No simple groups of order 66 or 144.

[@johnbailey8103]

Order 66? Definitely should've waited for tomorrow for Star Wars Day... But an early May the 4th be with you

[@goodplacetostop2973]

0:16 Ok, great. It’s always good to try new ideas 28:27 Good Place To Stop

[@martinnyberg9295]

Absolutely! Short lectures on advanced math once a week? I’ll be there. 👍🏼

[@f5673-t1h]

A generalization of the first solution is that if you have a group of order m*p^a (as given) where m < p, then it's not simple. This is because all divisors of m are below p, so you cannot have any divisor of np other than 1 that is congruent to 1 mod p.

[@jermeekable]

nice observation! i didnt catch that

[@alainrogez8485]

Other generalisation : if order of G is pq or pqr then this group is simple.

[@tomatrix7525]

Awesome. It’s nice to see higher-level math which may not have as wide an audience but is definitely interesting for a certain niche of viewers.

[@salim444]

at 17:43 the explaination is that if g in a sylowp and sylowq then the order of g divides p and q so it is 1

[@ImaginaryMdA]

I'd forgotten all of this. It feels good to be reminded.

[@carl3260]

12:25: should be “greater than” (which means it cannot be injective)

[@mcmanustony]

17 minutes in, had to recall that non identity elements can't be in both a Syl-2 and a Sly-3 subgroup because their order must divide 9 and 16. You may proceed..... :-)

[@ImaginaryMdA]

I had to pause here as well.

[@maxturgeon89]

Good catch, I was coming here to make that observation

[@emsihongg4868]

Reading your comment, I had to recall that the order of an element divides the order of its group, since an element generates a subgroup of its order to which one may apply Lagrange, whence this claim follows.

[@emsihongg4868]

And the cases (that the order of P_i \cap P_j is 1, 3, 9) follow from this argument as well.

[@technoguyx]

I'm prepping for a grad algebra exam so this will come in handy. Thanks for being so dedicated to your channel, been subscribed since the early days.

[@ZonaALG]

Execute order 66

[@User-gt1lu]

Or maybe order 69

[@malchar2]

Love to see these higher level problems on KZbin. It can be really hard to find accessible resources for petty much anything after calculus.

[@jermeekable]

YEAH! this the kinda stuff I was askin for! Well far more so! Im a mathematician turned programmer so I dont get high level proof math stuff to do anymore :(

[@MichaelPennMath]

This is good to hear. I have had this unlisted for months. I was worried about how it would be received.

[@jermeekable]

@@MichaelPennMath I obv cant speak for everyone but Id love to see more higher level stuff thrown in like this. Personally, Im game to see some Raw theorem proofs; example proving elliptic curves are isomorphic to tori. Ik that one might be a bit much to pack into one vid, but its just to illustrate my interests from your future content. I have posted a couple comments requesting this type content on older vids and have had others concur in response. One person said theyd like to see some topology proofs, which sounds good to me!

[@eulertoiler9774]

I'd love to see you do Deligne's proof of the Weil conjecture, just to see exactly where is a good place to stop with that.

[@jeffreycloete852]

Hi Prof Penn. .thanks. .again for sharing your time, energy and knowledge!

[@generalmathematics8499]

Claps for the the second one sir! (|G|=144). Constructing a homomorphism with non trivial kernel is really good idea for proving group is Not simple. Really impressed 🙏

[@OH-pc5jx]

Love this, keep ‘em coming

[@WhydYouChangeMyHandle]

As someone taking abstract algebra for like the fourth time, this is extraordinarily helpful. Thank you!

[@Zagszy]

This is a fantastic video. It clarified many uses of the Sylow Theorems and definitely helped me prep for my exam.

[@eukleidesk6759]

This was excellent. Really interesting. Thanks!

[@samsonblack]

Nice! I still remember the homework exercise from Algebra: for each composite number n less than 168 (excluding 60), prove there are no simple groups of order n. And, yeah, 144 is one of the most difficult orders.

[@gnarlybonesful]

Very excited for upper level maths. As someone with a math degree I've always been disappointed by the dearth of quality high level lectures on the internet.

[@Walczyk]

21:55 i'm guessing we skip 16 for an obvious reason as well?

[@emsihongg4868]

its because 3 does not divide 16, but there is an element of order 3 inside N. Furthermore, 24 and 48 are left out because they are not divided by 9, but P_i is in N. Therefore, leaving these out is required by Lagrange.

[@kyleneilson1457]

Two cool general facts about simple groups that imply the results from this video: For a non-abelain simple group with order n, n must 1. be not square free and 2. have more than 2 prime factors

[@thephysicistcuber175]

Why does 1 have to hold?

[@matron9936]

I mean, simply use the classification of all finite simple groups :3

[@Grassmpl]

For your case 3, the first subcase can't happen since 16 not cong to 1 mod 9.

[@thephysicistcuber175]

Yeah, but what matters is that it's congruent to 1 mod 3.

[@NoahPrentice]

I always thought that 144 was greater than 24. Learn something new every day.

[@issacodegard8501]

21:40 which are also multiples of 9

[@emsihongg4868]

Is the following valid for the kernel at 28:17 ? The kernel of \phi cannot be the whole group, since this would imply g x_i \in x_i N for all g. But one might choose g = x_j x_i^{-1} with i eq j. This violates the disjoint union.

[@Alan-zf2tt]

This is good - for so many good reasons

[@jimallysonnevado3973]

In the last part should phi be proven well-defined first? Does te representative of each coset matter?

[@joeo3377]

The representatives don't matter, as any element from each coset can serve as the representative. However, once chosen, the representatives are taken to be fixed. This means that if you pick different coset representatives than I do, you could end up with a different homomorphism than I will, but we will both end up with a homomorphism. You definition of φ includes the information about the coset representatives chosen, and that ensures the function is well-defined.

[@thephysicistcuber175]

12:23 ah yes, 144

[@pawebielinski4903]

Can't wait for another advanced one!

[@Sam_on_YouTube]

Basically, this video starts at the level of the last math class I ever took in college (17 years ago) and then gets harder from there. I think this one might be above my head. We'll see.

[@Sam_on_YouTube]

Yep, got most of the way through explaining what the question is and I'm lost. I remember the basic concepts of group theory, but nothing beyond that. I'll catch the next one.

[@Walczyk]

18:18 why must it be 3? this is probably a dumb question but why must the intersection be a factor of 9?

[@emsihongg4868]

If you pick any element in one of the Syl_3 subgroups, it must have order 1, 3, or 9, since the order of an element divides the order of its group (Lagrange). Thus, if you assume that there is a non-identity element in the intersection subgroup, it has order 3 or 9 (identity is unique), whence the order of this element divides the order of the intersection subgroup (Lagrange). Thus, the intersection subgroup should have order 3 or 9.

[@half_pixel]

21:45 couldn't N also have order 16, 24, or 48?

[@half_pixel]

Oh I see, it needs to also be divisible by 9 since it contains P_i

[@minerscale]

By the classification of finite simple groups, check the list. Oop! There are no simple groups of order 66 or 144. In all seriousness though great video :)

[@lucashoffses9019]

21:47 why can’t it be 16, 24, or 48? I’m guessing it has to be divisible by 9, but why?

[@emsihongg4868]

By Lagrange, because N contains a subgroup of order 9.

[@nothingbutmathproofs7150]

Dr Penn, I am a bit confused. You say that you want to show that there are no simple groups of order 66 yet you are looking at subgroups whose orders are all less than 66. How does that help? What am I missing? Thanks for your time.

[@aoehler1]

really liked this video!

[@s4623]

this video looks familiar... did you do another one of this problem before?

[@Grassmpl]

Please do it for order 540. It's so difficult.

[@gustavnilssongisleskog3145]

"Commander Penn... the time has come. Execute order 66"

[@Tehom1]

Does KZbin not let you input g⁻¹P₂g=P₁ ? I used usefulwebtool.com/math-keyboard for that.

[@FaerieDragonZook]

Why can't a normalizer be of size 16? 16 > 15

[@ferax_aqua]

N_G(P) is a supergroup of both P_i and P_j so its size must be a multiple of 9. Hence it cannot be 16, 24 or 48

[@Happy_Abe]

How is 144 less than 24

[@jermeekable]

depends on ur modulus XP

[@schweinmachtbree1013]

@@jermeekable actually the integers modulo n cannot be made into a linearly ordered set, checkmate :P

[@DepFromDiscord]

Hello people who watched this video when it was released. Hello Michael.

[@pbj4184]

Hello

[@pbj4184]

Hey you should post the timestamp just so that we have Good Place To Stop beat

[@DepFromDiscord]

@@pbj4184 the time is 8:53 AM PST, The date is 14/2/21.

[@DepFromDiscord]

@@pbj4184 let’s wait for the viewers reactions when they realize we can time travel

[@pbj4184]

@@DepFromDiscord No no I meant the Good Place To Stop timestamp. Are you a new viewer?

[@ZetaGirlPower]

If you continue on some of this group theory - I understand A(5) is the smallest non-abelian simple group, but I don't know why. Everytime I've read into it it doesn't make sense. Can you develop this on the channel to explain? It would be a great deep dive into a complex problem.

[@jermeekable]

bump

[@kyleneilson1457]

Two things are true about the possible orders of non-abelian simple groups: 1. Their order must have at least 3 prime factors (Burnside's Theorem) 2. Their order cannot be square free 60 is the smallest integer with these properties, and it is the size of A(5)

[@monzurrahman8307]

It's quite complex to prove. You have to go through all groups of order 2-59 to show that none of them is nonabelian simple, and then show that Alt(5) is uniquely nonabelian simple, up to isomorphism. Although the proof itself is not too difficult, building up the theory to rigorously prove the necessary results can take a while.

[@ZetaGirlPower]

@@monzurrahman8307 Yes this was what I've seen when I've tried to dive into it. Some of the groups are easy to dismiss as simple, others not so much. Thanks for understanding the issues I've seen with it.

[@monzurrahman8307]

Are you an undergrad? You will probably encounter the proof in a 3rd year group theory course. When we proved it, it was at the end of a 10 week course. I could possibly share the notes with you, but it would take a lot of patience to understand the proof completely.

[@kouverbingham5997]

please make a video about how we dont know whether pi+e is rational or not, and how pathetic that is

[@thephysicistcuber175]

#herebeforepublic.

[@maxzaputovich1412]

Order 66... Sounds familiar

[@PubicGore]

Cool! More advanced problems coming in more frequently!

[@MichaelPennMath]

If I continue to make some more advanced stuff, everyone can help out by making sure to watch the “popular” content as well so I can reach as many people as possible.

[@andreisupervloguri8058]

Wow

[@alnitaka]

I call 66 an elephant number. This is because it has an elephant factor. An elephant factor d of N is one that is greater than the square root of N, meaning that d is greater than all of the other factors of N multiplied together. It is the elephant in the living room of factors of N. In the case of 66, 11 is an elephant factor. No group with order an elephant number can be simple, because there can be only one Sylow d-group. Of interest here is the sequence A189712 in the Online Encyclopedia of Integer Sequences (oeis.org) which list all numbers m such that for each prime p that divides m, there is a k(p) such that k(p) == 1 (mod p), k(p) divides m evenly, and m divides k(p)!/2 evenly. The order of a simple group has to be one of these numbers. I call these Sylowpass numbers. 144 is Sylowpass. 66 isn't.

[@KJ-zs7pi]

Unlisted 😆

[@sugarfrosted2005]

You managed to not say pylow

[@davidepancotto7742]

Bruh, currently there are 66 comments.

[@imacup5047]

Now many of you will think how my comment is so early *Isn't It?*

[@Sam_on_YouTube]

KZbin keeps advertising feminine products to me. I am a man. My daughters also use this account, but are not quite old enough yet to need such things. Maybe KZbin is confusing my age and their gender? Maybe KZbin is making a gender stereotyped assumption about my taste? I don't know. But I find it funny.

[@elijahflynt3217]

really dude, are you going to go through like all of the math or just stop here. cuz this is a good place to stop. and that d be a good place to stop.

[@elijahflynt3217]

im just playin man, just playin, keep going on.... ..... i ran out of money to buy books...... damnit man

[@jrmc9222]

Execute order 66