Quotient Groups and Homomorphic Images | Abstract Algebra

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  • Опубліковано 11 гру 2024

КОМЕНТАРІ • 14

  • @MrCoreyTexas
    @MrCoreyTexas 4 місяці тому

    Here's what I didn't understand at first, about 9:40 you talk about adding the coset +1 to +3 and saying it would equal the coset +4, but if you add the elements shown, you get (-16,-6,4,14,24). So you're missing the -1 and 9 which are elements of +4. But you'd be adding every element in +1 to +3, and that would generate the -1 and 9, which is kind of an insane notion (adding infinitely many integers to infinitely integers to get a list of infinite integers)

    • @JPL454
      @JPL454 Місяць тому

      When you consider cyclic notation, you have to think about all possible multiples, so when you do +1 + +3 you can't just look at each column and add the numbers. You get the -1 with -4+3(which both belong to their respective +...) and 9 by adding 6 and 3. I think that you need to abstract a little bit from the common sum notation and just think of those as Z mod n (multiples of n)

  • @sasaknoob
    @sasaknoob Рік тому +1

    Nice Explanation bro

  • @shambo9807
    @shambo9807 Рік тому +1

    Where were these videos when I was struggling with Abstract Algebra?????????! Where was Gondor when my GPA fell?
    (Thanks tho, because I still need to understand it for set theory apparently. Maybe I can learn to not hate it.)

  • @wenzhang365
    @wenzhang365 4 місяці тому

    Excellent lecture. Nice hair style, BTW. 😀

  • @archiem654
    @archiem654 Рік тому +1

    Quotient group is set of all cosets. But wouldnt the set of cosets be the group itself? Since all cosets of a subgroup are disjoint and all of them will contain all of the elements of group itself....?

    • @awesome10pickles32
      @awesome10pickles32 Рік тому +1

      Good question! If H is a normal subgroup, than the set of all cosets of H, G/H, is a set of sets of group elements, not a set of group elements! Instead of the binary operation on elements of G, the quotient group G/H has a binary operation on cosets of H, so G and G/H are different groups.

    • @talastra
      @talastra 11 місяців тому

      Let's see if I've actually learned anything. I think couldn't this question be answered by saying: the set of all cosets (fully) partitions G?

    • @musicaltaco6803
      @musicaltaco6803 9 місяців тому

      If G is abelian, then G/H is G. The normal subgroup, N, is equal to G ifand only if G is Abelian. (I am taking algebra now, so I could be wrong.)

  • @cuneytkaymak4997
    @cuneytkaymak4997 Рік тому +1

    What exactly is a homomorphic image? And f:G -> G/H is a surjective but non injective mapping right? Im thinking about Z to Z/5z. Z/5z has 5 elements. 1,6,11,16…. are mapped to 1 and so on.

    • @talastra
      @talastra 11 місяців тому

      Look back in the series to find the videos that define homomorphic images.
      Maybe start here: ua-cam.com/video/rJpu22jMeIY/v-deo.html&ab_channel=WrathofMath

  • @James-eo2wh
    @James-eo2wh 3 місяці тому

    can you do the video on examples of quotient groups? thanks

    • @WrathofMath
      @WrathofMath  3 місяці тому

      Sure! Here's an early link you can use as soon as the video finishes processing: ua-cam.com/video/nR4bJb29ahU/v-deo.html
      Perhaps you wanted a more computational set of examples - I will do some of those also at some point - but I hope you'll find this video useful as well.