 We say a group is simple if it has no non-trivial proper normal subgroups Recall that every group has at least two normal subgroups The group itself is normal inside of itself, but you also have the trivial group Which is gonna be a normal subgroup. So these ones are guaranteed But what if there was no other normal subgroups inside of G? That's what we mean by a simple group and why do we call them simple groups that when you get more and more into Group theory you start to realize that simple groups are essentially the atomic building blocks of Every other group that is we can create larger groups by understanding What the simple groups are and again? That's that's conversations we could have in more detailed setting here But simple groups again really are the most fundamental of all types of groups Every group can really be built up from the simple groups and so there's been a great effort very recently in the In the history of abstract algebra to classify all finite simple groups the bulk of this work took place in the 1990s and early 2000s For which the the families of finite simple groups were classified This would include cyclic groups, alternating groups, the projective special linear groups There's the sporadic simple groups. There's simple groups of various lead types We won't we won't go into all of those here But I just want you to be aware that that classifying and identifying simple groups is a very important branch of modern-day group theory So in this video, I want to prove two things about simple groups first. I want to consider. What are the Abelian Finite simple groups it turns out that one's pretty easy and then I'm going to give you an example of a non-Abelian Simple group particularly the alternate group that won't take a little bit more to develop So suppose that G is an Abelian group then we're going to prove that G is simple if and only if G is cyclic Cyclic of prime order So it's just two directions, right? So we'll assume that G is Abelian in all cases, right? So let's first assume Well, let me let me first say the following right what we've learned previously about Abelian groups Is that to be a normal group right you want that? The left and right cosets are the same or another way of saying it is that Your your subgroup is closed under conjugation. Well in an Abelian group conjugation doesn't do anything, right? Because you have G X G inverse well since it's commutative You just get G G inverse X. This will just become X right conjugation doesn't move anything around in particular left and right Cosets are the same thing in every Abelian group So therefore every group every subgroup of Abelian group is normal And so the only way that an Abelian group could be simple is if it has no proper non-trivial subgroups that is the only subgroups are going to be the group itself and The trivial subgroup so if G is simple and you have some element X inside of G, right? So it's a non-trivial group so X is some non-identity element, right? Then take the cyclic subgroup generated by X that would be a non-trivial group if it's simple that would then force It to be improper that is it's the whole group so this shows that G would have to then be a cyclic group if it's Abelian and simple and in fact by Essentially the well since it's a cyclic group We've seen previously that for every divisor excuse me for a cyclic group every divisor of the order of the group Produces a subgroup, right? This is true for cyclic groups every divisor gives you a subgroup a unique subgroup up to that order and As such there can't none of these can exist, right? So the only divisors of the order of G would have to be one in itself, which would indicate that it's a prime number Of course the other possibilities. It's the trivial group itself, which technically would be the case We're not going to consider the trivial group to be simple much of the same reason why we don't consider one to be a prime number It needs to be something non trivial to be in consideration here If we want to think of the simple group is kind of like the primes of finite group theory So I guess it's the first direction that Abelian simple group is going to be cyclic in prime order If we go the other way around, right? If G has prime order then it necessarily is going to be cyclic and by Lagrange's theorem It'll have no non-trivial proper subgroups For which in that situation, right since it has no non-trivial proper subgroups In particularly has no non-trivial proper normal subgroups. And so therefore it's a simple group So like I said a very slick argument there that The that Abelian groups are simple if and only if they're prime Order which which implies that it would necessarily be a cyclic group of prime order the study of Simple non-Abelian groups is much much more difficult. Like I said, uh, it took decades for you know The best mathematicians in the field to be able to classify all of the non-Abelian simple groups And honestly the the paper if you count the the pages of all the collective papers that worked on it We're talking about hundreds probably thousands of pages It's it's it's still a chore the time of this recording here in the year 2020 It's still a chore just to kind of condense Condense the proof of these thousands of papers down to be something that could be taught, you know In a semester-long class right graduate course, of course So what about a non-Abelian simple group then? There is one that is within our grasp to provide it turns out we already are aware of some the alternating group, right? An it turns out that when n is greater than or equal to five then this is going to be a simple group A4 is not simple Because a4 has as it a normal subgroup the climb four group Since it's there as a normal group and then a3 technically is simple as well Because it's actually isomorphic to z3, which is a cyclic group of order three. So it's an abelian one. It's not it's A3 is an abelian alternating groups. It's sort of an exception there. So it's like Why is a4 different? What's so special of five? Why does it start working after five? and so what I'm going to do is collect some limas right now to Help us prepare for the proof that a5 and larger alternating groups are all simple non-Abelian groups So the first proposition here is actually this is something true for all Normal groups has nothing to do with the alternating group necessarily But if you have a group g and it has a normal subgroup n and then you have some other subgroup h Which may or may not be normal make no assumptions there Then if you take the intersection of n with h this will be a normal subgroup of h The intersection of subgroups is itself a subgroup, but we have to prove Which will actually prove that statement in this proof if you weren't aware of that But we'll also prove that in intersect h is closed under conjugation making it a normal subgroup of h So how do we first show that in intersect h is a subgroup of h, right? Well subgroups need three things. It needs to contain the identity Well as n is a subgroup it contains the identity as h is a subgroup it contains the identity So the intersection they both contain the identity so the intersection has it The intersection contains the identity It needs to be closed under multiplication So if we take two elements g and h inside of n intersect h Well, that means that g and h are contained inside of n and since n is a subgroup It'll be closed into the product g times h by similar reasoning, right? g and h will be contained inside of h Therefore since it's a subgroup it'll contain the product g times h So the intersection will contain the product so it's closed under multiplication Same basic argument for inverses right here that since n is a subgroup it'll contain the inverse of every element in there So if g is inside of n then g inverse will be inside of n Well, same thing for h, right? If h contains g then it'll contain g inverse because it's a subgroup and therefore the intersection Is can it will contain the inverse and so this right there shows that the intersection two subgroups is in fact a Subgroup of g right and in particular as the intersection is likewise a subset of h This would show you that intersect h is a subgroup of h as well Okay, so again that really just comes down to that subgroups are closed under the Under the operations of multiplication the identity and inverses, right? So intersections Are really nice for closure principles, right? Because if this set is closed with the property and this set's closed with the property Then the intersection will have that as well That's basically how we're going to show normality here using the operation of conjugation Remember we saw previously that a subgroup is normal in g If and only if it is closed under conjugation Or written in a slightly different way we get that for all g and g and for all n and n We have that g and g inverse is inside of n So normal subgroups are those closed under conjugation for which this conjugated element g could be any element Inside of g in particular. What if we restrict it to the attention that it is a It's the the things from h which of course is inside g. So that's perfectly legitimate here So let's take an arbitrary element g which which lives inside of n intersect h And take an arbitrary element of little of h living inside of h there Well, since n is a normal subgroup Then we have the property that h g h inverse will belong to n normality implies it's closed under conjugation So we get h g h inverse is inside of n. Well, okay But g is in h because it's inside the intersection of in with h and we also have that h belongs to h by assumption So since h is a subgroup The elements h g and h inverse are all inside of h. So their product is inside of h So we get that h g h inverse is inside n intersect h This then proves that h intersect h and intersect h is a normal subgroup of h All right, so let's get specific about the alternating group right here So our first lemma is that if n is greater than equal to three then an is generated by the set of three cycles So we've proven previously that the symmetric group s n right the s n is Is generated by the transpositions the two cycles and that's where the idea of like even permutations and odd permutations came from Because every permutation can be written as a product not necessarily a unique product But it could be written as a product of transpositions transpositions don't exist in a n because those are odd permutations And so we have to replace transpositions with three cycles But for a three in a larger we can prove that a three is generated by the three cycles That's a nice generating set so an You know since an consists only of even permutation It suffices to show that every pair of transpositions Is a product of three cycles right because every permutation to an is a product of an even amount of transpositions So that's that so those transpositions could come into pairs There's the first two then the next two then the next two then the next two etc and so when it comes to Transpositions right the nice thing is when you reverse their order, it's the same thing So every transpositions its own inverse so there's not a lot of combinations you get here when you talk about pairs of transpositions There's three possibilities There's the possibility that a the letters in the first transposition are identical to the letters in the second transpositions It's the same thing. So like ab times ab is the identity Um, okay. Sure. Um, the identity You know, maybe it goes without saying but if you're in a three you could take one two three And then times by its inverse one three two that'll give you the identity So the identity is a product of three cycles You could have your trans your transpositions your two cycles could have one letter in common The other two are distinct. So something like ab times ac Well, notice here that if you try to calculate this a goes to c C goes to a which then goes to b And then finally b goes to a so the product of those two transpositions is in fact a three cycle the three cycle ac b Okay, what if you have disjoint two cycles? So like ab and cd and those letters don't overlap whatsoever Well, turns out ab times cd is the product of three cycles here So notice if you take the product ac b times acd You're going to see that a goes to c c goes to b Great and then b goes to a that gives you a two cycle right there And then if I look at c c goes to d and then d goes to a and a goes to c We see this going on right here. So the the product of these Two three cycles does give us that the two two cycle and so since every pair of transpositions is a product of three cycles And elements of the alternating group those even permutations are generated by pairs of two cycles We see that the three cycles generate an as long as we have in grade equal to three But I thought you said grade equal to five. Where's the five coming to play? Well, that that's our next limit right here So let n be greater than or equal to five Then we're going to see that all three cycles are conjugate in an We proved previously that in the symmetric group Permutations are conjugates if and only if they have the same cycle type. So in particular in sn All three cycles are conjugates to each other Now that's not true in general right and the alternate well for the alternate group I mean because in a four for example in a four you actually have two different Consciously classes of three cycles like one two three Is one represents one class and then the others are the inverses one three two So these two inverses are not conjugates of each other and certainly that's because you don't have enough letters If you have five letters, then we can actually force conjugation between all three cycles. That's what we're going to prove right now So consider the two three cycles abc and x y and z Okay inside of an and kind of like the previous proof We're going to consider this we're going to we want to show that these things are conjugates That's not the similar part, but the similar part is we want to consider the overlap between their letters We have like abc and x y and z Um, do they share some of the letters or not right? Well, the first case is what if they share all the same letters What if like abc as a set is the same thing as x y and z? Although the order doesn't necessarily have to we don't know how they correspond right I'm not saying a equals x b equals y c equals z or anything like that. There's some correspondence But we do see there's only two possibilities right x y and z is either equal to abc or It's equal to acb. All right now in the first case if they're equal if x y and z is equal to abc Then of course they're conjugates of each other because they're equal Conjugation is equivalence relationship. So it's reflexive in particular if you take x y z And you conjugate by the identity, right? Then this will give you abc because abc is equal to x y z. All right, that's pretty simple Uh, so that gives us, you know, this observation right here. What about the other one? What if x y and z what if x y z is equal to acb in that situation? Okay So consider consider that possibility is when we have n greater than five is a three cycle Conjugate to its inverse now. This is this is the cool part, right? So if n is greater than equal to five take all the letters in play here one two three four up to n, right? The three cycle only involves three of them ab and c so that means if we take away ab and c there's at least Since since n is greater than equal to five. There's at least Two letters not involved inside of the three cycle abc or acb Um, choose any two of them. It doesn't matter which two you do but take at least take two of them We'll call them i and j and consider the following product bc times ij times abc times bcij inverse Oh boy, all right. That's kind of an interesting calculation there But let's let's actually run through the calculation bc times ij Times abc times we're going to take the inverse there. So we're going to get ij And bc which I really didn't need to turn the order around it's it's on the inverse because it's a two two cycle It's order two, but go through the calculation here. Let's see what happens to a So going through this a will go to b and then b goes to c All right, so recording that down we get that a goes to c now. What happens to c c goes to b b goes back to c and c goes to b so we have that b Um, what happens to b here b goes to c c goes to a and then so therefore wrapped it up. So, okay Yeah, we got we have acb going on here, but notice i and j They're not actually involved with abc here. So these things actually commute. So this thing is equal to bc times abc times bc Times ij times ij. So these things actually cancel out because they're inverses of each other All right, and so okay, so you see that the i and j basically didn't serve much of a purpose Why do we include that? Well abc if we conjugate it by bc bc is an odd permutation bc does not belong to an but aha bc Times ij that does belong to an because that's an even permutation Even though the ij is only present so that we can say it's an even permutation It's kind of a slick move there But this trick doesn't work in a four for example because there's not enough letters to get these superfluous Isons j is involved in here. So abc is conjugated to itself and it's conjugated to its inverse That was the possibility we considered if abc and the set abc and the set x yz Um coincided well if they don't coincide, right? What if they're not equal to each other? Well, then it turns out that at least you know if these two sets are not equal to each other One element doesn't belong to the other. So let's say like a doesn't belong to this set right here So x y and z none of those are equal to a but that would leave You know But there's only three elements in this set right here That also would imply that since these sets both have size three that one of the elements on the right hand side Wouldn't wouldn't match up with ab or c. So let's say that's x, right? So with the loss of generality we can say that a Does it does not coincide with x y and z and that x does not coincide with ab and z? All right So in particular the three cycle abc fixes a because x excuse me a abc fixes the element x because it does not belong to ab and z um And the the three cycle x y and z fixes The element a because a does not belong amongst x y and z right there Okay, so again since we have n is greater than or equal to five If we take away the letters abc and x right because those are all distinct letters Then you can take at least one more element i because there's at least five and consider the following product We have a x i times abc times a x i inverse three cycles or even permutations So this is a legitimate product to consider inside of a n or bigger right and what happens with the product here It's right over here a x i times abc Times i x a so if we go through and see what happens here what what happens to x x is going to go to a a goes to b And that's it. So we're going to get that x goes to b So then going through b goes to c and there's no other c So we're going to go c right here Then we go back to it again c goes to a and then a goes to x so that finishes off this Three cycle. Um, what about a and i right a goes to i but i goes to a so a is fixed And then i goes to x which x goes to i so i is fixed as well. So this does in fact give us the three cycle um x b and c so this shows us that abc is Conjugate to x bc, but it's like what i want abc to be conjugate to x y z what happens there Well, we're going to recurse on what we just did here. So notice that we got What we what we have so far is that we have that abc Is conjugate to x bc. So we basically switched the first letter We switched from a to x or remember x and a where these letters that didn't match up So now reconsider what's going on right now. So consider the set x b c versus x y z Okay, do these sets disagree if they do repeat this process. So let's say they're like b and y Are now the things that don't agree between the two sets If that's the case, then you can basically conjugate and you can replace x bc with x y c right and then you check again Do these things match up if? um, if they don't like if c and z don't agree with each other Then we can conjugate one more time and we get that x y and z right here So since conjugation is an equivalence relationship transitivity We can just do a couple steps to get from abc to x y z. So that's a possibility if they never match up But it could be at some point they match up, right What if the set x bc is equal to x y z? Well, that was the case we considered up here, right? In which case then we can force conjugation there and so whichever path one chooses eventually this will terminate With abc being a conjugate to x y and z So all three cycles in a n are conjugates as long as n is greater than or equal to five So that shows a lot of significance on y five y y five you need to have enough space for three cycles to wiggle in terms of conjugation here and then our last lemma, uh, let n be greater or equal to five right and suppose we have a normal subgroup of a n If this normal subgroup n contains a three cycle then in fact n is all of a n and why is that? Well, normal subgroups are closed under conjugation That's a fact of being a normal subgroup by the previous lemma We saw that all three cycles are conjugates to each other So if n contains A three cycle then it has all of the three cycles because they're all conjugates and normal subgroups are closed under conjugation But but in the first lemma of this slide right here If you have all the three cycles and you generate all of a n and therefore n equals a n For which then we see that for every normal subgroup of a n it either is trivial Um Or if it contains a three cycle it's everything right? So now now the thing is we have to prove is that every normal subgroup of a n proper subgroup Every non-trivial subgroup. Excuse me. Um, we'll contain a three cycle. That's that's our plan of attack right here So we're going to first do this for a five So here's the proof a five is a simple group Let n be a non-trivial normal subgroup of a five So again by the previous lemma if n contains a three cycle then we're done Because clearly if you have a non-trivial subgroup it contains something if that contains A three cycle then boom it produces all of a five and Thus it proves that a five is a simple group because it doesn't have any non-trivial proper normal subgroups So what are the possibilities right well in a five? There's only three types of uh permutations that live inside of a five Right. Oh, I mean, I guess there's the identity but let's exclude that for a moment Um, if you're a non-identity permutation, you're either a three cycle You're a two two cycle or you're a five cycle. Those are the only options Um inside of a five now. So sigma since it's non-trivial There's gotta be something that's not the identity inside of n. Well, the first possibility is it's a three cycle But like I said, if it contains a three cycle You know, it's done, right? So so let's not worry about that one Um, the next one then would be is if we have a two two cycle And so without the loss of generality we can assume, you know up to relabeling these things are like one two three One two three four one two three four five, you know, I'm very clever of my label in there But you know up to relabeling it's it's equivalent to something like that, right So sigma is equal to the two two cycle one two and three four Well, let's consider the element tau, which is one two and three five, right? That's an element that belongs to a five It's even so consider the product tau sigma tau inverse. So this would look like one two three five Then sigma remember was one two three and four and then tau since it's a two two cycle it's its own inverse You're just going to get one two Three five again, so we're taking the conjugate of this thing and so let's see what happens if you take one one goes to two two goes to one and So let's see one goes to two two goes to one and then one goes to two Like so That's going to give one goes to two. All right. Let me erase my marks there So we got one goes to two. So then you're going to see Honestly, I mean if you look at this thing you can commute things around. Let me say it that way So, I mean like you have a one two one two one two those all those can commute with each other two of them We're going to cancel out actually And so you're going to end up with a one two great. Then you're going to have this three five three four Three five that can't commute with each other because there's an overlapping three But you'll see that three goes to five and then five goes to three. So three is actually fixed So three is actually left alone and then we're going to see that probably four goes to five, right? So five goes to three three goes to four and so you're going to see four Four and five going on right here. So that's that's the conjugate tau sigma tau inverse now since Since sigma right Is that's our normal subgroup right if you conjugate by anything in the alternating group you get back to n Because normal subgroups are closed under conjugation, right? And so also what comes into play a lot are in this argument. It's going to be commutators, right? So let's consider let's consider right here the commutator tau sigma Which the the commutator here is just going to look like Tau sigma tau inverse which we did a moment ago at this element then times that by sigma inverse So let's finish up that argument here that calculation. I should say tau sigma. This is going to look like one two Four five and then we're going to multiply that by sigma inverse, which is itself just sigma So you get one two three four Again the one two's actually are disjoint so that you can move them around so they're actually going to cancel each other out And then you're left with four five times three four Which I want you to convince yourself. You're going to get three goes to four four goes to five And then you're going to see that five goes to four and then four goes to three So this is a three cycle. Okay, uh, this is the three cycle three five four and so therefore the second case Gets us a three cycle. So that normal subgroup is all of a five Um, so the last possibility is what if we have a five cycle? Let's say one two three four five Um, so let's consider the commutator in this situation. So we're going to take tau sigma Right, so consider This product so tau we're going to say is just be one three two That belongs to the alternating group right sigma is one two three four five Then we're going to get that's tau inverse which is one two three And so taking the inverse we're going to get one five four three two like so And so let's see what happens here. Where does one go one goes to five Five goes back to one and then one goes to three Uh, how about three three goes to two two goes to three three goes to four Like so, uh, we then get four goes to three Three goes to one one goes to two and then two goes back to one So those all wrap around so we get one three four And then lastly, uh, like us we should figure out what happens to two now Two goes to one one goes to two two goes to three and three goes back to two So two was fixed that also means That five will have to be fixed. So we see that this equals the three cycle one three four So which case that n contains a three cycle and therefore it is By the previous lemma it's going to have to be um all of a five this this proves that a five is a simple group For bigger an Right, so that is n is greater than equal to five. We're going to proceed by induction Where the a five case we just did a moment ago is our base case So for our induction hypothesis that if we you know, let's say we've proven that a five is simple up to Um the number k that is we haven't proven k yet. We've proven the five six seven eight all the way up to k minus one Let's suppose those are simple groups. We're going to induct upon that Let n be a non-trivial normal subgroup of a k and let sigma be some non-trivial element in here Our same goal is the same we want to use sigma to create a three cycle If we can find a three cycle inside of n then that means n is all of a k So suppose first of all there's two possibilities We're going to consider suppose that sigma has a fixed point that is there's some number You know a sigma i that equals i now up to relabeling of the numbers here We can actually suppose that this fixed point is the last number k right because again up to relabel It doesn't make much of a difference. So because of that because sigma fixes k We can essentially identify sigma with a permutation on k minus one many letters Okay, and so this is how the induction is going to get into play here. So a k minus one We know is simple by our inductive hypothesis right here By a previous lemma we proved in this video if you take a normal subgroup and you intersect it with a subgroup This will be a normal subgroup of that group right there Now n minus a k minus one is a non-trivial group because n contains sigma Which is non-trivial and sigma right sigma belongs To this intersection where sigma is not the identity. So this thing is non-trivial right here But by the lemma it's a subgroup of a k minus one a normal subgroup right And so since a k minus one is simple by our inductive hypothesis It must be that n minus a k minus one And since it's a non-trivial Since it's a non-trivial normal subgroup of a simple group. It's got an equal a k minus one So that tells us that this normal subgroup n contains all of n a k minus one But wait a second this alternating group on k minus one letters. It contains a three cycle So by the being about a boom n contains a three cycle and so n is all of a k great By by a previous lemma right there. So if it has a fixed point then It's going to be the whole it'll have to be it has to be all of a k there So now we're going to suppose right suppose sigma doesn't have any fixed points going on right there So in that situation in particular that means there exist some letters i and j inside of our set x right where x remember is one two Skip a few all the way up to n right here And so we have to have that There's no fixed point. So these letters i and j right Sigma send i to j and j is not equal to i that's what that should say clearly j equals itself that would that would be ridiculous So sigma i equals j but j doesn't equal i that's what we have right there now Let tau be a three cycle such that tau Of i is equal to i but tau of j is not equal to j So tau will fix i But it won't fix j and this can happen again because we have at least five letters And so consider the product of these things So remember sigma since i to j But tau fixes i so if you look at Sigma tau of i will tau since i to i so this will give you sigma of i but sigma of i will send you to j right But uh, excuse me sigma sigma of i will send you to j But j is not what tau of j does right and then sigma Sends i to j so we can we can factor this here. So notice here. What happens sigma tau and Sigma tau and tau sigma do different things to the number i Therefore sigma tau is not the same permutation as tau sigma These elements don't commute with each other. And so if we let alpha be their commutator Sigma tau sigma inverse tau inverse this cannot be the identity Right because if it is the identity then the two elements would commute with each other Which is not the case that would violate this this statement right here So alpha is some non trivial permutation that measures the commutator of sigma and tau right here Well since n is a normal subgroup and since sigma is inside of n That implies that alpha is inside of n because after all you take tau sigma inverse tau inverse Sigma inverse is inside of n because it's a subgroup tau sigma inverse tau inverse is inside of n because it's a normal subgroup It's closed under conjugation and since since tau sigma inverse tau inverse is in n and so is sigma their product will be an n So hence their commutator will be in there and so alpha this commutator belongs to n Okay, now look at look at sigma tau sigma inverse for a moment though, right? So if you factor alpha in a different way, right? So we factor it as a sigma times tau sigma inverse tau inverse, but you could also factor alpha in the following way it's sigma tau sigma inverse times tau inverse All right, so sigma tau inverse that's a conjugate of a three cycle In which case as this isn't this is a permutation. We'll have the same cycle structure So since tau was a three cycle sigma tau inverse must also be a three cycle And since tau again was a three cycle its inverse is also a three cycle So when you look at this factorization of alpha we see that it's a product of two three cycles Now those those three cycles could overlap in terms of their units. Maybe they do. Maybe they don't it depends but Worst case scenario, right? We have something like abc Times x y and z which again some of the letters between ab and c could overlap with the letters x y and z I can't make much of a statement there at the moment But worst case scenario alpha involves at most six letters of the alphabet In which case that tells us that alpha will fix k minus six letters Okay So if k is greater than six Then alpha has a fixed element and we actually can revert back to the fact that oh wait If if you have a fixed element then you then a n will equal ak, right Okay, so if if k involves more than six letters Well, we don't necessarily know that right we we've already taken care of five and now we're in an inductive hypothesis We this actually is a possibility we'll have to consider but also If there's any overlap between these elements right here If there's any overlap then in fact we can improve this bound If you know if they just shared one letter in common Let's say like a equals x or something like that then it turns out there's only five letters involved So we get five mine We then we get k minus five and then we can improve this mark to be like oh as long as k is greater than five Which we already took care of five. So we know that's the case so what i'm trying to say here is that if If k is greater than five or if a is a product of non distinct three cycles Then then that means alpha has a fixed point and the above argument applies So there's only one remaining case we have to consider So we have to consider the case where alpha is a product of disjoint three cycles in a six And which without the loss of generality as we can up to relabel and we can assume alpha Is just the the three three cycle one two three and four five six Okay, so let's let's play around with the commutator between alpha and the three cycle two three four We're going to call that beta for a moment. So consider this Commutator we're going to take alpha times beta times alpha inverse times beta inverse This is an element that lives inside of n right alpha belongs to n Here is a conjugate of alpha using elements from a six So since normal subgroups are closed under conjugation, this will belong to n's products closed Uh, the the product will be inside of n as well So notice throughout this argument we use the fact that normal subgroups are closed under conjugation so many times To you know show why conjugates and commutators belong to this normal subgroup So if we look at alpha beta alpha of inverse beta inverse, we're going to end up with one two three times four five six Then we times that by beta which is two three four we times that then by alpha inverse That's going to be one three two And four six five and then we times that by beta inverse which is two four three And so what happens in this calculation right here? So I claim this is going to be the five cycle One two three four five. So let's let's see how that works out. So one Goes to three three goes to four four goes to five And so that gives us that one goes to five Then we're going to get five goes to four four goes to two two goes to three So far so good three goes to four Excuse me three goes to two two goes to one one goes to two And then next we're going to get two goes to four four goes to six six goes to four Like so and then four goes to three three goes to two two goes to three And three goes to once that finishes off the cycle And as we're at a six then six necessarily has to be fixed. So that validates the argument we have right here So we see that this five cycle one five three two four belongs to end right here now This commutator alpha beta right it's a five cycle inside of a six that means the element six itself is fixed by The element alpha beta and so then the above argument applies basically what I'm saying is that there's always In this normal subgroup n there's always some element that Will fix some point and that fixed point we can then argue Then the normal subgroup has to be all of a k so in this last case We see that n equals a six and that considers every case Therefore n is equal to a k by induction and this then proves to us That the alternating group when it has five or more letters is always a simple group There's no non trivial proper normal subgroups in that group