 Welcome back everyone our hunt for non-Abelian simple groups is almost out of close We know that a five is a non-Abelian simple group of order 60 And we claim that that's in fact the smallest order for which you can have a non-Abelian simple group We know that 60 is gonna work. There are three other contenders though that we have to throw out We've done every number from one through 60 and now we're left with the dreaded multiples of 12 We already took care of 12 mind you, but we're now left with 24 36 and 48 multiples of 12 are problematic for the following reason 12 just by itself Factors as two squared times three. So if you're looking at seal off two subgroups, you could have one or three Those so threes a possibility, but when you're looking at seal off three subgroups, you could have one two or four Remember this number has to be one mod three so four works there So when you look at 12 the way that two and three interact with each other It actually there's possibility or you can have multiple seal off Subgroups and therefore they're not necessarily normal So we have to look for other subgroups that are going to be guaranteed to be normal Of course if other primes come into play like five or seven you can interact with those But you'll notice that 24 36 and 48 they all have they all have the form Where we have a power of two We have a power of three and in particular we have at least two Two's and so thus getting a multiple 12. These ones are a little bit harder because of those observations But in this video, we're gonna rule out 24 36 and 48. Let's start with the smallest of them 24 24 factors as two cubes times three So if we start to try to count how many seal off subgroups could we have well like I said a moment ago with regard to two We could have one or three if you have one seal off two subgroup. It's normal We want to find simple groups. So we're not going to do that So you could then get three So we're gonna have Three seal off two subgroups Okay, so that's what this observation right here is now if we were considering the three groups, right? because we have we have eight your possibilities are one two Four and eight which of these numbers are going to be one mod three Well, you get one and four if you had a unique seal off subgroup of or of just one of them Then that would be normal. So really we're in the situation where we're considering we have three Subgroups of order two and excuse me of order eight and we have four subgroups of order three and in past videos You try to count elements and then show that oh you either counted too many elements or you have to have unique seal off subgroup 24 gets a little bit more confused in that regard. So we're gonna have to take a different approach to that Because after all one of the things hard about this one is that if you look at if you look at the two the seal off Two subgroups. What type of groups are there of order eight? Sure. There's some Abelian groups, you know z eight z four crosses z two The elementary Abelian group z two cubed, right? It could be one of those But you also have some non Abelian groups now you have like the dihedral group you have the quaternion group and so Which of them which one is it right? There's five possibilities just for those And you know if you look at D4 for example, it has five elements of order two It's got two elements of order four Q8 has six elements of order four one element for two Z two cubed has seven elements of order two z eight of course has two elements actually four elements of order eight and Two elements of order four is that right? You know and so it's like if we're trying to count elements of order two four eight It gets a lot harder because there's so many possibilities So we can't rely necessarily on the isomorphism type of the seal off to subgroup Which we've often done with our past examples because at least one of the seal off subgroups was cyclic I mean in this situation Because of the divisor three here We do know that the seal off three subgroups are gonna be cyclic of order three So we could try to count elements there like okay We have four of them, right? So there's gonna be four Subgroups they intersect trivially so they each offer an element of they offer three elements excuse me two elements of order three So we end up with four times two which is eight. So you have eight elements there but When it comes to the remaining ones, you know 24 take away eight is 12 You you might be like how can you have multiple seal off two subgroups if they're order eight? Well, their intersection could actually be quite large So it's a little harder counting elements, so I want to offer an alternative approach to dealing with 24 So what we're gonna do here is we are gonna take two distinct seal off two subgroups their groups of order eight I want you to consider their intersections. What could there? What could they be? So at a very important formula that we've proven in after down to our one is the following if I take the set HK which remember HK this is just the set where we take all the possible products of some h times some k Where h is in h and k is in k like so This is just a subset of the group. There are some situations for which this itself is a group But we don't necessarily know if that's happening here. We did prove that this set always Satisfies this relationship here that the card Nality of the set is equal to the fraction Where we take the order of the subgroup h times it by the order of the subgroup k and divide that by the order of The intersection of h and k so what we know is the following h and k is both Seal off to subgroups. They have order 8 so you get 8 times 8 which is 64 Okay, so what are the possibilities? Well HK is a subgroup that divide it's a subgroup of h So the possibilities for the order of h intersect k you're gonna get one two Four and eight well, they're distinct so it can't be eight Right because if h intersect k had order eight then it would be a to also be k that would make them equal That's not the case If your order was one right then we get 64 divided by one But this is this is sitting inside a group of order 24. That's too many elements. So you can't get one Right and if you did two then 64 divided by two is 32 which is Still bigger than 24. That's too many elements. So if the order of the intersection was less than two That'd be too many elements. So it turns out the intersection of any two seal off two subgroups has to be four in that situation and So Start this again in three two one As such We know Something about the size here. Okay, so this intersection is exactly four and so then when we consider the size of HK Well, sure, you're gonna get 64 divided by four. That's equal to 16. That does fit inside of 24 So that's not a problem here. I'm gonna I'm gonna change gears a little bit I'm gonna then look at h intersect k as a subgroup of h h has order eight H intersect k has order four. So the index is two The same is also true if you look at h intersect k as a subgroup of k its index in k is equal to two Now we've proven previously that every group of index two So I should say every subgroup of index two is normal inside of that group Therefore eights intersect k is a normal subgroup of h and it's also a normal subgroup of k I'm not saying h intersect k is a normal subgroup of g But it is a normal subgroup in h. It's a normal subgroup in k. And so now I want to look at the normalizer I want to look at the normalizer of h intersect k inside of g right because what's the normalizer again? the normalizer is gonna be the set of all elements g inside of g such that g H intersect k is equal to h intersect k g in other words The normalizer is a set of elements of the group that normalizes the the group in question right here the subgroup in question In particular there are some things we always know which are inside the normalizer the normalizer always includes the group in question It always includes The center of the group because that centralizes everything does it normalizes everything And so typically if there are if there are central elements that means the normalizer grew It's not just h intersect k But in general we don't actually know what the centralizer of this group looks like at the moment But what we can say is that the normalizer is going to be the largest It's the largest subgroup It's the largest subgroup of g for which h intersect k is normal inside of it What that means for us here is that if we find a subgroup of g because these are both subgroups of g If we find a subgroup of g for which h intersect k is normal in that subgroup That means that subgroup belongs to the normalizer So the normalizer of h intersect k contains h it contains k Therefore it takes it also contains any combination of elements from h and k in particular the set H intersect k is inside the normalizer as we observed earlier the The set because we don't even know if it's a subgroup the set h intersect k contain 16 elements All right, which means that the order of the subgroup n of h intersect k is at least 16 So the order of n h intersect k here It's at least 16 But we also know that this order by Lagrange's theorem because this is a subgroup of g This order divides the order of g which is 24. All right, so we need a divisor of 24. That's at least 16 right and The only number that's gonna work by Lagrange's theorem is 24 So we actually get that the order of The normalizer is 24 that means since the normalizer is the order 24 That means the normalizer of h intersect k is in fact g Which tells us that interest h intersect g is normal inside of g So we found a subgroup of order 4 that's normal and g So while we couldn't guarantee the seal off two subgroups were normal the intersection of any two seal off Two subgroups isn't necessarily normal inside this group So this exercise illustrates the power of taking normalizers using normalizers and Intersections of seal off subgroups can be useful to find a normal subgroup therefore. There is no group of order 24, which is simple So 24 is now off our list. Let's let's move and look at the last two here I guess I should have 60 here. We know 60 is gonna work So I'm really considering what are the numbers that don't work, but I'll put 60 back on the list Sure, we know 60 is gonna work. What about 36 or 48? Let's now tackle 48 Okay 48 has the form two to the fourth times three So the same thing kind of happens here. What are our possibilities? We get that inch in sub two How many how many two subgroups do we have? It's either one or three if it's one then We're not simple because we have a normal seal off subgroup So it's gotta be three if we're looking for a simple one if you're looking for three There's actually a couple more possibilities now devisors of 16 or two four eight One two four eight and sixteen Two doesn't work eight doesn't work, but you could have four you could have sixteen having sixteen is a lot You're gonna figure out that doesn't work, but four is kind of in this Goldilocks zone. It's not too hot It's not too cold. It's not too big. It's not too small We could perhaps get away with four seal off three subgroups and so by counting just elements It's gonna get problematic Now we could look at the normalizer at the intersection of seal off Subgroups just like we did with 24 that same argument would work right here by changing the appropriate parts And I'll leave it as an exercise to the viewer to do exactly that What I want to do right now is actually provide an alternative argument That would then that's useful, right? So there's you know, it's not just a one a one horse One trick pony. I think that the phrase goes right The not one maneuver takes care of all of these these ad hoc arguments depend on the factorization So while the previous argument does work in 48 I want to give us another argument to show why there cannot be a simple group of order 48 Like I mentioned just above when we consider the seal off two subgroups There's either one or three of them by the third theorem of seal off if it were equal to one Then this problem would be trivial because you'd have a unique normal seal off two subgroup So clearly we have to consider the possibility where the number of seal off two subgroups is three Okay, now we've used seal off threes We've used the third seal off theorem so much when we do these calculations We also are using the first one a lot because we that gives us the existence of p-subgroups in particular seal off p-subgroups Number two often gets neglected here But I should mention we're using all the time because we have only one seal off p-subgroup That has to be normal. That's a consequence of the second theorem because they're conjugates of each other. Okay So what I want to do is now consider that conjugation action and We're going to treat it much more importantly right now because we have three seal off two subgroups G acts on the Seal off two subgroups by conjugation that seal off second theorem But that that action has to be non trivial because they're all conjugates of each other And so if we consider the conjugation action of G on the set of seal off two subgroups call that X for a moment this is a group action and by Applying the strong Kayleigh's theorem that actually provides for us a homomorphism from the group G Into the symmetric group on X But there's three seal off three subgroups here And so this sub this symmetric group sx is essentially just s3 up to isomorphism Now this map right here cannot be trivial, right? The orbit structure cannot just be like the first subgroup the second subgroup the third subgroup That's an option Not for this one though because they they have to be all together the orbit for us the seal off subgroups is all three of them together Like so so we don't have a trivial homomorphism We don't have a trivial action. This tells us that the kernel of this homomorphism is not all of G Now let me let me kind of give you a hint on where we're going here If you have a group homomorphism its kernel is always a normal subgroup I'm trying to show that we're not simple every group has normal subgroups Everyone because you have the trivial subgroup is normal and the whole group is normal So that when you're trying to show that something's not simple you have to produce a non-trivial Proper normal subgroup aha, so I'm gonna try to argue that the kernel of this Homomorphism is a non-trivial proper normal subgroup. So it's proper, right? It's not all of G Why is it not trivial? Okay. Well, like I said a moment ago sx here is really just s3 uptie Symmorphism in particular the order of this group is six three factorial and So as we map as we map G into a group of order six the image of this group Has to divide six So in particular this can't be a one-to-one map if the kernel was trivial That would mean the map is one-to-one in which case then there's an isomorphic copy of G inside of sx But we can't isomorphically put a group of 48 inside of a group of order six. It's too small The co-domain is therefore some things have to map to the same place the pigeonhole principle here And as such in particular the identity is gonna have there's gonna be more than one element that maps onto the identity The kernel of this map is in fact Tribule excuse me. It's non-trivial. So the kernel is not everything. It's not nothing It's something and so therefore the kernel of this homomorphism is a non-trivial proper normal subgroup and hence G Is not simple. I really like this argument using the group action here The conjugation action of the group on its seal off in this case seal off two subgroups Because you can play around with this when the index is small like, you know The symmetric group in general gets really really really big, right? But if you look at small symmetric groups like s3 has only order 6 s4 has only S4 only has order 24. So if the number of seal off subgroups is small You can play around with the kernel of this conjugation action and you can make something work with that as well So I leave it as a I leave it as an exercise to the viewer here to convince yourself You can do the same thing with 36 that if you look at the conjugation action The kernel of that conjugation action has to be a proper non-trivial subgroup, which is normal And therefore there are no subgroups of order 36 either. So that then finishes our list, right? using these conjugation actions or by using normalizers we then can take out 24 36 and 48 and so that leaves that the only the only Order left for a non-abelian simple group. It's gonna be 60 I should say we're only we're looking at possible orders less than 60 There of course are not a billion simple groups of order larger than 60 take a 6 for example It's order is larger than 60 But this then proves that you cannot have a non-abelian simple group of order less than 60 for which a 5 then fills the bill the You know the nature of course a vacuum, right? There's no argument that prevents 60 from working because it actually works. There is someone who took on 60 Now mentally there's no other Non-abelian simple group of order 60 other than a 5, but that's not an argument We're gonna provide in this video So this was a pretty fun journey we hunted through and looked at every possible order now We can prove that there is no non-abelian simple group of order less than 60 60 being obtained by the alternating group before we end this video I want to present to you the following super awesome theorem that is way beyond our ability to prove With regard to what we've developed right now in this lecture series here So this is often referred to as the odd order theorem. It's due to Fett and Thompson I hope I spelled their names correctly if not tell me in the comments Anyways, it tells us that every finite simple non-abelian group must have even order things like 60 Which is an even number. You can't have an odd ordered Simple group and be non-abelian the only ones available are the cyclic ones of prime order and so the fact that Even orders are a lot harder than odds. It's sort of a very awesome very curious thing I'm presenting it on the screen right now because it's relevant to our discussion of simple non-abelian groups that we considered in the previous Well in this video and all the previous ones as well I'm presenting it for its sake of appeal But I want you to be aware that the proof of this goes way beyond the techniques We've been using with the seal off theorems and and normalizers and group actions and such It's way beyond the scope of where we are right now. So if you're a student trying to prove a problem about finite simple non-abelian groups They're not existence or something like in a homework problem or an exam I would highly encourage you not to use this theorem because your professor probably won't think it's very acceptable, right? I mean since it's if I'm your professor, then I wouldn't let you do it because we haven't earned this one yet Our mathematical theory hasn't developed this far yet. It's a cool theorem. Just want to throw it out to there just spaz you but Be aware it's off limits when it comes to proving, you know, homework questions and like a undergraduate abstract underclass or like a first-year Graduate level abstract underclass because it's just it's it's awesome, but it's just too hard for we are where we are right now So with that, you know, thanks to make it at the end of this journey for our hunt for Non-Abelian simple groups turns out a five was the smallest one there was if you learn things about finite simple Groups in this video or any of the videos in these lectures, right? Lectures 10 and 9 actually are together as a two-part lecture If you learned anything give those videos a like subscribe to the channel if you want to see If you want to see more videos like this in the future And as always if you have any questions about anything you see in my videos Feel free to post your questions in the comments below and I'll answer them as soon as I can have a good day everyone. Bye