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Thread: well-defined proof

  1. #1
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    well-defined proof

    $\displaystyle A$, $\displaystyle A^{*}$, $\displaystyle B$, $\displaystyle B^{*}$ be four subgroups of a group $\displaystyle G$ with $\displaystyle A \vartriangleleft A^{*}$ and $\displaystyle B \vartriangleleft B^{*}$.
    Let $\displaystyle D=(A^{*} \cap B)(A \cap B^{*})$.
    I have shown the following:
    (1) $\displaystyle B \vartriangleleft B(A^{*} \cap B^{*})$
    (2) $\displaystyle (A^{*} \cap B) \vartriangleleft (A^{*} \cap B^{*})$
    (3) $\displaystyle (A \cap B^{*}) \vartriangleleft (A^{*} \cap B^{*})$
    (4) $\displaystyle D \vartriangleleft (A^{*} \cap B^{*})$

    If $\displaystyle x \in B(A^{*} \cap B^{*})$, then $\displaystyle x=bc$ for $\displaystyle b \in B$ and $\displaystyle c\in (A^{*} \cap B^{*})$;
    define $\displaystyle f:B(A^{*} \cap B^{*}) \rightarrow (A^{*} \cap B^{*})/D$ by $\displaystyle f(x)=Dc$.

    I'm facing problem to show that $\displaystyle f$ is well-defined.
    I started with $\displaystyle x_{1}=x_{2}$;
    implies that $\displaystyle b_{1}c_{1}=b_{2}c_{2}$. ----------($\displaystyle *$)
    I'm stuck here.
    I know I need to get $\displaystyle c_{1}c_{2}^{-1} \in D$; ------------($\displaystyle **$)
    implies that $\displaystyle Dc_{1}=Dc_{2}$;
    finally $\displaystyle f(x_{1})=f(x_{2})$.

    How to continue from ($\displaystyle *$) to ($\displaystyle **$)?
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  2. #2
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    I have another question.

    How to show that $\displaystyle Ker\:f=B(A \cap B^{*})$ ?
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