Quotient Groups Part 2

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  • Опубліковано 18 гру 2024

КОМЕНТАРІ • 11

  • @jacksoncrook9732
    @jacksoncrook9732 7 років тому +9

    thanks for these vids. I love the color.

  • @miss.needle7176
    @miss.needle7176 3 роки тому +1

    In beginning, I thought expressions are too long for me but finally, I am happy to learn something without confusion. Thanks. (videos still so long but... =) )

  • @wwinger1
    @wwinger1 6 років тому +2

    really great, greetings from germany

  • @jeromepatoux9719
    @jeromepatoux9719 5 років тому

    Hi, around 9:20, you mentioned that the composition law would not be well defined if N was not a normal subgroup. Why does using a normal subgroup ensure that the composition law is well defined? Where do we use the properties of a normal subgroup that help ensuring this?

    • @tekaaable
      @tekaaable 5 років тому

      This is mentioned in part 1: this is because when N is a normal subgroup the left and right coset are the same, so there is only one way to partition out G

    • @rtg_onefourtwoeightfiveseven
      @rtg_onefourtwoeightfiveseven 4 роки тому

      @@tekaaable Is there an issue with extending the definition of quotient group to include non-normal subgroups and specify that the partitions are made using (say) the left coset? Or does that break something down the line?

    • @rtg_onefourtwoeightfiveseven
      @rtg_onefourtwoeightfiveseven 4 роки тому +1

      @@tekaaable Wait, never mind, just hit 8:30.

    • @ANDROIDPOSTMORTEM
      @ANDROIDPOSTMORTEM 2 роки тому +1

      @@rtg_onefourtwoeightfiveseven Normal Subgroup is defined in term of stable conjugation of the elements of subgroup.
      Means a•n•a-¹ will also be an element of subgroup N, where n is an element of the group and n is an element of subgroup N.
      And look carefully that he used this stable conjugation property of Normal Group to prove that the defined composition is well defined. Check that he used (b-¹•n•b).
      For Non-Normal groups this property isn't true therefore we can't have a stable conjugation.

  • @ANDROIDPOSTMORTEM
    @ANDROIDPOSTMORTEM 2 роки тому

    I stopped the video and proved the same like this...
    (a•b) bar
    = Coset of a•b
    = Coset of (a•b)•n
    = Coset of n'•(a•b•n) [ because it's a Normal Group ]
    = Coset of n'•a•b•n
    = Coset of (n'•a)•(b•n)
    = Coset of a'•b' [ because n'•a is an element of Coset of a and b•n is an element of b ]
    = Coset of a' • Coset of b'
    = a' bar • b' bar
    Therefore, for all element of Coset of a and b the composition law is well defined.