One of the few aspects of KZread that improves with time is the growing audience for content like this. It has always been a minuscule percentage of users, but with the growth of the user base there’s enough viewership to reward and motivate content creators like this. I’ll never take this for granted.

@DjordjeRomanic

3 жыл бұрын

Well said!

@antilogis6204

3 жыл бұрын

There is a broader viewership, consequently resulting in more motivation for any uploader but as you said, by force. If you think about all the crappy clickbaity content which is being uploaded over time just for the sake of some views, fame, (easy 'money'?) and it's only discouraging, so in terms of percentages I wouldn't risk it one bit. But all in all, I agree with you, good quality content can only enrichen the whole platform.

@mygirldarby

3 жыл бұрын

Anyone who was around when the internet first became available to the public knows that this kind of content was all you could get (in more rudementary, non-video form). The "average" person on the internet in the early-mid 90's was very educated and polite. It was a joy amd an honor to have experienced it.

@devanshuyelurkar1017

2 жыл бұрын

Why have you removed your learning maths video, please reupload

@hueheeuuehueuheuheuhueheeu1245

4 ай бұрын

plot twist: they are the same user base in every math channel

I agree with the sentiment that perhaps too large of a jump was taken for the relationship between the tiling and the lack of formula. I followed everything before, and that step came seemingly out of nowhere. I couldn't believe the video was already over, made me feel like something was missing.

@KemonoFren

2 жыл бұрын

There is a lot to go through to explain the specifics that a video like this can't cover. It would be fully covered in a book on Abstract Algebra.

@MA-bm9jz

2 жыл бұрын

Read about field extensions, any abstract algebra book covers this, of course maybe you lack the pre requisites

@muesk3

2 жыл бұрын

To my commentors: of course some books on the topic would cover this. That is not the point. I was saying that previous videos on this channel never made such leaps, and I could always follow them without being a proper mathematician. It was just some feedback towards Aleph 0. I still very much enjoy this video and the channel.

@leonardeuler6170

2 жыл бұрын

Yeah I agree, I could follow the general idea before the last section. But the last section was too much of a jump, took some time to see , that he is talking about tiling by subgroups and counting the numbers.

@TheZenytram

Жыл бұрын

@@KemonoFren so this video is useless, it didnt showed the Insolvability of the quintic, it explained a bit of Galois group theory and then jump to conclusion. The same is as if it was a physics channel explaining a bit the atmosphere and by the end saying "bc the sky is blue the grass is green".

8:00 "In general, if a polynomial is solvable by radicals the number of tiles is a prime number." But why is this the case? This is kind of a conceptual gap in the video that left me dissatisfied. Great video otherwise!

@N9199

3 жыл бұрын

It has to do with the fact that in some sense every finite abelian group is a product of groups like Z/qZ for some q=p^k, and in some pretty specific sense this means that the number of tiles is a prime number.

@Entropize1

3 жыл бұрын

There is alot of work that goes into the insolvability of the quintic if you want a complete proof. This is essentially a culminating theorem in a first semester of Galois Theory. I recommend Milne's free online notes on Galois Theory after you've taken/learned basic group and ring theory.

@voiduniverse5563

3 жыл бұрын

Maybe this is given as a fact because he has difficulty proving it concisely or intuitively in this video. In textbook it is stated that a finite group is solvable if and only if all its composition factors are groups of prime order. If he needs to explain the fact, more details have to be involved.

@Sam_on_YouTube

3 жыл бұрын

I had the same issue.

@FranFerioli

3 жыл бұрын

This stuff has never been easy...

Nice explanation, although I'm slightly confused since you didn't explain everything. Where do the subgroup chains come from and what do they represent? I think you should show an example with concrete polynomials. And also, although this is probably a more advanced topic, I'd like to know why exactly do non-prime numbers mean that it's not solvable, since it seems that the entire proof hinges upon this single assumption.

@mrtaurho8846

3 жыл бұрын

Solvability of groups is defined in terms of some, well, let's call them "decomposition sequences". We take the (finite) group and break it apart into smaller and smaller pieces (the pieces being appropriately chosen subgroups contained in each other) until we hit rock-bottom, i.e. the trivial group. The group being solvable is then some conditions on how exactly this decomposition into smaller pieces takes place. As it turns out (and this is in fact not so trivial) if we have a solvable group we can always take such a "decomposition sequence" and refine it (by inserting some intermediate subgroups inbetween) until every "transition number" is prime. The converse is straightforward (if all "transition numbers" are already prime there's nothing to do) and hence a group is sovable if and only if no non-prime "transition numbers" appear in the "decomposition sequence". The details are bit more delicate (as every rigorous treatment of Galois Theory is) but I hope this conveys the basic idea.

@Tsskyx

3 жыл бұрын

@@mrtaurho8846 So like, if non-primes still appear in the chain after the first decomposition, then it's not solvable?

@mrtaurho8846

3 жыл бұрын

@@Tsskyx No, only if can't get rid of them via interated refinements. So, you can start with, say, the whole group as "decomposition sequence" and then insert some suitable (not any kind of subgroup suffices here) subgroup to get a new refined sequence. If now all "transition numbers" are prime we're happy; otherwise repeat. But there are groups (notably the A_n for n>4 from the video) which just don't have any suitable subgroups to insert. Hence we get stuck at a non-prime "transition number". It's a bit hard to break down without going deep down the mathematical rabbit hole.

@Tsskyx

3 жыл бұрын

@@mrtaurho8846 I see

@yikeslikes4457

3 жыл бұрын

@@ikarienator you misunderstood the question

When this channel explains something I already know, somehow it makes me know it better

Dude. I was reading the section about galois theory literally today in Modern Abstract Algebra and I didn't get it at all lol... Anyway, thank you so much for these videos, you truly are a hidden gem of youtube

@Brien831

2 жыл бұрын

@@AmourLearning I recommend richard borcherds videos.

The "why" you never get a prime tiling beyond 4 seems just as mysterious as to why there's no general formula for beyond quintics.

@gustavopauznermezzovilla4833

3 жыл бұрын

Actually, there is a very "algebrecity" of Sₙ for n ⩾ 5, that wraps this part out. I will copy an argument from my Galois course. Suposse that there are G₀, G₁, ..., Gₘ such that {e} = G₀ ◃G₁ ◃ ... ◃ Gₘ = Sₙ and every Gᵢ₊₁ --- Gᵢ have prime order, i.e., Gᵢ₊₁/Gᵢ is a cyclic group with prime order. Some algebra 1 theory tells us that these quotient group have the comutation property necessarily (it is abelian). Therefore, for any a,b ∈ Gᵢ₊₁, it would aGᵢ bGᵢ = bGᵢ aGᵢ and so a⁻¹b⁻¹ab ∈ Gᵢ. The element a⁻¹b⁻¹ab = [a,b] it is called the commutator, and we had seen that it transfers all the elements in Gᵢ₊₁ to Gᵢ. Okay, thats some generic bullsheet, and now we will see "why" the n ⩾ 5 its important. All those groups are subgroups of Sₙ, i.e., a set o permutations. Lets find out some dirty permutations. Chosse a,b,c,d,e ∈ {1,2,3,...,n} distinct (which can be done cause n ⩾ 5). After a lot good homework, the triplets factor as (a b c) = (d b a) ⁻¹ (a e c)⁻¹ (d b a) (a e c) = [ (d b a) , (a e c) ] Therefore, every triplet is an commutator (which we saw that trasnfers every element one step down. Using those "stepness" property on each group, every triplet in Sₙ should be in the trivial group G₀ = {e}, which is INSANE. WTF ALGEBRA DUDE?! IS THAT ALL ABOUT? SOME MIRACULOUS PROPERTIES OF AN RANDOM NUMBER? yep.

@rockinroggenrola7277

2 жыл бұрын

So, in the subgroup chain that was shown in the video, each of the subgroups creates a chain of "simple" factor groups. Factor groups are groups that you get when you "divide" successive groups in the subgroup change. Simple groups are groups that don't have any normal subgroups, a special type of subgroup. In these subgroup chains, each factor group must be simple. In the case of n = 2, S2 only has one subgroup, the subgroup of just the identity element. The factor group S2/{e} is a simple group. In the case of n = 3, S3 has a normal subgroup, namely C3, which is the cyclic group of order 3. When you take the factor group S3/C3, you get a simple group, and C3 can then be decomposed to the group with just the identity. However, at n >= 5, Sn can first be decomposed to the group An, which is the set of even permutations. However, An turns out to be simple for n >= 5, so the only subgroup which may come after it in the subgroup chain is the subgroup with just the identity. Therefore, you don't get a prime tiling from Sn to {e}. Instead, you get n!/2, which is not prime for n >= 5.

@georgecantu856

2 жыл бұрын

@@gustavopauznermezzovilla4833 so basically when n=5, Gn has multiple identities?

@artey6671

Жыл бұрын

@@gustavopauznermezzovilla4833 Dude, that's crazy. It's well-known that solvability is equivalent to the commutator chain becoming trivial, but I didn't know you can use that to easily prove that S_5 is not solvable. However, there is a stronger statement than this, namely that A_n is simple for n\ge5, where that argument doesn't work. So that cool proof is sadly not as useful as I had hoped.

Bro, I hadn't checked your channel in a while and I notice there are two videos missing: -How to learn pure mathematics -Math isn't ready to solve this problem What happened to those videos:(? I really loved them.

@nahblue

2 жыл бұрын

How to learn pure mathematics is here: kzread.info/dash/bejne/mKNhw86wYtSTcrg.html It's unlisted.

@joshuajoshua46

2 жыл бұрын

@@nahblue Why did he unlist them?

I wish they had your video during my Abstract Algebra days at Uni. The "roadmap" of a mathematics course at this level is generally obscured by the cramming of fundamental concepts into our heads, not how those concepts have built the map. Thank you for filling in a fundamental part of the map.

What happened to your video guide to self studying mathematics? It was super helpful.

@shalinisharma3542

2 жыл бұрын

I know bro I am searching for it

there's nothing like the feeling of discovering a new maths channel. great content you're severely under-appreciated

Great job Aleph! Been waiting for this for a while

@arindambhattacharyya5593

3 жыл бұрын

Hello there

@shohamsen8986

3 жыл бұрын

I like what u did there. Treating Aleph like his first name and Noll like his second.😂

I've seen this topic a bunch of times but never going deep enough and without this level of clarity. Massive thanks.

You had a video on books to read for people who want to learn college mathematics and it had a link to an open letter for calculus, can you share it again please? Or anyone else who passes bu here and know what I am talking about.

@thirdwave1777

2 жыл бұрын

looking for the same. please post here if you find.

@everlastingideas8625

2 жыл бұрын

@@thirdwave1777 I am still looking for it but yes, I ll post it here if I find it sure. Please do the same if you can.

I just want you to know that I was genuinely happy when I saw this in my abo feed.

Okay man I have to say that I've watched and read a fair amount of introductory content to Groups, Abstract Algebra and Field Extensions and this is by far the most comprehensive one. It is amazing how clearly you present the ideas, an statement of how organized you have the topic in your own mind, truly amazing! Thanks!

@harriehausenman8623

3 жыл бұрын

Me too! I try to tackle these ideas for a while and this kind of 'quick but still precise' recap helps a lot.

@cufflink44

Жыл бұрын

Gonzalo Christobal Well said.

👏👏👏 just made sense of a full course on Galois Theory full of formalizations of automorphisms and company and empty of intuition. Great video.

Why have you removed your learning undergraduate maths video, pleas reupload

i see a lot of people in the comments here who find this explanation unsatisfying, and this is certainly a good instinct to have! the reason is however because to actually understand this result you need a courseful of context, and a 10 minute youtube video can only do so much. i don't think this video was ever supposed to be a comprehensive explanation, but just an introduction to the topic c: btw the way the music loops is killing me

Great video! Thanks

I've not taken any analytic algebra courses and still able to follow along with the basics thanks to your great visuals and explanations.

I'm looking forward for when you'll have millions of subscribers. Great presentation, unique aesthetics and great didactics ! perfect!

Your channel rocks! Especially liked the self-study video. Thank you!

@shalinisharma3542

2 жыл бұрын

Where that video is I can't find it

A Cambridge University math course that comprehensively explains Galois theory so elegantly in 10 Mins. Wow thank you so much. I feel cheated by my university that could never really explain anything. We were simply following procedures.

In case anybody is interested, there's also a really nice elementary argument (no Galois theory) for the statement about a general formula due to Vladimir Arnold, which I just put up a video about here: kzread.info/dash/bejne/dId82Jt-nM6Xfbg.html Some other resources in the links there too :)

@MichaelRothwell1

2 жыл бұрын

This video by @not all wrong is amazing. Full proof, no gaps. Highly recommended! It also explains exactly how you get you get the decomposition series of a group - by taking as your normal subgroup the commutator subgroup at each stage. This obviously gives an abelian quotient, since you have set each commutator (ghg^-1h^-1) equal to the identity element.

Although there may be a detail or two that you did not explain, still you break the discussion down to the crux of the matter and thus make the discussion accessible to the uninitiated. That is inspiring! Now I really want to learn the details. Thank you 😊

I love to explore the mathematics this way! Thank you so much.

I have no idea whats going on but, I just love ur video's.

Great job!! I always wondered how that worked out. As I specialized in functional analysis and probability theory, I never got the chance to dive deeper into this stuff. Thank you!

You, sir, are another potentially big player on the mathematical KZread scene, which is clear from the quality of your content and the fact that you have almost 70K subscribers. Keep up the great content, you are doing the right thing. I like the way you present the content, reminds me of ASAP Science.

Many thanks. I did a course on Galois Theory in the third year of my maths degree, 40 years ago. I can't recall much of it, but this video has brought bits of it back to me. This has whetted my appetite to read up on it.

You just read my mind wow I started learning about this topic since 2 days. thanks for the great explanation and quality.

Been waiting for your video for a long time!

Another totally amazing video. Huge thanks.

Really enjoying your series of videos. Thank you.

Brillantissime vulgarisation ! En 10 minutes seulement les grandes ligne de la preuve et ses éléments de la théorie de Galois. Merci !

@grossetalain1234

Жыл бұрын

Tout à fait d accord. Seuls les anglo saxons arrivent à faire cela.

Love your videos! A recent problem set of mine involved proving that A_n is simple for n>4 and this is making me excited to learn Galois theory this spring

@harriehausenman8623

3 жыл бұрын

Just don't get into any duels ;-)

Very nice presentation of a difficult concept. Keep up good work.

great video as always, I'm looking forward to taking abstract algebra next year and this is an awesome preview thank you

I'm going to be studying field extensions and Galois Theory next week so this video is extremely helpful.

You used to have a great video for people who wanted to study pure maths with a syllabus suggested you tube videos and books where did it go I would love to see it again or at least have a list of the videos and books thanks

Just wanna say, love your videos Aleph! This one was especially interesting and cool, really excited for when I get to group theory in Uni :)

Two tiny nitpicks: First, at 2:31 I think it's misleading to write it as at every step a n-th roots is adjoined. This is not necessarily the case as one of your examples illustrates too. Might've been better to use n_i instead to indicate that the degree of the extension is not necessarily the same at each step. Second, I agree with the comment by @diribigal that the phrase "there is no general formula" is not exactly precise and might cause some confusion. For example, X^5-2 is a sovable polynomial of degree 5 as the Galois group is not all of S_5 but some proper solvable subgroup. However, at 8:25 you write "Gal(general degree n polynomial)=S_n" which, IMO, only adds to the confusion: the general polynomial meant here is in some variables t_1,...,t_n with no algebraic relation between them and not any polynomial over, say, Q. I know that it's hard to be this precise and still concise at the same time so I thought it might be worth adding a comment. Last but not least: I really enjoy your videos and I'm looking forward to more!

@harriehausenman8623

3 жыл бұрын

good point

@pineberryfox

2 жыл бұрын

nth root is fine actually. the square root of 2 is the sixth root of (2^3)=8 and the cube-root of something is a sixth-root of the square of that thing. in general, for finitely many adjunctions n is the lcm of the individuals. imo adding the subscripts would at least somewhat hamper readability, and because it's not necessary i'd rather them be left out

Very good 👍

Finally, someone can explain this quintic mystery 👏. Thank you!

Marvellous work!

The most impressive thing is that Galois came up with most of his ideas and wrote everything just a couple of days before he died. At the time he was only 20.

@gianfrancodue1476

3 жыл бұрын

Well, Hardy used to say no good Mathematics got discovered by someone over 50, most likely 30 too. He got a point, somehow

Great intuitive video, taking my second semester abstract alg class rn and this is spot on

really nice Video! I quite like the long end card (like 20sec), because i gives time to rethink everything you said, before having to interrupt the autoplay

Brilliant video!!

Thank you for the video. But, to be honest, the last part of the video was not comprehensible for me. I hope someone would be able to clarify it to me.

You do a great job.

Thanks a ton....scratching my head to understand this concept from last 4 years . Now i can make sense.

Very nice explanation. Thank you for this excellent work!

The quality of your content is exquisite! Keep going with this tempo! Btw,why did you delete the video on Hodge Conjecture ?

Great video!

A very nice presentation! Thanks for referring to my book "Galois Theory for Beginners".

I love your videos !

I loved Galois Theory. I wish I could continue this path, but the other mathematical subjects were too hard for me and could not pursue getting my Ph.D. I ended up taking 2 yeas self studying abstract algebra just to understand this proof. It was worth it.

@artey6671

Жыл бұрын

It's my understanding that you basically need to be a genius to get a PhD. If this proof gives you so much trouble, then maybe a PhD isn't the right thing for you.

Excellent. Thank you. Could not be explained better.

this is beautifully done.

I love this. The explanation was simple enough to understand as a Physics student.

Great work! I love watching these videos

Thank you so much for making this material, you're amazing 🙂👍

Please provide a full course in calculus cause you really really explain well

I am really amazed by how mathematicians and engineers are incapable of giving conceptual explanations of some fundamental ideas in their areas, but you did it!! you made me understand what solvable by radicals means, in a simple and insightful way; many thanks! the most useful part for me was the beginning, when you explained we had to add radicals step by step; I know the Insolvability of the Quintic isn't in contradiction with the Fundamental Theorem of Algebra, but then what is a root when it is not given in radicals?! this is also a thing which I think is fundamental but I can't find a good insightful nutshell explanation of;

beautifully explained !!

Totally enjoyed this video!

Congrats. This is the most understandable and intuitive presentation of the Galois Correspondence and quintic insolubility. Contents of books are a torture to assimilate. Other lectures (lecture series) can't see the forest for the trees. Thank you

amazing work

I had a surprisingly difficult time finding someone bold enough to make the simple, explicit claim that nested radicals correspond to field extensions. You'd think that would be easier to find, but I had a heck of a time finding it. Didn't see it on math stack exchange, and darn sure didn't see it in grad level textbooks because they never explain anything clearly. Didn't find it until seeing it in your video. Thank you very much.

So hyped for this one 💯💯💯

That's the summary of Abel's, Galiois work. Brilliant

Nicely explained

I really like the style of your videos, with the handwritten illustrations. What do you use to edit and create your videos? Also, thanks for sharing your mathematical knowledge!

New upload :)))) your videos are amazing

Great stuff as usual, but I still don't completely get it (logical, though - it's ten minutes, and I have only the most basic rudiments of Abstract Algebra).

@Cardgames4children

3 жыл бұрын

Don't worry, it's just not possible to completely "get it" in 10 minutes. Galois theory is the culminating end result of what turned out to be a 300 year effort by various mathematicians to answer one question: Does there exist a "quintic formula" and above? The quadratic formula was known in ancient times (albeit in word form, as symbolism didn't exist yet!), the cubic formula was found around the 1530s, and the quartic formula shortly after around 1545. But what about polynomials with higher degree? Around the early 1800's, mathematics has advanced enough where the time was "just right". And in the 1820's, the matter had been settled by two young geniuses, independently of one another. Abel proved in particular the impossibility of solving the quintic in radicals, while Galois took it a step further, settling the matter once and for all by figuring out exactly when polynomials ARE solvable in radicals. What is truely remarkable is that not only did Galois manage to do that, but he did it when he was 19 or so, and he didn't even have the field concept yet! He was the first to recognize the importance of groups (his terminology, in fact!). He must have possesed immensely incredible insight, having figured out this new angle of attack into an old problem. It's one thing to try and understand something with all the modern cushioning and years of something being established, but a completely different thing to even invent it yourself, and forging the difficult path into that unknown. But even in its modern formulization, Galois theory rests on a bunch of abstract algebra (mostly group and field theory) And it's STILL quite a difficult thing to learn! A lot of texts don't even show you the historical way things were done in the early days of Galois theory, via symmetric functions and whatnot. That's how Galois himself has done it, and I feel a lot of intuition might be lost by not showing these origins.

@MrAlRats

3 жыл бұрын

@@Cardgames4children I believe the reason, that most people are unable to appreciate the beauty of mathematics, is that we are taught it in a way that makes it impossible for most people to see for ourselves the reasoning involved in producing the mathematical results that we are taught. We have no intuition for why something is true or why a certain technique is able to solve a problem. If the origins of mathematical concepts and how to make an attempt at a proof of mathematical statements were taught alongside the things which are required to solve problems, then most people would have a better appreciation for mathematics.

@Cardgames4children

3 жыл бұрын

@@MrAlRats totally agree. While I appreciate the beauty of mathematics itself and the ideas / concepts it contains, I also really like the beauty of how those things fit together or why theorems are true. Seeing the thread of logic tie different ideas together at each step, then clearly reaching the final statement ie the theorem you're proving. It's fun to sit back and just marvel how closely-knit the concepts are. One of my favorites is showing why a field extension F(a) has a dimension equal to the degree of the minimal polynomial p(x) of a. Let it be n. F(a) is the set of all polynomials in a with coefficients in F: F(a) = {A(a) : A(x) € F[x]} By the division algorithm, we can write A(x) = q(x)*p(x) + r(x), where deg(r(x)) Thus any element of F(a) can be written in the form r(a), where r(x) is a polynomial of degree n-1: f_0 + f_1 * a + ... + f_n-1 * a^(n-1). Is the set {1, a, a^2, ..., a^(n-1)} independent? If it were NOT, then some of the above f_i would be nonzero, and so a would be the root of a polynomial of degree n-1, this is less that it's minimal polynomial degree and hence not possible. Hence all the f_i are 0, and so {1, a, a^2, ..., a^(n-1)} are independent. Therefore {1, a, a^2, ..., a^(n-1)} forms a basis of length n, and so F(a) has dimension n. What I love here is when the any-old-normal polynomial f_0 + f_1 * a + ... + f_n-1 * a^(n-1), viewed as a single object, is now taken, not as a polynomial, but as a linear combination of F-elements with the vectors {1, a, a^2, ..., a^(n-1)}. It was such a slight of hand that I think this proof is amazing, and shows the beauty of mathematics!

Wonderful. Thank you.

Pretty informative and understandable for distilling a whole undergrad course into 10 minutes

Your videos as awesome, very didatic. I would love to see a video about harmonic analysis, for instance about the philosophy of the Fourier transform or something about the Calderón-Zygmund operators

Really great video and channel!

There seem to be some steps missing. And I don't mean the step of having taken Modern Abstract Algebra. Why is it S4--2->A4--3->V4--2->C2--2->e and not S4--2->A4--12->e? Or, alternatively, why is it not something like S5--2->A5--5->W4-->3->V4--2->C2---2->e? Why is the 60 of irreducible complexity when the 12 was not?

@mrtaurho8846

3 жыл бұрын

As A5 is simple whatever W4 denotes it won't be a normal subgroup of A5. However, solvability requires that every (sub)group in the composition series is a normal subgroup of the the (sub)group immediately preceding it and that the quotients are all abelian. Hence, S4→A4→e is not enough as A4 is non-abelian and S5→A5→W4→V4→C2→e doesn't work as W4 is not normal in A5.

@PvblivsAelivs

2 жыл бұрын

@@mrtaurho8846 Hmm. I attempted a comment last night and it appears not to have gone through. I somehow didn't see the reply until then. I am largely self-taught and have not previously encountered the terminology of a "normal subgroup." However, I think I can surmise that it merely means that it is closed under the group operation. I have seen the term "abelian." But "commutative" is more commonly used, the definition of "abelian" is only that it is commutative, and we are hopefully trying to make this more accessible, not less. The question of why it was not S4 -2->A4-12->e asks why the quartic or biquadratic (I've seen both terms) does not fail the way the quintic does (12 not being prime or even a power of a prime.) And a big part of this gap is that the video does not explain what An is (the set of even permutations of a set with n elements, A2 being the identity permutation.) In preparing my reply to you (twice) I managed to work out what V4 (as mentioned in the video) is. It does help. And I note that your criterion does not match the author's -- at least in a technical sense. V4 is commutative (your criterion) so you would stop there. But it does not have a prime number of elements. (I need to be careful with my words here. I almost said it doesn't have a prime order. It does, 2.) The author breaks it down to C2, which is done wholly arbitrarily. When you tell me "whatever W4 denotes it won't be a normal subgroup of A5" you aren't really adding anything. I am not able to deduce the meaning when you call A5 "simple." I don't recognize the terminology and, perhaps more importantly, the context does not give me sufficient information.

@mrtaurho8846

2 жыл бұрын

@John Undefined You're missing some fundemental terminology commonly used (AFAIK) in any book on Abstract Algebra. This is fine as I only used it for brevity but let me expand. Normal Subgroup: A normal subgroup is not merely closed under the group operation (this is a subgroup) but satisfies a further technical condition. This condition states the the subgroup is mapped onto itself by conjugation (multiplying from the left by g and at the same time from the right by the inverse of g). Note that not every subgroup is normal in general, although this is the case for abelian groups. Abelian Groups: Indeed, abelian groups are simply commutative groups and this is actually the more common term in the literature. Commutative is more commonly used to describe a property which binary operations (roughly, something taking two inputs and producing one output) may or may not have. So, a group with a commutative group operation is called abelian by convention. Quartic≠Biquadratic: In general, an equation of the form ax⁴+bx³+cx²+dx+e=0 is called a quartic equation while a biquadratic equation is of the special form ax⁴+cx²+e=0, i.e. only the squares of squares appear (so bi-quadratics). V4 in A4: V4 is the Klein Four Group which is a (commutative) subgroup of A4, but not a normal subgroup. The precise definition of these decompositions, however, requires normal subgroups of which there are less. The fact that V4 is abelian is completely irrelevant for whether or not V4 a normal subgroup of A4. Moreover, I did not claimed the criterion being that V4 is abelian but rather that the quotient groups at each step have to be abelian. This is somethinh quite different! Also, breaking down V4 into C2 is the only possible way of further refining this decomposition to get prime order transitions (V4-e has a transition of order 4, which is not prime, while V4-C2-e has two transitions both of order 2, which is prime). And C2 is also the only non-trivial subgroup of V4. Simple Groups: A simple group is a group whose only normal subgroups are the trivial subgroup and the whole group considered as subgroup. If the original group is non-abelian and simple it won't be solvable. That's the crux about A5 (and hence S5). At this point I would advise you to read up the precise definition of solvability. Starting from there you might benefit from learning the necessary preliminaries for understanding the concepts properly. As long as we keep hand-waving some things I don't think it will get clear what I'm saying. If you mind, I can recommend the book by Judson (Abstract Algebra: Theory and Applications) which is freely avaiable online. Hope that helps :)

@PvblivsAelivs

2 жыл бұрын

@@mrtaurho8846 "Quartic≠Biquadratic" I have seen the quartic called the biquadratic and seen the terms used interchangeably. But I not trying to get caught up in the terminology. I am trying to communicate.

Another knockout! No fat but still clear and satisfying

Watching after an abstract algebra class is a whole new perspective

This video is the nice introduction.

Thank you very much, very informative video. Since your handle is Aleph 0, I'd love to hear your take on Cantor's countable/uncountable infinities (I'm not proponent of the idea).

Hi.. you post great videos. I was wondering if u can suggest any book(s) on your previous video regarding Navier-Stokes Equation (how it is used in fluid dynamics)?

Youre amazing

This isn't what I was looking for but thanks for that lesson.

I think there was a video of yours in which you gave your recommendation of mathematics books. I can't find it on youtube, I've been trying to for a while. What happened to that video?

if I remember correctly you had a video about tensors and how tensors are related to the direct summ or product or something like this. did you took it down or did i just dreamed about it because i cant find it anywhere. And you do a realy good job keep it up we all appreciate your work.

MANY THANKS.

I hope you do a video on V.I. Arnold's proof of the insolvability of the quintic. It might be easier for viewers to follow.

THANKS!

Great video. I would like to ask how do we find the equation that the roots must verify

Does this have to do with the fact that the alternating groups with n \geq 5 are simple?

@jeremywang2471

3 жыл бұрын

Yes, when n\geq 5, you get a composition series S_n>A_n>{e}, and Jordan-Holder says that you’ll never find any other composition series, so S_n is not solvable.

Awesome! (notifications ganggg)

Thankyou so much

Thanks❤️

I think he made a video on how to learn pure mathematics on your own , but I can't find his video . Does anyone remember it ?