take it to divide this number here. So Z over 6Z has exactly four subgroups. It's got the subgroup Z over 6Z itself which is say generated by some element G. So it consists of well I guess we could just take G to be one. So it consists of 0, 1, 2, 3, 4 and 5 and it's got a subgroup isomorphic to Z over 2Z which is generated by 0 and spanned by 1 and 0 and 3 and it's got a subgroup Z modulo 3Z generated consisting of the element 0, 2 and 4 and finally it's got a subgroup the trivial group just consisting of the element 0. Now for the group S3 it's almost as easy to find the subgroups. First of all we've got the group S3 itself and you remember this has six elements which are the identity then we can switch 1, 2 or 2, 3 or 3, 1 or we can rotate 1, 2, 3 or 1, 3, 2. So we're labeling the elements of the triangle as 1, 2, 3. Now the order of any subgroup must divide the order of S3 which is 6 so all subgroups have ordered 2 or 3 or 1 which makes them really easy to work out. So we've got a subgroup of order 3 consisting of 1, 1, 3, 2 and 1, 2, 3 and we've got three subgroups of order 2 so we've 1, 1, 2, 1, 1, 3, 1, 2, 3 and we have the trivial group so all together it has exactly 1, 2, 3, 4, 5, 6 subgroups and nothing terribly exciting is happening because there are no, there is yet no inclusions between subgroups other than the trivial subgroup and the whole group but we'll see some more complicated examples later. And what I want to discuss is the following problem. Suppose that H is a subgroup of G. Can we form a group G over H, a sort of quotient group and the idea of the quotient group is you sort of force all elements of H to be trivial. So we've seen some examples of this in number theory. If you take the group Z and you force all multiples of N to be trivial then you get a sort of quotient group of the integers mod N which is a group of order N so this is a quotient of Z by a subgroup NZ. And how can we do this in general? Well we should, we should get an exact sequence 1 goes to H goes to G goes to G over H goes to 1. So this means G should have a homomorphism onto this quotient group and the kernel should be exactly H. Well you can see from this that the only serious possibility we can do is to take the elements of G over H to be the set of left cosets of H. So G acts on this quotient group by multiplication and we saw that any set that G acts transitively on is the set of cosets of something. So G over H has to be the set of left cosets of H if it's anything at all. So the question is can we make this into a group? So is this a group and I mean is it a group in a natural way? I mean you can obviously define some sort of stupid group structure on it but we want a sensible group structure on this. So let's look at two cosets. We've got a coset G1H and a coset G2H and what is their product going to be? So let's multiply these cosets together and the only thing we could possibly set it equal to is G1G2H. So I'll put a question mark there because we haven't yet. There's a slight problem here. Is this well defined? So the problem is the coset G1 of H might also be a coset G3 of H for some other G3. So if we change G1 to G3 without changing this coset then do we change this coset here and that's a tricky problem. So let's have a look. So if we put G3 equals G1H then G3G2H should be equal to G1G2H because G3H is the same coset as G1H so the product with G2H should be the same. So is that true? Well for that we want G1HG2H equals G1G2H. Just remind you that an expression like this means there's a set of elements that are the form G2 times something in H. So is this true? Well this is equal to G1G2 times G2H G2 to minus 1 HG2 times H. So we want this to be G1G2 times something in H and you can see that will hold provided this is in H. So we've got the following question, is this element here in H? And if it's always in H then we've got a well-defined multiplication on cosets and it's very easy to check that if this multiplication is well defined then it's associative and has an identity in inverse forms of groups. So the big question is does this hold? And in fact you can see the following conditions are equivalent. First of all H is the kernel of sum map from G to any group X. Secondly GHG to the minus 1 equals H for all G in G. It doesn't really matter whether we have this as G and this is G to the minus 1 or this is G to the minus 1 or this is G. So this just means all elements you can get by taking H all elements of H multiplying on the left by G and on the right by G inverse. Thirdly well if GHG to the minus 1 equals H that's the same as saying G times H equals H times G for all G in G and this is the same as saying left cosets are the same as right cosets. And these are also the same as saying that a quotient group G over H exists defined by the construction above. So it's very easy to check that all these conditions are the same I'll just leave this as a sort of exercise for people to do. I've done a couple of cases of it and the others are easy. And if H satisfies any of these five equivalent conditions we say that H is a normal subgroup. This being a good example this will subgroup subgroup of mathematicians using very unimaginative terminology. The word normal in mathematics is grossly overused as an adjective and this is one of the dozen or so places in which it's used. So the obvious question is which subgroups are normal? Well obviously if G is a billion then it's trivial to check that all subgroups are normal because if it's a billion we can just sort of switch G to the other side of H. And the first example of a non-Abelian group we've had so far is the group S3 of order six. So if we're going to be looking for some examples of normal, non-normal subgroups this is the obvious place to look. So let's look at which subgroups of S3 are normal. Well first of all one and the whole of S3 are obviously normal and the same is true for any group. The trivial group and the whole group are trivially normal subgroups. What about the group 1, 1, 2, 3, 1, 3, 2? Well this group is index 2. So you remember the index of a subgroup is the number of cosets. Well did I mean left cosets or right cosets? Well it turns out it doesn't really matter because the number of left cosets is the same as the number of right cosets and to see this we note that we have a map from left cosets. If we've got a left coset gh then this goes to a right coset just by taking the inverse of every element here. So we take gh to the minus 1 by which I mean the inverses of all elements in this set and this is just h to the minus 1g to minus 1 which is hg to the minus 1. You should remember that when you take inverses you have to swap the order so ab to minus 1 is b to the minus 1, a to the minus 1. So there's a natural one to one map between left cosets and right cosets just by taking the inverse of every element of the coset. So the number of left cosets is the same as the number of right cosets and it's just called the index. So this is one of the few cases when we don't need to worry about whether we're talking about left or right cosets. So anyway you've got this subgroup of index 2 and all subgroups of index 2 are normal and this is easy to see because the only cosets so if the subgroup is h are h and things not in h because it has index 2, one left coset is h, the other left coset must be everything not in h and that's exactly the same as the right cosets. So left cosets are the same as right cosets so any subgroup of index 2 must be normal in particular this one. Of course we could just check directly but later it will be useful to know that subgroups of index 2 are always normal. So let's look at the subgroups of index 3 or order 2. Well let's look at the subgroup generated by containing elements 1 and 1 2. So this is order 2 index 3 and this is not normal. So let's check this explicitly. Let's find its left cosets and find its right cosets. So we've got 1 1 2 3 1 3 2 and I'm going to write out the elements here as 1 1 2 1 3 2 3. So here are the six elements of the group s3 and let's find the left cosets of 1 and 1 2. Well there's one obvious left coset which is these two elements here because the subgroup itself is a left coset of itself. So for the next left coset we want to work out what is the left coset of this. So we have to look at the elements 1 2 3 times 1 or 1 2 and 1 2 3 times 1 is obviously just 1 2 3 and 1 2 3 times 1 2 well it takes 1 2 3 times 1 2. Well if it acts on the element 1 this element takes 1 to 2 and this element takes 2 to 3. So this takes 1 to 3 and if we apply this element to 2 it takes 2 to 1 and then this one takes 1 to 2 so 2 goes to 2 and obviously 3 goes to 1. So this element is equal to 1 3. So we get a coset here and it's pretty obvious what the third coset is going to be it's going to be this. So here are the three left cosets of 1 2. Now let's work out the right cosets. So the right cosets I will do in blue so there's one obvious right coset which is this and then we've got to work out the right coset of 1 2 3 and the right coset is going to be 1 2 3 1 which is 1 2 3 and 1 2 3 1 2 and now again we have to work out what this does. So this element takes if we apply this to 3 this takes 3 to 1 and then this takes 1 to 2. So 3 goes to 2 if we apply this to 2 this takes 2 to 3 and that's fixed. So 2 and 3 are swapped and it must take 1 to itself. So this is equal to 2 3. So the right coset contains these two elements and the remaining right coset must be the two things the left over. So we can see the three left cosets are these red things and the three right cosets are these blue things and the subgroup is not normal because the left cosets are not the same as right cosets and there's no really honest way of making the set of three left cosets into a group. I mean obviously you can make it into a group because it's a set with three elements and there are groups of three elements but there's I mean that there are stupid ways to make it into a group but there are no nice ways to make it into a group. Well that's one group what about the other groups? Well we don't really need to do the other ones because if h is a subgroup so is ghg to the minus 1 and we can easily check this we can just check it's closed so if we've got some element ga g to minus 1 and we multiply it by gb g to the minus 1 this is just equal to ga b g to the minus 1 so it's closed under multiplication. This is called the conjugate of h by g and if you think about it a bit you will see that this is really an action of group g on the set of subgroups so gh on the set of subgroups h and the action is given by g of h this is the I'm defining an action and this is defined to be ghg to the minus 1 so this is the definition of the action of g on h. So let's work out what this action looks like for the subgroups of the group s3 so there are remember there were six groups there was s3 there was the group generated by one two these pointy brackets around an element or a set of elements often mean the group generated by those elements in other words the smallest subgroup contained by them so we've got one three and two three and we've got this element one one two three one three two and let's work out what the conjugates are well first of all if a subgroup is normal then the group acts trivial on it so it just maps s3 to itself and maps this group to itself and it maps one to itself. However if we take the subgroup one two and act on it by one three so we take one three sorry one two one three one three minus one you can calculate that this is actually the element two three so if we conjugate this subgroup by the element one three we get the subgroup two three and similarly if we conjugate two three by by one two we get one three and so and so all these three elements form an orbit under the action of g on the group so the orbits kind of look like this so these three subgroups are all conjugate the conjugate means you can get from one to the other by this action and conjugation of the group g is a sort of is actually a symmetry of g we notice that g a b g to minus one is g a g to minus one g b g to minus one so in other words conjugation by g is actually a symmetry of the group and anything you can say about something is sort of still true if if you conjugate by it in particular if this group is not normal then its conjugate is also not normal and this conjugate is not normal so these three groups are all not normal once we check one conjugate is not normal we don't need to bother with the other since they're all um equivalent under a symmetry of the group and you can see that under this action of g on the set of groups the normal subgroups are just the subgroups that are fixed by g meaning that every element of g maps that group to itself so so here we have g acting on a set of six elements it fixes these three elements which are the normal subgroups and it acts transitively on this set of three elements which are therefore non-normal subgroups um so um if g has a subgroup h then g is sort of composed um so suppose g has a normal subgroup h so if g has a normal subgroup h then we have this exact sequence one goes to h goes to g goes to g over h and if h is not one or g then these groups will be smaller than g so these are usually smaller than g so the idea is we can sort of break g up into into groups that we hope are smaller and therefore easier to understand and um um if we're lucky we can reduce properties of g to properties of these two groups um I'll just emphasize yet again that if we know h and if we know the normal subgroup h and the quotient group g over h this does not determine g just a reminder from last lecture that we've got these sequences nought goes to z over two z goes to z over four z goes to z over two z goes to one so nought and nought goes to z over two z goes to z over two z times z over two z goes to z over two z goes to nought so so we have two different groups they both have the same or an isomorphic subgroup z over two z and the quotients are isomorphic but that doesn't imply these two groups are isomorphic so there's a complicated problem of trying to work out what possibilities there are for g provided you know these two groups here we'll be giving an example of that um in a two or three lectures time when we discuss groups of order eight um well we haven't what we haven't yet done is classify the groups of order six so we will do that next lecture