 This lecture is part of an online mathematics course on group theory and we'll be covering the groups of order 12 using the Zilof theorems that we did last lecture to help classify them So let's start by writing down all the groups of order 12 We can think of So first of all, there's a cyclic group of order 12 then we can take cyclic group of order 6 Times the cyclic group of order 2 or we can take a cyclic group of order 3 times 2 cyclic groups of order 2 We've got the dihedral group of order 12 Then we can take the Symmetric group of order 6 on 3 points and multiply that by the group of order 2 and then Then we can take binary Cyclic group of Order 12 you remember a binary cyclic group means we take a group in SO3 of the reels That's all rotations and using quaternions. There's a map from the group of unit quaternions Onto the group of rotations and we can take the inverse image of that So if we take a cyclic group of order 6 here, it's inverse image will award a 12 here Similarly, we can form the binary Dihedral group of order 12 where we take a Dihedral group of order 6 and take its inverse image Then we can take a lot of semi direct products So we can take the group Z modulo 3 Z and take a semi direct product with Z modulo 2 Z times Z modulo 2 Z Or we can take a group Z modulo 3 Z and Take a semi direct product with the group Z modulo 4 Z which can act on the group with three elements We can also take rotations Of a tetrahedron Which is order 12 We can take the alternating group a4 that I'll explain in a moment And there's another semi direct product Z modulo 2 Z times Z modulo 2 Z and this is Acted on this is a symmetry of order 3 so we can take a semi direct product of Z modulo 3 Z So let's just Recall what the alternating group is the alternating group a4 is a subgroup of the symmetric group So the symmetric group acts on four variables w, x, y and z And we can look at w minus x w minus y w minus c x minus y x minus c y minus c and if you act on this polynomial here by a permutation it it changes either to itself or to minus itself because It will permute all these factors and some of the factors will end up multiplied by minus one So if f is equal to this a4 is the permutations fixing f and Since any permutation maps this either to f or to minus f We see that a4 is index 2 and s4 so as order 12 So we've got 1 2 3 4 5 6 7 8 9 10 11 12 groups of order 12 Well, there are in fact only five groups of order 12 and it turns out that many of these groups are the same in fact all the groups on The same row are isomorphic to each other. So these are isomorphic These are all isomorphic These are isomorphic. These are isomorphic These are isomorphic It's not too difficult to check all these isomorphisms so So we want to show that any group of order 12 is isomorphic to one of these five groups and Now it will be very useful to recall the seal of theorems so any group G of order 12 has Seal of subgroups of orders three and four and the number so the number of three seal of subgroups three seal of subgroups is One mod three by one of the seal of theorems and it divides the order of the group So the number of seal of three subgroups is either one or four and the number of seal of two subgroups must again be One mod two and it must divide the order of the group. So it's either one or three So What we're going to do is first look at the case when the number of seal of Three subgroups is one so assume One seal of three subgroup Then it must be normal Because if it wasn't normal there would it would have a conjugate different from itself So we see that G is a semi direct product of a seal of three subgroup with a Seal of four subgroup and now we have four cases because The seal of four subgroup can either be z modulo four z or z modulo two z times z modulo Two z And it must act on the seal of three subgroup the seal of three subgroup must be z over three z And this only is two automorphisms Which are one mapping everything either to one or minus one So an action of a seal of four subgroup on a seal of three subgroup must be a homomorphism of one of these groups to a group of Order two and it can either be a trivial homomorphism or a non-trivial homomorphism And this one has one non-trivial homomorphism and this is three non-trivial homomorphisms But they're all really the same Opto automorphisms of this group. So we have four cases to consider The seal of two subgroup Can be z modulo four z or z modulo two z times z modulo two z and the action on Z modulo three z can be trivial or non-trivial and All we have to do is to figure out what we get in each of these four cases So z modulo four z with trivial actions giving us the cyclic group of order 12 Z modulo four z acting on z modulo three z with a non-trivial action is giving us some semi-direct product like this Which as we said is actually also isomorphic to the binary Dihedral group of order 12 Trivial actioners of this group on s3 is just going to give us a product So we get z modulo six z times z modulo two z Which is of course the same as z modulo three z times z modulo two z times z modulo two z And finally if we have a non-trivial action of this on z modulo three z What we're getting is z modulo two z times Group s3 or equivalently the group d12 So this gives the four cases when there is a normal subgroup of order three now suppose So now suppose the subgroup of order three Is not normal then the number of conjugates You remember it's one mod three and it has to divide 12 so it must have four conjugates so So we've got four subgroups s1 s2 s3 and s4 of order three and No two of these subgroups kind of any element in common other than the identity because if they had an element in common They would be the same so this So the number of elements of order three Well, we have two elements of order three in each of these subgroups and these four element These eight elements are all distinct so this leaves so there are four elements of G not of order three So they must form the seal of two subgroup Because we know it has a seal of two subgroup of order four and the seal of two subgroup can't contain elements of order three So it must consist of all the elements left over So it must be normal Because the only possibility is it contain is it consists of these four elements that are not of order three So our group is a semi direct product for a group of order four by a group of order three So it's either Z modulo 4 Z semi direct product Z modulo 3 Z or Z modulo 2 Z times Z modulo 2 Z Semi direct product Z modulo 3 Z Because these the only two groups of order four and this case here is not possible because The group of order four has no Automorphisms of order three so we can't form an interesting semi direct product except by making this act trivially in which case there would only be one subgroup of order three which Contradicts our assumption the subgroup of order three isn't normal. So the only possible case is this and We notice that this group here has an automorphism of order three In fact, we've got two automorphisms of order three because you can just permute the The you can just do a cyclic permutation of the three elements that aren't the identity So we get exactly one group that is a semi direct product of the client for group by group of order three And you can easily check this is isomorphic to a for and it's isomorphic to The other one was the rotations of a tetrahedron In the easy way to see this is to notice that we've shown this is the only group with more than two elements of order Three that is order 12 and these two groups both of more than two elements of order three and they have order 12 So they're all they're all the same so That classifies the groups of order 12 and we're now going to look at the groups in a bit more detail And so what we're going to do is just look at the subgroups and see what happens Let's start with the easy cases for Z modulo 12 Z And it's very easy to write down its subgroups because the subgroups are just going to be indexed by Devices of 12 so we get subgroups of order one two four three six and 12 so the the letters of subgroups looks like this where I've just written for each subgroup. I've just written down down its order and and Z Modulo 2 Z times Z modulo 6 Z is not much harder to do Again, these groups are both a billion which makes it easy to work out their subgroups So it's got three subgroups of order two all contained in the subgroup of order four And it's got a subgroup of order three and this is contained in three subgroups of order six and This is contained in the subgroup of order 12 and each subgroup of order six contains Subgroup of order two, so we also get lines like that So that does the abelian ones Abelian groups tend not to be very exciting the non-Abelian groups are a little bit trickier to do So let's do rotations of a tetrahedron Or we can think of this as a semi direct product of The group of order four by the group of order two. Well, it's got three subgroups of order two and it's got a normal subgroup of order four It may not be immediately obvious what the subgroup normal subgroup of order four of rotations of the tetrahedron is but it consists of The three you see it's got three elements of order two which consists of rotating the tetrahedron like that Or like that or whatever the third one is and these form a group of order four in terms of permutations the group of order four contains the identity permutation and The three permutations of this form So these form a little normal subgroup of order four of the group of rotations of the tetrahedron Well, it's also got four subgroups of order three and that's about it really So it's lattice of subgroups looks like this where you can now see the only normal subgroups are This one this one and this one Now let's look at the binary dihedral group so this contains A subgroup of order three And the quotient by the group of order three is cyclic so it contains The only groups containing it to order six and order 12. It's also got a unique subgroup of order two However, it's got three subgroups of order four So it's lattice of subgroups ends up looking a bit like this And binary dihedral group. There's not really very much interesting to say about it And finally we'll do the group Dihedral group of order 12 and try and pick out its subgroups and this starts to get a bit complicated and first of all it's got Subgroup of order one and it's got seven subgroups of order two Because if we look at a hexagon It's got three conjugacy classes of subgroups because you remember we can Reflect in these three lines, which gives us three subgroups and we can reflect in These three lines Which gives us these three subgroups and then we can rotate by 180 degrees which gives us yet another subgroup of order order two It's also got a subgroup of order three consisting of just rotations and It's got three subgroups of order six So one of these subgroups Contains these three The other subgroup of order six Contains these three and these two subgroups are isomorphic to the symmetric group on three elements And it's got one subgroup of order six isomorphic to z modulo 6z Which contains the final Final element of order two Next we've got three subgroups of order four and You can see one of these subgroups of order four by drawing this orange rectangle here and just looking at the Automorphisms fixing this rectangle and there are three more rectangles you can draw So there are there are three of these and this subgroup of order four Contains one of these green infolutions and one of the blue ones and there are two other subgroups of order four similarly contain Groups like that one contains that and let's do that one containing that and These are all contained in a big group of order 12 Um So we can now pick out the normal subgroups Well any subgroup of index two is normal and the whole group is normal and the trivial group is normal and also this group must be normal because there's only one subgroup of order three and This group is normal because it's in the center. So the red ones are the normal subgroups and The remaining subgroups you can see they form Conjugacy classes of three subgroups So that pretty much does the groups of order 12 and Next lecture we will probably do automorphisms of cyclic groups and classify the groups of order 15