I agree about the presentation being repetitive, not to mention somewhat slow-paced...which is exactly how I prefer a lecture. Some would call me a bad listener, but to me it's because my mind is the most active while I witness something being carried out before me, so my mind tends to wander....albeit often in the same direction that the lecture is moving while I'm not listening. The repetitiveness lets me re-synchronize with the lecture quite easily. And the slow pace means I probably won't need to rewind. Stay the course, sir. Your style is much appreciated. (By me, anyhow.)
One advantage of this presentation style is that the video itself isn't needed for following the material. For example, I think a vision impaired, mathematically inclined person could easily follow this presentation.
yeah others can just watch it on a faster speed setting. It's great that he goes over every little detail as there's always something you won't know as an individual that's different to what another person doesn't know
@@sparky213thousand That is true, I find the videos quite slow as I already have spent time in group theory and generally skip some parts of them by there are times when I just listen to the videos without watching them and then every word seems important. I use to do that when running and I don't want to listen to music. In that case is very helpful that the videos have this pace. In any case it is true than only now I can say that I understand group theory.
Thank you so much. This was so helpful. I am grateful for all your videos. You are nice to listen to, very thorough, detailed, patient and kind. You helped me learn.
I often watch the whole advertisement before your video starts, ON PURPOSE, even if I'm not interested in the advertised product, because you definitely deserve to be paid for your work. (Ahem, everyone else.) THANK YOU.
I had no idea that Ben1994 would not get rewarded for his efforts until I read the post herein from Padraic Ceallach. From henceforth, when watching any of Ben1994's videos, I'll retain the advertisement to ensure that the latter party gets rewarded financially.As usual, Ben1994's videos are EXCELLENT. None of the three textbooks (Fraileigh, Rotman, etc.) that I have provide incisive explanations of Group Action as well as Ben1994. It appears to me, as an amateur maths enthusiast, that most mathematicians are unable to explain the subject in a simple manner. Ben1994 is an exception!!!
Very clear. I now understand the underlying basis for group actions in 1 x 30 min video. I have spent hours reading a book and it was making no sense. Thank you.
The question which occurred to me is: Can property 2 (the identity property) be false when property 1 (the associative-law-like property) is true? If not, then property 1 implies property 2, so you might as well only have property 1. However, property 2 is not redundant. Consider f:GxA->A where there exists e in A such that for all g in G, a in A we have g.a=e (i.e. every single action gives us element 'e' as the answer - the most boring composition table imaginable.) This satisfies property 1 but not property 2.
Thank you very much ! I'm French and your video really helped me understand group actions although english isn't my first language : your explanations are so clear ! Keep it up !
Is the F: GxG -> G function the same one that forms the group and therefore prop. 1 is essentially redundant (being already included in the definition of the group itself)?
I should probably finish the video before asking my question but the set A does not need to be a subset or anything of the sort, right? E.g. say G = {1,2} and A = {3,4}, let f: (g,a) -> (a) then I can say something like, f(1,3) -> 3, f(1,4) -> 4, f(2,3) -> 4, and f(2,4) -> 3. Then my function 'f' is a valid group action?
On condition (2) for the mapping F to be a group action. In the composition table, row titles are elements of G cross A , right? So, there should not be a row title consisting in element e ( = identity) alone . I mean, since the domain of mapping F is G cross A, the first row should be the row dedicated to an ordered pair, not to the identity element alone . Or do I miss something?
At 5:25 , your "definition" is incomplete. You also need to condition that f(e, a) = a and f(g1 * g2, a) = f(g2,f(g1, a)) where * is the binary operation on G and e is the identity of the group (G, *).
These videos are well done, thorough and comprehensive. However, I have one suggestion regarding the colors. I think that using colors all of the time is overdone. In some circumstances they can be helpful, but that should be the exception, not the rule.
But it's not technically associative since it is dealing with two _different_ operations (multiplication within the group and "scalar" multiplication of the group on the set). Associativity only ever deals with _one_ operation.
@@MuffinsAPlenty Right Muffins, it is the kind of structure you would need to formalize scalar multiplication, or composition of linear maps on vector space theory, it is a generalization of associativity, and it is highly nontrivial.
also this is very abstract stuff, there's no way to to open up this Russian doll type problem without taking it slowly and clearly. everything in one, two or three steps removed, go any faster, my head melts and I lose it.
