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Math Help - cardinality proof

  1. #1
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    cardinality proof

    Let G be a group and let A \leq G be a subgroup. If g \in G, then A^g \subseteq G is defined as

    A^g=\{a^g | a \in A \} where a^g=g^{-1}ag \in G.

    Show that the sets A and A^g have the same cardinality.
    Hint: sets A and A^g have the same cardinality if and only if there exists an injective and surjective map f: A \to A^g.

    Attempt: So, in order to show that the mapping f is injective I have to show that

    f(a)=f(a^g)=f(g^{-1}ag) \iff a = a^g (for a \in A and a^g \in A^g).
    But how can I deduce that when I don't know what the function f is?
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  2. #2
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    You don't. You have to define a specific f and show that that f is both injective and surjective. If you simply assert the existance of such an f, you are assuming what you want to prove.

    Since A^g= {g^{-1}ag |g\in G} how about trying f(a)= g^{-1}ag?
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    You don't. You have to define a specific f and show that that f is both injective and surjective. If you simply assert the existance of such an f, you are assuming what you want to prove.

    Since A^g= {g^{-1}ag |g\in G} how about trying f(a)= g^{-1}ag?
    To show that the mapping is injective:

    f(a)=f(a^g) \iff g^{-1}ag=g^{-1}a^gg
    \iff agg^{-1} = a^g gg^{-1}
    \iff a=a^g

    Is this correct?

    To show that the mapping is surjective:

    For any a^g \in A^g let a=g^{-1}a^gg (since a^g = g^{-1}ag).

    f(a)=g^{-1}(g^{-1}a^gg)g

    = g^{-1}a^gg = a^g

    Is it right?
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