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Representation of a finite group

  1. Aug 8, 2011 #1

    syj

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    1. The problem statement, all variables and given/known data

    Prove that a representation of a finite group G is faithful if and only if its image is isomorphic to G.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 8, 2011 #2

    micromass

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    What did you try already?? If you show us where you're stuck, then we'll know where to help...
     
  4. Aug 9, 2011 #3

    syj

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    I am not very eloquent when it comes to proofs.
    So I am just going to lay out what I know.

    Let the representation be noted as F, and the image of G'
    if F is a faithful representation then ker{F}={1G}

    Can I conclude then by the first isomorphism theorem that G is isomorphic to G'?





    I know that for an "if and only if" proof there are two directions. If I can get the first direction of the proof, I can easily get the other direction.
     
  5. Aug 9, 2011 #4

    micromass

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    Indeed, the first isomorphism theorem does the trick!! :smile:
     
  6. Aug 9, 2011 #5

    syj

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    Ok, so is this enough:

    If f is faithful then ker{f}={1G}
    therefore by the first isomorphism theorem, G[itex]\cong[/itex]G'

    If G[itex]\cong[/itex]G' then by the first isomorphism theorem ker{f}={1G}
    therefore by the definition of a faithful representataion, f is faithful.

    it seems so plain.
    lol.
    too plain to be complete.
    but if it is, i am one happy girl ;)
     
  7. Aug 9, 2011 #6

    micromass

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    This is true (but only for finite groups), but you might want to explain in some more detail.

    The rest is ok!
     
  8. Aug 9, 2011 #7

    syj

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    can you please explain how i should expand further?
    I am told that G is finite in the question.
    thanks
     
  9. Aug 9, 2011 #8

    micromass

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    Well, you know that

    [tex]G\cong G/\ker(\phi)[/tex]

    Why does that imply that [itex]\ker(\phi)=\{1\}[/itex] ??

    Think of the order...
     
  10. Aug 9, 2011 #9

    syj

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    ok,
    am i making sense here:

    a corollary to the first isomorphism theorem says:

    |G:ker([itex]\varphi[/itex]|=|[itex]\varphi[/itex](G)|

    from this can I conclude:
    |[itex]\frac{G}{ker(\varphi)}[/itex]|=|G'|

    and then conclude:
    ker([itex]\varphi[/itex])={1G}
     
  11. Aug 10, 2011 #10

    micromass

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    Indeed, that works!! :smile:
     
  12. Aug 10, 2011 #11

    syj

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    wooo hoooo !!!!
    i am the happiest girl in the world!!
    until the next proof comes my way ... at which time I shall bug u some more!
    thanks so much.
     
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