Group theory 4: Lagrange's theorem

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  • Опубликовано: 14 дек 2024

Комментарии • 18

  • @AndrewLi1235
    @AndrewLi1235 3 года назад +18

    This is the first time I've seen the use of orbits for the discussion of cosets. Really appreciate it!

  • @jackozeehakkjuz
    @jackozeehakkjuz 4 года назад +22

    At 25:00 it should be 2¹¹-2 is divisible by 11.

    • @avi123
      @avi123 4 года назад +14

      Or 2¹⁰-1 is divisible by 11

    • @PunmasterSTP
      @PunmasterSTP 4 месяца назад

      That was a good catch.

  • @julesjacobs1
    @julesjacobs1 4 года назад +21

    Your lectures are amazing! I love that you explain the big picture.

  • @mikeywatts366
    @mikeywatts366 Год назад +1

    At 6:30, why is the identity element necessarily in H? That is, if G acts on a set S, then what makes 1s=s? I guess, what does it mean for a group G to act on a set S? I interpret that to mean an arbitrary function f: (G x S) -> S. I guess I'm misinterpreting the definition of "acts on"?

    • @mikeywatts366
      @mikeywatts366 Год назад

      A "group action" is a well-defined concept with two axioms, one of which is that axiom 1s=s.

    • @mishaerementchouk
      @mishaerementchouk 10 месяцев назад

      Good question. It allowed me to explicate that feeling of discomfort related to using just “a set”. In representing the group action as a function f: G x S -> S, function f must respect the group structure, for instance, f(g, f(h, s)) = f(gh, s). Hence f(1, f(1, s)) = f(1, s), so, instead of the group action on “some” S, we can always consider the action on its subset S’ = f(1, S). Now, for all s in S’, we indeed have f(1, s) = s.

    • @simonreiff3889
      @simonreiff3889 4 месяца назад

      You basically have it. Recall a group action is a binary operation, call it ° : G x S --> S, where G is a group (say a set G endowed with a group product *), and S is a set (not necessarily a group), satisfying e ° s = s (left identity with e being the identity in G) and (g * h) ° s = g * (h ° s) (compatibility of the group product, * , and the left action of G on S, °). The axioms of a group action look a lot like the axioms for a group, but notice that the identity of a group is always unique (prove this by taking e and e' as identities of G and then show they must be equal), but it may be possible to find many elements g in G, not just e, such that the group action g ° s = s. Note that this happens all the time with FUNCTIONS. For instance, cos x = 1 for many values of x, not just x = 0.
      For the above reason, I suggest thinking of the group action ° as an abstraction of a function--precisely the word you used, correctly in my view--while thinking of the group product * as an abstraction of a mathematical operation. Whereas x*y might be the abstraction for x+y, xy, or any other binary operation, think of ° as taking a group's elements as functions on a set, so write g ° s = g(s). We obviously don't "mix" composition of functions and operations, but rather must keep them separate. The f(s) notation is precisely why we frequently consider and work with LEFT actions (without loss of generality) because it would be strange to write a function as (s)g.
      Now consider the simplest function of all, the identity function f(s) = s, where s is in S (only a set), and f is an element of G (a group), and let's see what happens if we compose f with another function g (also an element in G) where the group product * is defined to be composition of functions. In our notation above, we get g ° (f ° s) = g ° s = (g * e) ° s = (g * f) ° s. (Check this!) Notice that the group product only happens on the LEFT, before ending with a ° operation on an element of s. In other words, the simplest function (group action) is the action of the identity 1 on s, and we want 1 ° s = s (where 1 * g = g * 1 = g for all g in G). We also get a nice compatibility between the group product and the group action even though we cannot "mix" the two any more than we can mix g ° f with f(x). Keeping the group action and group product explicitly different notationally helped me a lot to understand what was happening. It follows once you sort this out what orbits and fixed points are, and then what the number of elements in each should be, and so on. Hopefully this helps you too!

  • @netrapture
    @netrapture 3 года назад +1

    I like the check on coset disjointness starting at 15:35, but couldn't we say at 16:07 "for some h_1, h_2" rather than for "some g_1, g_2" and at 16:09 we could say "show g_1.H = g_2.H" rather than "assuming g_1.H = g_2.H"

  • @sgssergio
    @sgssergio 2 года назад +2

    The numbers of cosets is called index too and denoted by [G:H]?

  • @SitichokeAmnuanpol
    @SitichokeAmnuanpol 22 дня назад

    Great lecture, thank you.

  • @PunmasterSTP
    @PunmasterSTP 4 месяца назад +1

    Coset? More like "Cool information with which we'll be all set!" 👍

  • @conradmorris8881
    @conradmorris8881 Год назад

    excellent thank you

  • @Anony1176
    @Anony1176 Год назад

    Sir it will be our great great help if you kindly please upload videos of differential equation special function

  • @shaneyaw4542
    @shaneyaw4542 2 года назад +1

    This lecture was poorly planned and executed. The video neglects important topics, hardly clarifies abstract concepts, and the written notes are almost incomprehensible. It would have been worthwhile for the lecturer to slow down and clearly define the mathematical concepts, especially if mathematical rigor is to be expected from the student.

    • @griggpev
      @griggpev 8 месяцев назад

      refer to what he said in the beginning of the first lecture. the pacing is fine for what he planned. for a graduate student looking for a refresher, this is fine, and the brevity enhances clarity in that case.