Dihedral Groups -- Abstract Algebra 4

[@petterfriberger3539]

At 24:12 Michael says "commutativity" but he means associativity.

[@Nickle314]

I was getting my 8 year old to do this group the other day with some cut out triangles. I'll have to show her this when she gets home from school.

[@quantumgaming9180]

Sounds nice, hope your child will like to explore this subject by herself

[@Nickle314]

@@quantumgaming9180 She quite good. Other observation from these groups. 1. r * r is another rotation so the rotations form a subgroup. 2. r * s is a reflection 3. s * s is a rotation. Two reflections make a rotation. That's interesting. If we take D4, there are 4! combinations or 24. However the size of D4 is 8. What's going on with the missing 16 elements from the permulation set? On a separate note, my brother teaches maths. He had an 11 year old who was investigating logs with complex bases. That child did very well on a maths Olypiad exam. As he says, I have to work out what to do with her next year. ...

[@Anonymous-cw4yd]

@@Nickle314 An arbitrary permutation is not a symmetry. In a symmetry of n-gon, let say vertex 1 ends up on i, you have two choices for vertex 2: it is either in position i+1 or in i-1. If it is in position i+1, then 3 must be in position i+2, and so on. If 2 is in position i-1, then 3 must be in position i-2, and so on. So knowing where vertex 1 and 2 ends up, completely determine the symmetry. As you can see we have to add additional condition on permutation to describe a symmetry. That additional condition is "adjacency".

[@Nickle314]

@@Anonymous-cw4yd Exactly. For n>3, you get twists.

[@Nickle314]

@@Anonymous-cw4yd On separate issue teaching young kids. Group theory, they can touch by rotating shapes etc. Another example, Mobius band, colouring the edges, the sides, cutting down the middle, cuting 2/3rds of the way in. All maths, all real. Another example, exponents, adding to multiply them. Show that 2^3 * 2^1 = 2^4. Work though to check the answers. Then ask, what's 2^0, and see if they can work it out. Another one that interests them is infinity and an infintesimal. The infinite hotel they like as well So I pity her first senior maths teacher when she starts talking Aleph 1. :-)

[@curtiswfranks]

That dihedral group diagram for the regular 4-gons would look really good in three dimensions with each arrangement of vertex labels forming a vertex of a cube-shaped diagram.

[@matheusjahnke8643]

23:04 "That actually follows from the things we have on the board" rs=sr² r³=s²=e Starting with rs=sr²... apply r after everything rrs=rsr² r²s=(rs)r² r²s=(sr²)r² r²s=sr⁴=sr³r¹=ser=sr Indeed... it does follow

[@simonreiff3889]

D_1 and D_2 are the only abelian dihedral groups. Note that for all elements r and s in any abelian group, rs=sr, but on the other hand and for any dihedral group, rs=sr^(n-1) where r and s are a counterclockwise rotation by one click and a reflection, respectively.. Hence, if a group is both dihedral and abelian, we have rs=sr=sr^(n-1); left-cancelling from the rightmost equation yields r=r^(n-1); multiplying both sides by r yields r^2=r^n=e; and by inspection we see that for n=1 and n=2, we have D_n is abelian. I believe it is possible (but tedious) to complete the proof by showing that for all D_n (n>2), D_n is not abelian, by using induction on n, starting with a base case of n=3 (proven in the video since rs does not equal sr in D_3), then assuming as the induction hypothesis that D_k is not abelian (k>=3) and showing that that assumption implies that D_(k+1) likewise is not abelian. The much easier way to prove that D_1 and D_2 are the only abelian dihedral groups for all n greater than or equal to 3, however, simply is to notice that a rotation of two clicks in the counterclockwise direction from the starting position of a regular n-gon (say the position occupying "1" between "2" and "n-1"), or a rotation of two clicks in the clockwise direction from the starting position for that matter, never go "far enough" in D_n (n>2) to transport us from the starting point all the way back to the starting point solely through rotations in a single direction. Hence, r^2 cannot equal e for D_n (n>2), imolying that rs does not equal sr for all r, s € D_n (n>2), and thus proving that the dihedral group D_n (n>2) cannot be abelian.

[@TheOiseau]

That's all very well, but how do you draw the required objects in part (b) of the question? How does one draw a regular 2-gon or 1-gon? A straight line? ^_^

[@simonreiff3889]

@@TheOiseau D_1 is isomorphic to Z_2, and consists of two points. There is one axis of symmetry given by the line that perpendicularly bisects the line segment connecting the two points. D_2 is isomorphic to the Klein 4-group Z_2 x Z_2, i.e., 2 pairs of 2 associated points. There are several ways to visualize this group and its axes of symmetry. One way is as a cross (i.e., horizontal and vertical axes collinear with the line segments connecting the pairs of points). Another is as two overlapping but oriented line segments, one in the AB direction and one in the BA direction (this version does look a lot like D_1).

[@TaviZeir]

Another simple object for D_2 is a non-square rectangle :)

[@iGeen7]

​@@TheOiseau2-gon must contain 2 edges (by definition) - you can colour them by 2 different colors. 1-gon must have 1 edge which you can draw as a loop, but this loop should have an arrow on one end.

[@schweinmachtbree1013]

24:58 that isn't quite right because the relations need to be only in terms of i and j, so you can't use "-". you can fix this by adding "-1" to the generators, in which case you need to express that it's a central element (commutes with everything), for which it is sufficient for it to commute with the generators, and hence you get

[@iabervon]

I think what he should have used was ji=i^3j, which is just in terms of i and j. On the other hand, we haven't seen enough yet to know whether we also need i^2=j^2 or ji=ij^3 or something like that.

[@yf-n7710]

25:27 Does this work? I feel like there's an implicit assumption of the commutativity of -1 here. Wouldn't "ji = i^3j = ij^3" be a more full definition? You don't even need any -1 at all. I checked, and you can prove with my version that i^2 = j^2, and it follows from there that i^2 is commutative.

[@yoav613]

Abstract algebra is very interesting!

[@martin6276]

Superb, thank you! I think your teaching style (and blackboard skills) have become even more outstanding!

[@abdelatielazzouzi6373]

thanks for your lesson I could understand the symetry groups

[@nahblue]

Why counter clockwise rotation for R? I understand it must just be a convention, but it's unexpected to me.

[@matematicacommarcospaulo]

I liked a lot your maps for D3 and D4

[@cunningham.s_law]

can you do groups of specific polynomials like orthogonal polynomials

[@foobargorch]

I've always been confused about why the vertices of the polygon are unlabeled when studying rotation, if there aren't any labels to distinguish the points then subject to say a single rotation isn't the resulting object identical to the input? In other words why is labeling the vertices of the triangle not something essential but the opposite?

[@foobargorch]

is just because the labels are not essential to the definitions of the group elements themselves as operations on points in a space?

[@schweinmachtbree1013]

The elements of a dihedral group are not different vertex configurations of an n-gon; the elements are _functions_ -- also called transformations -- of the plane which leave the n-gon fixed (not fixed point-wise or vertex-wise of course, because only the identity transformation does that; letting the polygon be P ⊂ *R* ^2 and letting the transformation be f, what I mean is f(P) = P). Since the elements are functions, the group operation is just function composition. So labelling the vertices is just done in order to describe what the different transformations do. For example for the symmetry group of the triangle P = Δ with vertices A,B,C ϵ Δ, the picture Michael draws for the vertical rotation s_A at 5:38 signifies that it's the transformation such that A↦A, B↦C, and C↦B.

[@iabervon]

We're studying "symmetries", which are the operations we can perform on an object and get something identical. In particular, we're looking for the operations that would give different outputs from something like the letter G, but give the same output for a regular polygon with n sides. For example, if we did an operation to an equilateral triangle pointing up and got one pointing down, this operation is not a member of D3. So we label the vertices in order to show that, if every spot on the object was distinguishable, we'd know we did something, but if they're not labelled, we get the same shape. Alternatively, we could put a G in the middle of the triangle, and find all the ways we can transform the figure so the triangle is the same but the G is different.

[@foobargorch]

these answers actually left me a bit more confused than before... i guess this is because *the* dihedral group is the abstraction connecting a whole bunch of equivalent structures where the concrete elements as functions have concretely different types. on math stack exchange, "Precise definition of Dihedral Group" has an answer which makes perfect sense to me: "The Dihedral group is the automorphism group of the cycle graph. Formally, the cycle graph Cn (for n≥3) has vertex set [n]={1,…,n}", which also makes it clear how it's a sub group of the symmetric group too. but this seems to imply the labeling is essential! that said, if i think of the elements of the group as functions from points on the plane to points on the plane, as the 2nd reply suggests, then i can see how restricting this to mappings whose domain and range are the vertices of the polygon results in an object that is equivalent to the graph formulation, since it's just an arbitrary numbering of the cartesian points, and the points along the edges are fixed by the edges and this function being a graph automorphism, and symmetry as graph isomorphism is very intuitive for me as well. the 1st reply, otoh seems to suggest that this is also equivalent to functions on *sets* of points if i understand correctly? that i still find very confusing since sets are unordered, and i don't understand how symmetries would be distinguished from the identity using that model...

[@iabervon]

@@foobargorch Yeah, there's a question we haven't gotten to of what makes two groups "the same group", and many of the same groups come up in graph theory, rigid transformations, permutations, etc. You can think of the rigid transformation definition as starting with all rigid transformations and composition as the operation, and then, for any subset of R^2 S, taking the transformations such that f(S)=S (using the "image of a set" definition of f(S): f(S)={f(s)|s in S}). This gives you Dn for a regular n-gon, but when you pick other subsets of R^2, like regular tilings, you get groups that don't come out of finite graph theory, while graph isomorphisms will give you groups that rigid transformations don't give you.

[@TimHaloun]

So in this group, the elements aren't actually the positions of anything, but the operations of relabeling in and of themselves. And so the group action is the composition of relabelings? I guess that makes them fundamentally different from the earlier examples where if felt like all the elements were distinct things like standalone values i.e. just exemplar numbers or matrices. I feel the video really could have benefited from calling out this big distinction!

[@willemesterhuyse2547]

Has anyone come across a mathematical proof requiring some other formula from Principia Mathematica (other than syllogism)?