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Isomorphism of quotient group

  1. Apr 4, 2013 #1
    How can one prove that for homomorphism [itex] G \xrightarrow{\rho} G' [/itex] and H as kernel of homomorphism, quotient group G/H is isomorphic to G'?

    Thanks.
     
  2. jcsd
  3. Apr 4, 2013 #2

    Ben Niehoff

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    Think about what happens to H under the mapping, and how this relates to the definition of G/H.
     
  4. Apr 4, 2013 #3
    I understand that it suffices to prove that [itex] Ker[G/H \xrightarrow{\rho} G'] =e [/itex]. How do you prove this?
     
  5. Apr 5, 2013 #4

    jbunniii

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    This is only true if ##\rho## is surjective. In general, ##G/\ker(\rho)## is isomorphic to ##\text{image}(\rho)##, which may be a proper subgroup of ##G'##.

    Hint: you need to find an isomorphism ##\phi : G/H \to \text{image}(\rho)##. Every element of ##G/H## is a coset of the form ##g + H## for some ##g \in G##. What is the most natural way to define ##\phi(g+H)## as an element of ##\text{image}(\rho)##?
     
  6. Apr 5, 2013 #5

    jbunniii

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    But you already defined ##\rho## as a map from ##G## to ##G'##. It can't also be a map from ##G/H## to ##G'##.
     
  7. Apr 5, 2013 #6
    right, sorry, [itex] \rho' [/itex]
     
  8. Apr 5, 2013 #7
    I don't know, how?
     
  9. Apr 5, 2013 #8

    jbunniii

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    What is an element of ##\text{image}(\rho)## which is in some way related to ##g##?
     
  10. Apr 5, 2013 #9

    micromass

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    Make an educated guess and prove that your guess is correct.
     
  11. Apr 5, 2013 #10
    By image(rho) you mean the subgroup of G' onto which G is mapped?
     
  12. Apr 5, 2013 #11

    jbunniii

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    Yes, that's right.
     
  13. Apr 5, 2013 #12
    Still confused.
     
  14. Apr 5, 2013 #13

    micromass

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    So ##\rho## takes in an element of ##G## and gives you an element of ##G^\prime##.

    You want to define some map ##G/H\rightarrow G^\prime##. are given the point ##g+H##. You know ##g\in G##. You want to use ##\rho## somehow to get a point in ##G^\prime##. What are you going to try?
     
  15. Apr 5, 2013 #14
    bundle all ##\rho##s to get a ##\rho'##?
     
  16. Apr 5, 2013 #15

    jbunniii

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    You are overthinking this. Given an element ##g\in G##, and the mapping ##\rho: G \to G'##, can you write down a formula for an element of ##G'## which involves ##g## and ##\rho##?
     
  17. Apr 5, 2013 #16
    ##\rho(g)##
     
  18. Apr 5, 2013 #17

    jbunniii

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    Yes, that's what you need. So now how can you define ##\phi : G/H \to G'##?
    $$\phi(g+H) = ???$$
     
  19. Apr 5, 2013 #18
    ##\phi(g+H)=\rho(g)## but how do you prove the isomorphism?
     
  20. Apr 5, 2013 #19

    jbunniii

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    OK good, you have found the right mapping, that's half the battle.

    Now you need to prove the following:

    1. ##\phi## is well defined. This is important because you defined the map ##\phi## in terms of a particular element, ##g##, of the coset ##g + H##. But that coset may have other elements. You need to show that ##\phi## gives you the same result if you choose a different element ##g_1 \in g + H##.

    2. ##\phi## is an injection. i.e. ##\ker(\phi) = \{e\}##.

    3. ##\phi## is a surjection.

    4. ##\phi## is a homomorphism.

    I recommend starting with 1, so you can be sure you understand why the definition of ##\phi## makes sense. Formally, you need to show that if ##g + H = g_1 + H##, then ##\phi(g+H) = \phi(g_1 + H)##.
     
  21. Apr 5, 2013 #20
    thanks, let me think and report back. this is an important thing to thoroughly understand.
     
    Last edited: Apr 5, 2013
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