 In this video we're going to continue our hunt for non-Abelian simple groups. We know there is a non-Abelian simple group of order 60. It's in fact a five, the alternating group. And we're trying to argue that there is not one smaller than that, no non-Abelian simple group of course. So in previous videos we've cut our list down to the possible orders of 24, 30, 36, 42, 48 and 60. We can't remove 60 because we know it works, but we're going to shave off a few more in this lecture and in particular by the end of this video we're going to take off 30 and 42. All right, so let's start with 30. 30 is different than any of the other ones we've considered so far because 30 is the product of three distinct primes, two, three and five. There's no repetition, but we have three distinct primes. So I want us to consider what are the possibilities here? So if we think of a Silov five subgroup, how many are there? By Silov's third theorem, we know that this has to divide six because six is the five prime part. That is if we take away the factors of five, what's left over six, two times three. Silov's third theorem says the number of five subgroups will divide this number, so there's six. We also know that in five it's congruent to one mod five. Like so, so we have to look at the divisors of six, one, two, three and six. Which of those are divisive, which of those are one mod five, not two and three, you're left with one and six. Now, if there was a unique Silov five subgroup, it would be normal and therefore that group would not be simple. So we don't have to worry about that case. So we're then gonna consider, okay, what if we have exactly five, excuse me, what if we have six Silov five subgroups? All right, so they're not unique, so we get six of them, excuse me. Now, because they're each order five, each of these Silov subgroups has to be order five. This, the Silov five subgroups have to be Z five itself, which is gonna contain the identity, right? The identity, and then you're gonna have four elements of order five. It's a cyclic group after all, something like this. Then when you consider, if you had two different Silov five subgroups, like if P and Q are Silov five subgroups of G, then because of order considerations using Lagrange's theorem, you're gonna get that P intersect Q is equal to one. That's the only possible order of their intersection. So these subgroups only contain the identity together. And so each Silov five subgroup contributes five elements of order five to the pot here. So if I take six, the number of subgroups we have, five minus one, the number of elements of order five there, you get six times four, which is 24. We have 24 elements of the group, which are order five. Now we started with a group of order 30, right? 30 take away 24 is equal to six. We have then six remaining elements of the group, which are not elements of order five. One of those elements is the identity. Some of them have to be group, have to be order two. Some of them have to be order three. So so far this argument is very similar to what we did when we considered 56. But at this moment, we're not quite there yet because we don't know the group necessarily has a subgroup of order six. So we have to still consider what's happening here. Now let's consider the three subgroups, the Silov three subgroups. If there is a unique Silov three subgroup, it's necessarily normal. So therefore we're not simple. So okay, can we have non-normal Silov three subgroups? We'd have to have multiple ones. Again, by using Silov's third theorem, we get that in three, it has to divide, in this case, 10, right? All right, well, that's kind of interesting there. So because where did 10 come from? Well, we took 30 divided by three has to be 10, but it also has to be congruent to one mod three. So what are our possibilities? One would work, two doesn't work, five, no, doesn't work. Oh, but 10, 10 could work, right? 10 is nine plus one. So if there was only one three subgroup, it would be a normal subgroup, we're not normal. So we have to assume we have 10, we have to assume we have 10 subgroups of order three. But then the same argument applies. Each of these Silov subgroups has to be order three. I mean, they're order three and therefore they have to be cyclic of order three. They will contain the identity, some other element, and then it's square. Both of these elements, of course, are elements of order three and the intersection of any two Silov three subgroups is gonna be trivial. So between the 10 subgroups, each of them will contribute two elements of order three. So that gives us 20 elements total, but we only have six elements left. So there's a deficit of 14 elements and I haven't even considered the elements of order two yet, the Silov two subgroup. So we've already got a contradiction that we can't have multiple Silov five subgroups and multiple Silov three subgroups. That requires too many elements here. In which case this then tells us that every group of order 30 must have a unique Silov five subgroup or a unique Silov three subgroup. Therefore, in either case, it can't be simple because one of the Silov subgroups will then be normal. I want you to convince yourself that the same arguments could be adapted for the number 42, which is two times three times seven, which we try to provide the details very quickly here. The idea in this situation is even better if you have n equals seven, right? This has to, excuse me, divide six, but also we need that n seven is congruent to one mod seven. So it's actually easier than 30, right? Because divisors six are one, two, three and six. The only one that's congruent to one mod seven is one. So if you're a group of order 42, then you have to have a unique Silov seven subgroup. And so I'm gonna leave it as an exercise to the viewer here to strengthen this pattern here. If you take n to equal p times q times r, where p, q and r are three distinct primes, then we could say that no group, no group of order n is simple. Oftentimes you get something like the following where maybe one of the primes is too big so that there's no opportunity to have multiple Silov r groups in that situation, like with 42, but it could also be the case where you have 30 where it's like, okay, yeah, this other, where's my factorization of 30? There you are. It could be that when considering five, the other number is big enough to have multiple. And then when you look at a different prime like three, their product is big enough to have multiples. But when you start considering all of them, you're gonna get too many elements in the end. So 30 and 42 are off our list and we'll consider the remaining ones in our next video.