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Thread: abstract

  1. March 16th 2009, 11:50 AM #1
    pila0688
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    Exclamation abstract

    Assume the following are isomorphic:

    Prove |G| = |H|

    Prove "phi"(eG) = eH.

    Prove "phi"(a^n) = ("phi"(a))^n.
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  2. March 16th 2009, 01:56 PM #2
    SimonM
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    Let \phi be an isomophism between G and H.

    \phi is a bijection, what does that tell you about |G| and |H|?

    \phi(e_G) = \phi(e_G^2) = \phi(e_G)\phi(e_G) = \phi(e_G)^2

    What does this tell you about \phi(e_G)?

    Can you induct on a similar expression that I've used above?
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