 Welcome back to the lecture series Math 4230 abstract algebra 2 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. As we can see on the screen, lecture 9 is entitled, Seelof Theory and Simple Groups. What do Seelof subgroups have to do with simple groups? Well, the short answer is the following. If you have a Seelof P subgroup that's unique, then it has to be normal. Additionally, if it's not unique, then it can't be normal. This is a consequence of Seelof's second theorem. Also, Seelof's third theorem gives us an ability to count the number of Seelof P subgroups. So by studying the Seelof P subgroups, we can make statements about the presence of non-normal subgroups or the presence, more importantly, of normal subgroups. And in fact, Seelof theory is very useful in showing that certain groups cannot be simple because they guarantee perhaps the existence of a normal subgroup in those situations. In addition to Seelof subgroups, when we consider things like simple groups, we might make arguments about centers, commutator subgroups, or other guaranteed normal subgroups. We often look also at normalizers of Seelof subgroups as those can be useful in this discussion. So this video was the lecture nine, I should say, is the first of a two-part lecture where really are trying to consider simple groups of small order. In this video, we're actually going to consider groups of order P times Q. So if the order of the group G is the product of two primes, it turns out we can actually make a complete classification of those types of groups. In particular, we're gonna see that they cannot be simple groups. And so that's what we're going to do. So the first thing in that goal is actually presenting a lemma. So this lemma we could have done some time ago, we just haven't needed it until now. Suppose G is a group, and H and K are normal subgroups of G, such that H intersect K is equal to one, that is, it's the trivial subgroup, and the product H times K is equal to G. If those three conditions are satisfied, then it turns out that G is, in fact, isomorphic to H cross K. And our strategy of proof here is we're gonna argue that G is the direct, the internal direct product of H and G, excuse me, H and K in this situation, for which to be an internal direct product, we need that the Frobenius product of the two sets, H and K is the whole group, we need their intersection is trivial. We also need that they centralize each other. In other words, for all elements H inside of H and all elements K inside of the subgroup K, we have that HK equals KH. And so that's gonna be the plan of attack we take here. So we have to argue that the normality of these two subgroups forces the elements centralizing each other. Now, before we jump into that, I do want to kind of mention what does this have to do with groups of order P times Q? Well, the idea here is that by Silov's first theorem, we're guaranteed subgroups, you know, these these maximal P subgroups of maximal Q subgroup, the Silov subgroups, right? For which as their orders are different powers of primes, Silov subgroups for different primes necessarily intersect each other in a trivial manner. Okay, if you have only two prime divisors, then it means the product of two Silov subgroups of different primes will be the whole group. So yeah, if you have only two prime divisors, then Silov's first theorem guarantees these two conditions. Now, like we mentioned before, if the Silov subgroups are unique, they'll be normal. And so under many situations, if you only have two prime divisors, we pretty much have to have a direct product. Of course, there's other options, but that's that's what we're considering in this situation. All right. So why do the elements of H and the elements of K centralize each other? I'm not saying H is abelian, I'm not saying K is abelian, but I'm saying that our assumptions guarantee that they will commute with each other. Well, since H is a normal subgroup, it's closed under conjugation. So in particular, if I take the element H inside of H and conjugate it by an element from K, that has to fall back inside of H since it's a normal subgroup. So let's say that K, H, K inverse is equal to H prime for some other element, H prime and H. Well, similar argument applies to K. It's a normal subgroup. It's closed under conjugation. So if I take the element K here, if you conjugate it by the element H inverse, notice the inverse is on the other side now. This one had K and then K inverse. This one has H inverse and H. That's still okay. I mean, it's still a conjugate of K. Because K is a normal subgroup, this falls inside the subgroup K. And therefore, there's some element inside of the subgroup K called K prime, so that H inverse K, H is equal to K prime. Okay. So now I want you to consider these products right here. So take this first equation. If I times both sides of the equation on the right by K, you're going to get that K, H is equal to H prime K, K like so. And then if we take this equation right here, if we multiply on the left of on both sides of the equations on the left by H, we're going to get that K, H is equal to H, K prime. And clearly K, H is equal to each other. So let me slide down a little bit. So the observations here is that, of course, K, H is equal to H prime K, K, H is equal to H, K prime. Clearly K, H is equal to itself. So in particular, these two things are equal to each other. That's what we want right here. We get that H, K prime is equal to H prime K, like so. Now again, playing with this equation, if we multiply both sides of the equation by H prime inverse, H prime inverse, right, that gives us H prime inverse H, K prime is equal to K, we're then going to multiply both sides of the equation by K prime inverse on the right, K prime inverse, so that these cancel out. And then we're left with the equation that H prime inverse is equal to K, K prime inverse. Of course, if you take the inverses, I did a little bit different here in my notes, if you take the inverse of this, you actually get instead the equation H inverse H prime is equal to K prime K inverse equivalent equation. They imply each other. And so in particular, what I'm trying to say here is we have right here this H inverse H prime. That those are elements of H, right? H and H prime are inside of H therefore H inverse is inside of H, their product is going to be inside of H. This element right here belongs to H. But then when we look at K prime K inverse, same argument, that's going to be an element of K. So this element is called X for a moment. X can be factors H inverse H prime, it could be factors K prime times K inverse. Though that's an element in H, it's an element in K, therefore this element belongs to H intersect K. But by assumption, this is trivial. So X is equal to the identity. Well, if H inverse H prime is equal to the identity, we can times both sides on the left by H. And we get that H and H prime are actually the same element. Right same thing here. K prime times K inverse has to equal the identity. So if you times on the right by K, you end up with K equals K prime. And so therefore these elements are the same thing. So come up to the original equation, well, not the original equation, but come up to this equation. We have that K H equals H K prime, but K prime is equal to K. So this gives us that K H equals H K. And as K and H were arbitrary elements of the two subgroups, we see that they centralize each other. And therefore by a previous theorem, we've proven that this would be an internal direct product. Therefore, G is in fact isomorphic to the direct product of H and K. Alright, so with that limit out of the way, now let's talk about groups of order P times Q, where in this discussion, P and Q are distinct primes. We can assume that Q is the bigger prime than P. I mean, clearly one's bigger than the other. So let's just say Q is the bigger one. In case it ever matters. And so let's now consider G to be a group of order P times Q. My claim in this theorem is that G necessarily has a normal subgroup of order Q by Sealoff theory. If that subgroup is normal, then it was actually a unique Sealoff Q subgroup. Because after all, every Sealoff Q subgroup of G here is going to have order Q, right? And if it's unique, then they will have to be normal by the Sealoff theorems. So in particular, no group of order P times Q can be simple because it has a non trivial normal subgroup. And then furthermore, if it turns out that Q is not congruent to one mod P, then in fact, in that situation, G must be a cyclic group. So it's not just it's not simple. It's it's not. I mean, it has to be abelian has to be cyclic, we can say a lot about those things. Alright, so for the sake of this proof, let the number n represent the number of Sealoff Q subgroups. By Sealoff's third theorem, we know that n divides the order of G and has to be congruent to one mod Q. Well, since n is congruent to one mod Q, it actually we can refine that a little better. It's not just that n divides G, it actually has to divide the portion of G that's not divisible by Q. In particular, it has to divide G divided by Q, which means n divides P. But P, it's itself a prime number. There's not a lot of divisors you can use in that situation. P only has two divisors, one and P. So n is either one or P. If n is equal to one, then we know it's going to have to be we have a unique Sealoff Q subgroup, which is normal by Sealoff's second theorem. Okay, so that's our goal. So we have to rule out the possibility of Q right here. But remember, excuse me, get the rule out the possibility of P. P is less than Q. So if n was equal to P, that means that n is congruent to P, but not congruent to one mod Q. Because and why is that? Well, because P is too small, right? If you're if n is congruent to one mod P, it's going to look like one, one plus Q, one plus two Q, one plus three Q, etc. But P sits in between one and one plus Q, it's not large enough to possibly reduce down mod Q. So if n is equal to P, it can't be congruent to one mod Q because Q is bigger than P. So the only other possibility is n equals one. And then like we mentioned before, by Sealoff's second theorem, if you only have one Sealoff Q subgroup, you know, it's unique, then it actually has to be normal. It can't be a it's a the Q subgroup, it has order Q. So it's a proper subgroup of G. It's non trivial, because its order is Q. And so we have a proper non trivial nor subgroup that implies that this group G is not simple. So that's the first statement here. There is no simple group of order P times Q. Well, let's add the other assumption. Let's now suppose that Q is not congruent to one mod P. In this situation, I then claim that G is a cyclic group. Let K be the number of Sealoff P subgroups of G. Well, by Sealoff's third theorem, right, we get that K has to divide Q, and we get that K has to be congruent to one mod P. Well, since K divides Q and Q is also a prime number, just like before, K has to equal one or Q. But by assumption, by assumption here, it can't be Q, because Q is not congruent to one mod P, which if K was equal to Q, that would be a problem, right? So that enforces that K is equal to one as well. So we get a unique Sealoff P subgroup. Alright, so under our second assumption here, our group G has a unique has a normal Q subgroup, it has a normal P subgroup Sealoff subgroups for both of them. Well, because the Sealoff P subgroup has order P, it has to be isomorphic to ZP. Every group of prime order is cyclic, so it's isomorphic to ZP. The Sealoff Q subgroup because it has order Q has to be isomorphic to ZQ in that situation. I should also mention that we satisfy the conditions of lemma 1521, which we did earlier in this video, right? We have these unique Sealoff subgroups. So H is basically ZP, K is equal to ZQ. Both of these are going to be normal because they're unique Sealoff P and Q subgroups respectively. Their entire the product of these two has to equal G because if we consider their orders, these are both equal to PQ. The product HK is not necessarily a subgroup in general, but it is a subset of Q, but it's a subset that obtains the cardinality of the whole set. So because the product HK is has size cardinality PQ, it's got to be all of G. So we get that HK is equal to G. And then also considering orders H intersect K is trivial, because the intersection here by Lagrange's theorem, we need a divisor, a common divisor of P and Q as they're both primes that has to be one. So the conditions for lemma 1521 were satisfied. So we see that G is in fact the direct product of ZP with ZQ. But as we studied finite abelian groups back in abstract algebra one, we know that the direct product of two cyclic groups of co prime order is itself a cyclic group. So G is isomorphic to ZP, ZPQ in that situation. So if you have a group of order PQ, it can't be simple. And a lot of the times it is cyclic. I want us to look at some examples of this. Okay, consider a group of order 15. Notice 15 factors as three times five, it's a product of two primes, it's can't be a simple group. And notice that five is not congruent to one mod three five is actually congruent to two. So by theorem 22, which is we just finished proving any group of order 15 has to be cyclic. Therefore, there is only one group of order 15 up to isomorphism. And it is the cyclic group. So notice we've done with just now we have a classification theorem, we can then say something about groups of order 15. There's only one of them. It's the cyclic group. Alright, I want to then look at a different different order. Let's look at 14. It's also a product of two primes. This is often referred to as a semi prime, right? It's a product of two primes, these are very important. And you know, not just an algebra but a number theory cryptography. Let's take a product of two and seven this time. Well, notice that in this situation, clearly seven is too big. So you're going to have to have a normal seal off seven subgroup. That was the whole that was the main argument of the previous theorem, right? Seven is too big. But we could have multiple seal off two subgroups, right? So if n sub two is the number of two subgroups, it has to be congruent to one, right, has to be congruent to one mod two. But that just means n two is an odd number. It has to also divide seven. So we have as a possibility in two equals one or seven. So you could maybe have seven seal off two subgroups, each of those seal off subgroups has to look like Z two, just like the unique normal seal off seven subgroup has to be Z seven. So is there a way we can combine these together? That isn't necessarily cyclic? Maybe it's not even a billion. Because by the finite, the fundamental theorem of finite a billion groups, the only abelian group of order 14 is Z two cross Z seven, which of course, by reason, earlier reasoning in the same video, that would have to be Z 14. So the only abelian group of order 14 is the cyclic one, but could we get a non abelian one? Well, if we have different seal off seal off two subgroups, they're not normal, which means you can't be abelian. So yeah, that's that's the case we're trying to consider right now. So let's suppose we have this unique seal off seven group, but we have seven seal off two subgroups and all of these seal off groups, of course, are cyclic order two and order seven respectively. So I want us to consider the elements of order seven for a moment, because after all, Z seven, it contains the identity, but the other six elements of that group are necessarily going to be elements of order seven. These are elements of order seven. Of course, we have the identity right here. I also want us to consider the seven seal off two subgroups. Each of those subgroups contains exactly two elements. One is the identity, one is an element of order two. So if we go through all of these seven seal off two subgroups, each of them contributes an element of order two. And so we have seven elements of order of order two. Excuse me. Let me look like make that look more like a two. So I want you to consider the elements we have so far. We have one identity. We have six elements of order seven. We have seven elements of order two. And if we add those together, one plus six plus seven, that's 14. So under these assumptions where we have one seal off seven subgroup and we have seven seal off two subgroups, that accounts for all 14 elements. So there can't be anything else in this group. There's only 14 elements. Okay. So let's call one of those elements of order seven, let's call it R, and let's call one of those elements of order two. Let's call that, let's call that S. Okay. And so let P be a seal off seven subgroup. It's necessarily normal. And so if we were to take the element R, right, which lives inside of P, and we conjugate it by S, because P is a normal subgroup, it's closed under conjugation. So SR, S inverse belongs to P. But as P is a cyclic group, SR, S inverse belonging to P means that it's inside of capital P. There's some power of A so that SR, S inverse is equal to, to R to the A. Again, for some integer, some integer A right there, which of course, it's unique up to modulo seven. All right. Now what if we conjugate it again, right? If we take R and you conjugate it by S, then you conjugate it by S again. I want to remember that S is an element of order two. So S times S is just the identity. S inverse times S inverse is just the identity. So if you conjugate R twice by S, you just get back R. But if you conjugate it once, you get RA. And if you conjugate it again, you're going to get A again because this is a power of R. And so this is going to give you R to the A squared, like so. All right. So R raised to the power A squared gives you back R to the first. And so because of this, how, what does this say about the power? The power here is going to have to be, we're going to have to have that A is co-prime to seven. A belongs to seven, Z seven star. I mean, that's what we get from this relationship here, which of course, that as a group, Z seven star is a multiplicative group. It is, it's a cyclic group of order six. And so it's isomorphic to the additive group Z six in that situation. But in particular, is in particular, we also know that A has to have order two inside of Z seven star because of this relationship, right? That's what we're trying to get at here. And so there's only one element of order two inside of Z seven star. That's actually the element A six, A is equal to six, that is, or negative one, if you prefer. So when you take R and you conjugate it by seven, you actually get R to the sixth or maybe better we look at it as R to the negative one. Now you're going to start to see why I labeled the things as R and S in this situation, right? Because R is an element of order seven, S is an element of order two, and when you conjugate R by S, you get R inverse. That sounds like a group we've seen before, but that's right. Our group is none other than the dihedral group D seven because D seven is uniquely characterized by these relationships. D seven, we can write as a group generated by two elements, R and S, R satisfies the relationship that R to the seventh is the identity, S satisfies the relationship that our S squared is equal to the identity. And finally, there's a commutator relationship that S, R, S inverse is equal to R inverse, like so that uniquely defines D seven. So if we have a non-Abelian group of order 14, it's got to be D seven. And so if it's Abelian, it's going to be Z 14. If it's non-Abelian, it's going to be D seven. That's pretty cool that when we looked at groups of order 14, we found only two of them. We eliminated any other possibility using our applications of Seeloff's three theorems. That is, we use these Seeloff seven groups and the Seeloff two groups. I want to point out that the previous example really generalizes to the following theorem. Let P be an odd prime, so it's any prime except for two. Then, there are only two groups of order two times P up to isomorphism. And that's going to be the cyclic group and the dihedral group, because the exact same argument will apply to each of those situations. In particular, two P is a semi-prime. So by the previous theorem that we did, we know that there is a net, there's automatically a unique Seeloff P subgroup. It'll be normal. It can't be simple. If the second case doesn't apply, because of course two, excuse me, P is congruent to one mod two, because P is an odd number. And so there is a possibility that if you have a unique two subgroup, you're going to get Z two P. If you have non-unique Z two subgroups, then you're going to have P many of them. And then, like we discussed before, that's going to lead to the dihedral group. So it generalizes perfectly in that situation. All right, so to finish this video, I want to then consider, well, what about the general case where we have a semi-prime type P times Q, but it's not necessarily two. Like, what if both of them are odd numbers? So let's consider that one for a little bit further. Well, because of the theorem we proved in this video, if Q is not congruent to one, it has to be cyclic. And so what happens if Q is congruent to one? Well, kind of like we considered earlier with the dihedral group, like again, you have basically the two options, right? If you have unique Q groups and P groups, then it's cyclic. You're always going to have a unique Q group, because we're still considering Q to be larger here. But if you have a non-unique P group, that is to say, because the number of Silov P subgroups is equal to Q, that would happen only if Q is congruent to one mod P. Then it turns out you can construct a non-billion group that generalizes the dihedral group we saw in the previous example. We'll take two generators, and for the lack of better labels we'll call them R and S. We're going to have that R cubed is equal to the identity. We're going to have that S to the P is equal to the identity. And then like likewise, finally, we're going to have that R, excuse me, S R S inverse is equal to R to the A, where A is some element of Z Q star. In particular, the order of that element A is equal to P. But since Q is a prime number, this group right here is congruent to Z P, excuse me, Z Q minus one, like so for which it has a it has a unique subgroup of order P, because we are assuming by this assumption right here this is equivalent to saying that P divides Q minus one, right? This as a cyclic group has a unique subgroup of order P. And so basically A has to be a generator of that subgroup. And it really doesn't matter which one you choose because choosing a different generator up to isomorphism is not going to make any difference on this group whatsoever. That is this group, this group right here will be isomorphic to it, okay? And so that's what happens here that when you conjugate R by S, you get a power of R. And that power of R is an element of order P in the automorphism group. Again, that's something to be significant of this right here. This Z Q star, this is the automorphism group of Z Q, like so. And this leads to the idea of what's called a semi-direct product, that this semi-direct product Z Q, semi-direct product. So it's like an X, but there's this little line in here. It's like an amalgamation of the direct product symbol with the normal symbol here, the normal subgroup. Because in this situation Z Q is always the normal subgroup of the whole group here. In this case we have a cyclic group, semi-direct product, a cyclic group. It's a lot easier to describe in that situation. But we take everything over here, everything over here. But how do things between the two groups interact with each other? With an internal direct product, they just centralize each other, just commute with each other. But if you unravel this right here, this is saying that S R is equal to S, excuse me, R A S. So to commute R with S, you look at an automorph of R A to the R. And so in general, in general, if you have, if you want to take the semi-direct product of H with K right here, what you need is an automorphism. You have some type of homomorphism, in fact, from K into the automorphism group of H. And so then you have the relationship that if you have H, excuse me, if you have K times H. So K is something in K right here. H is something over here. Then this becomes phi of K of H times K right here, which this can be a little bit confusing right here, because phi of K is an automorphism on H. And so therefore, this is the automorphism actually acting on H. Because of this notation, it's often written in the following form. You have H to the K times K, like so. So what does this mean right here? This means the image of H under the automorphism associated to the element K, for which when you have a cyclic group, it's always a power like this. So H will map, excuse me, r maps to r to the A. And so this gives us a complete characterization of all of groups of order P times Q. If you have a group of order P times Q, you get only two subgroups up to isomorphism. You get the cyclic group, and if there's an automorphism available, you can get a semi-direct product, such as the dihedral group. But in general, if there's no, if there's no automorphism available, then you only get a cyclic group, like in the case of a group of order 15. These semi-direct products are very important when studying finite groups. And of course, we showed up in this situation, showed that groups of order P times Q can sometimes have a semi-direct product, such as the dihedral group. We can generalize essentially the dihedral group using this idea of a semi- direct product when you have a cyclic group, semi-direct product, a cyclic group. This is what's known as a split metacyclic group, in case you like to hear jargon like that. Metacyclic is suggesting it's actually the product of two, of two cyclic groups, and the split here is actually suggesting it's a semi-direct product of two cyclic groups. For a group of order P times Q, you always, you only get this or the direct product. But I should also mention the semi-direct products which generalizes this idea of a direct product. We've seen in the past, like in an algebra one, when we talked about the Euclidean and the affine groups for certain matrix symmetries, those were in fact or semi-direct products there as well.