1. ## 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 ($\displaystyle **$)?

2. I have another question.

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