 This lecture is part of an online math course on group theory and will be about outer automorphisms, especially outer automorphisms of symmetric groups. So we'll start by recalling what automorphisms of a group look like. First of all, we have inner automorphisms. The inner automorphisms are automorphisms that take x to gxg to the minus one for some g in the group g. And this is trivial if g is in the centre of g. So this just means centre because that's just the same as saying x is equal to gxg to the minus one, so g would have to commute with everything. So the group of inner automorphisms is just the quotient of g by the centre of g. And there may or may not be other automorphisms you can't get like this. And the quotient of all automorphisms by inner automorphisms is called the outer automorphism group. So we get an exact sequence. The centre of g is contained in g and this maps to all automorphisms of g and this maps to the outer automorphisms of g and this maps on to one. So this quotient here is just the inner automorphisms of g. And what we want to do is to calculate the group of all automorphisms of g and as we generally know what g in the centre are the problem is to find this outer automorphism group. So for a simple example if g is say z over 2z squared the centre of g is equal to g so the group of inner automorphisms is trivial but the outer automorphism group of g which is equal to the automorphism group of g is equal to all the automorphisms of this group which is just the symmetric group s3 because you can just permute for three non-identity elements. Now for most groups the automorphisms of a group g are usually obvious whatever this means. It means they're usually easy to find for example what does obvious mean? Suppose we take the group the projective special linear group over a finite field. So this means determinant one matrices that are n by n projective means you're quotient out by the centre and this is the finite field of order q and then we have the following obvious automorphisms we can have inner automorphisms. We can have automorphisms we can conjugate by elements of gln of fq. So this includes the inner automorphisms but it also includes some other automorphisms for instance we could conjugate by this matrix here and this might give us a new automorphism. Thirdly we can have field automorphisms so if we've got a map from the field q that will give an automorphism of this group. Finally we can just map g to the transpose of its inverse and you can easily check that that also gives an automorphism of the group and the full group of automorphisms of this group is generated by these obvious ones and for most simple groups something similar happens you write down the obvious automorphisms you can think of and that gives all of them. So for the alternating group a n the obvious automorphisms is just the group the symmetric group s n which has a n as a normal subgroup index two. So the outer automorphism and for most n if n is not equal to 6 this gives all automorphisms of a n so the outer automorphism group of a n is equal to z over 2z for n not equal to 6. A6 is a bit different in this case we will see that there are some extra automorphisms in fact the outer automorphism group of a6 will turn out to be z over 2z times z over 2z. So this is a very unusual example of an unexpected outer automorphism of a simple group. These outer automorphisms turn up in some of the sporadic groups for example there's a sporadic mature group called m11 acting on 11 points and the subgroup fixing a point is order 720 and is made of a6 together with a sort of none one of these non-standard outer automorphisms of it. So let's see why a6 has outer automorphisms well it'll be slightly simpler to do outer automorphisms of s6 the symmetric group. The symmetric group is a little bit easier to handle than the alternating group and if you've got an outer automorphism group of s6 that's going to give you an outer automorphism of its normal subgroup a6 so that doesn't really matter. So let's see why s6 has an outer automorphism. Well we start by observing the s5 as a subgroup of index 6 so this would be order 20 because s5 has order 120 and that's s5 not s6 I'm working with s5 for the moment and this subgroup is easy it's just the Frobenius group of order 20 that we discussed earlier that it consists of the automorphisms of the finite field of order 5 of the form x goes to ax plus b where b is in the field of 5 elements and a is in the field of 5 elements and a is not equal 0. So this group here obviously is order 20 because there are 5 possibilities here and 4 possibilities there. Well if s5 has a subgroup of index 6 of a group means the group acts on the set of 6 cosets of that group so we get this gives a homomorphism s5 to s6 so s5 acts transitively on a set of 6 elements well you may think that's not a big deal it's obvious that s6 contains subgroups s5 because it contains the subgroup s5 fix in your point but this is a sort of unusual s5 and s6 so this gives an s5 containing s6 not fixing a point so you see s6 has 6 obvious subgroups s5 fixing points but what we're saying is there are some more copies of s5 not fixing a point in fact s6 has 6 plus 6 subgroups s5 normally sn it usually has just n subgroups sn minus 1 but the group s6 is funny because it contains more of these subgroups than you would expect well in particular s6 has a transitive s5 subgroup of index 6 now just as s5 having a subgroup of index 6 meant that we got a homomorphism from s5 to s6 if we've got a subgroup of index 6 of s6 this gives a homomorphism from s6 to s6 acting on the 6 cosets of this subgroup and if you think about it a bit you can see this isn't actually in an automorphism because the subgroup fixing a point of this action is one of these funny s5s rather than an s5 fixing a point so this gives an abstract argument showing that s6 has an outer automorphism well what we would like to do is write down an outer automorphism explicitly so I mean the trouble with this abstract argument is it's hard to get a picture of what this outer automorphism is actually doing so let's recall that s6 is generated by 5 elements of order 2 with 5 transpositions satisfying the following relations first of all each of these elements if I call these a1, a2, a3, a4 and a5 each of them satisfies ai squared is equal to 1 and aiaj squared equals 1 if i not joined to j and aiaj cubed equals 1 if i is joined to j remember we had this coxida diagram giving presentation for s6 and we're going to find an automorphism by finding 5 other elements b1, b2, b3, b4, b5 with the same relations so we want b1 squared to have ordered 2 and so on and then we can get a homomorphism of s6 to s6 by mapping ai to bi so this homomorphism will be given like that and this homomorphism will not be inner because the bi will not be transpositions so you recall a transposition just exchanges 2 elements and any inner automorphism has to take transpositions to transpositions so each bi will be of the form it will be a product of 3 different transpositions so we should think about what these elements look like so if we've got 6 points and we've got a product of 3 transpositions oops sorry I didn't mean to join those let me start again if we've got a product of 3 transpositions it might do something like this and if we've got another product of 3 transpositions it might be related like that alternatively it might be related like this so here it might do this and it might exchange these two and do the same as that so it might have one transposition in common with the blue one in which case the others must be arranged somehow like this or it might have none in common in which case you can check it must be arranged like this and in this case the product has ordered 3 so the product has ordered 3 here and 2 here or of course the 2 transpositions might be the same in which case the 2 elements might be the same in which case their product just has ordered 1 so what we want to do is to find 5 of these involutions of this shape with the following property that any 2 that are joined by a line in a diagram like this are related like that otherwise they're related like that so let's see if we can do that so my first one is going to be 1, 2, 3, 4, 5, 6 and now my second one must be a product of 3 transpositions none of which is these ones so I can try 2, 3, 3, 4, 5, 6, 1 the next one must have no transpositions in common with this but 1 transposition in common with this so I could take 1, 2 so it's in common with that and then 3, 5, 4, 6 the next one must have a transposition in common with each of these but none in common with these and by looking around a bit I can take 3, 4 which is that one times 6, 1 which is that one times 2, 5 and now it has nothing in common with that the final one has to have a transposition from each of these and none in common with that so I can take 1, 2, 4, 5, 6, 3 and now I can write down my explicit automorphism of S6 I'm going to take 1 transposition 1, 2 to here 2, 3 to here 3, 4 to here 4, 5 to here and 5, 6 to here so this is an explicit out of automorphism of a group S3 if you want you can draw little pictures of these so if I do my 6 points like this then I'm going to colour all the transpositions so I might call this number these 1, 2, 3, 4, 5, 6 and then the yellow one is going to look like this but its image will look like this so 1, 2, 3, 4, 5, 6 so its image looks like this and then 2, 3 might be this one here so 2, 3 looks like that and its image here will be the operation that takes 2 to 3, 4 to 5 and 6 to 1 so we go on like this 3, 4 looks like this and it goes 1, 2, 3, 5, 4, 6 next one exchanges 4 and 5 and goes to 3, 4, 6, 1, 2, 5 and the final one if I can find this in colour there it is exchanges 5 and 6 and exchanges 1 and 2 4 and 5, 6 and 3 so what this means is the transposition exchanging these two is mapped under the outer automorphism to the element exchanging these two, these two and these two so this is an actual picture of an outer automorphism of S6 well now you can ask do other symmetric groups have other outer automorphisms and the answer is no they don't so let's look at the automorphism group of S6 sorry not of S6 of SN for other values of N and first of all we can notice that suppose we have an outer automorphism sigma so suppose sigma takes transpositions to transpositions then sigma is inner and the idea is we can reconstruct the points from transpositions we can reconstruct the endpoints from the transpositions at least if N is not equal to 3 we can do it if N equals 3 as well but that takes a bit more work and you see this as follows suppose we've got these endpoints here wherever they are what you can do is you can look at all the transpositions fixing one point and this is a cluster of N-1 transpositions such that the product of any 2 has order 3 and if you think about it a bit you'll see that the only way to get a product of N-1 transpositions such that the product of any 2 has order 3 is to either pick a point like this and join it to all the others or if N equals 3 you can do this so N equals 3 is a bit of a special case so it doesn't quite work for N equals 3 so if N is not equal to 3 you can reconstruct the points by knowing the transpositions so any automorphism that takes transpositions to transpositions also acts on the points on the endpoints that NS acts on and must therefore be an automorphism of SN so the question is are there automorphisms that don't preserve transpositions so automorphism is not preserving transpositions you remember the outer automorphism we found for S6 did not preserve transpositions because it took each transposition to a product of 3 transpositions well let's take a look at what the other automorphisms the other involutions look like so let's look at S1, S2, S3, S4, S5 and S6 and we can ask how many transpositions are there well the number of transpositions is N, N minus 1 over 2 so the number is going to be 0, 1, 3, 6, 10, 15, 21 and 28 what about products of two transpositions well here the number is N, N minus 1, N minus 2, N minus 3 because we can pick these four points but then we've got to divide by 2 times 2 times 2 factorial because we can swap each pair of points and permute these so the number of these we get is there are none here there are none here there are 3 for S4, 15 for S5, 45, 105, 210 what about products of 3 transpositions so again we do N minus 1 up to N minus 5 and we divide it by 2 times 2 times 2 times 3 factorial and the number we get is 0, 0, 0, 0 until we get to 6 and then we have 15 here 105 and then we get some large number 4, 20 I believe and now you notice we have this funny fact that the number of transpositions is the same as the number of this sort of involution and obviously if an automorphism takes transpositions to some other conjugacy class these conjugacy classes must have the same number of elements and this happens for S6 on the other hand if we take more transpositions and ask how many there are well we just get 0, 0, 0, 0, 0, 105 and you see S8 there's no other conjugacy class of 28 transpositions there are no outer automorphisms of S8 and similarly for S7 or S5 or S4 or S3 the outer automorphism of S6 depends on this funny coincidence that these two numbers are the same and the point is these numbers are always bigger than n times n minus 1 over 2 if n is greater than 6 that's quite easy to check because we've just got this formula for it you can see the number of them grows until it reaches some point and then it starts decreasing again but even when you get right to the end the number is always bigger than this number here except when s e n equals 6 that's quite an easy calculation so for example if n is even the number is going to be n n minus 1 down to times 1 divided by n over 2 factorial times 2 to the n over 2 and it's an easy estimate that I'm feeling too lazy to do to show that this is bigger than n times n minus 1 over 2 if n is greater than 6 so that shows that the outer automorphism of S6 has order at most 2 and we've shown it as order at least 2 so it has ordered 2 and it shows that the outer automorphism groups of all other symmetric groups are just one so that completes the description of the outer automorphism groups of S6 next lecture we're going to stop looking at finite groups quite so much and discuss free groups