Graphing the Groups of Units

[@ThisIsAYoutubeAccountAsd]

5:48 "Why spend a few hours making a chart, if I could spend a few weeks writing a program to do it for me?" XKCD 974

[@macchiato_1881]

I think this should be a mandatory introductory video to abstract algebra. This is so good

[@CasualGraph]

0:50 "This is a group because the multiplication table follows a nice sudoku rule." And associativity, without that the rule described only guarantees you have a quasigroup.

[@hobbified]

Which, if you want to continue the sudoku treatment, just means that the table has symmetry over the main diagonal.

[@CasualGraph]

@@hobbified That's commutativity. Associativity is a(bc)=(ab)c, not ab=ba. There exist commutative quasigroups that are not associative.

[@kevon217]

this was super illuminating. very serendipitous stumbling on this vid (thank you youtube algo gods) because I’ve recently been trying to further understand the properties of 4 choose 2 and 24.

[@Aleixus]

20:35 the fact that I was correct in guessing 120 and all of the rests (840, 9240, 120120...) made me happy... knowing about primes was useful for once lol

[@darqed]

this video is really cool, a gem found by the algorithm with less than 1k views

[@Guys-s5v]

look again

[@darqed]

@@Guys-s5v yeah but i've never seen this channel ever before and when it got recommended it had below 1000 views

[@MichaelDarrow-tr1mn]

you should make it so when you scroll manually it scrolls based off of where your mouse is rather than where the origin is

[@MichaelDarrow-tr1mn]

this becomes a big problem when you accidentally scroll all the way out and when you scroll in almost the entirety of the graph is drifting downward, away from the origin

[@TheGrayCuber]

Thanks for the suggestion! I just published a change such that the scrolling uses mouse location

[@schweinmachtbree1013]

At 0:49 you said 'this is a group ⇐ the multiplication table has every number appearing once in every row and column', but what is actually true is 'this is a group ⇒ the multiplication table has every number appearing once in every row and column'. this property of a multiplication table actually defines what is called a quasigroup, and a quasigroup with identity is called a loop. Niche math alert: just as associative rings have groups of units, certain non-associative rings have loops of units, namely so-called alternative rings, which weaken associativity from "∀x,y,z : (xy)z = x(yz)" to "∀x,y : (xx)y = x(xy) and x(yy) = (xy)y and x(yx) = (xy)x", and their loops of units are so-called Moufang loops because they satisfy the "Moufang identities", which are also weakened versions of associativity. The most well-known example of this is the octonions, which are a non-associative alternative ring, and its units -- the non-zero octonions -- are a Moufang loop. This does not generalise to higher-dimensional hypercomplex numbers because the sedenions and so on are not alternative rings, only so-called flexible rings, which satisfy just "∀x,y : x(yx) = (xy)x" (this implies so-called power-associativity: "∀x : x(xx) = (xx)x").

[@TheGrayCuber]

Thanks for pointing this out, and adding the niche math! I didn't mean this very rigorously, just to give an idea of what a group is to someone who is not familiar.

[@Higgsinophysics]

Those few weeks paid off! It's beautiful

[@djsmeguk]

Fascinating. The sequence 7,21,56,168,504,1736 does not exist in OEIS! The minimum sizes are OEIS A272590, though they start with 2 for a 1 cycle.

[@elijahberegovsky8957]

I think this should be submitted to OEIS then!

[@djsmeguk]

@@elijahberegovsky8957 Agree, but that should be up to the video maker, since they seem to have an idea how to generate the sequence in general. I could only give a description.

[@elijahberegovsky8957]

@djsmeguk I guess, we should wait until the next video comes out, and either ask them to submit the sequence, or more likely submit it ourselves with the links to the videos

[@TheGrayCuber]

I will submit the sequence along with the next video, which will give more information about how to find the terms!

[@elijahberegovsky8957]

@@TheGrayCuber Woohoo! One more fun sequence to my favourite encyclopedia!

[@veztron]

I think it would be nice to see this graph (quotient isomorphisms) layed out using a directed acyclic graph layout algorithm!

[@ShockRox]

7:37 * Poe the cat appears *

[@RubyPiec]

Thank you for posting the best part of the video

[@tcaDNAp]

There's a graphic novel called Prime Suspects about the mysterious connections between primes and groups

[@nmay231]

Perhaps you already included a reference to the previous video that I had missed, or I missed some very important details about the relations between unit groups and cycles, but I was completely lost until I realized this was a sequel. I think a mention of the previous video would have been very useful for me. EDIT: It's also almost 1 am, so I'm sure that has a factor to my lack of understand lol

[@RuleAndLine]

This is so cool. Gonna go try and prove the formulas used, I wasn't familiar with them

[@charlottedarroch]

The moduli n = 32, 80 and 96 have U_n with canonical cycle structures C_2xC_8, C_2xC_4xC_4 and C_2xC_2xC_8 respectively, and are the only moduli with these structures for U_n. The moduli with unique structure for U_n and with n

[@charlottedarroch]

After observing that all these moduli are divisible by 8, I wondered if there was a modulus n which is not divisible by 8, but has unique structure for U_n. Clearly n must be divisible by 4, for otherwise n is odd, or twice an odd number. But the video already covered that U_m ≅ U_(2m) if m is odd. So I started considering when the structure of U_n is non-unique for n = 4m and m odd, and observed the following isomorphisms: If n = 4m with m coprime to 2 and 3, then U_n = U_(4m) ≅ U_4 x U_m ≅ C_2 x U_m ≅ U_3 x U_m ≅ U_(3m). If n = 12m with m coprime to 2 and 3, then U_n = U_(12m) ≅ U_4 x U_3 x U_m ≅ C_2 x C_2 x U_m ≅ U_8 x U_m ≅ U_(8m). If n = 36m with m coprime to 2, 3 and 7, then U_n = U_(36m) ≅ U_4 x U_9 x U_m ≅ C_2 x C_6 x U_m ≅ U_3 x U_7 x U_m ≅ U_(21m). However, I then considered n = 252 = 2^2*3^2*7, which has U_n ≅ C_2 x C_6 x C_6. One can show that if U_n has canonical cycle structure C_2 x C_6 x C_6, then n|5040. One can then check all cases and determine that the only divisor n of 5040 with U_n ≅ C_2 x C_6 x C_6 is n = 252, so indeed n = 252 is a modulus with unique structure for U_n, with n not divisible by 8, and by the above isomorphisms is in fact the least such modulus.

[@charlottedarroch]

I ended up writing some code to find the moduli n up to 1000 which have unique structure for U_n. They are n = 24, 32, 80, 96, 120, 128, 160, 168, 240, 252, 256, 264, 324, 384, 400, 408, 416, 456, 480, 504, 512, 544, 552, 640, 648, 672, 696, 768, 840, 928.

[@AbideByReason]

Really nicely explained video!

[@isaacjuggling]

It's great that your standard representation works well for this video, but it's not the most intuitive in my opinion. Isn't it the case that for every unit group we could take as its representation the product of the groups C_p, where p is prime, or is a power of a prime? Of course there could be more than one instance of given C_p, but then we can write it as C_p^n where n is number of apprentices of C_p. Reducing groups to this form would consist only in decomposing the group index into prime factors, and checking if something is a subgroup also seems to be quite simple.

[@TheGrayCuber]

This is a great point, and I think the prime powers representation is more useful in a lot of cases. I favored this 'standard representation' in the video for two reasons: it is shorter to write which helps the graph be less messy, and it helped us notice the simple 8, 24, 120, 840 sequence first, before the 7, 21, 56 that you get from the prime power representation

[@lapiscarrot]

I may be misunderstanding but if Z68400≈=Z16×Z9×Z25×Z19 so that U68400≈=Z2×Z4×Z2×Z3×Z5×Z4×Z2×Z9≈=Z2³×Z4²×Z3×Z9×Z5, wouldn't that make the representation 2(5,2)×3(2,1)×5(1) since, as a product of prime power cyclic groups, there are two factors of the group which are powers of 3 with exponent at least 1? I'm assuming the nth term in the tuple for each prime p is equivalently described as the number of such factors with base p and power at least n.

[@MooImABunny]

11:46 I definitely agree, U2 is, in fact, quite boring

[@ejejej9200]

Awesome!

[@muskyoxes]

Step three: graph the cycles. Okay, first we need a physics engine...

[@brromo]

The program breaks at around ~825 or if you click to add too fast

[@guigui0246]

No, it depends on your computer/browser I went to above 1k with auto-add without breaking (only lag) I broke it at 526 with 10-15 clicks/second

[@guigui0246]

And with autoclick at 100/second it breaks as 35

[@TheGrayCuber]

I've published some changed that will help with this fast clicking. It still breaks at some point but it does seem to go faster than before

[@胡玮涛]

❤❤

[@alexdaguy9626]

osugame be like:

[@writerightmathnation9481]

Pedagogically, the phrase “doesn’t have an isomorphism” is problematic. Every group is isomorphic to itself, and every group is isomorphic to a subgroup of some larger group. You should use a phraseology that incorporates these facts via a relativization to your context. You may feel that a phrase such as “has no nontrivial unit group isomorphisms” is cumbersome or is too long, but I don’t think so.

[@caspermadlener4191]

There is a really nice representation of abelian groups I came up with, but I assume it already exists. For U₆₈₄₀₀, the representation is 2(5,2)+3(1,1)+5(1) The direct product of two abelian groups will take the direct sum of the representations, you can only add corresponding prime numbers. And for cyclic groups, of order pⁿ, the representation is p(1,1,...,1), with n ones. You can add an arbitrary amount after this. The first number after the p for a group is the dimension of the elements of order dividing p, and two sum of the first two numbers is the space dividing p². And the best thing: The amount of homomorphisms from p(a,c,e,...) to p(b,d,f,...) is pᵃᵇ⁺ᶜᵈ⁺ᵉᶠ⁺··· This last formula is what makes this representation the natural one. A similar version works as a replacement for the normal form of a matrix.