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Math Help - Normal subgroup

  1. #1
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    Normal subgroup

    Let G be a group, and let H \leq G

    (That means H is the normal subgroup of G, right?)

    a)Show that aHa^{-1} is also a subgroup of G.

    My proof:

    Suppose that x,y is in G, so that xHx^{-1}, yHy^{-1} \in aHa^{-1}

    Since H is normal in G, pick Hx^{-1}=x_{1}H, yH=Hy_{1},x_{1},y_{1} \in G

    (xHx^{-1})(yHy^{-1}) = x(x_{1}H)(Hy_{1}y^{-1}) = xx_{1}Hy_{1}y^{-1}

    Now, (xHx^{-1})^{-1}=x^{-1}Hx \in aHa^{-1}

    Is this right?

    b) If H is finite of order k, then aHa^{-1} is also finite of order k.
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  2. #2
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    Do you know this theorem?
    A set H in a group G is a subgroup iff a \in H\;\& \;b \in H \Rightarrow \quad ab^{ - 1}  \in H?

    Suppose that x \in aHa^{ - 1} \;\& \;y \in aHa^{ - 1}  \Rightarrow \left( {\exists c \in H} \right)\left( {\exists d \in H} \right)\left[ {x = aca^{ - 1}  \wedge y = ada^{ - 1} } \right].

    What can be said about xy^{ - 1}?
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  3. #3
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    Quote Originally Posted by tttcomrader View Post

    b) If H is finite of order k, then aHa^{-1} is also finite of order k.
    \{ a_1, ... , a_n \}
    Let the above be the elements of H.
    Then,
    \{aa_1a^{-1},...,aa_na^{-1} \}
    Are all distint.
    Because otherwise if,
    aa_ia^{-1} = aa_ja^{-1}
    This implies,
    a_i = a_j.
    Which is an impossibility for distinct elements.
    -----
    Here is a useful theorem to know.

    Theorem: Let H be a subgroup of G the following are equivalent.
    1)For all g\in G and h\in H implies ghg^{-1} \in H.
    2)For all g\in G we have gHg^{-1}.
    3)The left and right cosets coincide, i.e. aH = Ha.

    All these 3 are equivalent formulations for "normal subgroup". My favorite is #1.
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