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Take a regular octagon, like an American stop sign. It has a bunch of symmetries... and this is a picture of all its symmetries!
You can do nothing to it: that operation is called 𝐼. You can rotate it an eighth of a turn clockwise: that's called 𝑟. You can do that twice: that's called 𝑟². And so on, up to 𝑟⁷. When you rotate your octagon an eighth of a turn 8 times it's back to where it started, so
𝑟⁸ = 𝐼
But you can also flip the sign over, say along the vertical axis. Let's call that operation 𝑠. If you flip it over twice it's back to where it started, so
𝑠² = 𝐼
There's another equation that's less obvious. Rotate your octagon an eighth of a turn clockwise, flip it over along the vertical axis, and then rotate it an eighth of a turn clockwise again. This is the same as just flipping it over: the two rotations cancel out! So
𝑟𝑠𝑟 = 𝑠
(The picture states this equation another way: 𝑟𝑠𝑟𝑠 = 𝐼, or (𝑟𝑠)²=𝐼 for short. It's equivalent.)
The picture shows all the different operations you can get by doing 𝑟 and 𝑠. There are only 16, namely these:
𝐼, 𝑟, 𝑟², 𝑟³, 𝑟⁴, 𝑟⁵, 𝑟⁶, 𝑟⁷
and the 8 operations you get by rotating any amount and then flipping:
𝑠, 𝑟𝑠, 𝑟²𝑠, 𝑟³𝑠, 𝑟⁴𝑠, 𝑟⁵𝑠, 𝑟⁶𝑠, 𝑟⁷𝑠
In the picture, the arrows show what happens when you do another rotation 𝑟: for example there's an arrow from 𝑟⁶ to 𝑟⁷, or less obviously from 𝑟𝑠 to 𝑠 because
𝑟𝑠𝑟 = 𝑠
The green lines show what happens when you do 𝑠.
Finally, the little numbers say how many times you have to do an operation to get back to where you started! For example, if you rotate an octagon a quarter turn that's 𝑟². If you do this 4 times you get back where you started, so the number next to 𝑟² is 4.
(1/2)
Now I'll say the same stuff like a mathematician: we saw that the dihedral group with 16 elements, also called 𝐷₁₆, has a presentation with generators 𝑟,𝑠 and relations
𝑟⁸ = 𝑠² = (𝑟𝑠)² = 𝐼,
We also saw its Cayley diagram, labeling each group element by its order.
Quicker, but fewer people will understand!
Now for something a bit less known. The group we just saw has an evil twin, another group with 16 elements, called the 'quasidihedral group'. Only one of the relations is different: now we have
𝑠𝑟𝑠 = 𝑟³
This makes the Cayley diagram look like an 8-pointed star inside an octagon!
I heard about this group from @Ianagol, and I instantly looked it up on Wikipedia, where they have these nice pictures:
https://en.wikipedia.org/wiki/Quasidihedral_group
In fact most finite groups have a size that's a power of two! So there are a *lot* of different groups with 16 elements - namely, 14 of them. So, if you were stuck on a desert island, you could have fun figuring out what they all are, and drawing the Cayley diagrams of all 14. In fact if the world keeps going to hell, I might go to a desert island and do just that.
(2/2)
@johncarlosbaez @Ianagol "ost finite groups"?
@bstacey @johncarlosbaez @Ianagol maybe it was “most”. Interestingly enough both of these groups are non-isomorphic semi-direct products of the cyclic group in eight elements (C_8) and the cyclic group in two elements (C_2) as evidenced by the relation srs=r^b (b=3 for the quasidihedral and b=7 for the dihedral)
Whenever I see someone put the numbers 3 and 7 (or 2, 4 and 8) in a sentence and John is in sight I get overwhelmed by suspicion. I also have a strong feeling that, in this case, it is just a coincidence. But there are no coincidences, aren't there? What are you up to, John?
On anotter mather, why are there so many groups with order a power of two? Your construction, @mathematicalsynesthesia, connecting the two examples, clearly contributes to make groups having a factor of two. This can be generalized. Given any group G, we can add an involution J and stick G to the other side of J. but we can shuffle it G'=s(G) a before glueing them together. What kinds of shufflings s are allowed? Surely G must be isomorphic to G', unless we're very lucky.
Of course, instead of an involution J we might use a cyclic group like C³. There are so many (groups with order being) powers of two, because those C³ based things grow the order so much faster that we don't see them until much later. And by then the power-two groups have amassed even more specimen.
@RefurioAnachro wrote: "What are you up to, John?"
I was discussing the McGee graph shown below, and its 32-element symmetry group, which I call the McGee group:
https://mathstodon.xyz/@johncarlosbaez/114162305599970513
@Ianagol noticed that it contains an interesting 16-element subgroup, which is called a quasihedral or semidihedral group:
"Deleting the red edges (and erasing the red vertices), one gets the Cayley graph for the semidihedral group of order 16 with respect to the generator of the cyclic group Z8 and the involution automorphism of Z8 (in the Cayley graph, one may draw one green edge for each involution instead of two directed edges). This group is a subgroup of the McGee group. https://groupprops.subwiki.org/wiki/Semidihedral_group:SD16"
Since the Wikipedia pictures of this 16-element subgroup is so nice, I decided to explain it in a simple way.
@RefurioAnachro wrote: "On anotter mather, why are there so many groups with order a power of two?"
It's about you build up finite groups from finite simple groups; since the group with 2 elements is so small there are lots of ways to build up groups from it. For more try this:
• Tom Leinster, Almost all of the first 50 Billion groups have order 1024, https://golem.ph.utexas.edu/category/2012/11/almost_all_of_the_first_50_bil.html
Now, that's a nice discussion (thanks)! Where Allen Knutson suggested the book Groups of order 2^n with n<=6, which is for sale for some puny $350, or you can download it here instead (not quite in $350 quality, but still):
That books seems to be another classic I have never looked at. I miss Allen. Is he writing on Category Zulip?
> since the group with 2 elements is so small
Anyways, yes, that seems to align with my intuition which... I didn't express in my comment earlier.
> there are lots of ways to build up groups from [C²].
Still I don't have a good intuition how to build all these groups. Are all 14 groups of order 16 built from C²-s? Building means product? Then C^4 already isn't a product of C²-s. Wait, we also get semidirect products, the dihedral groups are semidirect products. Which is a product with a twist I called "s" in my previous post.
Okay, I'll have to think a bit about what's allowed to build groups. Why is that even news to me. So I tried to cheat and looked up small groups on Wikipedia:
https://en.wikipedia.org/wiki/List_of_small_groups
And there I learned about cycle graphs. They are pretty! But they also look much less symmetrical than Cayley diagrams. Which is a bit unsatisfying. A cycle graph for a group made of some C^n x C² looks like an n-cycle and n 2-cycle pins sticking out. The two groups you showed have the same cycle graph! Here's some more info about cycle graphs:
https://en.wikipedia.org/wiki/Cycle_graph_%28algebra%29
Just took a little sip, and bam, my maths addiction kicks in with full force. Damn you, @johncarlosbaez ;^)
I like this remark by Mark Meckes:
> complexity increases a lot when the multiplicities of prime factors rises. Notably log2(2000)>10, but log3(2000)<7 and log5(2000)<5.
https://golem.ph.utexas.edu/category/2012/11/almost_all_of_the_first_50_bil.html
That's just before you solved by reference:
> the number of groups of order p^n […] is p^(2n³/27+O(n^8/3))
@RefurioAnachro - Note that it's still NOT PROVED that the fraction of groups of order ≤n whose order is a power of 2 approaches 1 as n→∞. It's not so easy to prove, though I'm sure it's true.
@RefurioAnachro wrote:
"Still I don't have a good intuition how to build all these groups. Are all 14 groups of order 16 built from C²-s?"
Yes. By the way, everyone calls the cyclic group of order n C₂, with a subscript.
"Building means product?"
No. I wouldn't have said "building" if there was a phrase like "product" that you already know that expresses the concept.
"Then C₄ already isn't a product of C₂'s"
Right. C₄ is an 'extension' of C₂ by C₂. You can build all finite groups from finite simple groups by iterated extensions. Products are the simplest sort of extensions. To fuel your math addiction:
@RefurioAnachro wrote: "I miss Allen. Is he writing on Category Zulip?"
No, he's not a category theorist at heart. But he emails me and also posts comments on my Azimuth blog and the n-Category Cafe.
Thanks! I'm sure I can find an occasion to say something as an excuse to wave at him.
Sometimes I get sentimental about the days. I met most of the people from back then at some place or other, but lost contact to a few. And some of those left a mark, so I miss them. In fact, Allen's paper on juggling pops up from time to time, because the juggling enthusiasts around me (I'm not one of them) keep discovering it (I might have helped once or twice).
@RefurioAnachro - yes, the "good old days" of the internet appeal to me. Back then I didn't think about how many people were reading my stuff, just the friends I met. Twitter permanently changed me by making me think about numbers of people.
@raph was just saying BlueSky is better than Mastodon, so I've been thinking about this again. Luckily many of my friends are here.
@johncarlosbaez @RefurioAnachro My position is a bit more nuanced than just "better". Bluesky has a better user experience and better moderation, so is poised for significant growth and will become the main gathering place for short comments. It also has *some* protection against hostile takeover, but doesn't have Mastodon's free software roots, and it's also US based. Those are reasons I'm still here and haven't embraced Bluesky fully.
@johncarlosbaez @RefurioAnachro @raph In sci.physics.research I think I enjoyed knowing this was a small bunch of people - it felt special - although I recognise there are some negative aspects to that too.
@dpiponi @johncarlosbaez @RefurioAnachro @raph Do you remember Ludwig Plutonium…? even then the SNR was not 1…
High SNR is the reason I like mathstodon. I suspect the coverage on Bluesky is more complete but at the cost of lower SNR; they’re just operating further along the ROC curve…
@Tom_Drummond @johncarlosbaez @RefurioAnachro @raph And Alexander Abian and Jack Sarfatti (who pops up in the fascinating book How the Hippies Saved Physics).
@mathematicalsynesthesia @bstacey - yes, I was not talking about "ghost finite groups". 
Very nice fact about semidirect products! Here's another fun fact from Wikipedia: if n is a power of 2 bigger than 8, there are exactly four isomorphism classes of nonabelian groups of order n which have a cyclic subgroup of index 2. There's a dihedral group, a semidihedral group (also called a quasidihedral group), a generalized quaternion group, and one other:
@johncarlosbaez @mathematicalsynesthesia Or "original soundtrack finite groups".
@johncarlosbaez My first thought is to wonder what is a "quasi-octagon", ie. what shape has this thing as its symmetry group?
@julesh @johncarlosbaez it's got two (Galois-conjugate) faithful 2D irreps over C:https://people.maths.bris.ac.uk/~matyd/GroupNames/1/SD16.html. A generic orbit there contains 16 points, so will give you a shape, but I have no idea what it is.
A subgroup of order 2 acts as -I on those reps, so each orbit consists of 8 points and their antipodes.
The reps seem to genuinely be complex (bc the characters include sqrt(-2)), so you probably really have to think of it as 16 points in C^2, which might be harder to visualize than R^2.
@julesh @johncarlosbaez otoh re: my last comment. The complex characters of A5 are both restrictions of the same S5 irrep, which is real.
@julesh @johncarlosbaez that site also gives the actual matrices of a faithful 2D irrep in characteristic 3, where one could actually compute and play with an orbit.
@joshuagrochow - another approach is to avoid linear representations and instead seek subgroups H of the 16-element quasidihedral group G, since these give actions of G on the finite sets G/H. To satisfy @julesh we want H to have 2 elements, so that G/H is an 8-element set.
And lo! There's an obvious 2-element subgroup H ⊆ G consisting of 𝐼 and 𝑠. So I hereby declare that with this choice, G/H is Jules' "quasi-octagon".
It's an 8 element set, and I think it has 8 rotational symmetries like an ordinary octagon, but when you "flip it over", its elements get permuted in an exotic way.
@johncarlosbaez @julesh I was trying to see it in a vector space so the "geometry" of the "figure" would be obvious. Notice that r^4 also generates an order 2 subgroup that's not conjugate to s, so that's a different 8-point G-set
@joshuagrochow - great, that one could be "better".
@johncarlosbaez I mean, r^4 is central, and acts as -1 in those irreps, so the 8 points of that G-set arise as a generic orbit in the projective space of either of the faithful 2D irreps. (cc @julesh)
@johncarlosbaez @julesh OTOH, s acts with trace 0 in those 2D irreps, so it has eigenvalues 1 and -1. The 1-eigenval corresponds to a nonzero fixed point of s, so there is a (unique-up-to-scaling, non-generic) orbit of size 8 in those vector spaces that realizes the G-set you said, G/<s>.