Many things in this video are WRONG! Duplicating @sbares 's comment: Gal(x^7-2, Q) is not C7 as you claim, it is a group of order 42, as is seen by the tower of extensions Split(x^7-2,Q) = Q(2^(1/7), exp(i2pi/7)) > Q(2^(1/7)) > Q where here ">" means "extension of" (sorry, can't type TeX in youtube comments lol). The minimal polynomial of 2^(1/7) over Q is x^7 - 2, so this extension is of order 7, and the minimal polynomial of exp(i2pi/7) over Q(2^(1/7)) is the seventh cyclotomic polynomial, so the order here is phi(7) = 6. Thus |Gal(x^7, Q)| = [Split(x^7-2,Q) : Q] = 7*6 = 42. There's a similar problem with (x^7 - 1)(x^5 - 1). The minimal polynomial of a seventh (or fifth) root of unity over Q is not x^7 - 1 (or x^5 - 1), it is the seventh (or fifth) cyclotomic polynomial. It is not too hard to see that the splitting field here is the 35'th cyclotomic field, so the Galois group is in fact (Z/35Z)*, which is isomorphic to C12 x C2 Finally, the example x^5 - 2x + 1 factors as (x-1)(x^4 + x^3 + x^2 + x - 1) so the Galois group is not S5, it has to be a subgroup of S4 (In fact, Sage tells me it should be the whole S4).
There are some errors here. At 7:20 you say that the Galois group of x^7 - 2 over the rationals is cyclic, but it's not. It's dihedral with an order of 14. At 8:10 you make a similar error for (x^7 - 1)(x^5 - 1). The orders are 6 and 4. At 12:39 you mention that the Galois group of x^5 - 2x + 1 is S5. But it has a root of 1 and is reducible to a linear and a quartic, for which there exists a formula. It's therefore S4.
I overall agree with your quite useful comment, but the Galois group G of x^7-2 is not dihedral with order 14. Indeed, the splitting field of x^7-2 is generated by the (real) 7th root of 2, along with the (all complex, except 1 of course) 7th roots of unity, whose Galois group is (Z/7Z)*=Z/6Z=C6. The degree of the extension (and so the order of the G) is therefore 7*6=42. Moreover, if you call r2 the real 7th-root of 2 and z=exp(2ipi/7), then Q(r2,z)/Q(z) is Galois with degree 7, so has Galois group C7. As the intermediate extension Q(z)/Q is Galois, the subgroup C7 is normal in G=Gal(Q(r2,z)/Q), with factor group Gal(Q(z)/Q)=(Z/7Z)*=C6. Now, by the Schur-Zassenhaus theorem, as 6 and 7 are relatively prime, the group G splits as a semidirect product of C7 by C6. In particular, it cannot by dihedral. By the way, the same arguments apply for polynomials of the form x^p-n where p is an odd prime and n is an integer different from -1,0,1: its Galois group is a semi-direct product of Cp by C(p-1), the latter acting on Cp through the isomorphism C(p-1)=Aut(Cp). [sorry for the long comment]
Well, if the École Polytechnique's examiners weren't so shitty and his father hadn't comitted suicide, maybe Galois wouldn't have gone to that duel. Anyway, it was definitely NOT just a matter of thinking that a duel was necessary and dying. While Galois was probably a genius, a bunch of mediocre people have a lot more shameful or questionable behaviors or words than dying in a duel and still are recognised are geniuses. Dying in a duel is actually very honorable.
I don't know how much mileage I'd get out of this if I didn't have an undergrad course in modern algebra, but from perspective of someone who knows what groups and fields are, but never encountered Galois groups before and their relation to polynomials, this is fantastic. I just wish it'd push just a little bit further into why S2, S3, S4 can always be decomposed into a product of cyclic groups, even if just as visualization of some special cases.
6:23 I'm afraid that the Galois group of X^7-2 consist of 42 elements, not 7 (I believe it is equal to the semi direct product of the cyclic C_7 and C_6) . Apart from that, great video!
Overall I'm always happy to see more math content creators on youtube, and I'm excited to see future videos from you. Most of the Galois groups are actually calculated incorrectly here, and those kinds of details should really be corrected/verified before creating a video like this. Ignoring that, here's a few notes and nits, though (reading this over I'm worried that I'm coming off too harshly but I promise I did enjoy the video and would like to see more :) ) One big question I'm left with after seeing this is - who is your intended audience? Curious high schoolers? Undergraduate math majors? Undergraduate non-math STEM majors? Any curious undergraduates? Graduate math students/Post graduates in mathematics? My guess is that it is meant for curious high schoolers or undergraduate non-math STEM majors (particularly because of the commuting functions analogy and notation), perhaps with some math majors as well. I think this question needs to be addressed quite carefully, because it will address the question of how much rigor your videos require, which I think was probably the primary weak point of the video (and is imo the point that most math videos on youtube struggle with, even for big channels like numberphile). I'll try pointing out stuff as I see it through the video: 0:00-0:50 I think this is a solid intro, motivations for the subject are definitely clear. I think it's a *little* disingenuous to say that we'll "answer the question today using Galois theory", since it's really more of taking a peek at the theory that needs to be developed in order to answer the question, but all good intros are probably a little disingenuous in a clickbait-y way, so I think this is fine. 0:51-2:34 I think this is pretty good, and a fine introduction into the idea of field extensions. However, I think it could be a *little* clearer about the finite operations thing. 1+sqrt(2) is an example, but it's also probably worth showing something like 3*sqrt(2)+(4-sqrt(2))/(5+sqrt(2)). It's also somewhat nonobvious that this ends up being equivalent to the set x+y*sqrt(2) for rationals x, y (division being the nonobvious part). If the intention is to just briefly wave Galois theory in front of the audience, then omitting such details is probably fine, but it's worth at least pointing to the parts that you are handwaving over to acknowledge that they have been handwaved (textbooks do this with the classic "(Why?)" inserted mid paragraph). 2:35-3:50 I think this is good; perhaps the idea of extending Q by the roots of any arbitrary polynomial was glossed over a little too quickly given how central the concept ends up being to the rest of the video (and the topic generally). 3:51-5:33 This part is fine, but it feels a little unclear as to what purpose it is serving in the overall video. I imagine it's trying to grow some intuition about how finite cyclic groups work when your elements are functions wrt repeated compositions, but this only feels like it is showing this connection as someone who has already seen it. It is not super clear to me whether an uninitiated student would be growing this intuition by watching this section. 5:34-5:50 Alright this is probably the biggest handwave of the entire video. I think building up the notion of what exactly the "symmetries" of an equation means is quite involved, and is not accomplished just by looking at the sqrt(2) -> -sqrt(2) example. In your defense, I think many textbooks also use this example and pretty much only this example, but it really is too complicated a concept to glean from just this example. There is a lot about field extensions and automorphisms that is being omitted here, and the viewer probably should be aware of this omission. Also, the notion of a "group" is kind of just introduced without any definition. 5:51-6:40 So, as other commenters have mentioned, this is actually not accurate. One natural question an attentive student might have is, "Aren't there 7!=5040 ways to map the roots of this equation to each other? Why do we only care about the transformations that take 1->2, 2->3, etc.?" This also gets a little more muddled since we are extending Q by the 7th root of 2, in addition to a primitive 7th root of unity. Using f(x)=x^7-1 here was probably better. 6:41-7:40 I think this is good. Minor nit: I think the numbers on the dial should probably be filled in with white or something, it can be a bit hard to read sometimes with the lighting. 7:41-8:17 I like the combination lock analogy, but technically the galois groups of these are incorrect. x^7-1 has a galois group of Z_6 and x^5-1 Z_4. Z_7 x Z_5 would have still been cyclic, btw. 8:18-10:19 I think this is okay, but this is one of the areas where you are probably shifting audience levels. Knowing that function composition doesn't always commute is pretty standard for math majors, perhaps not obvious for high schoolers and should be known to at least a good amount of STEM undergraduates. Prior to this point the video seemed good for all three audiences, but here it's appealing a bit much to one demographic and perhaps spending a disproportionate amount of time on it. In general I do like the clothes-wearing analogy for function commutativity. I think personally I would have liked this point in the video to justify why exactly these automorphisms (ie symmetries) commute rather than learning about what commutativity is. The rest of the video is fine apart from the galois group computation errors, I think the transition at 11:54 is a little awkward since you go through an example where a group isn't abelian and then talk about groups where you can construct them from cyclic extensions, which are necessarily abelian, without any word like "however" or "on the other hand" so it feels like you're talking about the same thing. The chaining of combination locks was good, and I think it captures the notion of direct products of cyclic groups well. Other minor nit is that the music is not loud enough to add much to the video but probably not quiet enough to be totally ignored. Overall I think the video is good; since it's your first, it's natural that there will be some feedback. At the end of the day I'm just a random dude with some feedback. My algebra is not super strong so I may have made mistakes in this comment as well. I hope you keep the spirit of this video and continue to make more, looking forward to seeing what your channel provides! :)
It's very good that you commented on some of the more severe mistakes in this video. It appears that the creator confused the Galois group of a field extension and of a polynomial equation (which may differ, when only adjoining one root and not using, say, primitive elements) at some points, or rather didn't check his computations. There is a relevant post on reddit ( www.reddit.com/r/math/comments/kk7cde/galois_theory_explained_visually_the_best/? ) also talking about the problems. (Note: I really loved the video but I think the number of mistakes is problematic)
I am doing an MSc. in maths and found this video fantastic. Certainly won't make up for your maths course and hours of drilling exercises, but it's nice to see this stuff explained in normal non-convoluted language for once.
@@pauls.2451 I agree that it is nice to provide some "disillusionment" from how scary higher level math seems, but that kind of thing needs to be done with care. If you stray too far from rigor and too close to "wow" factors you end up with some problematic videos like numberphile's infamous disastrous -1/12 video, which has forever plagued the way many laypeople see math. Much of the best parts of this video build some intuition about cyclic groups and their direct products, but give a bit of incorrect intuition about what exactly Galois groups are. Don't get me wrong, I love the fact that someone is taking on the challenge of explaining Galois theory to laypeople, because I think this type of thing can really get people excited to see what math is really all about. But it requires a great deal of attention to detail to make sure that you are really conveying what should be conveyed, rather than what people will digest easily at the expense of precision. Btw, if you or anyone else is looking for a layman friendly book to get into the subject, I've worked through quite a bit of "Field Theory and its Classical Problems" by Hadlock and it's really quite good. No prior algebra knowledge assumed.
wrt? With regards to? Honestly, if you to write a thousand word youtub comment and decide to be cute by abbreviating three words; I kind of don’t respect you. Jmoymmv.
Dear Aman, Assalamu Alaikum, in case you are a Muslim and "Greetings," in case you are not. You seem to know so much about the Galois theory. Why do you not make some of your own videos. I hope I can have email correspondence with you and ask questions about this subject. I cannot give my e address explicitly. it will be erased so: ali.jamily1at g.come. thank you
Oh damn, this is really good. The combination lock idea is brilliant. The way a radical creates a cyclic group is also why the scale we use in music works the way it does. Since we use twelve tone equal temperment, each note is 2^1/12 apart. Once you stack 12 together you double the frequency. The 12 possible pitches form a cyclic group symmetry.
That doesn't have much to do with radicals though. If anything it's a coincidence related to auditory perception of frequency being logarithmic. Sure 440 Hz and 880 Hz are A4 and A5 but they are two different numbers and two different notes despite being harmonically similar. In the mathematical sense its the 12 complex 12th roots of 2 that form the cyclic group but when considered as complex frequencies they actually correspond to the phase of the oscillation, not the frequency. Hence why they can come back to where they start in a closed cycle. I hope that makes sense. I'm having a hard time explaining this as clearly as I'd like to.
Probably one of the most beautiful fields of mathematics I have come across... everything from the content of the field itself and how a single teenager needed nothing but a simple problem to completely revolutionize our understanding of the world. I am so grateful to study such content in the coming months... when people ask me what math and physics is like I tell them it is stranger than you can ever imagine.
I really liked the cosmic background music you put in here. I am sure Galois also had such a trip before the day he died when he was waiting for sun to come up and writing his proof. He had the same sparks and clashes in his mind that he felt that it was necessary that although nobody listened to what he needed to say, it was important that he expressed himself. He says in his notes, "sun is almost rising, I have to hurry up...."
Thank you for popularizing Galois theory! However, in addition to the mistake pointed out in the description, I think there's a mistake around 7:19 : the Galois group of the equation x^7 - 2 = 0 over the rationals is not cyclic but rather an extension of the cyclic group of order 6 permuting primitive 7th roots of unity by the cyclic group of order 7 that acts on 7th roots of 2 by multiplication by 7th roots of 1.
Yeah he didn't really describe how the galois group is obtained at all. It is assumed to preserve multiplication of numbers in the field and built up from there. Then in simpler terms if you have 7 roots of 2, the quotient of any two is a complex 7th root of 1, and these roots of unity must permute among themselves as elements in this field, in a way that fixes the trivial root 1. The one with the smallest complex argument (angle from positive real) generates the rest and is called primitive. Then a permutation in the galois group for x^7-2 is given by finding where this primitive root of 1 goes, and where the real 7th root of 2 goes. But it's still solvable as a group because, without too much group theory, we can work with the cyclic subgroup that fixes the primitive root and only permutes the roots of x^7-2 by multiplying by some fixed complex root 7th root of 1. A good way to think about it is that the naive way you'd want to permute the 7 roots of 2 is with a group having 7!= 5040 elements, highly nonabelian (though still solvable, that's besides the point). This roots of unity business is reducing the complexity to a small subgroup and proves solvability, it shouldn't be thought of as making it harder.
@@orangeguy5463 That the group S_7 of permutations of 7 elements is solvable is false! General degree n equation isn't solvable in radicals for n ≥ 5 precisely because S_n isn't solvable.
On RUclips, I’d rank you as one of the Top 3 explainers of dense mathematics. Galois theory was always presented as too abstract for beginner students yet this video gave me a good grasp of the basic tools this discipline offers. I look forward to watching more of your content!
@@mueezadam8438 Cool - thanks! I came across Aleph 0 very recently and then the RUclips algo also recommended this video. It's good to see these newer channels begin to rise up to the very high bar set by 3Blue1Brown
To be fair yt algorithm showed us this video with only one month delay. Usually it shows me the videos I want to see with 8 to 10 years of waiting. This time it did a great job
it would be nice to see a video that explains why solvability of the Galois group is necessary and/or sufficient for the solvability by radicals, that would count as "Galois theory explained"
Yeah I still don't know why this is true after watching this video. It seems that this is the most important part of the theory so it's strange not to explain it.
@@mimikal7548 obviously this comment is late, it is more an exercise in me trying to explain it. I think that, at least for sufficiency, the groups must be cyclic, so that when you apply more groups, you won't continuously be making algebraic equivalences that get you farther and farther away from the solution. it's like going into a funnel vs coming up from a funnel, it is easy one way and hard the other. (do not read this part, unless you have a better explanation - in review, this is a horrible and confusing explanation) It might be possible to think about it this way: all the information is contained within the equation and within the rules of mathematics. we can view symmetries as changing context. Consider a jigsaw puzzle, where you have the puzzle (we'll say it's like a line, in that there's one place to put a piece and a finite amount of jigsaw puzzles left). When you place a piece and it fits (let's say that's symmetric), you'll obviously see the jigsaw puzzle and say, yes, that's right, just like you can sometimes tell when solving a polynomial equation that you're getting off track and terms are growing larger and such. When you apply another symmetry, you preserve the "nature" of the equation, but change how it looks - just like when doing algebra, your 5th step is equivalent to your 1st step, because you do everything to both sides of the equal sign. So, if the equation can be seen as a galois group with cyclic component groups, then obviously, there is a solution. It's like backtracking through a maze, it is easier from the end to the start than start to end (NP vs P). By tracking the grammar of the math, while whatever format the problem is in might change, Galois theory can tell us if the overall problem has a solution, or if it doesn't. Which is pretty cool, it is like framing a painting in different frames. The style of the painting and the effect it has on your room might change, but the painting stays the same. Or, it is like learning. The subject you learn stays the same, you just need to change how you see it, until you finally grasp it. Obviously, changing how you see it is done automatically, by your brain, although you can see it at a psychological level if you want.
Just great. Took algebra (fields and groups) 40+ years ago: this was a pleasant refresher. And I like your general statement on swapping the study of an object for that of its symmetries: it's also what you do with symmetry groups in physics all the time.
I think there is a note of confusion during the segment on composition of transformations. If one maps an element of a set through a composition of transformations, then the transformation on the right side will be applied to the element first. During the segment where transformations were demonstrated via the wearing of clothing, the left transformation was applied first. However, this video was very helpful to me, and I am looking forward to more content! Thank you.
I’m just dumb senior in HS with little to zero knowledge in abstract algebra (I tried to study it myself but I was not able to grasp the abstraction) this gave me a vague sense about galois theory and motivated me to continue studying it!
Really liking how this channel presents stuff. Thanks for making this content, I am trying to self-teach math since I don't want to go back to school for a math degree.
I like your idea very much, but I think there are things that can be improved. 6:15 Can have some explanation on why we can't shuffle the roots arbitrarily. 8:12 This is a cyclic group! 9:00 The wrong order. 11:21 May also try to visualize it as two dials (in addition to what you already have)
@@__mrmino__ That's exactly what's missing. Galois groups are the maps that preserve addition and multiplication (as in the first example). Naturally these maps are permutations of roots, plus it's sufficient and easier to consider the roots. However just a random permutation might not be obtained from a member of the Galois group.
Studying this stuff in uni, obviously in greater details, but this gave me a better perspective on some things like cyclic groups. It’s awesome when a video that is understandable for someone who does not know the subject still helps complete the work of books and professors. Absolutely loved it!
Excellent video. You made an inherently complicated subject comprehensible by clear explanations and clever use of graphics - well done! I look forward to watching some of your other math videos.
Great video! 9:26 we have a minor mistake: \phi\circ\lambda is to first apply \lambda then \phi, but the audio takes it in the opposite direction. Hope this help!
Often more introductory explanations of group theory use left-to-right instead of right-to-left. There are a few group theory lectures on youtube that I've seen the same.
This is one of the best explanations of Galois theory I’ve seen. I’m physically exhausted by how many people I’ve shared it with my friends. It’s so intuitive that now I finally have a place to redirect people who are scared of Field Theory
As far as I can tell this is the only video this channel has so far, but it sure started out with a bang. I'm glad people post this kind of stuff on youtube because a lot of the literature out there is so much less accessible. Because of channels like this, I can eventually see a future where one day Galois theory will be just as accessible as calculus is today. Still challenging, but accessible. Maybe even at the high school level.
Thank you for this explanation. I've been studying group theory on and off for years, but I always stop short of diving into Galois theory, because it seems difficult to approach. But this gives me the motivation I was looking for.
Very nicely explaind. I also studied mathematics, but neglecting much algebra. What you tell and the way you tell is simple and easily to be understood, but for me a non familiar with algebra not familiar. Books on Galois theory often ignore this NON FAMILIAR but you ignore not. THANKS FOR GOOD DIDACTIC
Very nice and understandable explanation of the link between solvable polynomials and solvable Galois-Groups of polynomials. Thank you so much for showing the core ideas
Great! A clear explanation of what Galois theory aims to prove. It would be great if courses on this started with this overview, to give an idea of where they are going
Nice video. I like the "military maneuver" metaphor. People like moving from more "analytic" concepts (e.g., how a polynomial function behaves) to more "algebraic" terms (like groups here). The ideal algebraic thing would of course have been to obtain a general expression (or algorithm) for all the roots, but unfortunately we cannot fully win that "war".
I disagree, I think it's much cooler that we know that there are no solutions for n>4, it's one of the most surprising facts in math. If there was an algorithm for every polynomial, first, chances are that it would be so ugly that nobody would write it down for all n, second, it would be more convenient to solve a polynomial via Netwton method if one needs to know the root numerically.
@@strangeWaters Yes, there are typical textbook proofs that show that under the assumptions of smoothness (always the case for polynomials) and being "sufficiently close" to the root, Newton's method will converge. For simple roots, once being sufficiently close to the root, the convergence is quadratic, which is a very nice property of Newton's method. For multiple roots the situation is more complicated in practice. While standard fixed-point theory would still provide geometric convergence, the actual rate worsens as the multiplicity increases. So some special care is needed. Apart from these mathematical consequences of the multiplicity of roots (which hold even under perfect arithmetic), there are other numerical aspects. A multiple root suffers from ill-conditioning in the sense that there is larger margin of error due to amplified effects of round-off noise and the ways you evaluate the polynomial may affect the error more. After all, there are some reasons why people don't prefer polynomial root finding approaches even for things that seem natural candidates for that, like finding the eigenvalues of a matrix. In fact, in some cases (like orthogonal polynomials) one would opt for reformulating the root-finding problem as one of finding eigenvalues of a matrix. Interestingly, for complex polynomials, if you color the points in the complex plane according to which root is recovered by the Newton's method starting from that point and how many iteration were necessary (or if the method failed to converge), you end up plotting some fractals. Newton's method has intrigued a lot of the greatest minds, Kantorovich being one of them.
Great presentation I would also have mentioned that by the same token, this is how the complex field is constructed - by adjoining i to R, corresponding to the equation X^2 +1 = 0
i'm just a (former) homeless guy who used to sleep in the wet woods behind Berkeley..... (in a loincloth). i have no idea what you're talking about! but thank you for posting and trying.....
Sadly our education system doesn’t dedicate that much time to math profs so that they can explain these ideas in such details, the concepts in this video would be covered in 2 mins due to time restraints. A normal Galois theory class requires huge dedication to group theory and field theory then connecting them by Galois theory stuffed into 3 months, it’s excepted grad students and senior undergrad students will go out and watch videos like these on their own time to understand the big picture better.
It is because creating videos like these takes a huge amount of skill and time. Moreover, these lecturing skills and not (completely) the same as these skills.
That can't happen. First of all Professors teaching courses at this level are not only teaching you the subject but preparing you to be a mathematician. That means they need to teach the material on a much more fundamental level so that the student can do more of the learning independently through exercises and reading. Moreover making one of these videos takes days or weeks of planning, scripting, and editing. Professors do not have that kind of time to prepare an entire semesters worth of material. Regardless, introductory videos like these are not exhaustive and are absolutely not a substitute for learning the subject from a textbook or a class.
bm3253 Math does not change often. So if they take the time to produce something like this for critical concepts they could reuse it every semester. Any extensions and exceptions can then be supplemented for completeness. Making mathematical concepts more mysterious and vague does not produce better mathematicians.
@@nikeshsolanki9542 this is true , people assume theirs proffessor can create such video or at least combine the effect of this graphical animations but they cant . Also , we may note that there is a way that proffesors teach well, they need to incooporate thoose materials in class within time restrains beacause it lays the foundations and speak to our brains differently ,moreover, we know that learning visually also is a great method to enhance our understanting and it's also found in studies .
I got a bit confused... I went to Wolfram and it seems that x^5 -2x + 1 can actually factored in a first degree and a quartic so at least the two separe factored pieces can be solved by radicals. BTW the factorization provided was (x-1)(x^4 + x^3 + x^2 + x - 1)=0. Thanks for the video! I think for the first time I really understood it! PS. x^5 +2x + 1 seems to be a different story as as in this case Wolfram had to 'cheat' by using hypergeometric functions.
As you noted x^5-2x+1 is indeed solvable by radicals. Better might've been either x^5-x-1 or x^5+2x+1 as you suggested (I think both should work). However, sadly there are a few more mistakes in the video when it comes to computing the Galois groups. Edit: I think I found the source of the wrong equation www.google.com/amp/s/rohilprasad.wordpress.com/2015/12/17/constructing-a-polynomial-with-galois-group-s5/amp/
This was a great intro to Galois theory. Different and better than I’ve seen before. I hope you’ll do a whole series. I’d love to see how these concepts evolve into Lie groups and to solutions of physical problems.
The Galois group for x^7-2 isn’t the cyclic group with 6 or 7 elements, like it says in the video. It is for x^7-1 but not x^7-2. The rotation symmetry where the seventh root of 2 gets multiplied by the seventh root of unity does generate a cyclic group but there are other symmetries too. Let r be a primitive seventh root of unity, then r can get mapped to r^2 (and r^2 goes to r^4 etc.). That generates another cyclic group with 6 elements and these two cyclic groups combine together to give another group with 42 elements. You could write the presentation for this Galois group as I’m not sure what that group would be called though.
Yes I agree f(x) = x^7 − 2 The 7 zeros are 2^(1/7)* k^j where k is the 7th root of 1, and j=0,...,6. Splitting field: F = Q[2^(1/7),k]. The minimum polynomial of 2^(1/7) is x^7−2 So |Q[2^(1/7)]:Q| = 7. The minimum polynomial for k over Q is x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 & over Q[2^(1/7)] it is the same. |F:Q| = |F:Q[2^(1/7)]| * |Q[2^(1/7)]:Q| = 6 * 7 = 42. The Galois group has order 42. AMs can be. 2^(1/7) -> 2^(1/7)*k^j, for j=0,...,6, and k -> k^j, for j=1,...,6. Giving 42 combinations, which all happen. These two subgroups are C_7 & C_6 but do not commute.
Awesome video, very clear! You didn't go into much depth, but you hinted at enough terms and theorems to allow one to five deeper based on the video. Very nice!
@11:51 equation can be factored into (x"2 plus 1)(x"3 plus 2) equal 0. The solutions exist: i, -i and -cube root (2). But shown is a non-abelian transformation. But the equation can be factored. Confused.
Wow I have little understanding of math past a high school level but I was able to understand(more or less) what you were presenting. You have a gift for presentation I hope you continue to make videos.
I really liked this video. You are doing a great job. I heard about Galois theory from many people but never actually knew what it is. This video explained everything in a nutshell.
More than interesting this part of Galois theory...is the first time for me to see and understand these type of equations that you explain in the video, so really true, because is affected in the same part with Complex numbers, how is the another part that we need to know too, and the another point the variation of the other types of funtions that you have done in the example how interchange the differents between between the different results that are interpreted in the numbers as they are not abelian numbers, in this case it is expressed as non-abelian numbers that result in a given difference between each of them by indicating that they are different solutions and that they are not equal. Given in the cases of the mathematical equations that he explains in the video, thank you teacher, kind regards from Cancun, Mexico.
Great video, but some examples are wrong. The Galois group of x⁷ - 2 is not C₇. The degree of the splitting field over Q would be 42, as adjoining the real seventh root of 2 gives you a subfield of the reals, after which you need to adjoin a primitive 7th root of unity, whose minimal polynomial over Q(⁷√2) is the same as the one over Q using the tower law. The Galois group is actually a semi direct product of C₆ and C₇. If you change the polynomial to x⁷-1, you get C₆ as the Galois group. The Galois group of (x⁷-1)(x⁵-1) is not a direct product of C₅ and C₇. The degree of its splitting field is (I think) 24 (adjoin a primitive 5th root of unity then a 7th root, showing that the minimal polynomial of the 7th root over the intermediate extension has degree 6 might need some work idk). If it is 24, then the Galois group is the direct product of C₄ and C₆.
I would love to know the details about the 7 solutions to that 7th-degree polynomial, and the corresponding symmetries (42 symmetries according to another comment, not the 7 mentioned in the video)
Think about building it up from two extensions, first adding a complex seventh root of unity, say ζ, then adding the (real) seventh root of 2. The Galois group for Q(ζ)/Q consists of automorphisms which fix the real line and permute the remaining 7th roots of unity. Not by rotating, though, because that wouldn't fix the real line. An example of a map which generates these automorphisms sends ζ to ζ^3 (the number you raise to has to be a primitive root in order to generate the whole group). Raising ζ to the 3rd power repeatedly gives ζ^3, ζ^2, ζ^6, ζ^4, ζ^5, ζ. In terms of permutations, if each number is a power on ζ, then the permutations for the group are 0123456, 0362514, 0246135, 0654321, 0415263, 0531642. This is cyclic of order 6, though I know it's a slightly complicated order for the elements to rotate through. Each number is multiplied by 3 each time, mod 7. Anyway, next, this field needs to be extended by adding the (real) 7th root of 2. The 7th roots of unity are already present, so just adding the real 7th root of 2 covers all the complex ones as well, which are just a 7th root of unity times the real 7th root of 2. The automorphisms here (over Q(ζ)) will be generated by the map which sends the 7th root of 2 to ζ times the 7th root of 2, so in this case, it is just rotating the the 7th roots of 2 by 2π/7 in the complex plane. The reason why other permutations don't lead to automorphisms is because the roots of unity being present prevent non-rotational ones from being valid. When you put these two extensions together, you get a group of order 42 with these order 6 and 7 subgroups, but this is not an abelian group. It has 6 elements of order 7, 14 elements of order 6, 14 elements of order 3, and 7 elements of order 2, and of course one element of order 1 :) It's also sometimes known as F_7. And it's solveable, since it was made from two cyclic extensions. I don't know if that really helps to explain exactly what it is, but you can manually compute the automorphisms if you want.
Thanks Seán! I can't say I have a complete understanding, but your explanation certainly helped. Along the way, I learned about complex roots of unity and automorphisms.
Interesting video that explains a highly complicated subject in simple ways. Looking forward to a video that explains why the equation is solvable by radicals if and only if the corresponding group is solvable.
I hadan assignment related to the question that why quintic equations are not solvable by radicals...this video really helped me❤️thanks for mking this amazing video.😊
8:10 The wheels will have 6 and 4 edges. Not 7 and 5. As x=1 is a root of each of the two factors of the polynomial. Also, your symmetry composition goes the opposite way of convention. XoY usually means apply first Y, then X.
Correct regarding the convention on composition. Not sure I see regarding the dials. Cycling through all 5th roots of 1 will give us legitimate field automorphisms. There are five of them and they form a cyclic group that is a subgroup of the Galois group for the 12th degree polynomial. But if so, then according to Lagrange's theorem the group order divides by 5. But if our group is a direct product of C4 and C6 then its order is 24 and does not divide by 5. Thoughts?
@@mathvisualized355 One of the fifth roots of 1 is 1. The polynomial x^5-1 has 1 as root. It can be rewritten as (x-1)(x^4+x^3+x^2+x+1). The second factor here yields a Galois group of order 4, which shuffles the 4 non-1 fifth roots around.
Also, x^7 - 2 does not have a cyclic galois group of order p. It has a galois group of order p(p-1). The group is isomorphic to all group of invertible matrices of the form (a b) (0 1)
Super awesome, I am just thinking guys, let me know your ideas. A funny property of this is that, for example in case you have three polygons, when solving the equation by flipping instead of with the lock, that polygon must have a degree equal to the number of polygons you would need for the lock, assuming that in the example given of x to the five and x to the seven, two polygons is the maximum amount of polygons necessary to solve by locking. From this we could already know if an equation is solvable by radicals. It would be great to write a proof based upon this, for example by proving that you cannot take a polynomial and split it into a number of parentheses (I forgot how to call this in math) that are equal or greater than the degree of such polynomial, if the degree is larger than 4.
I enjoyed this video. Galois Theory was lectured badly on my course and this was before the internet. Never got my head round it. This video makes things clearer. However, it is not correct to say that the example given, a 35!order group, is not cyclic. It is cyclic of order 35. If you take your a and b of order 5 and 7 respectively, their product has order 35 and generate the group.
Really enjoyed this. Have you thought about making a video about permutation groups and cyclic groups specifically? Maybe a proof of 5+ elements not being able to decompose into cyclic groups? That is a very fun thing to show visually
But apart from the combination that he says between group theory and body theory is another point in which Galois explains as a functional method given as abelian numbers and non-abelian numbers ... in fact, for me since I have been studying mathematics for quite some time back, I feel that even oneself must still continue to seem because this area of mathematics is quite extensive, thank you teacher.
Hi, I have watched your video 5 times so far. It is refreshing and very very original. Thanks you for sharing this wonderful way to simplify the Galois Theory. I have been interested on the Quintic and the solvability of polynomials all my life. A lot of work to put all the essence on the theory in a 15 minutes video. You are a genius ! How did you do it ? What software do you use ? The result is amazing... Bravo ! Can you please put some more information about yourself ?
This is a rather deep question about the structure of the symmetric groups. Ultimately, the question boils down to the structure of a particular subgroup for S_k, which we denote A_k. A_k is the alternating group on k letters. It consists of all of the "even" permutations. Permutations can be classified as either even or odd, depending on whether it has an even or odd number of inversions. As it turns out, the even permutations by themselves form a group! (The odd permutations do _not_ form a group.) But the alternating group has a very important structure within the symmetric group - it's a _normal_ subgroup. This is the type of subgroup we need in order to do that "extension" idea talked about in the video in a meaningful way. What needs to be Abelian is not the subgroup A_k, but rather the _quotient_ group where we collapse S_k by forcing all of A_k down to a single element - by treating any even permutation as if it were the identity permutation, the do-nothing permutation. And it turns out that if you collapse S_k along A_k, you do get an Abelian group (in fact, a cyclic group)! However, A_k is what's called a "simple" group for every k _except_ k = 4. Simple here is referring to a "building up" sort of sense. Simple groups are the basic building blocks for this "extension" notion. You can't build up simple groups from smaller groups. So A_k for k >= 5 can't be broken down further, and it's far too complicated to be Abelian itself. For k < 4, A_k is too small not to be cyclic, and all cyclic groups are simple! But here's the difference - since A_k is itself cyclic, this works for solvability! For k = 4, we actually get something quite interesting. A_4 is definitely not cyclic or even Abelian. Far from it. However, A_4 has a nontrivial normal subgroup, the Klein-4 group. In general, for k >= 4, A_k has the Klein-4 subgroup living inside of it (for larger k, we get more and more copies of Klein-4 living inside of it). But 4 is pretty much a special number here. Permuting 4 things is too few things for there to be multiple copies of Klein-4. The uniqueness of Klein-4 in A_4 combined with the size of A_4 forces this Klein-4 subgroup to be normal in A_4. Then the collapse of A_4 along its Klein-4 subgroup is too small not to be Abelian (indeed, too small not to be cyclic). Moreover, the Klein-4 subgroup itself is too small not to be Abelian. So it still ends up working, but we're right at the cusp of it working - look how much more work we had to do to even realize A_4 as solvable compared to A_3, for instance. These alternating groups get a _lot_ bigger _very_ fast. Once you hit 5 things to permute, A_5 has more than enough room to be complicated enough not to be solvable. I don't know that there's any nice way to make it easier to digest than this, unfortunately. The alternating groups A_k are extremely complicated. If it were easy to understand A_k for every k, it would actually be easy to understand all finite groups. But this isn't the case. Understanding groups is hard.
Which book of Lawvere are you refering to? I'm only aware of one at the moment (the one on Set Theory together with Rosebrugh) which I did not read (yet!).
Hey there, I love your video! It explains Galois Theory incredibly easily and smoothly, and in a way that people who are not interested in math can understand it. I'd love to contribute subtitles in English and German, so I can show this video to my friends. If you're interested, hit me up with a DM so I can submit them to you, once they're done! Have a great day and thank you for this gem :)
wow I'm glad the youtube algorithm showed me this hidden gem. I like your presentation and style.
Looking forward to seeing your next video!
A bit slow in my opinion but yeah... (The RUclips algorithm really seems to've liked this video compared to other math ones....)
+1
I'm sorry but what the fuck how do you only have one video???
To me it is the first time I got an inkling of what abstract algebra actually is.
Many things in this video are WRONG! Duplicating @sbares 's comment:
Gal(x^7-2, Q) is not C7 as you claim, it is a group of order 42, as is seen by the tower of extensions
Split(x^7-2,Q) = Q(2^(1/7), exp(i2pi/7)) > Q(2^(1/7)) > Q
where here ">" means "extension of" (sorry, can't type TeX in youtube comments lol).
The minimal polynomial of 2^(1/7) over Q is x^7 - 2, so this extension is of order 7, and the minimal polynomial of exp(i2pi/7) over Q(2^(1/7)) is the seventh cyclotomic polynomial, so the order here is phi(7) = 6. Thus
|Gal(x^7, Q)| = [Split(x^7-2,Q) : Q] = 7*6 = 42.
There's a similar problem with (x^7 - 1)(x^5 - 1). The minimal polynomial of a seventh (or fifth) root of unity over Q is not x^7 - 1 (or x^5 - 1), it is the seventh (or fifth) cyclotomic polynomial. It is not too hard to see that the splitting field here is the 35'th cyclotomic field, so the Galois group is in fact (Z/35Z)*, which is isomorphic to C12 x C2
Finally, the example x^5 - 2x + 1 factors as
(x-1)(x^4 + x^3 + x^2 + x - 1)
so the Galois group is not S5, it has to be a subgroup of S4 (In fact, Sage tells me it should be the whole S4).
There are some errors here.
At 7:20 you say that the Galois group of x^7 - 2 over the rationals is cyclic, but it's not. It's dihedral with an order of 14.
At 8:10 you make a similar error for (x^7 - 1)(x^5 - 1). The orders are 6 and 4.
At 12:39 you mention that the Galois group of x^5 - 2x + 1 is S5. But it has a root of 1 and is reducible to a linear and a quartic, for which there exists a formula. It's therefore S4.
i wish this comment rises to further top
This comment is true
also a 2d reflection is not equivalent to a 180° rotation
I overall agree with your quite useful comment, but the Galois group G of x^7-2 is not dihedral with order 14. Indeed, the splitting field of x^7-2 is generated by the (real) 7th root of 2, along with the (all complex, except 1 of course) 7th roots of unity, whose Galois group is (Z/7Z)*=Z/6Z=C6. The degree of the extension (and so the order of the G) is therefore 7*6=42. Moreover, if you call r2 the real 7th-root of 2 and z=exp(2ipi/7), then Q(r2,z)/Q(z) is Galois with degree 7, so has Galois group C7. As the intermediate extension Q(z)/Q is Galois, the subgroup C7 is normal in G=Gal(Q(r2,z)/Q), with factor group Gal(Q(z)/Q)=(Z/7Z)*=C6. Now, by the Schur-Zassenhaus theorem, as 6 and 7 are relatively prime, the group G splits as a semidirect product of C7 by C6. In particular, it cannot by dihedral.
By the way, the same arguments apply for polynomials of the form x^p-n where p is an odd prime and n is an integer different from -1,0,1: its Galois group is a semi-direct product of Cp by C(p-1), the latter acting on Cp through the isomorphism C(p-1)=Aut(Cp). [sorry for the long comment]
I came looking for just such a set of comments. 🙂
And then the genius Galois thought it was necessary to have a duel, and died.
Imagine if he’d thought, “Bugger it, I’ll write this up tomorrow night. I’m going to bed.” ...
Exactly! , this was too bad we lost also in this dual of his .
I want to live in the universe where Galois won the duel and survived.
Well, if the École Polytechnique's examiners weren't so shitty and his father hadn't comitted suicide, maybe Galois wouldn't have gone to that duel. Anyway, it was definitely NOT just a matter of thinking that a duel was necessary and dying.
While Galois was probably a genius, a bunch of mediocre people have a lot more shameful or questionable behaviors or words than dying in a duel and still are recognised are geniuses. Dying in a duel is actually very honorable.
Galois was a hothead.
I don't know how much mileage I'd get out of this if I didn't have an undergrad course in modern algebra, but from perspective of someone who knows what groups and fields are, but never encountered Galois groups before and their relation to polynomials, this is fantastic. I just wish it'd push just a little bit further into why S2, S3, S4 can always be decomposed into a product of cyclic groups, even if just as visualization of some special cases.
6:23 I'm afraid that the Galois group of X^7-2 consist of 42 elements, not 7 (I believe it is equal to the semi direct product of the cyclic C_7 and C_6) . Apart from that, great video!
Yes, it's true.
probably a typo of x^7-1 :'(
@@JustinLe no because in that case is an extension of degree 6 lol. By the way it's a good video, no one really noticed it 😆😆
If the roots of unity are included in the base field its still right. Which I assume since that is what ables us to use rotations.
@@David-km2ie yes i was assuming the base field was Q obviously! But whatever.. if everyone understood the right context shame on me 🙈
Overall I'm always happy to see more math content creators on youtube, and I'm excited to see future videos from you. Most of the Galois groups are actually calculated incorrectly here, and those kinds of details should really be corrected/verified before creating a video like this. Ignoring that, here's a few notes and nits, though (reading this over I'm worried that I'm coming off too harshly but I promise I did enjoy the video and would like to see more :) )
One big question I'm left with after seeing this is - who is your intended audience? Curious high schoolers? Undergraduate math majors? Undergraduate non-math STEM majors? Any curious undergraduates? Graduate math students/Post graduates in mathematics?
My guess is that it is meant for curious high schoolers or undergraduate non-math STEM majors (particularly because of the commuting functions analogy and notation), perhaps with some math majors as well. I think this question needs to be addressed quite carefully, because it will address the question of how much rigor your videos require, which I think was probably the primary weak point of the video (and is imo the point that most math videos on youtube struggle with, even for big channels like numberphile).
I'll try pointing out stuff as I see it through the video:
0:00-0:50 I think this is a solid intro, motivations for the subject are definitely clear. I think it's a *little* disingenuous to say that we'll "answer the question today using Galois theory", since it's really more of taking a peek at the theory that needs to be developed in order to answer the question, but all good intros are probably a little disingenuous in a clickbait-y way, so I think this is fine.
0:51-2:34 I think this is pretty good, and a fine introduction into the idea of field extensions. However, I think it could be a *little* clearer about the finite operations thing. 1+sqrt(2) is an example, but it's also probably worth showing something like 3*sqrt(2)+(4-sqrt(2))/(5+sqrt(2)). It's also somewhat nonobvious that this ends up being equivalent to the set x+y*sqrt(2) for rationals x, y (division being the nonobvious part). If the intention is to just briefly wave Galois theory in front of the audience, then omitting such details is probably fine, but it's worth at least pointing to the parts that you are handwaving over to acknowledge that they have been handwaved (textbooks do this with the classic "(Why?)" inserted mid paragraph).
2:35-3:50 I think this is good; perhaps the idea of extending Q by the roots of any arbitrary polynomial was glossed over a little too quickly given how central the concept ends up being to the rest of the video (and the topic generally).
3:51-5:33 This part is fine, but it feels a little unclear as to what purpose it is serving in the overall video. I imagine it's trying to grow some intuition about how finite cyclic groups work when your elements are functions wrt repeated compositions, but this only feels like it is showing this connection as someone who has already seen it. It is not super clear to me whether an uninitiated student would be growing this intuition by watching this section.
5:34-5:50 Alright this is probably the biggest handwave of the entire video. I think building up the notion of what exactly the "symmetries" of an equation means is quite involved, and is not accomplished just by looking at the sqrt(2) -> -sqrt(2) example. In your defense, I think many textbooks also use this example and pretty much only this example, but it really is too complicated a concept to glean from just this example. There is a lot about field extensions and automorphisms that is being omitted here, and the viewer probably should be aware of this omission. Also, the notion of a "group" is kind of just introduced without any definition.
5:51-6:40 So, as other commenters have mentioned, this is actually not accurate. One natural question an attentive student might have is, "Aren't there 7!=5040 ways to map the roots of this equation to each other? Why do we only care about the transformations that take 1->2, 2->3, etc.?" This also gets a little more muddled since we are extending Q by the 7th root of 2, in addition to a primitive 7th root of unity. Using f(x)=x^7-1 here was probably better.
6:41-7:40 I think this is good. Minor nit: I think the numbers on the dial should probably be filled in with white or something, it can be a bit hard to read sometimes with the lighting.
7:41-8:17 I like the combination lock analogy, but technically the galois groups of these are incorrect. x^7-1 has a galois group of Z_6 and x^5-1 Z_4. Z_7 x Z_5 would have still been cyclic, btw.
8:18-10:19 I think this is okay, but this is one of the areas where you are probably shifting audience levels. Knowing that function composition doesn't always commute is pretty standard for math majors, perhaps not obvious for high schoolers and should be known to at least a good amount of STEM undergraduates. Prior to this point the video seemed good for all three audiences, but here it's appealing a bit much to one demographic and perhaps spending a disproportionate amount of time on it. In general I do like the clothes-wearing analogy for function commutativity. I think personally I would have liked this point in the video to justify why exactly these automorphisms (ie symmetries) commute rather than learning about what commutativity is.
The rest of the video is fine apart from the galois group computation errors, I think the transition at 11:54 is a little awkward since you go through an example where a group isn't abelian and then talk about groups where you can construct them from cyclic extensions, which are necessarily abelian, without any word like "however" or "on the other hand" so it feels like you're talking about the same thing.
The chaining of combination locks was good, and I think it captures the notion of direct products of cyclic groups well.
Other minor nit is that the music is not loud enough to add much to the video but probably not quiet enough to be totally ignored.
Overall I think the video is good; since it's your first, it's natural that there will be some feedback. At the end of the day I'm just a random dude with some feedback. My algebra is not super strong so I may have made mistakes in this comment as well. I hope you keep the spirit of this video and continue to make more, looking forward to seeing what your channel provides! :)
It's very good that you commented on some of the more severe mistakes in this video. It appears that the creator confused the Galois group of a field extension and of a polynomial equation (which may differ, when only adjoining one root and not using, say, primitive elements) at some points, or rather didn't check his computations.
There is a relevant post on reddit ( www.reddit.com/r/math/comments/kk7cde/galois_theory_explained_visually_the_best/? ) also talking about the problems.
(Note: I really loved the video but I think the number of mistakes is problematic)
I am doing an MSc. in maths and found this video fantastic. Certainly won't make up for your maths course and hours of drilling exercises, but it's nice to see this stuff explained in normal non-convoluted language for once.
@@pauls.2451 I agree that it is nice to provide some "disillusionment" from how scary higher level math seems, but that kind of thing needs to be done with care. If you stray too far from rigor and too close to "wow" factors you end up with some problematic videos like numberphile's infamous disastrous -1/12 video, which has forever plagued the way many laypeople see math. Much of the best parts of this video build some intuition about cyclic groups and their direct products, but give a bit of incorrect intuition about what exactly Galois groups are.
Don't get me wrong, I love the fact that someone is taking on the challenge of explaining Galois theory to laypeople, because I think this type of thing can really get people excited to see what math is really all about. But it requires a great deal of attention to detail to make sure that you are really conveying what should be conveyed, rather than what people will digest easily at the expense of precision.
Btw, if you or anyone else is looking for a layman friendly book to get into the subject, I've worked through quite a bit of "Field Theory and its Classical Problems" by Hadlock and it's really quite good. No prior algebra knowledge assumed.
wrt?
With regards to?
Honestly, if you to write a thousand word youtub comment and decide to be cute by abbreviating three words; I kind of don’t respect you.
Jmoymmv.
Dear Aman, Assalamu Alaikum, in case you are a Muslim and "Greetings," in case you are not.
You seem to know so much about the Galois theory. Why do you not make some of your own videos. I hope I can have email correspondence with you and ask questions about this subject. I cannot give my e address explicitly. it will be erased so:
ali.jamily1at g.come. thank you
Oh damn, this is really good. The combination lock idea is brilliant. The way a radical creates a cyclic group is also why the scale we use in music works the way it does. Since we use twelve tone equal temperment, each note is 2^1/12 apart. Once you stack 12 together you double the frequency. The 12 possible pitches form a cyclic group symmetry.
That doesn't have much to do with radicals though. If anything it's a coincidence related to auditory perception of frequency being logarithmic. Sure 440 Hz and 880 Hz are A4 and A5 but they are two different numbers and two different notes despite being harmonically similar. In the mathematical sense its the 12 complex 12th roots of 2 that form the cyclic group but when considered as complex frequencies they actually correspond to the phase of the oscillation, not the frequency. Hence why they can come back to where they start in a closed cycle. I hope that makes sense. I'm having a hard time explaining this as clearly as I'd like to.
the cyclic visualization is really helpful.
Probably one of the most beautiful fields of mathematics I have come across... everything from the content of the field itself and how a single teenager needed nothing but a simple problem to completely revolutionize our understanding of the world. I am so grateful to study such content in the coming months... when people ask me what math and physics is like I tell them it is stranger than you can ever imagine.
I really liked the cosmic background music you put in here. I am sure Galois also had such a trip before the day he died when he was waiting for sun to come up and writing his proof. He had the same sparks and clashes in his mind that he felt that it was necessary that although nobody listened to what he needed to say, it was important that he expressed himself. He says in his notes,
"sun is almost rising, I have to hurry up...."
Thank you for popularizing Galois theory! However, in addition to the mistake pointed out in the description, I think there's a mistake around 7:19 : the Galois group of the equation x^7 - 2 = 0 over the rationals is not cyclic but rather an extension of the cyclic group of order 6 permuting primitive 7th roots of unity by the cyclic group of order 7 that acts on 7th roots of 2 by multiplication by 7th roots of 1.
Yeah he didn't really describe how the galois group is obtained at all. It is assumed to preserve multiplication of numbers in the field and built up from there.
Then in simpler terms if you have 7 roots of 2, the quotient of any two is a complex 7th root of 1, and these roots of unity must permute among themselves as elements in this field, in a way that fixes the trivial root 1. The one with the smallest complex argument (angle from positive real) generates the rest and is called primitive. Then a permutation in the galois group for x^7-2 is given by finding where this primitive root of 1 goes, and where the real 7th root of 2 goes. But it's still solvable as a group because, without too much group theory, we can work with the cyclic subgroup that fixes the primitive root and only permutes the roots of x^7-2 by multiplying by some fixed complex root 7th root of 1.
A good way to think about it is that the naive way you'd want to permute the 7 roots of 2 is with a group having 7!= 5040 elements, highly nonabelian (though still solvable, that's besides the point). This roots of unity business is reducing the complexity to a small subgroup and proves solvability, it shouldn't be thought of as making it harder.
@@orangeguy5463 That the group S_7 of permutations of 7 elements is solvable is false! General degree n equation isn't solvable in radicals for n ≥ 5 precisely because S_n isn't solvable.
x⁷-2=0
x⁷=2
x=2^(¹÷⁷)
As a Ph.D. in Math, I didn't even know this! Thanks!
Oups, this is typically studied at Bachelor level ;-)
@@Jooolse well, yeah, keep learning! lol
@@Jooolse depends what modules you choose and what uni you study at. You could ignore algebra courses and focus on Analysis instead.
I may never understand Galois Groups but this video has already helped me get closer than I have ever gotten to understanding them.
Clear and simple. Thank you. It's is so much easier to dig in deeper when one has a clear overview like this.
On RUclips, I’d rank you as one of the Top 3 explainers of dense mathematics. Galois theory was always presented as too abstract for beginner students yet this video gave me a good grasp of the basic tools this discipline offers.
I look forward to watching more of your content!
I'd be curious to know what the other two top channels are in your ranking :)
@@aiwen6942 Aleph Null and of course, Grant Sanderson!
@@mueezadam8438 Cool - thanks! I came across Aleph 0 very recently and then the RUclips algo also recommended this video. It's good to see these newer channels begin to rise up to the very high bar set by 3Blue1Brown
Bruh. This explained everything SO MUCH BETTER. Wish the youtube algorithm showed me this earlier.
To be fair yt algorithm showed us this video with only one month delay. Usually it shows me the videos I want to see with 8 to 10 years of waiting. This time it did a great job
Bro the channel is just a month old chill
Bril. Don’t say bruh.
It makes me feel that I’m listening to an idiot.
it would be nice to see a video that explains why solvability of the Galois group is necessary and/or sufficient for the solvability by radicals, that would count as "Galois theory explained"
Yeah I still don't know why this is true after watching this video. It seems that this is the most important part of the theory so it's strange not to explain it.
@@mimikal7548 obviously this comment is late, it is more an exercise in me trying to explain it. I think that, at least for sufficiency, the groups must be cyclic, so that when you apply more groups, you won't continuously be making algebraic equivalences that get you farther and farther away from the solution. it's like going into a funnel vs coming up from a funnel, it is easy one way and hard the other.
(do not read this part, unless you have a better explanation - in review, this is a horrible and confusing explanation)
It might be possible to think about it this way: all the information is contained within the equation and within the rules of mathematics. we can view symmetries as changing context. Consider a jigsaw puzzle, where you have the puzzle (we'll say it's like a line, in that there's one place to put a piece and a finite amount of jigsaw puzzles left). When you place a piece and it fits (let's say that's symmetric), you'll obviously see the jigsaw puzzle and say, yes, that's right, just like you can sometimes tell when solving a polynomial equation that you're getting off track and terms are growing larger and such. When you apply another symmetry, you preserve the "nature" of the equation, but change how it looks - just like when doing algebra, your 5th step is equivalent to your 1st step, because you do everything to both sides of the equal sign. So, if the equation can be seen as a galois group with cyclic component groups, then obviously, there is a solution. It's like backtracking through a maze, it is easier from the end to the start than start to end (NP vs P). By tracking the grammar of the math, while whatever format the problem is in might change, Galois theory can tell us if the overall problem has a solution, or if it doesn't. Which is pretty cool, it is like framing a painting in different frames. The style of the painting and the effect it has on your room might change, but the painting stays the same. Or, it is like learning. The subject you learn stays the same, you just need to change how you see it, until you finally grasp it. Obviously, changing how you see it is done automatically, by your brain, although you can see it at a psychological level if you want.
Just great. Took algebra (fields and groups) 40+ years ago: this was a pleasant refresher. And I like your general statement on swapping the study of an object for that of its symmetries: it's also what you do with symmetry groups in physics all the time.
Amazing explanation
I think there is a note of confusion during the segment on composition of transformations. If one maps an element of a set through a composition of transformations, then the transformation on the right side will be applied to the element first. During the segment where transformations were demonstrated via the wearing of clothing, the left transformation was applied first.
However, this video was very helpful to me, and I am looking forward to more content! Thank you.
I’m just dumb senior in HS with little to zero knowledge in abstract algebra (I tried to study it myself but I was not able to grasp the abstraction) this gave me a vague sense about galois theory and motivated me to continue studying it!
simple and direct , easy but interesting subject EXCELLENT JOB.
Really liking how this channel presents stuff. Thanks for making this content, I am trying to self-teach math since I don't want to go back to school for a math degree.
I like your idea very much, but I think there are things that can be improved.
6:15 Can have some explanation on why we can't shuffle the roots arbitrarily.
8:12 This is a cyclic group!
9:00 The wrong order.
11:21 May also try to visualize it as two dials (in addition to what you already have)
@@__mrmino__ That's exactly what's missing. Galois groups are the maps that preserve addition and multiplication (as in the first example). Naturally these maps are permutations of roots, plus it's sufficient and easier to consider the roots. However just a random permutation might not be obtained from a member of the Galois group.
Studying this stuff in uni, obviously in greater details, but this gave me a better perspective on some things like cyclic groups. It’s awesome when a video that is understandable for someone who does not know the subject still helps complete the work of books and professors. Absolutely loved it!
Very good work sir, I work every day with Galois groupes, if I had to explain it to someone that doesn't know maths, I would do something like that.
Nice. Keep up the good work. Algebraic Topology and Algebraic Geometry concepts away you, my friend.
Excellent video. You made an inherently complicated subject comprehensible by clear explanations and clever use of graphics - well done! I look forward to watching some of your other math videos.
Great video! 9:26 we have a minor mistake: \phi\circ\lambda is to first apply \lambda then \phi, but the audio takes it in the opposite direction. Hope this help!
Often more introductory explanations of group theory use left-to-right instead of right-to-left. There are a few group theory lectures on youtube that I've seen the same.
It definitely depends on which book you use, but yes, usually the transformation is applied right-to-left because it's like functions sorta.
Excellent! Understanding the essence of Galois Theory in 15 minutes. Worth every second!
Galois theory was one of my very favorite units what must be 10 years ago now. Thank you for this beautiful video - a fun tour down memory lane
This is one of the best explanations of Galois theory I’ve seen. I’m physically exhausted by how many people I’ve shared it with my friends.
It’s so intuitive that now I finally have a place to redirect people who are scared of Field Theory
As far as I can tell this is the only video this channel has so far, but it sure started out with a bang. I'm glad people post this kind of stuff on youtube because a lot of the literature out there is so much less accessible. Because of channels like this, I can eventually see a future where one day Galois theory will be just as accessible as calculus is today. Still challenging, but accessible. Maybe even at the high school level.
Thank you for this explanation. I've been studying group theory on and off for years, but I always stop short of diving into Galois theory, because it seems difficult to approach. But this gives me the motivation I was looking for.
I love the background music, it is so soothing that I felt like I was peacefully dying in sleep.
Very nicely explaind. I also studied mathematics, but neglecting much algebra. What you tell and the way you tell is simple and easily to be understood, but for me a non familiar with algebra not familiar. Books on Galois theory often ignore this NON FAMILIAR but you ignore not. THANKS FOR GOOD DIDACTIC
Very nice and understandable explanation of the link between solvable polynomials and solvable Galois-Groups of polynomials. Thank you so much for showing the core ideas
Great! A clear explanation of what Galois theory aims to prove.
It would be great if courses on this started with this overview, to give an idea of where they are going
9:48 Superman needs to see this.
This is becoming my favorite maths channel please make more videos!
Very interesting, cleared up a lot of question marks concerning motivation left after taking a Galois theory class
Am I the only non math enthusiast here who had no idea what he was talking about half the time, but still watched cause math's interesting as fork?
More interesting than spork, even
sem
It's even better than sex 😂
Take abstract algebra
Matrix/vector operations visualized like this would be amazing, such as projections, dot-products...
Look at 3Blue1Brown Essence of Linear Algebra videos.
This is a lovely video. In fact, I don't think I've ever seen solvability by radicals explained so clearly and concisely. Thank you!
I've been looking for this video for a long time. You managed to keep all the juice with the right amount of definition. Thanks!!
Nice video. I like the "military maneuver" metaphor. People like moving from more "analytic" concepts (e.g., how a polynomial function behaves) to more "algebraic" terms (like groups here). The ideal algebraic thing would of course have been to obtain a general expression (or algorithm) for all the roots, but unfortunately we cannot fully win that "war".
I disagree, I think it's much cooler that we know that there are no solutions for n>4, it's one of the most surprising facts in math.
If there was an algorithm for every polynomial, first, chances are that it would be so ugly that nobody would write it down for all n, second, it would be more convenient to solve a polynomial via Netwton method if one needs to know the root numerically.
Question: is Newton's method guaranteed to converge? Do we have a proof that it "solves" all polynomials, in some approximate sense?
@@strangeWaters You need to be "close enough" to the root. So you kinda have to plot the polynomial first to see roughly where the roots are.
@@strangeWaters Yes, there are typical textbook proofs that show that under the assumptions of smoothness (always the case for polynomials) and being "sufficiently close" to the root, Newton's method will converge. For simple roots, once being sufficiently close to the root, the convergence is quadratic, which is a very nice property of Newton's method. For multiple roots the situation is more complicated in practice. While standard fixed-point theory would still provide geometric convergence, the actual rate worsens as the multiplicity increases. So some special care is needed. Apart from these mathematical consequences of the multiplicity of roots (which hold even under perfect arithmetic), there are other numerical aspects. A multiple root suffers from ill-conditioning in the sense that there is larger margin of error due to amplified effects of round-off noise and the ways you evaluate the polynomial may affect the error more. After all, there are some reasons why people don't prefer polynomial root finding approaches even for things that seem natural candidates for that, like finding the eigenvalues of a matrix. In fact, in some cases (like orthogonal polynomials) one would opt for reformulating the root-finding problem as one of finding eigenvalues of a matrix. Interestingly, for complex polynomials, if you color the points in the complex plane according to which root is recovered by the Newton's method starting from that point and how many iteration were necessary (or if the method failed to converge), you end up plotting some fractals. Newton's method has intrigued a lot of the greatest minds, Kantorovich being one of them.
@@strangeWaters SO, B~₩◇▪︎•○OS
a.
I've heard of this topic before but your approach in explaining with such visualization is very well crafted. Thanks
This is gold. I already took field theory class but this vid really helped me understanding galois theory.
I am anxiously awaiting part II.
R u single
what means "a simetry of a given (arbitrary) ecuation" which was refered to at @5:38?
Thanks so much for putting this so simply. Now I kind of get the motivation of using a derived series in the definition for solvable groups.
Great presentation
I would also have mentioned that by the same token, this is how the complex field is constructed - by adjoining i to R, corresponding to the equation X^2 +1 = 0
i'm just a (former) homeless guy who used to sleep in the wet woods behind Berkeley..... (in a loincloth). i have no idea what you're talking about! but thank you for posting and trying.....
I always wonder why Professors dont explain things like guys on RUclips, like you thank you for this insight in Algebra.
Sadly our education system doesn’t dedicate that much time to math profs so that they can explain these ideas in such details, the concepts in this video would be covered in 2 mins due to time restraints. A normal Galois theory class requires huge dedication to group theory and field theory then connecting them by Galois theory stuffed into 3 months, it’s excepted grad students and senior undergrad students will go out and watch videos like these on their own time to understand the big picture better.
It is because creating videos like these takes a huge amount of skill and time. Moreover, these lecturing skills and not (completely) the same as these skills.
That can't happen. First of all Professors teaching courses at this level are not only teaching you the subject but preparing you to be a mathematician. That means they need to teach the material on a much more fundamental level so that the student can do more of the learning independently through exercises and reading.
Moreover making one of these videos takes days or weeks of planning, scripting, and editing. Professors do not have that kind of time to prepare an entire semesters worth of material.
Regardless, introductory videos like these are not exhaustive and are absolutely not a substitute for learning the subject from a textbook or a class.
bm3253 Math does not change often. So if they take the time to produce something like this for critical concepts they could reuse it every semester. Any extensions and exceptions can then be supplemented for completeness. Making mathematical concepts more mysterious and vague does not produce better mathematicians.
@@nikeshsolanki9542 this is true , people assume theirs proffessor can create such video or at least combine the effect of this graphical animations but they cant .
Also , we may note that there is a way that proffesors teach well, they need to incooporate thoose materials in class within time restrains beacause it lays the foundations and speak to our brains differently ,moreover, we know that learning visually also is a great method to enhance our understanting and it's also found in studies .
I got a bit confused... I went to Wolfram and it seems that x^5 -2x + 1 can actually factored in a first degree and a quartic so at least the two separe factored pieces can be solved by radicals. BTW the factorization provided was (x-1)(x^4 + x^3 + x^2 + x - 1)=0. Thanks for the video! I think for the first time I really understood it!
PS. x^5 +2x + 1 seems to be a different story as as in this case Wolfram had to 'cheat' by using hypergeometric functions.
As you noted x^5-2x+1 is indeed solvable by radicals. Better might've been either x^5-x-1 or x^5+2x+1 as you suggested (I think both should work). However, sadly there are a few more mistakes in the video when it comes to computing the Galois groups.
Edit: I think I found the source of the wrong equation www.google.com/amp/s/rohilprasad.wordpress.com/2015/12/17/constructing-a-polynomial-with-galois-group-s5/amp/
Great explanation, you are going to be big on youtube one day!
Really good class. Build up from simplicity and comprehensive examples. Love it!
This is a very, very beautiful video. This is how you spoonfeed and it’s wonderful of you to have put this together, thank you.
This was a great intro to Galois theory. Different and better than I’ve seen before. I hope you’ll do a whole series. I’d love to see how these concepts evolve into Lie groups and to solutions of physical problems.
Wow, what a video
A giant introduction, that gives vague understanding of a problem, but gives no proves on why it works
The Galois group for x^7-2 isn’t the cyclic group with 6 or 7 elements, like it says in the video. It is for x^7-1 but not x^7-2.
The rotation symmetry where the seventh root of 2 gets multiplied by the seventh root of unity does generate a cyclic group but there are other symmetries too. Let r be a primitive seventh root of unity, then r can get mapped to r^2 (and r^2 goes to r^4 etc.). That generates another cyclic group with 6 elements and these two cyclic groups combine together to give another group with 42 elements.
You could write the presentation for this Galois group as
I’m not sure what that group would be called though.
Yes I agree
f(x) = x^7 − 2
The 7 zeros are 2^(1/7)* k^j where k is the 7th root of 1, and j=0,...,6.
Splitting field: F = Q[2^(1/7),k].
The minimum polynomial of 2^(1/7) is x^7−2
So |Q[2^(1/7)]:Q| = 7.
The minimum polynomial for k over Q is x^6 + x^5 + x^4 + x^3 + x^2 + x + 1
& over Q[2^(1/7)] it is the same.
|F:Q| = |F:Q[2^(1/7)]| * |Q[2^(1/7)]:Q| = 6 * 7 = 42.
The Galois group has order 42.
AMs can be.
2^(1/7) -> 2^(1/7)*k^j, for j=0,...,6,
and k -> k^j, for j=1,...,6.
Giving 42 combinations, which all happen.
These two subgroups are C_7 & C_6 but do not commute.
I didn't realize the BGM at frist but soon I found what made me feel thrilled...
Awesome video, very clear! You didn't go into much depth, but you hinted at enough terms and theorems to allow one to five deeper based on the video. Very nice!
@11:51 equation can be factored into
(x"2 plus 1)(x"3 plus 2) equal 0. The solutions exist: i, -i and -cube root (2).
But shown is a non-abelian transformation. But the equation can be factored. Confused.
Wow I have little understanding of math past a high school level but I was able to understand(more or less) what you were presenting. You have a gift for presentation I hope you continue to make videos.
I really liked this video. You are doing a great job. I heard about Galois theory from many people but never actually knew what it is. This video explained everything in a nutshell.
Looking for more videos... was really disappointed to see only one video 😅. Excellent video!!! Thank you! 👏👏👏
I'm getting in on the ground floor. Looking forward to your 100k-subscriber special!
Great video! Loved the dial visual of cyclic group extensions.
Nice Video👍 very intuitive.
This was a terrific introduction. Great job
More than interesting this part of Galois theory...is the first time for me to see and understand these type of equations that you explain in the video, so really true, because is affected in the same part with Complex numbers, how is the another part that we need to know too, and the another point the variation of the other types of funtions that you have done in the example how interchange the differents between between the different results that are interpreted in the numbers as they are not abelian numbers, in this case it is expressed as non-abelian numbers that result in a given difference between each of them by indicating that they are different solutions and that they are not equal. Given in the cases of the mathematical equations that he explains in the video, thank you teacher, kind regards from Cancun, Mexico.
Your channel will be famous. Keep at it my friend.
Great video, but some examples are wrong.
The Galois group of x⁷ - 2 is not C₇. The degree of the splitting field over Q would be 42, as adjoining the real seventh root of 2 gives you a subfield of the reals, after which you need to adjoin a primitive 7th root of unity, whose minimal polynomial over Q(⁷√2) is the same as the one over Q using the tower law. The Galois group is actually a semi direct product of C₆ and C₇. If you change the polynomial to x⁷-1, you get C₆ as the Galois group.
The Galois group of (x⁷-1)(x⁵-1) is not a direct product of C₅ and C₇. The degree of its splitting field is (I think) 24 (adjoin a primitive 5th root of unity then a 7th root, showing that the minimal polynomial of the 7th root over the intermediate extension has degree 6 might need some work idk). If it is 24, then the Galois group is the direct product of C₄ and C₆.
I would love to know the details about the 7 solutions to that 7th-degree polynomial, and the corresponding symmetries (42 symmetries according to another comment, not the 7 mentioned in the video)
6:14
7 solutions, but 42 symmetries
@@BChen7 corrected, thanks
Think about building it up from two extensions, first adding a complex seventh root of unity, say ζ, then adding the (real) seventh root of 2. The Galois group for Q(ζ)/Q consists of automorphisms which fix the real line and permute the remaining 7th roots of unity. Not by rotating, though, because that wouldn't fix the real line. An example of a map which generates these automorphisms sends ζ to ζ^3 (the number you raise to has to be a primitive root in order to generate the whole group). Raising ζ to the 3rd power repeatedly gives ζ^3, ζ^2, ζ^6, ζ^4, ζ^5, ζ. In terms of permutations, if each number is a power on ζ, then the permutations for the group are 0123456, 0362514, 0246135, 0654321, 0415263, 0531642. This is cyclic of order 6, though I know it's a slightly complicated order for the elements to rotate through. Each number is multiplied by 3 each time, mod 7.
Anyway, next, this field needs to be extended by adding the (real) 7th root of 2. The 7th roots of unity are already present, so just adding the real 7th root of 2 covers all the complex ones as well, which are just a 7th root of unity times the real 7th root of 2. The automorphisms here (over Q(ζ)) will be generated by the map which sends the 7th root of 2 to ζ times the 7th root of 2, so in this case, it is just rotating the the 7th roots of 2 by 2π/7 in the complex plane. The reason why other permutations don't lead to automorphisms is because the roots of unity being present prevent non-rotational ones from being valid.
When you put these two extensions together, you get a group of order 42 with these order 6 and 7 subgroups, but this is not an abelian group. It has 6 elements of order 7, 14 elements of order 6, 14 elements of order 3, and 7 elements of order 2, and of course one element of order 1 :) It's also sometimes known as F_7. And it's solveable, since it was made from two cyclic extensions. I don't know if that really helps to explain exactly what it is, but you can manually compute the automorphisms if you want.
Thanks Seán! I can't say I have a complete understanding, but your explanation certainly helped. Along the way, I learned about complex roots of unity and automorphisms.
Loved the video. It ties together so many of the ideas I had floating around loose.
These are useful for error correcting codes, but coefficients of polynomials are only 0 or 1. Galois fields of GF (2^3), GF(2^4) etc.
Interesting video that explains a highly complicated subject in simple ways. Looking forward to a video that explains why the equation is solvable by radicals if and only if the corresponding group is solvable.
Excellent presentation, great pace and animation.
This passed WAY over my head
spare change, bro?
I would love a follow-up explaining why only cyclic groups are solvable with radicals.
I'm taking rings and fields right now and this is a fun watch.
I hadan assignment related to the question that why quintic equations are not solvable by radicals...this video really helped me❤️thanks for mking this amazing video.😊
8:10 The wheels will have 6 and 4 edges. Not 7 and 5. As x=1 is a root of each of the two factors of the polynomial.
Also, your symmetry composition goes the opposite way of convention. XoY usually means apply first Y, then X.
Correct regarding the convention on composition. Not sure I see regarding the dials. Cycling through all 5th roots of 1 will give us legitimate field automorphisms. There are five of them and they form a cyclic group that is a subgroup of the Galois group for the 12th degree polynomial. But if so, then according to Lagrange's theorem the group order divides by 5. But if our group is a direct product of C4 and C6 then its order is 24 and does not divide by 5. Thoughts?
@@mathvisualized355 One of the fifth roots of 1 is 1. The polynomial x^5-1 has 1 as root. It can be rewritten as (x-1)(x^4+x^3+x^2+x+1). The second factor here yields a Galois group of order 4, which shuffles the 4 non-1 fifth roots around.
@@MasterHigure Argh! You're right... I'm gonna have to publish a correction to this. Thank you for pointing me to it.
I included it in the top of the description, for the lack of a better solution at this point. Thanks again for pointing it out, MasterHigure.
Also, x^7 - 2 does not have a cyclic galois group of order p. It has a galois group of order p(p-1).
The group is isomorphic to all group of invertible matrices of the form
(a b)
(0 1)
Super awesome, I am just thinking guys, let me know your ideas. A funny property of this is that, for example in case you have three polygons, when solving the equation by flipping instead of with the lock, that polygon must have a degree equal to the number of polygons you would need for the lock, assuming that in the example given of x to the five and x to the seven, two polygons is the maximum amount of polygons necessary to solve by locking. From this we could already know if an equation is solvable by radicals. It would be great to write a proof based upon this, for example by proving that you cannot take a polynomial and split it into a number of parentheses (I forgot how to call this in math) that are equal or greater than the degree of such polynomial, if the degree is larger than 4.
I love this, thank you so much for clearing up all doubts I had about studying Group Theory. You've rightfully earned a subscriber!
I enjoyed this video. Galois Theory was lectured badly on my course and this was before the internet. Never got my head round it. This video makes things clearer. However, it is not correct to say that the example given, a 35!order group, is not cyclic. It is cyclic of order 35. If you take your a and b of order 5 and 7 respectively, their product has order 35 and generate the group.
Really enjoyed this. Have you thought about making a video about permutation groups and cyclic groups specifically? Maybe a proof of 5+ elements not being able to decompose into cyclic groups? That is a very fun thing to show visually
But apart from the combination that he says between group theory and body theory is another point in which Galois explains as a functional method given as abelian numbers and non-abelian numbers ... in fact, for me since I have been studying mathematics for quite some time back, I feel that even oneself must still continue to seem because this area of mathematics is quite extensive, thank you teacher.
Thank you I'm now interested in math too, but come from physics. Need those intuitive explainations. 🎉
This is a really great video on Galois group theory!
Im here before this channel blows up
Looking forward to further videos.
Really helpful and amazing video . keep making more visual video
Very nicely explained.
Hi, I have watched your video 5 times so far. It is refreshing and very very original.
Thanks you for sharing this wonderful way to simplify the Galois Theory.
I have been interested on the Quintic and the solvability of polynomials all my life.
A lot of work to put all the essence on the theory in a 15 minutes video.
You are a genius !
How did you do it ?
What software do you use ?
The result is amazing...
Bravo !
Can you please put some more information about yourself ?
Why is it that S_5 cannot be built as a series of cyclic extensions? What changes between S_4 and S_5? Why does it start at 5?
This is a rather deep question about the structure of the symmetric groups. Ultimately, the question boils down to the structure of a particular subgroup for S_k, which we denote A_k. A_k is the alternating group on k letters. It consists of all of the "even" permutations. Permutations can be classified as either even or odd, depending on whether it has an even or odd number of inversions. As it turns out, the even permutations by themselves form a group! (The odd permutations do _not_ form a group.) But the alternating group has a very important structure within the symmetric group - it's a _normal_ subgroup. This is the type of subgroup we need in order to do that "extension" idea talked about in the video in a meaningful way. What needs to be Abelian is not the subgroup A_k, but rather the _quotient_ group where we collapse S_k by forcing all of A_k down to a single element - by treating any even permutation as if it were the identity permutation, the do-nothing permutation. And it turns out that if you collapse S_k along A_k, you do get an Abelian group (in fact, a cyclic group)!
However, A_k is what's called a "simple" group for every k _except_ k = 4. Simple here is referring to a "building up" sort of sense. Simple groups are the basic building blocks for this "extension" notion. You can't build up simple groups from smaller groups. So A_k for k >= 5 can't be broken down further, and it's far too complicated to be Abelian itself.
For k < 4, A_k is too small not to be cyclic, and all cyclic groups are simple! But here's the difference - since A_k is itself cyclic, this works for solvability!
For k = 4, we actually get something quite interesting. A_4 is definitely not cyclic or even Abelian. Far from it. However, A_4 has a nontrivial normal subgroup, the Klein-4 group. In general, for k >= 4, A_k has the Klein-4 subgroup living inside of it (for larger k, we get more and more copies of Klein-4 living inside of it). But 4 is pretty much a special number here. Permuting 4 things is too few things for there to be multiple copies of Klein-4. The uniqueness of Klein-4 in A_4 combined with the size of A_4 forces this Klein-4 subgroup to be normal in A_4. Then the collapse of A_4 along its Klein-4 subgroup is too small not to be Abelian (indeed, too small not to be cyclic). Moreover, the Klein-4 subgroup itself is too small not to be Abelian. So it still ends up working, but we're right at the cusp of it working - look how much more work we had to do to even realize A_4 as solvable compared to A_3, for instance. These alternating groups get a _lot_ bigger _very_ fast. Once you hit 5 things to permute, A_5 has more than enough room to be complicated enough not to be solvable.
I don't know that there's any nice way to make it easier to digest than this, unfortunately. The alternating groups A_k are extremely complicated. If it were easy to understand A_k for every k, it would actually be easy to understand all finite groups. But this isn't the case. Understanding groups is hard.
What a video! Haven't seen something so clear and didactic since Lawvere book
Which book of Lawvere are you refering to? I'm only aware of one at the moment (the one on Set Theory together with Rosebrugh) which I did not read (yet!).
Hey there, I love your video! It explains Galois Theory incredibly easily and smoothly, and in a way that people who are not interested in math can understand it. I'd love to contribute subtitles in English and German, so I can show this video to my friends. If you're interested, hit me up with a DM so I can submit them to you, once they're done! Have a great day and thank you for this gem :)
Why is it that a cyclical galois group implies solvable by radicals?
This is the absolute best explanation, thank you :) !