302.3A: Review of Homomorphisms

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  • Опубликовано: 18 дек 2024

Комментарии • 34

  • @PunmasterSTP
    @PunmasterSTP 4 месяца назад

    Looks like these videos are morphin' into a great playlist!

  • @Hythloday71
    @Hythloday71 7 лет назад +2

    What is the character of the structure that is preserved by homomorphisms generally ? Could it be said that an 'isomorphism' preserves 'information' in the physics sense of preserving distinctions ? Or is that what is generally done with homomorphisms ? Thinking about it, the particular cases of onto and 1-1 just serving to inform about the complete structure accountability ? There seems also something perhaps 'order' preserving about a homomorphism ?

    • @MatthewSalomone
      @MatthewSalomone  7 лет назад +4

      All types of homomorphisms "preserve" the algebraic operation of the groups they connect, but as you say, unless they are one-to-one they lose information in the sense of not being uniquely reversible. Isomorphisms, on the other hand, preserve everything about the groups they connect except possibly the "names" of the elements. We don't look at order in this series of videos, but that would be an extra consideration on top of isomorphism (that would, e.g. distinguish the isomorphisms f(x) = x and f(x) = -x on the integers).

    • @alojzybabel4153
      @alojzybabel4153 4 года назад

      @@MatthewSalomone Is there any _systematic_ way (that is, an _efficient algorithm_ ) for comparing two groups in order to check whether they have in fact the same structure and differ only by the labels? Let's say, given their multiplication tables?

    • @MatthewSalomone
      @MatthewSalomone  4 года назад +1

      Alojzy Bąbel Depends. For finite groups of order n, sure - you could just test all permutations of one Cayley table against the other - order O(n!). You may be able to do better with linear algebra, by testing whether there exists a permutation matrix P such that PG1 = G2. But if you move away from Cayley tables (which are very explicit descriptions of groups) to presentations (more minimal descriptions using generators and relations) then no, there's no efficient algorithm: en.wikipedia.org/wiki/Group_isomorphism_problem

  • @ravishkumar5756
    @ravishkumar5756 7 лет назад +1

    Your illustration is so good sir.......

  • @hebrewwolf6540
    @hebrewwolf6540 4 года назад

    01:04 I assume you meant to say "composition of the images" ?

  • @00bonobo
    @00bonobo 10 лет назад +5

    Many thanks. I love your videos. I feel very confortable learning with you about a subject that I am very interested in.

  • @AnthonyCasadonte
    @AnthonyCasadonte 12 лет назад +1

    I see how the function you chose "3^n" works between addition and multiplication operations given the rules of exponents, so that is one way to get an intuitive sense of why you chose that particular function as the homomorphism. But, why in particular the base 3? I see how it would not work with 2, but would it work with 5? In general, for a group of integers with addition and the multiplicative group of integers for another modulus, I see why exponentiation makes sense but what about the base?

  • @AnthonyCasadonte
    @AnthonyCasadonte 12 лет назад +1

    For the automorphism example, saying how phi is its own inverse makes me think of whether we can have a group of homomorphisms with the binary operation of composition. The identity is the trivial homomorphism and in this case the inverses are themselves. Can we think of this?

  • @bonbonpony
    @bonbonpony 6 лет назад +2

    Are those morphisms the same as in category theory? Because they sound oddly familiar :q

  • @keerthanakeerthu2256
    @keerthanakeerthu2256 6 лет назад

    Thank u so much... It's helpful to take class .. very neat explanation. . I like it. .. thank u thank u... It cleared my all doubts in homomorphisms. .

  • @bclan6937
    @bclan6937 6 лет назад

    Very well done. And clear.

  • @ouafieddinenaciri3783
    @ouafieddinenaciri3783 10 лет назад

    Hi Anthony
    If we choose 2^n it works, I mean we obtain an homomorphism.
    All that we lose is : that is neither a monomorphism nor an epimophism.

  • @RARa12812
    @RARa12812 2 года назад

    At 5 12 why one side is addition and other side is multiplication 1 +2 on one side 2 ×3 on other side

  • @lemyul
    @lemyul 5 лет назад +1

    thanks salmon

  • @TARINunit9
    @TARINunit9 6 лет назад +3

    1:10
    Thank you so damn much

    • @Taricus
      @Taricus 5 лет назад

      I was thinking the same thing at the same exact point. He flat out said exactly what it was, directly! ^_^

  • @ravishkumar8422
    @ravishkumar8422 2 года назад

    Z2*Z2 to S4 is not homorphism because order of image should divide order of preimage. 4 doesnot divide 2. order of (12)(34) is 4 and order of (1,1) is 2. and 4 doesnt divide 2, there fore it is not homorphism, I mean (1,1) doesnot goes to (12)(34). Am I right?

  • @yugandhar59
    @yugandhar59 8 лет назад +1

    nice explanation

  • @sanjursan
    @sanjursan 10 лет назад

    Good Vid!! More fun to watch Liliana de Castro, but more and better info here. Thank you for this vid.

  • @wdobni
    @wdobni 6 лет назад

    as may be....but it seems like a very tortured way to get through a morning......so much of pure math is gaming and tortured nomenclature.....and at the end of the mental puzzle the puzzle pieces go back in the box......
    its all very inventive but also stark and cold and lifeless like a hunk of granite....it all seems to lead nowhere...like an elaborate roadsign in a desert that has no road and no towns

    • @Taricus
      @Taricus 5 лет назад +1

      You can use it if someone says, "Hey, I came up with a new operation!" and if you work it out, it may just be some regular old operation that everyone already knows, but in disguise. It may look new, but you can show that the algebra is just a different-looking form of the same algebra.

  • @muhammadrustamzada
    @muhammadrustamzada 8 лет назад

    Angel :D Thanks, you saved my life :D

  • @RURALSCHOOLTALENT
    @RURALSCHOOLTALENT 11 лет назад

    show that the group l 1 a l such that a €z is iso morphic to Z

  • @manimegalai7695
    @manimegalai7695 7 лет назад

    very useful

  • @prof.mmahboobkhanraja9563
    @prof.mmahboobkhanraja9563 8 лет назад

    Lot of thanks i got it

  • @HanzAlbertNguyen
    @HanzAlbertNguyen 10 лет назад +1

    you are doing good presentation, but you are talking as if you are robbing, slow downnnnnnn , mannn!

    • @sanjursan
      @sanjursan 10 лет назад +7

      Hmm, I actually like the speed just fine.

    • @m0ng00se4
      @m0ng00se4 10 лет назад +3

      sanjursan i agree, i can always rewind or pause if i find he's going too fast
      it's a massive quality of life improvement to not watch the cosets be written in real time

    • @beback_
      @beback_ 7 лет назад +2

      You can try and interleave his lectures with those of Ben Garside, who presents super slowly. Go to Ben when Matt is being too fast, and to Matt when Ben is being to slow.

    • @stapleman007
      @stapleman007 2 года назад +1

      This is a review of Abstract Algebra I, not a full explanation. Also, there are pause buttons on YTube.