What is a group? -- Abstract Linear Algebra 3

แชร์
ฝัง
  • เผยแพร่เมื่อ 30 ต.ค. 2024

ความคิดเห็น • 53

  • @kevinnguyen4055
    @kevinnguyen4055 2 ปีที่แล้ว +20

    Reminder to self:
    A magma is a set M with an operation, M x M -> M.
    A semigroup is a magma that is associative.
    A monoid is a semigroup with identity.
    A group is a monoid with inverses.

  • @iabervon
    @iabervon 2 ปีที่แล้ว +42

    In 2020, with all the lockdowns, Norway had to introduce the Non-Abel Prize, because so many mathematicians weren't commuting.

  • @kaasci
    @kaasci 2 ปีที่แล้ว +9

    I've got spare time so I've started teaching myself from Linear Algebra Done Right over the weekend (this is my second pass, and a friend recommended it), these videos could not come at a better time in all honesty.

    • @josheichholz1184
      @josheichholz1184 2 ปีที่แล้ว +1

      I’m taking a linear algebra course that’s heavily relying on Linear Algebra Done Right, so this series is also amazing for me

  • @abishekpuri5684
    @abishekpuri5684 2 ปีที่แล้ว +5

    Warmup Question 2:
    By induction
    n=1 then observable its true
    given for n=k, (a^-1ba)^(k+1) = (a^-1ba)^(k)*(a^-1ba) = (a^-1b^(k)a)*(a^-1ba) = (a^-1b^(k+1)a)

  • @abishekpuri5684
    @abishekpuri5684 2 ปีที่แล้ว +6

    Warmup Question 1:
    e=aa=aea=abba=(ab)(ba) so ab and ba are inverses. But we know inverse of any element is itself due to aa=e and inverses are unique (proven in video) so it means ab must equal ba

  • @derekpoon5308
    @derekpoon5308 2 ปีที่แล้ว +3

    Warm up Q1:
    ab = (bb)ab(aa) = b(ba)(ba)a = bea = ba

  • @DirtShaker
    @DirtShaker 2 ปีที่แล้ว +4

    10:54 ... "my operation will be addition" ;-)

  • @Boringpenguin
    @Boringpenguin 2 ปีที่แล้ว +1

    7:18 and a monad is just a monoid in the category of endofunctors

  • @abrahammekonnen
    @abrahammekonnen 2 ปีที่แล้ว +1

    11:00 your operation is multiplication not addition (again being a pendant because you did write down multiplication).

  • @scramah195
    @scramah195 ปีที่แล้ว

    Note that for the final proof, 0*0=0 does NOT serve as a counter example...because 0 is NOT a member of the classic structure associated with multiplication, which is R/{0}...and, implicitly, these propositions assume that 'a' is a member of said group.

  • @abrahammekonnen
    @abrahammekonnen 2 ปีที่แล้ว +3

    3:16 You say -5, but write -2.(-2 is correct just being pedantic).

  • @johnwroblewski4773
    @johnwroblewski4773 2 ปีที่แล้ว +2

    Just a note that the link for Linear Algebra Done Right is incorrect. It points to the first book in the list.

  • @BethKjos
    @BethKjos 2 ปีที่แล้ว

    @7:30 Do we get left- and right-monoids from left- and right-identity elements? Does this begin to hint at commutation?

  • @malawigw
    @malawigw 2 ปีที่แล้ว +1

    Will all the other course videos from main channel be included on this channel eventually?

  • @goodplacetostop2973
    @goodplacetostop2973 2 ปีที่แล้ว +3

    31:02

  • @matematicacommarcospaulo
    @matematicacommarcospaulo 2 ปีที่แล้ว

    In Brazilian Portuguese we do not have an equivalent name for magma. We simply say that the set is closed under the operation (o conjunto é fechado para a operação)

    • @schweinmachtbree1013
      @schweinmachtbree1013 2 ปีที่แล้ว

      good, I think the term magma is rubbish and unnecessary xD (there is the alternative term "binar" though which I do like)

    • @matematicacommarcospaulo
      @matematicacommarcospaulo 2 ปีที่แล้ว +1

      @@schweinmachtbree1013 in Portuguese we can say operação binária

  • @schweinmachtbree1013
    @schweinmachtbree1013 2 ปีที่แล้ว +2

    24:17 I think you assumed what you were trying to prove in this part of the proposition. The correct proof is that _a_ ^{-1} being the inverse of _a_ means by definition that _a a_ ^{-1} = _e_ = _a_ ^{-1} _a,_ but we can flip the equations around to give _a_ ^{-1} _a_ = _e_ = _a a_ ^{-1}, which by definition means that _a_ is an inverse of _a_ ^{-1}, so since inverses are unique we thus have ( _a_ ^{-1} )^{-1} = _a._

    • @schweinmachtbree1013
      @schweinmachtbree1013 2 ปีที่แล้ว +1

      oh wait no your proof is totally fine. just consider my comment as an alternate proof then, oops

    • @ahzong3544
      @ahzong3544 2 ปีที่แล้ว

      @@schweinmachtbree1013 how is his proof correct? can you please elaborate

    • @schweinmachtbree1013
      @schweinmachtbree1013 2 ปีที่แล้ว +2

      @@ahzong3544 the second equals sign follows by using e = x^{-1}x with x being a^{-1}

    • @ahzong3544
      @ahzong3544 2 ปีที่แล้ว

      @@schweinmachtbree1013 I see now, thank you so much

  • @emanuelegilio8801
    @emanuelegilio8801 2 ปีที่แล้ว

    9:34 Isn't it enough to say that there is only the reverse left (or right)?

  • @JaykTheJackal
    @JaykTheJackal ปีที่แล้ว

    presumably function composition among invertible functions can also be said to be a group?

    • @JaykTheJackal
      @JaykTheJackal ปีที่แล้ว

      I suppose it would have to be invertible functions f: X -> X

    • @JaykTheJackal
      @JaykTheJackal ปีที่แล้ว

      Wait, is that just the symmetric groups?

  • @ttrss
    @ttrss ปีที่แล้ว

    What is the group operation for the symmetric group? Is it composition of permutations?

    • @galus_anonimus
      @galus_anonimus 9 หลายเดือนก่อน

      10 months later:
      Yes, it's the most probable option; set of all permutations with the operation of composition is in fact a group and prof. Penn referenced the inverses as probably the inverse functions, because he said that they are implied by the bijectivity of permutations.
      Of course inverse functions are inverse element for operation of composition.
      Hope it helps

  • @christianorlandosilvaforer3451
    @christianorlandosilvaforer3451 2 ปีที่แล้ว

    i got some kind of confusion... the neutral element is unique.. but as a part of the set it has to match an inverse even so? .. and if the inverse is itself could be rigth? i found this a little bit contradictory

    • @Mutlauch
      @Mutlauch 2 ปีที่แล้ว +1

      The neutral element is the identity and from that follows, that it is its own inverse. But this relationship is not equivalent, so you can have an element that is its own inverse but does not equal the identity.
      Ex: G* := {1, -1, i, -i}
      1 is the identity, thus its own inverse
      -1 is its own inverse but not the identity

  • @StanleyDevastating
    @StanleyDevastating 2 ปีที่แล้ว +2

    @28:05 I think the permutation (1 2 3)^2 = (3 1 2) not (1 3 2)?

    • @hawkeyeplank
      @hawkeyeplank 2 ปีที่แล้ว +4

      It took me like half an hour to figure that one out, it is actually 132, remember to write the starting number and the first step of your permutation and you should get it

    • @ratandmonkey2982
      @ratandmonkey2982 2 ปีที่แล้ว +2

      took me at least a half hour too. Another vote for (1 3 2) . Tricky stuff.
      Check out
      th-cam.com/video/UI-SWiyaArE/w-d-xo.html
      very carefully.
      Another good video:
      th-cam.com/video/MpKG6FmcIHk/w-d-xo.html

    • @paulshin4649
      @paulshin4649 2 ปีที่แล้ว +1

      Also note that (3 1 2) is in fact the same as (1 2 3). Since in a group we have a^2 = a if and only if a is the identity (last result in the video), and since (1 2 3) is not the identity of the group S_3 (or any other symmetric group on more than 3 letters), we could not have (1 2 3)^2 = (1 2 3). (Just another way to check the correctness of the calculation.)

    • @StanleyDevastating
      @StanleyDevastating 2 ปีที่แล้ว

      @@paulshin4649 good point! Yes I agree with all of you, coming back and looking at this I was evaluating cycles incorrectly.

  • @abrahammekonnen
    @abrahammekonnen 2 ปีที่แล้ว

    12:17 so the set is a group with respect to multiplication AND addition.
    Edit: Not Addition.

    • @scalesconfrey5739
      @scalesconfrey5739 ปีที่แล้ว +1

      No... just multiplication.
      The would-be-identity-element (namely 0) is not in the set. -1 + 1 = 0, but 0 is not in {1,-1,-i,i}.
      It isn't even a magma under addition, (1) + (i) = 1+i, which isn't even in the set. So the set isn't closed under +.

  • @jaredvv86
    @jaredvv86 2 ปีที่แล้ว

    Are your examples from a text book. I swear I have seen a few of these before

  • @abrahammekonnen
    @abrahammekonnen 2 ปีที่แล้ว

    9:49 In your definition of a group you essentially have 4 conditions: closed, associativity, identity, and inverse all with respect to a binary operation. But the standard definition of a group only has 3 conditions: Associativity, Identity, Inverse all with respect to a binary operation.
    Does that mean those 3 conditions imply closedness?

    • @AssemblyWizard
      @AssemblyWizard 2 ปีที่แล้ว +3

      No, the standard definition includes all 4, according to wikipedia

    • @lina31415
      @lina31415 2 ปีที่แล้ว +6

      well, the "closedness" is often omitted, because the definition of a group operation implicitly states the "closedness" condition (* : G x G ----> G), so explicitly adding it to the axioms is superfluous.

  • @prbprb2
    @prbprb2 ปีที่แล้ว

    Do you really need to use associativity to prove that inverses are unique? I am thinking it is not necessary...

  • @hawkeyeplank
    @hawkeyeplank 2 ปีที่แล้ว

    (Shoes and socks)(socks and shoes) very very funny ekekek. Also, we need LaTeX support in the comment section.

  • @artificialresearching4437
    @artificialresearching4437 2 ปีที่แล้ว

    I might be wrong, but this could work: th-cam.com/video/PvUrbpsXZLU/w-d-xo.html
    P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)

  • @braindead3201
    @braindead3201 2 ปีที่แล้ว +1

    How come you don’t include 0 in the natural numbers? I think it’s much easier, notation wise, if we reserve Z+ for the positive integers and we let N be the set of nonnegative integers.

    • @goodplacetostop2973
      @goodplacetostop2973 2 ปีที่แล้ว +8

      In this world nothing can be said to be certain, except death, taxes and math people arguing about 0 in ℕ.
      Personally I’m team « 0 in ℕ » but that’s the kind of debate people have strong opinions and that can go on forever. Michael thinks 0 is not in it, and he won’t change his mind because of TH-cam comments lol

  • @schweinmachtbree1013
    @schweinmachtbree1013 2 ปีที่แล้ว

    Not sure about the inclusion of magmas, semigroups and monoids in this course (at least introducing the terminology for all of them); seems like abstraction for abstraction's sake. and if you insist on breaking down groups in this way then it sort of obliges you to do the same thing for vector spaces later on, which would involve defining vector spaces as modules over a field.
    You _could_ justify the inclusion of semigroups/monoids by going super abstract and considering semimodules over semirings (a semiring being a ring without additive inverses, and a semimodule being an abelian semigroup equipped with a scalar multiplication from a semiring, satisfying the usual properties (a+b)v = ab + bv, a(v+w) = av + aw, and a(bv) = (ab)v). But "pulling the legs off the spider" in this way isn't really appropriate in a linear algebra course; semimodules will still form a category (that is, composition of morphisms will be associative and there will be identity morphisms) which is enriched over itself (that is, the set of morphisms between two semimodules will itself be a semimodule, although one will have to be careful will left, right, and two-sided semimodules when the semiring is non-commutative), but apart from these basic properties, the theory of (semi)modules is very different to the theory of vector spaces.

    • @ttrss
      @ttrss ปีที่แล้ว

      Ok buddy

  • @christianorlandosilvaforer3451
    @christianorlandosilvaforer3451 2 ปีที่แล้ว +1

    24:22 r u using what u want to prove ? mmmm