Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Correspondence Theorem in Group Theory

  1. Aug 17, 2013 #1
    Hello,

    I'm following the proof for this theorem in my textbook, and there is one part of it that I can't understand. Hopefully you can help me. Here is the part of the theorem and proof up to where I'm stuck:

    Let ##N## be a normal subgroup of a group ##G##. Then every subgroup of the quotient group ##G/N## is of the form ##H/N## for some subgroup ##H## of ##G## with ##H\le G##.

    ...

    Proof: Let ##H^*## be a subgroup of ##G/N## so that it consists of a certain set ##\{hN\}## of left cosets of ##N## in ##G##. We define the subset ##\beta(H^*)## of ##G## to be ##\{g\in G:gN\in H^*\}##. Then ##\beta(H^*)## clearly contains N and is a subgroup of G:

    [here follows demonstration that ##\beta(H^*)## is a subgroup of G]


    The portion in red is what I'm having trouble with. I don't see why it's immediately clear that ##N## is contained within ##\beta(H^*)##.

    Any help is appreciated.

    I first thought that perhaps ##H^*## is supposed to correspond to the ##H## of the hypothesis, but then I realised that cannot be true, since the elements of ##\beta(H^*)## would be precisely those of ##H## and the properties would be trivial. In particular, proving that ##\beta(H^*)## is a subgroup of ##G## would be pointless.
     
  2. jcsd
  3. Aug 18, 2013 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    Isn't the identity of ##H^*## exactly the coset ##N##?
     
  4. Aug 18, 2013 #3
    That's correct. So we know the identity element ##1## belongs to ##\beta(H*)##. We also know that ##N## is a subgroup, so ##1\in N##. Therefore ##\beta(H*)## contains at least one element of ##N##.
     
  5. Aug 18, 2013 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    For each [itex] g \in N ,\ gN = N[/itex] and [itex] N \in H^* [/itex], so [itex] g\in \beta(H^*) [/itex]
     
  6. Aug 19, 2013 #5
    Ah, well that clears that up. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Correspondence Theorem in Group Theory
  1. Group theory (Replies: 3)

  2. Group Theory (Replies: 6)

  3. Group theory (Replies: 1)

Loading...