Quotient Groups Part 2

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  • เผยแพร่เมื่อ 9 ก.พ. 2025
  • In this video we discuss the construction of quotient groups and the First Isomorphism Theorem.

ความคิดเห็น • 11

  • @jacksoncrook9732
    @jacksoncrook9732 7 ปีที่แล้ว +9

    thanks for these vids. I love the color.

  • @miss.needle7176
    @miss.needle7176 4 ปีที่แล้ว +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 ปีที่แล้ว +2

    really great, greetings from germany

  • @jeromepatoux9719
    @jeromepatoux9719 6 ปีที่แล้ว

    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 ปีที่แล้ว

      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 ปีที่แล้ว

      @@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 ปีที่แล้ว +1

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

    • @ANDROIDPOSTMORTEM
      @ANDROIDPOSTMORTEM 2 ปีที่แล้ว +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 ปีที่แล้ว

    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.