 Welcome back to our Lecture Series Math 42-30, Abstract Algebra 2 for students at the University. As usual, I'll be your professor today, Dr. Andrew Missildine. In Lecture 36, we are going to talk about the solvability of polynomial equations using the technique of radicals. And we're going to do that in the second half of this lecture that actually talk about solvability. What does it mean to solve a polynomial by radicals? In particular, why is the general Quintic equation impossible to solve by radicals? Now, in order to do that, we have to first develop the notion of a solvable group. With our lecture series, we actually could have done this a long, long time ago, right? I mean, if you follow along with the numbering of our lectures here, we're actually currently in Chapter 23. But the notion of a solvable group actually was presented back in Chapter 13 from Tom Judson's Abstract Algebra textbook for which our numbering is following there. We opted not to discuss it back in Abstract Algebra 1, that is Math 42-20, but it becomes a necessity now if we really want to understand the insolvability of the Quintic. We have to understand what it means for a group to be insolvable, which of course means we have to know what a solvable group is. And the terminology for a solvable group is somewhat curious, but it really, its name comes from this idea of solvability of polynomial equations by radicals. And so while the term might seem weird, it's going to turn out that a polynomial is solvable by radicals if and only if it's a Galois group. It's a solvable group. And so that's where the terminology comes from here. So what does it mean for a group to be solvable? Well, there's a few things we've got to say first. So given a group, we can define a so-called subnormal series of G. We say it has length N. If there's a sequence of subgroups, which we'll call HI, such that the following conditions happens. At the very bottom of our chain, we call that the trivial subgroup that's going to be H0. Then you have some subgroup H1, which contains H0, admittedly everyone contains H0, but in particular this containment is proper. H1 is not equal to H0. Then there's some other subgroup H2, that properly contains H1. Then there's some other subgroup H3 that properly contains H2. Again, all of these contaminants are proper. And you go up your chain until we get to Hn minus 1. This is a subgroup properly containing Hn minus 2. And then it is contained properly inside of Hn, which is the whole group itself. So when you look at this, we're going to have N plus 1 groups in this chain right here. The bottom is the trivial group. The top is the whole group. And you have all of these groups in the middle that are properly contained. There's never equality in that chain whatsoever. That's what's going to give us a series of groups, of course. Why do we call it subnormal? Well, we require that each group in this chain, when you compare neighbors in the series, HI has to be a normal subgroup of HI plus 1. So H0 is normal inside of H1. H1 is normal inside of H2. H2 is normal inside of H3, et cetera, et cetera. Hn minus 1 is normal inside of Hn. Now, there's a very important distinction that needs to be made here. When we talk about a subnormal series, we are not assuming that these subgroups are normal inside of G. No, no, no, no. We are saying they're normal inside of the next group. So H1 is not necessarily a normal subgroup of G. It will necessarily be a subgroup of G, but it's not necessarily a normal subgroup of G. It could be, but not necessarily. But what we do require is that H1 is a normal subgroup of H2. And that relativity there can make a difference there. Clearly, if H1 was normal inside of G, it will be normal inside of H2 and every other subgroup as well, but it's only required that it be a normal subgroup of the next group in the series. And that's why we refer to this as a subnormal series, that every group in the series is normal in the next group in the series, but not necessarily normal in the whole group. Now, if you do require that each subgroup in the series is normal inside of G, you refer to that as a normal series. And one can talk about normal series of groups, but for the notion of a solvable group, it's better to focus on subnormal. And so we only require that you're normal in the successor inside of that series. Now, we should mention that the subnormality condition here guarantees that the quotient group H1 out H1 minus 1 is well-defined for every pairing you have here. That is consecutive pairings. And these quotient groups are referred to as the factors of a subnormal series, okay? So that gives us what a subnormal series is. And again, we also have a notion of a normal series, but we're only going to focus on subnormal series in this situation. Now, if you start with a finite group, every finite group will have a maximal normal subgroup. That is, there's a largest normal subgroup proper inside of G. And so you can take that then to be the next one. But then that group, because it's a finite group, by induction, it would have a maximal subnormal series. And once you can glue those things together, what I'm trying to say here is that every finite group has a maximal subnormal series. What do we mean by maximal? We have a subnormal series for which we can't insert any new subgroups, that these are proper groups, right? They're proper subgroups of each other. There could be a large gap in between them. The fact that we say that a subnormal series is maximal means that we can't fit any other subnormal subgroup in between there. And this is true for finite groups. Now, for infinite groups, you can have an issue right here that these subnormal series might not have a maximal extension. You might not be able to enlarge it larger and bigger and bigger and bigger and bigger and bigger, right? Take the integers, for example, right? If you take the integers, you can then put inside of it 2z, which you can put inside of that 4z, for which you could then put inside of that 8z, right? And by definition, these subnormal series have to have finite length. And so you have to kind of terminate this process after a while, but you could always make it bigger by adding one more, right? You could always go one more, one more, one more. And so in the case of the integers, you can't ever construct a maximal subnormal series, okay? But finite groups, because of the induction, we can always construct a maximal subnormal series. A group, if it has a maximal subnormal series, we call that maximal subnormal series a composition series. Composition series is a maximal one, a maximal subnormal series. Now, if you have a normal series, if you have a maximal normal series, that's called a principal series, but like I said, we're not going to worry about normal series or subnormal series in this context, but I just want to throw out the vocabulary. The factors of a composition series, which is a maximal subnormal series, are called composition factors, and they actually measure something about that group. Now, if you have a composition series, so it's a maximal subnormal series, look at the composition factors, these quotient groups. Because the composition series is maximal, that means you can't fit in any more normal subgroups, that means if you take two consecutive groups inside of your composition series, like HI and HI-1, since it's maximal as a subnormal series, that means there doesn't exist any normal subgroup sitting between HI and HI-1. And so then by the correspondence theorem, if there's no normal subgroup sitting in between HI and HI-1, the quotient group has no proper non-trivial normal subgroups, and therefore it's a simple group in that situation. And so that's a very nice condition that we have a subnormal series as a composition series, if and only if all of its composition factors are, all the factors are simple groups. And this is actually how one can try to fix some of these. This is how one can try to fix these, the infinite group problem, like I said before, because after all, when you look at Z mod 2Z, that's a simple group of just Z2. If you take 2Z mod 4Z, that's a simple group of Z2. So in some essence, I can get this infinite subnormal series for which all the factors are simple. So one can get around the infinite group problem, at least for a billion groups. And I don't want to say too much about that. But for finite groups, be aware that you have a composition series if and only if all of these factors are simple, because again, HI mod out HI-1 will be simple if and only if HI-1 is a maximal normal subgroup. And if those are all maximal normal subgroups, that means you can't fit any other normal subgroups in this series. So you can check whether a subnormal series is composition or not by looking at these factors. These simple groups are not. All right. Let's see. We've commented on some of these things already. So HI-1 is a maximal normal subgroup if and only if there's no proper normal subgroups in between them, which would mean that the series is maximal in that situation. Okay. And then the simple factors of a composition series are called composition factors. All right. I've illustrated all of those things now. So let's look at a few examples of this. Let's take the dihedral group D4. So this is going to be the symmetries of the square. And consider the following what I claim is a subnormal series here. So start off with D4 and then look at this Klein 4 group that sits inside of it. So I'll actually indicate this. We have this Klein 4 group, V4 here, which is going to be generated by the 22-cycle 1234 and the 22-cycle 1324, like so. I claim that this is in fact normal inside of here, right? Because D4 is a group of order 8. V4 is a group of order 4. And therefore the index of this group, D4 with V4, has got to be 8 divided by, that's a horrible 8, 8 divided by 4, which is 2. Every subgroup of index 2 is normal. So we do get that this Klein 4 group is normal inside of D4. Okay. But we can make that same argument again. If we take this Klein 4 group, take your favorite 22-cycle in that Klein 4 group. So take, for example, 1234, all right? This is a proper subgroup because this now has order 2. This has order 4. And the same argument applies. Since we have a subgroup of order 2 inside of a subgroup of order 4, the index has to be 2 in that situation. Therefore, we get that this is a normal subgroup of the Klein 4 group. But by contrast, right, this is only a subnormal series. This group, this cyclic group, Z2, is not normal inside of D4. It's normal inside of the Klein 4 group, but it's not a normal subgroup of D4. That's the important thing about these subnormal series. And then of course, the trivial subgroup is normal. It's normal inside of D4. So it's normal inside of everything else. But again, this is a subgroup of index 2. So it's going to be normal. So we do have an example of a subnormal series right here. I'm just going to abbreviate it like this. We have D4, which then contains V4, which then contains Z2, which then contains, whoops, the trivial subgroup. So we have something like that. Now let's look at all the various composition factors. Because I actually claim this is a composition series. How do I know that? Well, because I can look at the factors right here. When we take D4 mod out by V4, this is a group of order 2. So it has to be isomorphic to the cyclic group of order 2. And by similar reason, V4 mod out Z2 and Z2 mod out 1, these are all, all three of these composition factors are groups of order 2. And there's only one group of order 2 up to isomorphism. So all three of these factors have to be isomorphic to Z2, which Z2 is a simple group. It's a simple abelian group, mind you, but it's a simple group. So this does give us in fact a composition series where our three factors are going to be Z2. So we got Z2 and Z2 and then Z2. Those are our three composition factors. All right, looking at the next example right here, let's look at a different composition series for D4. So for example, we have D4 and sitting inside of that is the subgroup generated by R. This would be our rotation by 90 degrees. As a group, this is the same thing as Z4. This is the cyclic group of order 4 because R has order 4. Again, by considering indices, right? D4 is order 8, V4 is order 2. So the index of this group, the subgroup is going to be 2. That makes it a normal subgroup. Then we can look at the subgroup of Z4 that's generated by R squared. This is going to look like Z2 because the 180-degree rotation has order 2. By the same reasoning as before, this will be normal in the larger group because it has index 2. Now, unlike the previous example, this subgroup, the subgroup generated by R2 is actually normal inside of D4. That's not required to be a normal, a subnormal series. But this is actually an example of a normal series because all of these subgroups are normal because after all, the subgroup generated by R2 is in fact the center of the dihedral group, which is a normal subgroup. And of course, the trivial subgroup is normal as well. So this is a composition series because for the same reasoning, if you take D4 mod V4, that's a group of order 2, so it's Z2. If you take V4 mod Z2 or Z2 mod 1, again, these are all groups of order 2, so they have to all be Z2. So in this situation, we have our composition series, which has D4 inside of it. You have a V4, which is a normal subgroup. And then you have, of course, a Z2 down here, and then you have the trivial group again. And then all of these composition factors are, again, Z2, Z2, and Z2. All right. So some things to mention here that the first composition series we have is subnormal but not a normal series. So this is not a principal series. But when you look at the second one, it actually does turn out that each of these groups here are normal subgroups of D4. So this composition series actually is a principal series since it's a maximal normal series, which is kind of a cool little observation there. Like I said, we're not going to say too much about principal series. I just want to bring it up. But the thing I really want to emphasize here is that if you look at the composition series for this one and you look at the composition series for this one, it's the exact same. You got Z2, Z2, Z2 in both situations. So the composition series didn't matter on the composition, excuse me, the composition factors didn't depend on the composition series. You got the exact same ones. We're going to see this pattern occurring over and over and over again. And one thing I want to mention is that by the Seeloff theorems that we developed previously, every P group is going to have a composition series that basically looks like the following, right? Where you take your P group, then by the Seeloff theory, there does exist subgroups of order. So let's say that the P group has order of P to the K. Then there's going to be subgroups of order of P to the K minus one. Those will have to be normal inside of the P group. But then there's going to be subgroups of order of P to the K minus two, which will have to be normal inside of the previous group, not the whole group. And then by induction, you can go downward. So for a P group, you will always, always get a composition series whose composition factors look like ZP times ZP times ZP times ZP times ZP times ZP. And you're going to get K copies of ZP. So we can predict the composition series. Well, at least I can say I can predict the composition factors for any P group. D4 is such an example. All right, let's look at a non-P group. Let's switch to the cyclic group of order 60. Z60, of course. And so let's see what happens here. Well, since it's an Abelian group, every subgroup is going to be normal inside of Z6. So every composition series we construct in this situation will in fact be a principal series, even though we don't need it to be a principal series. That's what happens for finite Abelian groups here. And so we can look at the subgroups in this situation. So if you take the subgroup of Z60 that's generated by three, this will give us a cyclic group of order 20, okay? Then inside of that, you can take the subgroup generated by 15. This will be a subgroup. So the subgroup generated by three is Z20. The subgroup generated by 15 is Z4. And then inside of that, you have the subgroup generated by 30, which will look like Z2 in that situation. So you have all these cyclic groups. I'm going to write those again. Z60, it contains inside of it is Z20. Sitting inside of that is Z4. Sitting inside of that is Z2. And sitting inside of that is the trivial subgroup. So in summary, this is our composition series. Normality is not at worry because we're an Abelian group. Why is it a composition series? We'll look at the factors. If we take Z60 and we mod out by Z20, then that gives us a group of order three. That has to be the cyclic group of order three. So we get this first composition factor of three. If you take Z4 and excuse me, if you take Z20 and mod out by Z4, that'll give you a cyclic group. All quotients of cyclic groups are cyclic. That'll give you a cyclic group whose order is five. So that's got to be Z5. That's our second composition factor. Notice because the orders of these cyclic groups are primes, that makes them simple groups. If you have Z4 mod out by Z2, that gives you a Z2. And then if you take Z2 mod out by the trivial subgroup, since this is an additive group, the trivial subgroup has a zero this time, Z2 mod out by zero gives you Z2. And this gives you the composition factors, Z3, Z5, Z2 and Z2. I want you to notice that three times five times two times two is in fact equal to 20. So this, it turns out, and we'll see, well, let's actually look at another example, right? What if we take a different composition series, Z60, where this time we look at the subgroup generated by two, which this, I'll use a different color to indicate this one. This one's gonna look like Z30. Then you're gonna take the subgroup generated by four, which is the same thing as, that's a cyclic group of order 15. Then you're gonna take the subgroup generated by 20, which is as a cyclic group Z3, and then you take the trivial subgroup in that situation. So what our composition series looks like this time, you should take Z60, which contains Z30, which contains Z15, which contains Z3, which then contains the identity in that situation. The composition factors, well, if you take a cyclic group of order 60, mod out by a cyclic group of order 30, that gives you a cyclic group of order two. If you take 30 and you take, and divide that by 15, you're gonna get two. So another Z2. If you take 15 and divide it by three, you're gonna get five. And then finally, if you take three, and you don't take out, you don't mod out by anything, you're gonna get Z3. The composition factors are, again, one and the same thing. It doesn't matter what the composition series looks like. The composition factors are in fact the same things. And notice the exact same things, two, three, five, and three. We could predict what these things are going to be, because when you take a cyclic group, its composition factors are all gonna be cyclic groups, and you're gonna take all the prime divisors of your number, with multiplicity, right, because two shows up twice. In fact, one can argue that for a finite abelian group, this is the principle that you're gonna get, that the composition factors for a finite abelian group are gonna be exactly those cyclic groups of prime order, where you take all the primes of the abelian group up to multiplicity. All right, and so one more example here. If you take, for example, the symmetric group of degree five, S5, a composition series would be the following. You take S5, which contains inside of it, A5, which contains inside of it, the trivial subgroup. That's a pretty short composition series, but it is in fact a composition series for the following reason. If you take S5 mod out by A5, since the degree of A5 is two, that's what makes a normal set of S5 there, since it's a degree two normal subgroup, the code should have to be Z2, okay? But on the other hand, if you take A5 mod out by one, that gives you back A5. So, as we've proven previously, A5 is a simple group. And so, as Z2 is simple and as A5 is simple, that gives you a composition series. And it turns out that up to equivalence, this is the only composition series for S5, because as we'll see in just a second, the composition factors are determined by the group itself. It doesn't matter what the composition series are. The composition factors are always one and the same thing. If you take Z5, you always get Z2 and A5. Now, how do I know that? Well, I'm actually going to state the Jordan-Holder theorem here, which says that 82 composition series of a group are isomorphic. What does it mean for two composition series to be isomorphic? It means that there's a one-to-one correspondence between the composition factors where that correspondence is an isomorphism preserving correspondence. Basically, the Jordan-Holder theorem tells us all composition series of a finite group produce the same composition factors. It's a unique characteristic of the group. We're not going to prove the Jordan-Holder theorem in this video. It's not beyond the scope of our lecture series. It's mostly just a timing thing. I want us to be aware that the composition factors are uniquely determined by the group itself. It doesn't matter which composition series you choose. You're always going to get the same series, the same composition factors, excuse me. The series can change. This then leads us to the topic for this video here, the titular topic. That is of a solvable group. We say that a finite group is solvable if it has a composition series with cyclic composition factors. With those cyclic factors, of course, we'll have to have prime order because an abelian group is simple if and only if it is of the form zp, where p is a prime. A finite group is solvable if and only if its composition series only involves cyclic composition factors. Looking at some of the examples we've already looked at, if you take z60, all of its composition factors will get z2, z2, z3, and z5. This is a solvable group because its composition factors are all cyclic. This will actually apply to any abelian group because of the fundamental theorem of finite abelian groups. Every abelian group can be written as a direct product of cyclic groups. One can argue that when you have a direct product of groups, their composition factors will then be just the union of the composition factors for each of the direct factors in the direct product. Basically mimicking the strategy we did with z60, every cyclic group will have cyclic composition factors and therefore every abelian group is solvable. We also looked at d4. d4 was a solvable group because its composition factors were z2, z2, and z2. We also made the argument that every p group will have a similar composition series and thus similar composition factors. If you have a group whose order is p to the k, then your composition factors are going to be zp, zp, zp, and you're going to do that k times. For a p group, the composition factors is always going to be zp up to some number. Those are cyclic simple groups and therefore p groups are always solvable as well. Those are two very important families of solvable groups. There's lots and lots of solvable groups. The important thing to mention here is that if you look at the symmetric group, Sn where n is greater than or equal to 5, that is not a solvable group because if you look at A5 or A6 or A7 or An for n larger, greater than 5 in that situation, that is a simple group which is not cyclic. In that situation, you have a composition series that looks like Sn which contains An which contains 1. Your composition factors will be z2 which is Sn mod of An and you're going to have An in that situation when n is greater than or equal to 5. Since the composition factors don't depend on the series, this is what the composition factors have to look like and therefore we get a non-solvable group because we have a composition factor that's not cyclic. It turns out this observation right here, the special number 5, is exactly why degree 5 polynomials, ak-quintics polynomials cannot be solved, cannot always be solved by radicals but we'll provide the details of that in the next video.