 This lecture is part of an online mathematics course on group theory and will be about extensions of groups. So to motivate this let's try to classify groups G of order 8. So we've done orders 1 to 7 in earlier lectures so this is the next case. Now the elements of G have order dividing 8 so they all have order 1, 2, 4 or 8 and we want to show there's an element of order 4. Well there isn't always. Suppose there are no elements of order 4 or 8 then all elements have order 2 and if all elements have order 2 well we've already classified groups such that all elements of order 1 or 2 and we saw that just products of groups of order 2 so this means G is equal to Z over 2 Z times Z over 2 Z times Z over 2 Z. So we may as well assume G has an element little G of order 4 it is order 8 we can just square it and get an element of order 4. So we look at the subgroup H which is all powers of G so H is order 4 and we notice H has index 2 so is normal in G. So let's draw a picture what we've got we've got an element 1 we've got a group Z over 4 Z which is isomorphic to H and this is a subgroup of G and the quotient is a group of order 2 and the only possibility is Z modulo 2 Z so we get an exact sequence like this this means Z over 4 Z is a subgroup of G or more precisely isomorphic to a subgroup of G and the quotient of G by this group is isomorphic to the group of order 2. So this is a typical extension problem an extension problem means we have an exact sequence like this where we know these two groups and want to figure out what this group is so we say G if we have a sequence 1 goes to A goes to G goes to B goes to 1 we say that G is is an extension of B by A actually we sometimes say G is an extension of A by B because it's almost impossible to remember which way around A and B are supposed to go but whatever so we now have this problem how do we classify all possible extensions of Z modulo 2 by Z modulo 4 so let's take a look at the possibilities so let's say that A contains an element A little A with A to the 4 equals 1 I've changed it from G to A so we can remember it's an A now we pick an element B in G mapping to an element of order 2 in B this doesn't mean the element B of G has ordered 2 it means its image in the group B has ordered 2 so let's look at the possibilities we know that B normalizes A so this means that B to the minus 1 AB must be a generator of the group A because A is a generator so it must be A or A cubed which is the same as A to the minus 1 because these are the only two elements in A of order 4 and B squared must be an element of A as it has image 1 in B so B squared is equal to 1 A A squared or A cubed now if B is equal to A cubed we just change A to its inverse which is A cubed so we can assume that B squared is equal to 1 A or A squared so here we've got three possibilities and here we've got two possibilities and finally of course we should remember we've got this equation A to the 4 equals 1 so here we've got three equations for A and B and there are all together six possibilities for what these three equations are so let's write them all out and see what we get so first of all we can have B to the minus 1 AB is equal to A or B to the minus 1 AB is equal to A to the minus 1 so I'm going to draw a big rectangle giving us all the possibilities and here we have B squared equals 1 here we have B squared equals A squared A and here we have B squared equals A squared so let's see what we get here well if B squared equals 1 this means that B actually generates a subgroup of order two so we get a semi direct product so if A and B commute we just get the semi the semi direct product is then a product so we get Z over 2 Z times Z over 4 Z as the only possibility so this is the group A and this is the group B if B AB to the minus 1 is A to the minus 1 we get the semi direct product Z over 4 Z semi direct product Z over 2 Z so the group has a normal subgroup of order 4 and it also has a group of order 2 acting non-trivially on this and it's pretty easy to see what we get we just get the group of all symmetries of a square so the element A is going to be just rotation so A does this so this is the element A with A to the 4 equals 1 and the element B well we can just take it to flip the square like this so B exchanges the two sides of the square with B squared equals 1 and you can check that B AB to the minus 1 is indeed A to the minus 1 so that gives the second possibility this case is really easy to deal with because it doesn't exist the point is that if B squared equals A this implies B actually commutes with A because A is just B squared so we can't B AB equals A to the minus 1 this case is quite easy because B commutes with A and B squared equals A so we get B to the 8 is equal to 1 so we just get the cyclic group of order 8 generated by B and we notice that A is the group generated by B squared so it consists of 1 B squared B to the 4 B to the 6 so that's again a group we've had before this case here turns out to be the case we've already got before because what we can do is if we put B a better not call it big B let's put C equals AB then we see that C squared is now equal to 1 and C to the minus 1 A C equals A and now we see that this case here is really the same as this case here so these are actually the same um we can in other words we've just picked it turns out that when we picked this element B we actually made a bad choice of element and we could have chosen a better element AB and would have got this presentation here so although this extension at first sight looked like a non-split extension that wasn't a semi-direct product we see that is in fact a disguised split extension so we say an extension is split if it's really a semi-direct product and this example shows it can be a bit tricky to tell whether an extension is split or not finally we get to this case here which is the most interesting one where we have two elements B to the four equals one because it's equal to A squared A to the four equals one and B to the minus one AB equals A to the minus one and the question the first question is is there any group of order eight um generated by elements with these properties you see it's not obvious because sometimes if we write down um properties of these elements there might be no group that satisfies them well this will in fact turn out to be the quaternion group denoted by Q of eight um this by the way I forgot to say was the dihedral group which we will be studying later and is called D8 so the first thing we've got to do is to show that that actually is a quaternion group so let's write it down explicitly well we can just take these elements we take A is equal to I minus I naught naught so this is a two by two matrix with I squared equals minus one and we take B to be the matrix naught one minus one naught and we can then check that A to the four equals one B to the four equals one and towards the relation B to the minus one AB equals A to the minus one so we found two matrices which satisfy these conditions we'd better check that these do actually generate a group of order eight because you see in this one we we could actually find a group satisfying these conditions but it would only turn out to a order of four not eight it kind of collapses a bit um well what we do is we look at the element C which is naught I naught and we look at the identity matrix which is one naught naught minus one and then G is going to be the eight elements plus or minus A plus or minus B plus or minus C and plus or minus one naught one and we can check that G is a group of order eight um so the quaternion group um really does exist um by the way the usual notation isn't to use AB and C for these elements but to use elements I J and K or sometimes big little i and little j and little k and then um and we denote um this element by one and minus one naught naught minus one by minus one and then we find I J and K satisfies these relations I squared equals J squared equals K squared equals I J K equals minus one and these are rather famous because they are there's they're the relations for the ring of quaternions that was discovered by Hamilton in the 19th century and he is rumoured to have been so pleased by this discovery that he carved these on the bridge somewhere um so um next lecture we're going to study the group of quaternions in more detail