 Continue in our search for non-Abelian simple groups. We have lowered our list just to the numbers 12, 24, 30, 36, 42, 48, 46, and 60. We do know there is a non-Abelian simple group of order 60, that's A5, but are there any smaller than that? In this video, we're going to take a look at the number 56 and show why there cannot be a simple group of order 56 and then try to make some generalizations from that if we can. So 56, the factorization here is significant, 56 factors as 2 cubed times 7. For which if there was a unique Seeloff 7 subgroup, then that Seeloff 7 subgroup would be normal and therefore the group would not be simple. If I'm looking for a simple one though, then I have to assume we have a non-unique Seeloff 7 subgroup. So by Seeloff's third theorem, we know that the number of seven groups is going to divide 8, in particular it's going to divide 56, take away the 7, but it also has to be congruent to 1 mod 7, like so. So if we look at the divisors of 8, 1, 2, 4, and 8, 2, 4 don't work. We'd have to get, that is they're not 1 mod 7. We don't want 1 because that would be a unique one. So what we then have decided is that if we have, if we're looking for a simple group right here, it can't have a normal subgroup and therefore it has to have 8 Seeloff, 8 Seeloff 7 subgroups. Okay, now as the subgroups, each of these 7 subgroups has order 7, they're all going to be isomorphic to Z7 itself. So imagine we consider two of these Seeloff 7 subgroups right here, two different ones, let's be clear about that. Consider the intersection for a moment, right? These are groups of order 7, they do have the identity in common, but they're not going to have anything else. By Lagrange's theorem, the intersection of two subgroups, let me rephrase that. The order of a subgroup divides the order of the group itself. So if we take the intersection of P and Q, which itself is a subgroup, it has to divide 7. If it were 7, that means P and Q were the same group, can't be the case. So their intersection has to be one, it's trivial. So none of these Seeloff 7 subgroups contain anything in common except for the identity. So each of them, each of them is going to contain 7 minus 1, a.k.a. 6 elements of order 7 that none of the other subgroups have. And by the assumption that it's not simple, right? Excuse me, the assumption that this is not normal, the Seeloff 7 subgroup, we have to have 8 of them. So we end up with 8 times 6 mini elements of order 7, which gives us 48 elements total. So notice here that if we take these 48 elements of order 7 and we subtract them from the total group, there is 56 total, 56 takeaway 48 is now 8. So there are 8 elements left in the group that are not order 7. Well let's consider the Seeloff 2 subgroup for a moment. 56 is equal to 2 cubed times 7. So a Seeloff 2 subgroup is going to contain 8 elements. Okay? So honestly, and we know by Seeloff's first theorem that there is a Seeloff 2 subgroup. There has to be at least one. It contains 8 elements and none of those elements are of order 7. So these 8 remaining elements that aren't order 7 have to be the 8 elements of this Seeloff 2 subgroup. It contains all of them. So by forcing us to have multiple Seeloff 7 subgroups, it then forces that we have a unique, we have a unique Seeloff 2 subgroup. So if you're a group of order 56, you either have a unique Seeloff 7 subgroup or you have a unique Seeloff 2 subgroup. You have to have one or the other. You could have both. Maybe you're like the cyclic group of order 56, but you're going to have to have one or the other. So you either have a unique Seeloff 7 subgroup or a unique Seeloff 2 subgroup. A unique Seeloff subgroup is always normal and therefore no group of order 58 can be simple. It's not possible because your 2 group or your 7 Seeloff subgroup will be unique. I want you to be able to convince yourself that if we change the appropriate parts, this exact same argument applies to 12 as well. That is, if you consider the 3 subgroups, your options are going to be N3 equals 1 or 4. If you have 1, it's unique. Therefore, it's normal. You're not simple. So the other options, you have 4. But as you have 4 subgroups of order 3 that each share the identity, 4 times 3 minus 1, this is going to give you 4 times 2. This is 8 elements. 12 take away 8 leaves you 4 because there's 8 elements of order 3. So there's 4 elements which are not order 3, but the Seeloff 2 subgroup itself is size 4. So the 4 remaining elements have to belong to the Seeloff 4 subgroup. So you either have a unique subgroup of order 3 or you have a unique subgroup of order 4. In either case, those Seeloff subgroups are normal. So there are no simple groups of order 12. And so I want to take those off of our list. We took away 12, we took away 56, and our list just got shorter.