Visual Group Theory, Lecture 2.3: Symmetric and alternating groups

Visual Group Theory, Lecture 2.3: Symmetric and alternating groups

In this lecture, we introduce the last two of our "5 families" of groups: (4) symmetric groups and (5) alternating groups. The symmetric group S_n is the group of all n! permutations of {1,...,n}. We see several different ways to describe permutations, though we primarily stick with the "cycle notation". The alternating group A_n is a subgroup of S_n, and it consists of the permutations that can be written as a product of an even number of transpositions. There are exactly five regular three-dimensional convex polyhedra (tetrahedron, cube, octahedron, dodecahedron, icosahedron), and the symmetry groups of each of these are symmetric or alternating groups.

[awazin]

Your series on group theory is absolutely brilliant :)

[wieslawpopielarski]

I think computing disjoin permutation should be explain in better way
in homework example
(1 2 3)(1 2 3 4)
1->2 and 2->3
(1 3
3->1 and 1->2
(1 3 2
2->3 and 3->4
(1 3 2 4
4->4 (there is no 4 in first transition) and 4->1 (so we close bracket
(1 3 2 4)

the second
(1 6)(1 2 4 5)(1 6 4 2 5 3)
1->6 and 6->6 and 6->4
(1 4
4->4 and 4->5 5->3
(1 4 3
3->3 3->3 3->1
(1 4 3)
2->2 2->4 4->2, so
(2 2) which simplifies to (2)
so (1 4 3)(2)
5->5 5->1 1->6
(5 6
6->1 1->2 2->5
(5 6)
so finally (1 4 3)(2)(5 6)

[hanyanglee]

15:40 Oh, this example reminds me of the time when I did speed solving of Rubik's cube.
20:10 Inside the "1." the second one, it's easy to get to (1 4 3). Then 3 goes back to 1. At this moment, the next part starts with the first unused index, which is very much alike when we memorise the cube in blinkfolded 3x3x3.

[dohduhdah]

At 19:59 / 29:49 in the video, it asks writing elements of S3 in "disjoint transpositions" and "disjoint adjacent transpositions". I don't understand the use of the word 'disjoint' here. Isn't the idea with disjoint cycles that they don't share any common elements? For instance, I'd expect (12)(34) to be an example of two disjoint transpositions and since they are disjoint, they commute and (12)(34) = (34)(12), but I'd expect (12)(23) *not* to be an example of disjoint transpositions because they don't commute (12)(23) is different from (23)(12). If I apply (12)(23) to [1, 2, 3] I would get [2, 3, 1], but when I apply (23)(12) to [1, 2, 3] I would get [3, 1, 2].

[Anonymous-pmqc]

Professor, I think you need to mention that your list of symmetry groups is for PROPER SYMMETRY GROUPS, not FULL SYMMETRY GROUPS. That is, the full symmetry group for the tetrahedron would actually be S4 instead of A4. This tripped me up quite a bit. I feel that you should mention the tetrahedron has symmetries over planes of reflection, that swap only 2 vertices. But that orientation-reversing isometries (like reflections, glide reflections and improper rotations) under which the figure is invariant, are not allowed in a proper symmetry group.

Anyhow, your course is great. But if you are looking to spruce it up a bit come around next time, this clarification would help the brighter students who get frustrated, quite a bit.

Cheers!

[qingzhenwu]

and what confused me more is, if i treat (12) as swapping object 1 and object 2, then i can still draw the same cayley diagram, and i can do a different mutiplication as (12)(23)=(123) instead of (132) when treat (12) as swaping objects at fixed position 1 and 2. So it really does not matter which way you treat it. But, i think the reason they chose the to mark fixed postion as 123 is to make it easily compare to mark fixed 3d coordinate axis so (12) and (23) can easily be corresponding to rotation and flip, etc

[Mrpallekuling]

For the most part you use Cayley graph (which is far more common), but sometimes you use Cayley diagram. I guess the two terms mean the same thing.

[ChuanChihChou]

29:00 Not the most natural choice of generators of A_5, not S_5! 😅

[qingzhenwu]

hi professor macauley, i am confused with mutiplication of cycle notiation . for example lets take S3 group. e is 123. if do (12), we sawp 1 and 2, it become 213. then do (23), we swap 2 and 3, then 213 become 312. And also we have (12)(23)=(132), if we let 1 goes to 3, 3 goes to 2, 2 goes to 1, then 123 become 231, which is different from 312 that get from do (12) then do (23) . So i think, either my mutiplcation is wrong, or what you mean by (12) is not swap the object 1 and object 2 but rather swap the objects at fixed position 1 and position 2

[godfreypigott]

Is the fact that Sn and Dn only correspond for n=3 the reason why the shortcut for calculating the determinant of a 3x3 matrix doesn't work for larger matrices?

[ekaptsv]

Honestly, the exercises at 19:55 still look quite hard to me even after I watched the preceding part of the lesson twice. It would be nice if you provide us with solutions for some of that problems.
Anyway, thank you for these inspiring lectures very much!

[simianomatlaldo]

2:52 no clue how you got to that at all

[Olga-btxl]

how to solve it? (16)(1245)(164253)=(431???) I am confused (

[smackronme]

Can someone help me with the solution for product of disjoint transpositions?