View source for Transcript/Why There's 'No' Quintic Formula (without Galois theory)

We all know and love the quadratic formula. It tells us how to solve quadratic equations and all we need to know is how to add, subtract, multiply, divide and take square roots. Just plug in the numbers and outcome two solutions.

Pretty neat. Maybe it's less familiar that there is also such a formula for the cubic equation. It's a bit longer, but it's basically the same deal.

You just take the coefficients of the polynomial equation equation, you add, subtract, multiply, divide, take some square roots, take some cube roots and it tells you what the answers are. I mean, it's a bit longer, it's not quite as useful. And okay, it's admittedly not mostly how people actually solve cubic equations in the real world, but it's still pretty cool that it exists.

And there's even one for the quartic formula and that one really is quite a bit longer, but it's still the same deal. It's built out of plus minus times, divide and taking roots. Expressions like this are called algebraic.

Unfortunately, not all patterns go on forever and this one actually stops at five. I want to talk about the Quintic equation. There is no Quintic formula, okay? So that's not strictly correct, although roll credits, it's not all wrong.

There's no algebraic quintic formula, so there's no way of solving a general quintic equation with arbitrary coefficients just using addition, subtraction, multiplication, division, and taking nvuts or to give them their much cooler name radicals involving the coefficients. And today we're going to see why using a really remarkable argument. Due to the amazing ######## #####, I promise we're not going to use anything more than a little bit of knowledge about complex numbers.

In particular, if you've heard of this subject before, you may have heard of Galwa theory being involved. We're not going to need to touch Galwa theory, although the way that we approach the problem does have some deep relations with Galwa theory and maybe we'll mention those a little bit later. Let's have a brief reminder of almost the only thing about complex numbers we're going to need throughout this entire video.

So when we're first thought about complex numbers, we usually think about them in terms of their real and imaginary parts. So some coordinate x and some coordinate y and therefore we can plot them in the complex plane as a point. But this representation makes it quite difficult, quite ugly really, to multiply complex numbers together.

It's much neater if we think about the product of complex numbers in polar form, this means thinking about the distance that the complex number lies from the origin r and the angle that the point makes with the real axis theta. So these are called the magnitude and phase most commonly of the complex number. So this polar form is the way that we're going to mostly think about complex numbers.

And the reason why it's really neat is, as I say, that it makes multiplication a lot easier so if we take two complex numbers, r E to the I theta and R primed, e to the I theta primed and we multiply them together, well, we can just regroup these terms, right? We can take the two magnitudes, those are just real numbers and multiply those together. So the new complex number just has magnitude r times R primed, the product of the old magnitudes. But the phases also collect together in a really neat way.

So E to the I theta times E to the I theta primed actually factorizes together as E to the I theta plus theta primed. In other words, the phases of complex numbers add up when we multiply them. And this fundamental fact is going to underlie pretty much everything else that we use in the rest of the video.

Actually, it's really only a special case of this that we'll use most of the time, which is just what happens if we square or cube or take some complex number to some power. And we can see from the way that this phase is added that if we compute Z squared, for example, r E to the I theta squared, well, the magnitude is just R squared and the phase gets doubled, right? It becomes E to the two I theta. Okay? So you can see this pattern will continue, right? So if I compute z cubed, for example, the angle it makes with the x axis will be three times the angle that z used to make with the x axis.

And that's basically all we're going to need to know about complex numbers. This result, by the way, is ######'# theorem. So I suppose the starting point for our study of polynomials ought really to be to find out whether there are any solutions and if so, how many.

And this is a very famous result, and it goes by the name the fundamental theorem of algebra. So it says that if we have an nth order polynomial equation in one variable, then it has n solutions. And okay, there's a couple of little caveats we need here.

Firstly, they might well be complex, otherwise it's really easy to write down a quadratic equation with no real solutions. The other slight caveat is that, well, actually when we write down a general polynomial, it might have what we call repeated roots. So for example, z squared minus two, z plus one, it's a second order polynomial.

So it should have two roots. But if we factorize it, we get it in the form z minus one squared equals zero, and the only solution to that are z equals one. But in some sense it has that solution twice, right? If we write it in this factorized form, it has two factors of z minus one.

And that's actually really what we mean by this whole theorem. It's that if we write down an nth order polynomial, some leading term that's nonvanishing, so A isn't ever zero, then we can always just factorize it out. So there's a factor of a outside the whole thing and then a product of n terms, each of which looks like z minus something, and those somethings are the solutions of the polynomial equation.

Okay, so let's stretch our legs into the complex plane to see a proof of this. So suppose that the biggest term in our equation is a z to the n. Since this is really what makes it an nth order polynomial.

Only if this term is present that it's nth instead of n minus oneTH order, it must be somehow that this term is the most important thing improving our result, that this polynomial is going to have n solutions. So where does this term really shine? Well, thinking in the complex plane, this term involves the highest power of z. So it's going to grow most rapidly as z gets bigger.

So the idea is we should look at really large values of z and then this polynomial will be really well approximated by just that term alone, a z to the n. So let's think about moving z around a very large circle. The idea is that, well, we want to think about where z has a large magnitude so that this term dominates, but we also want to make use of the whole complex plane somehow.

So going around in a circle sort of seems to make some sense. I guess as a starting point it's a good thing to play with. Okay, so let's do that.

We take a z to the n, or for example, taking n equals two, we'll just take a z squared and ask how that changes as we move z around a large circle. Well, firstly we're going to have to zoom out a bit because the magnitude of a z to the n is always going to be much bigger and even taking n equals two squaring a number makes it quite a lot larger. So okay, we can zoom out, we're going to fit that on our picture.

But the other thing we have to think about is what happens to the phase of the complex number. Now z squared, remember, has a phase which is always twice as large as the phase of z. So if z goes around a circle once, then z squared, well, its phase has to go up twice as fast and therefore it's actually going to make two complete circles around the origin in that time.

And that generalizes, right? If it was a z to the n, then it would be some large complex number which goes around the origin n whole times. I guess we need to remember that this isn't actually going to be exactly what the polynomial is. It's not going to be exactly a z to the end.

There's going to be some correction terms. So just for an explicit example, let's look at this quadratic polynomial and you can see that, okay, now as z varies around some large circle. The whole polynomial does indeed still go around the origin twice, but it's not a perfect circle anymore.

These other terms just slightly correct it. Okay, so now let's think about shrinking down the path that z follows as we do that the path that the polynomial follows is also going to shrink quite rapidly, and it's going to become a worse approximation to think of it as close to a circle that's traversed twice in this example. So the shape has changed a bit, but we still haven't got down to where the solutions actually lie.

So let's just start slowly shrinking it more. Okay, this is the point at which there's actually a zero, right? The polynomial passes through zero. So there's a solution, and we can see what that solution is, right? It's just z equals two.

And indeed that's on the curve of z values that we saw, okay? And if we keep shrinking a bit more, we'd find that there's another solution, magnitude one. And then if we keep shrinking it down, eventually the polynomial also shrinks down to a single point, right? Because once we've shrunk the z circle down, we're just evaluating the polynomial at z equals zero, and that has a definite value. Okay, so how could we have known for sure that we were going to come across some zeros when we did this process? Well, it's precisely because we start with a circle or not circle, some sort of path that wraps around the origin multiple times.

And as we shrink that down, it shrinks to a point. And while either that point is the origin, or in order to shrink down to a point, it has to have passed through the origin. This is the argument that there must be at least 10 of the polynomial, right? As we shrink down this complicated path that the polynomial follows, it must intersect the origin eventually.

And that proves that there's at least one solution to the polynomial, and that's actually enough to get the whole result. So if we think about what happens for our quadratic polynomial, we do this once we find a solution, we can factorize it out, right? And the remainder has to be zero because the polynomial has to vanish. So it must be possible to factorize out the z minus two term from this polynomial.

But what's left? Well, it's a polynomial of degree one less. And okay, in our case, it's just a linear polynomial, and obviously that has a zero. We can just see what it is, and then we factorize the polynomial and found our two solutions.

But this would work for a higher order polynomial too, right? If we started off with an Nth order polynomial, we find one solution, but we can factorize that solution out. It'll give us an n minus oneTH order polynomial. And now we can repeat the same argument again.

We'll have another solution, we can pull that out, and then another solution. We can pull that out and we can keep going until we get to this factorized form. And that proves the fundamental theorem of algebra.

Okay, so we know that our quintic, our fifth order polynomial, has five solutions somewhere in the complex plane. But what's the problem we're actually trying to solve? Well, we're trying to go from the set of coefficients to the set of solutions and just dividing through by the leading term A in the polynomial, because that just rescales the whole problem. It's not really interesting.

We realize that what we're trying to do is go from five coefficients, the BCDEF coefficients, to the five solutions z one through z five. Say this is the thing that our hoped for quintic formula would do, right? But we're going to claim that that doesn't exist. Of course, going the other way is dead easy.

If I want to tell you what the coefficients are, given the roots, well, I just expand the brackets, right? I just take this factorized form and I multiply it all out and I can just work out what all of the coefficients are. But going the other way is somehow harder. Okay, so do you notice anything about the way that we went from solutions to coefficients, which suggests that it might be harder to go back the other way? The thing which I notice is that if you swapped around two of the solutions, say z one and z four, then you'd still get the same polynomial, right? Because you just rearranged the terms in this factorized form.

It's the same expression. So Bcde and F are the same, but the set of solutions has been switched around. But that's fine, surely, right? Can't we just pick an order? Well, actually, one of the most interesting things about solving polynomial equations, and this idea is at the heart of a lot of other things to do with polynomial equations, including the traditional Galwa theory approach to this whole topic, is that there really isn't a natural way of ordering the solutions to a polynomial equation.

Let me show you what I mean. We'll study the more familiar quadratic case just to get some sense of what I mean by this. So let's think about the equation z squared minus W equals zero or just z squared equals W.

Quadratic equation has two solutions. And for example, if W is four, they're the two square roots of four, which are plus two and minus two. And maybe you will choose z one equal to two and z two equal to minus two.

The game we're going to play is I'm going to continuously change the equation and you're going to have to continuously change the solutions to the equation. So I'm going to do this by increasing the phase of W. Now, since z squared equals W and the phase of z squared changes at twice the rate of z, you're going to have to change the solutions while changing their phase at half the rate that I change the phase of w, right? So you'll have to choose this one to still be z one on the right and this one to be z two on the left.

So I'm going to keep doing that. Get round to W equals four I, and then all the way around to W equals minus four. And at this point, the two solutions are two I and minus two I, the two square roots of minus four.

But you have to choose z one equals two I and z two equals minus two I because, well, you started with particular choice of z one and z two and it's smoothly changed. So what am I now going to do? Can you see the trick? Trick is just I'm going to continuously change W all the way back to where it started, I'm going to change it back to being four, but I'm going to do that by completing a circle around the origin. When we're finished, what's going to happen? Well, z one is going to be equal to minus two and z two is going to be equal to plus two.

I've tricked you into reordering your set of solutions, the same set of solutions that we started with two and minus two, but you've chosen them in the other order. And that's what I mean when I say that you can't consistently choose an order for the solutions of this polynomial. What I mean is that by smoothly changing the equation, the solutions will change in a smooth way, but they won't necessarily go back to same ordering as you started with.

So this property means that the quadratic formula, which has to tell us what z one and z two are in terms of W, can't be a proper continuous function, right? Because it's multivalued in terms of w. By insisting that you change the solutions continuously, that you change z one continuously, I've proven that it can have two different values at w equals four, therefore it's not what we usually mean by a function. So this is an important idea.

So it's going to underlie a lot of the rest of what we're doing by going around the origin. This root, the square root that we're taking, the quadratic formula's output, if you like, does not get back to where it starts, even though all the coefficients inside that formula do get back to where they start. So this corresponds exactly to the fact that the quadratic formula has a square root in it.

We even write it with a plus minus sign to remind ourselves that there's multiple options for which square root we might have to take. And actually, if you think about it, we've proven that the quadratic equation, this particular quadratic equation in fact, cannot be solved using a formula that only involves addition, subtraction, multiplication and division mixed up with the coefficient w in some way. Because if it was of that form, then it couldn't be multivalued, right? None of those are multivalued operations.

So it would always just say, if you give me this w, then this is the value of z one. Okay? So since no such expression can have that weird multivalued property, it must be that the quadratic formula involves some additional ingredients. And it does, it involves square roots.

But if we are allowed to use any root we want, write down some crazy algebraic expression that involves nested fourth roots of 7th roots, added to fifth roots, and all sorts of garbage like that, how could we possibly argue that that couldn't reproduce multivalued nature of the solutions of polynomial equations? Well, it must be somehow, if this is a helpful way of approaching the problem at least, that algebraic expressions are somehow limited in how multivalued they can be, and that if we can show that a polynomial has solutions which are somehow more multivalued than any algebraic expression will have proven, that there's no quintic formula. Maybe that's the idea. So remember, we can permute the solutions of a polynomial in any way we want at all, and we will get back the same coefficients at the end of the day.

So our formula written in terms of those coefficients must be multivalued in such a way that it can reproduce the results of any process of exchanging the solutions of a polynomial. So we clearly need to understand roots better nth roots, if we want to show that they can't reproduce those arbitrary ways of mixing up the solutions of a polynomial. So we're going to have a little bit of a think about the nature of the multivaluedness of Nth roots.

Let's visualize the fourth roots of two. So the four solutions have the same magnitude, which is about 1.19, something like that what you'd get if you press the square root button a couple of times on your calculator.

There are three other choices, of course. One that's just the negative of that, one that's I times that, and one that's minus I times that. So these are the fourth roots of unity multiplied by 1.19,

right? They're the fourth roots of one, but made a bit bigger. They have phases, zero quarters, one quarter, two quarters, and three quarters of a full circle. The reason for that is, well, when we take a fourth power, they all get multiplied by four.

So they all become an integer number of multiples of two pi, which means that they're all real numbers when we put them to the fourth power. So this is a helpful way of thinking about the fact that they always end up being evenly spaced around some circle. Okay, anyway, so we're going to think about what happens as we smoothly change the quantity we're taking the fourth root of, and just as before, the phase of each root is also going to have to increase smoothly.

But now, at a quarter the rate of the thing that we're taking the fourth root of. So if we, for example, move a whole time around the origin, then you're simply going to end up rotating the four solutions one place around. And if we go around again, well, they rotate one place further and as we go around the origin the opposite direction, they just rotate back in the other direction.

And if we vary the quantity you were taking the root of in any other way, changing its magnitude, but without going around the origin, well, the roots just end up back where they started, right? So this is, as you might hope for, a kind of tame multivaluedness, right? We can predict the way that any nth root changes just by asking how many times the quantity inside the nth root goes around the origin. And we're going to avoid thinking about any path that takes the quantity inside the root to zero, because then all of these roots sort of merge together and lose their identity. So whenever we're choosing paths, just imagine that we're choosing paths which keep everything inside roots nonzero.

How can we use this neat property? This is probably the biggest idea in the video and it's really quite subtle, so it's worth taking some time to get your head around. Our starting point is going to be the argument we made about quadratic equations just a few minutes ago. So remember we claimed that the quadratic equation cannot be solved with just a rational function of the coefficients of that equation.

And the argument was just, well, imagine that there was such an expression and that it didn't involve any roots, then it would be a continuous and single valued function of those coefficients. Right? Okay. But then we thought about a particular example of a polynomial equation.

We said, let's think about an equation which has solutions plus two and minus two, and then let's smoothly switch those two solutions around. So if our formula for z one is continuous as a function of all of the things in this equation, well, then it must end up swapping to the other root. So if z one is plus two at the start, then it ends up being minus two.

But that's a problem, right? Because the equation actually at the end of this process has got back to where it started. Since all we've done is switch the roots around, the actual A, B and C coefficients didn't change. And therefore, since z one is a single valued function of all of those coefficients, it must also have got back to where it started, which is a contradiction.

And that's what guaranteed for us that the quadratic equation must have some sort of nth root in its solution. Okay, so that's all well and good, but we want to think about the Quintic equation. And let's think about how we might try and prove, for example, that you can't solve it using an algebraic expression that contains just a single root of Bcde and F.

And, well, I guess we'd start by doing the same thing, imagining that z one was such a function and then asking what properties it has so it would still be continuous. This nth root hasn't ruined that property, but the nth root does mean that it can be multivalued as a function of these coefficients. Okay, so then we have to make use of these properties.

The first part, I suppose, goes the same, right? So we just have some polynomial with five solutions which are whatever we want them to be, and then we can swap them around in whatever way we choose to. And all we have to do to make an argument like this work is just make sure that at the end of the day, z one has changed to z three. Perhaps, but then what about the other stage? Well, we need to guarantee that our formula for z one hasn't changed in order to get the contradiction and prove the theorem.

And that's where we want to somehow use this tameness of this route, right? We want to argue that even though an nth route could be multivalued for some reason, the clever way that we've chosen to swap these things around means that it isn't multivalued. Because if that's true, then we have the same tension between the fact that z one does change by continuity, but that on the other hand, it doesn't change because of some nice property of the function and then we get a contradiction and our result. So now we know what the right question is.

Question is, how do we swap the solution z I around without changing any nth roots of the coefficients bcde and F. So what permutations must leave an nth root of the coefficients unchanged? So I want you to imagine that you're given some you're going to take the nth root, say, of some complicated function involving plus minus times and divide some rational function of the coefficients of a quintic formula. And you want to find a cunning way of switching around the solutions to that corresponding polynomial such that that nth root doesn't change even though it's an nth root and can be multivalued.

You want to prove that there's a particular permutation which leaves that nth root unaltered at the end of it. Let's say we pick two of these points, z one and z two, and we just swap them around in some smooth way. And as we do that, the quantity R that we're going to take root of is going to change in some way and maybe it goes around the origin, for example.

That would be most interesting thing that it could do. It could go around the origin some number of times and pick up some total phase like two pi times some integer m. At the end of the swap, the coefficients are the same, so the value of R must be the same, but it can follow an interesting path as we're altering the actual solutions of that equation.

Right, okay. So we need to now make sure that the nth root of R doesn't change, but so far it might have changed, right? It can change by two pi m divided by N because it's an nth root. It's like a power one over n.

It can change at one nth, the speed of the quantity we're taking a root of. So what could we do? Well, the simplest thing we could do, I guess, is just exactly undo the permutation. We did the same permutation of z one, z two, but following the same path in reverse.

And as we do that, R is going to follow the same path also in reverse. And therefore it's going to pick up a phase which is just minus two pi M. And that kind of works, right? The total phase change of R is zero.

So the phase change of the nth roots is also zero because it's gone around the origin M times one way and then M times the other way. So the root doesn't actually change at the end of the day. And that's kind of neat.

But of course it hasn't helped us because the set of solutions has got back to where it started. So we've proven that the root isn't multivalued, but the set of solutions also hasn't changed. We haven't relabeled them.

So we don't need the nth root to be multivalued in order to reproduce this set of solutions. That hasn't helped us prove anything. But can you see how this idea of doing and then undoing a permutation could be slightly generalized to something which does help us because it doesn't actually leave the set of solutions untouched.

So think about that for a minute. Can you find a slightly less trivial thing that involves doing and undoing permutations? Did you get it? Suppose we do one permutation, pick up some phase like two pi M like before, and then we do a different permutation, we pick up a different phase in R, right? So the the R is going to follow a new path because we're changing a different set of solutions, different pair of solutions around, say. And we're going to wrap the origin maybe some different number of times as we do that.

But next what we do is undo the first permutation and then undo the second permutation. That involves following those two paths that are followed in reverse, right? Same permutation process, but done the opposite way around. Okay, now this is what's called a commutator doing A, doing B, undoing A, undoing B.

And you can see on the left hand side, it genuinely switches around the set of solutions, right? Even though in some sense we've done and undone every permutation by doing them in a clever order, we actually alter the ordering of the set of solutions. But now think about how the nth root of R. Has behaved well.

We accumulated some phases and then we accumulated the negative of those phases. And therefore the total phase change of R, total number of times it's wrapped around the origin, counting plus and minus signs for anticlockwise and clockwise directions is zero. The phase of R has net not changed at all, and therefore an nth root of R also doesn't change.

And this is exactly what we wanted. Right? We've managed to choose a way of switching around the set of solutions which does something interesting to R, but at the end of the day guarantees that the nth root of R gets back to where it started. And that is exactly what we need in order to move on with our argument.

So, as I mentioned, this is something called a commutator. You do one thing, call it sigma, you do another thing, call it tau, then you undo the first thing sigma inverse, and then you undo the second thing tau inverse. So do A, do B, undo A and do B.

This sort of operation is maybe most familiar to people who've played with Rubik's Cubes, because quite a common kind of move is actually a commutator move. It involves doing a rotation of one face, a rotation of another face, and then undoing the rotation of the first face and undoing the rotation of the second phase. And it's kind of useful for the same sorts of reasons it's useful for us here.

It changes some stuff, doesn't leave the cube completely unchanged, but it doesn't change too many things. It's very useful building block when you're trying to find interesting ways of permuting objects. So where have we got to? Well, actually, what we've done is prove that no cubic or quartic aquintic formula can exist that contains only basic algebra, addition, subtraction, multiplication, division and a single nth root, because we know how to change around the solutions such that they end up permuted around.

But the results of any possible formula that looks like this don't change around. So that's quite a beautiful result and quite a clever way of arguing it. But didn't we write down a cubic formula earlier? Yeah, we did, but it looks like this.

And notice it involves complicated expressions involving nested roots, and we haven't ruled those out yet. We have actually ruled out things that involve adding together or multiplying or dividing roots, because we can choose a path which individually makes each one of those roots get back to where it started. And then when we add, subtract, multiply and divide them, well, that doesn't help us build anything more multivalued.

So we've ruled out anything that doesn't involve nested roots. So why is it that nested routes somehow evade our argument? Well, think about what happened before. We firstly argued that expressions involving the coefficients get back to where they started as soon as we switch around a pair of solutions.

Any permutation of the set of solutions leaves an expression involving the coefficients unchanged, but doesn't necessarily leave an nth root of those coefficients unchanged. On the other hand, if we follow a commutator move, so that means the coefficients get back to where they started after sigma, get back to where they started after tau, and then we undo those with sigma inverse and sigma tau. And so the coefficients have actually been back to where they started, like four times.

That guarantees that an nth root of that expression gets back to where it started. But now that works for the inner route, but it doesn't work for the outer root. Because whilst the coefficients that were in the original equation, the B's and C's have been back to where they started four times in this clever commutator pattern, this nth root, the inner nth root only just got back to where it started.

Or at least that's all we can guarantee. At the end of the commutator, this nested route is the same as where it started. But that's just like a single path that wraps around the origin, maybe.

And then this outer route may be multivalued in terms of that. So the outer route can have changed at the end of our commutator. But this kind of suggests what we might like to do in order to prove that there isn't an expression involving nested roots, for example, for the Quintec, right? Where you need to rule out nested roots to prove that there's no quintic formula.

Can you see what might be a sensible idea? So we've shown that an expression involving the coefficients is single valued. An expression involving an nth root of those coefficients is single valued after a commutator. So an expression involving a root of a root of the coefficients is single valued.

If we follow a commutator of commutators, that is to say, imagine we do one commutator that leaves the inner route unchanged. Now we do a different commutator, which also leaves the inner route unchanged. Then we undo the first commutator, and then we undo the second commutator.

That's quite an elaborate process. But at the end of it, this outer route must have been left invariant by exactly the same argument as before. You just think about the inner route as being a quantity which gets back to where it started after each of these commutators.

And because it's a commutator of commutators, the root of that quantity must also get back to where it started. So that's quite cool, I think. But a commutator of commutators is quite complicated, right? You could write it out in terms of permutations, and it would be a real proper mess.

But okay, what can we do? We know that a commutator of commutators is going to leave expressions involving once nested roots unchanged, but there is such a formula for the cubic equation. So therefore it must be that all commutators of commutators, when we have three solutions to move around, are actually trivial, because the cubic formula will be single valued when we follow a commutator of commutators. Therefore, it must be that the set of solutions of the cubic equation get back to where they started.

So there's a little challenge for you. Can you prove that all commutators of commutators involving three points are trivial? They get the points back to where they started. Have a go at that.

It's quite a fun exercise. Do you want a hint? A hint would be to first show that all commutators of three points are a cyclic permutation. So whenever you do any commutator of any kinds of permutations of three points, you always just get a cyclic permutation that maps one to two to three and three to one or same thing in reverse or just leaves everything where it started.

Okay? So if you follow that hint, what you should find is that the possible three point permutations that you can write down, there's six of them. You should find that whenever you do one another, undo the first one, undo the second one, you get a cyclic permutation. We already showed that the commutator of two swaps z one, Z two and Z two, Z three.

For example, we know that that's a cyclic permutation because we saw it earlier. So you only need to check a few other cases, really in order to prove that first result. And then the other result is just well, what happens when we take a commutator of those cyclic permutations? Well, cyclic permutations actually commute with each other.

It doesn't matter what order I do cyclic permutations of three points in, it has the same result. So for example, if I commute them one way clockwise and then one way anticlockwise and then undo the first thing one way anticlockwise and then one way clockwise, everything ends up where it started, right? And if you just think about that, that's basically the only case that you need to worry about until all commutators of commutators just leave the set of three points exactly where they started. And that is why there can be a cubic formula with once nested roots, because commutators of commutators of three points don't actually change them.

By the way, if you do want a little bit of group theory and to make some connections with Galoir theory, this is where that starts becoming really clear. Because this is what's called the derived series of the group of permutations of three elements. So if you take all possible permutations three elements, you get this thing on the left that's the group S three itself.

The next thing along is what you get by doing commutators. So if you're only allowed to do commutators of elements and you can do a few commutators of elements if you want, then you find that you get another group, which is the first element in the derived series of group S three. And it happens to be the group of cyclic permutations C three.

And if you do that one more time, you ask, what can you build out of just commutators of commutators? Then you find there's only this trivial permutation. And this set of three groups is called the derived series. And this is exactly the sort of thing you worry about when you do Gala Theory to prove this same result.

But, okay, I promised no advanced Galawar theory, no group theory. So that's just a little note for the interested amongst you. Okay, so what's next on our journey? Well, there was a quartic formula, right, that had roots of roots of roots inside it.

So that suggests that there may be a couple more challenges I can give you, which are, firstly, to show that if you have four points, the four solutions of a quartet equation to mix up, then there is a commutator of commutators that mixes up those four points in some nontrivial way. The challenge is to find such a thing and that would rule out something that looked like the cubic formula, where you just have one level of nesting going on. So a root of a root that is indeed true.

Have a go. And the other thing to check is that there's no commutator of commutator of commutators which mixes up four points, and that allows this formula to exist, this quartet formula to exist with only one extra level of nesting. So, yeah, have a little bit of a play with drawing permutations of four points out on the piece of paper and see if you can prove those results.

What about the quintic? I claim that there's no way to express its solution using any small print finite number of nested roots. How could we possibly prove this result? Well, actually, all we need to do is prove that there are commutators of commutators of commutators of commutators that genuinely mix up, switch around a set of five solutions. If there are such arbitrarily nested commutators, then, well, if you give me a possible quintic formula that involves, say, nothing deeper than an M level nested root, then I can just look for commutators of commutators of commutators also M times nested.

And that will prove that your formula is not multivalued when I follow that complicated commutator pattern. But since there is such a commutator that does mix up the points, again, I've ruled out the possibility of such a quintic formula. So can we actually find commutators or commutators or commutators, et cetera, of five points, which genuinely mix up a set of five points? Yes, we can.

And I totally challenge you right now to go and have a go at arguing that, yes, there is a way of doing a commutator of commutator, a commutator of so on, of five points, which leaves them changed at the end of it. Okay, have a go. And there's a hint if you want it, which is to think just about three points at a time.

So in particular, suppose that you want to obtain a cyclic permutation of three points. I claim that it's possible if you have five points, not if you have fewer than five points, to obtain that cyclic permutation as the commutator of two other cyclic permutations of three points. So give that a try.

It's quite a satisfying result to derive, and it's just drawing out some pictures and working out how certain permutations act on set of five points. Go ahead. Okay.

I hope you had a go at that. I think it's a very fun, satisfying thing to see play out. Let's just illustrate it.

And I'm going to make it slightly neater just by ordering the points neatly. Okay, so that's better. Let's see how we can obtain a permutation of one, two, three cyclically around as a commutator.

So the first thing I'm going to do is take these points on the left hand side. I'm going to switch them around one place. And now I'm going to go over here, do something different.

I'm going to switch these three points around also one place, and it's a commutator. So I have to undo the first thing and then the second thing, which happily gets four and five back to where they started. And lo and behold, one, two, and three have been permuted one place.

And this is exactly the result I said I wanted. Right. I've now managed to obtain a cyclic permutation of three points as a commutator of cyclic permutations of three points.

And that's actually pretty much the last stage of the proof. All we need to do is just use this to build arbitrarily nested commutators. But can you see how I do that? I know how to write cyclic permutations of three points as commutators, but now, since they were commutators of cyclic permutations of three points, I know how to rewrite each of those commutators as a commutator itself.

So this sigma has as a representation as a commutator of cyclic permutations of three points, and so does Tau. And therefore, actually, I have a representation of this one, two, three permutation as a commutator of commutators. And I can just keep going, right.

I now have a way of finding a representation of a one, two, three cyclic permutation as an arbitrarily nested commutator, and therefore I can mix up the set of solutions whilst leaving an arbitrarily nested root completely untouched, and therefore there is no quintic formula. QED. I think that's a really beautiful argument.

Well done, #######. Good job. I mean, I always used to think you needed Galwa theory to prove this result, and it's a lot more machinery to develop to understand that proof, whereas this approach that just involves complex numbers, thinking about basic properties of those, and some clever reasoning about multivaluedness and continuity, somehow much more satisfying.

It's a little bit less powerful in some ways than Galwa theory, because Galva theory allows you to hold up a particular example of a polynomial and say this polynomial with these numeric coefficients doesn't have a solution in terms of radicals of nth roots of rational numbers. But our approach has its advantages too. It, for example, rules out adding any exponential, for example, into this some e to the z type function.

You can't have a quintic formula that involves nth roots, some basic algebra and e to the z because that's also a single valued continuous function. So by the same arguments, you can't build a quintic formula using it. But okay, whatever your tastes, I hope that you think this is a nice argument and it leads on to develop a bunch of the ideas that you sort of study in more advanced group theory in quite a satisfying way.

It really tells you why you should be thinking about commutators, what's interesting about them, what's special about them. And yeah, I think it's really valuable for that. I think maybe it's nice just to give you some additional maths words to drape around what we've looked at today.

What we've shown is that the permutation groups called the symmetric groups s one, S two, S three and S four which just the different ways you can permute one, two and three and four objects are all what's called solvable which means that taking commutators of commutators of commutators of commutators eventually you only get trivial operations from doing this. And that's not true for S five. It's not a solvable group.

And yeah, so there's instead you get stuck, right? You take commutators of commutators. But there's a whole bunch of different permutations of five points which can be written as arbitrarily nested commutators. It's not quite all of them.

So if you'd like to try something out, you could try showing that it's actually exactly half of all possible permutations of five points. You can realize by taking commutators of commutators, of commutators of commutators multiplying those together, you can only get half of the possible permutations of five points. Trying to work out how to describe which half you do and don't get is quite a nice challenge if you've not seen much group theory before.

But anyway, so you do get a group out of this. It's just a particular set of permutations with some structure, and it's called a five. The things which you can get using this approach.

And that's what's called a perfect group. So it's its own commutator subgroup. If you take that particular collection of permutations, then anything within it can be realized as a product of commutators of things in that group.

That's what we call a perfect group. Okay, so I think that's maybe enough jargon for one day. Okay, I hope you enjoyed hearing this argument.

I thought it was a beautiful argument when I first learned about it from a video and paper which I'll link to below. I encourage you to have a look at those. Yeah, it's just such a nice way of approaching this problem.

And so clever as an argument and quite hard to get your head around the first time. So if there's anything I could clarify or if I made any mistakes, please do let me know in the comments below and I'll hopefully come back to you quickly and be a little bit more helpful. But, yeah, please do like video subscribe if you enjoy this sort of content.

And see you soon. Goodbye.