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Thread: cardinality proof

  1. #1
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    cardinality proof

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

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

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

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

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

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

    Is this correct?

    To show that the mapping is surjective:

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

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

    $\displaystyle = g^{-1}a^gg = a^g$

    Is it right?
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