 Welcome back to our Lecture Serious Math 42-30, abstract algebra two for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. Lecture 10 is the second part of a two-part lecture, which we started obviously at number nine, for which we're trying to study non-Abelian simple groups. We made the observation that A5 is a non-Abelian simple group of order 60, and I made the claim that you can't get any other non-Abelian simple groups of a smaller order than A5. It's the smallest non-Abelian simple group. We were considering all numbers from one to 60, and in the last part of lecture nine, we ruled out everything except for the following on our list. 12, 18, 20, 24, 28, 30, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56, and then 60 of course is going to work. And so what we wanna do in this, in lecture 10, in multiple videos in lecture 10, is I want to rule out some of these remaining possibilities. So what we're gonna do right now is consider the number 28, okay? The factorization of the number plays a critical role here. 28 is two squared times seven. It's four times seven here. So by the Silov theorems here, if N7 is the number of Silov seven subgroups that G has if it's a group of order 28, then by the third theorem, this has to be a number that divides four, but also it has to divide four, but it also has to be one mod seven, which in this situation, you'll notice that four is smaller than seven, okay? And so the divisors of four are gonna be one, two, and four, one is one mod seven, two is not, because it's too small, right? It's less than seven, four is also not gonna work either. So what you see in the case of 27 is because four is so small, it doesn't have any divisors that are larger than seven. So in terms of reduction, modulus seven, the only one who could do it is one. And as such, every group of order 28 is gonna have to have a unique Silov seven subgroup, again, because the other factor is too small that it's that unique Silov seven subgroup is gonna have to be normal. And so this is not a simple group. So 28 is removed from our list. It is also true that we can remove 20 for the same reasoning here, right? If you have, if you consider a Silov five subgroup, your options are one, two, and five, excuse me, one, two, and four, but the only one that's congruent to one mod five would be one. So groups of order 20 have a unique Silov five subgroup. And by similar reasoning, we get that 44, which is four times 11, because four is too small, we get that there's gonna only be one 11 subgroup of 44. So that Silov subgroup necessarily has to be normal. So we can remove 20 from our list and we can remove 44 from our list as well. I also want to point your attention to, actually I think there's one more that's missing here. Aha, 52, you're on my list as well. I could see you hiding there. 52 is four times 13, 13 is way bigger than four. If 13, excuse me, if we need a number that's a divisor of four, one, two, and four again, that's congruent to one mod 13. Since this number is smaller than the other prime, we get that you can't have it. There's nothing else you could do here. So we can also move 52 from our list here. So I'm gonna leave it as an exercise to the viewer to prove the following statement right here, theorem. So imagine that N is a number that factors as P to the, P to the, what am I gonna call it? I'll use an N here. So we'll use a P to the M times a number K, such that, so P is prime, and we have that K is actually smaller than K is smaller than the prime itself, okay? So in that situation, I would then leave it to the viewer to prove, kind of mimicking what we're doing right here, prove that all groups of order N, all groups of order N, all groups of order N are in fact, not simple. And so I want you to be aware that this is the situation we're in so far. Like if you look at 28, this is our number K, this is our number P to the M, it's actually just P to the first, right? Seven is so much bigger than the rest of it, K, that there's just no other possibility. Same thing, five is bigger than four, 11's bigger than four, 13's bigger than four, okay? Now in this situation, and that's how we ruled out those numbers, but there's also the possibility that P could be repeated prime, like take the number 18, for example, 18 factors as two times three squared. The fact that three is bigger than everything else is gonna lead to a very similar situation. So I'm not gonna provide the details of this theorem. This is actually a student, this is a homework exercise for my student. It's a good exercise for the viewer here. But by proving this exercise right here, which is a generalization of what we're talking about right now, you can rule out the numbers 18. What else can we rule out? We can also rule out the number 50. 50 is two times five squared, five is bigger than two. We can rule out 54, right? 54 is two times three cubed, three is bigger than two, so we can get rid of that. So another number I also wanna take off our list is 45. Now 45 factors as three squared times five, for which this theorem doesn't apply to it directly because of course, five is not bigger than nine, nor do we have that three is bigger than five. So it doesn't apply directly, but this one's gonna be very similar to the examples we've done in this video right here. So I wanna leave this as an exercise as well. So as we continue on our search for non-Abelian simple groups, we're gonna take 45 off our list because again, those are gonna be left towards the viewer right here. Now the good news is we took a lot of things off our list in this video, but the bad news is we've gotten all of the easy ones now. The remaining ones, most of which are multiples of 12 will take a little bit more effort and so we'll approach some of those in the next few videos.