If $N \lhd G$ and $A \lhd B \le G$ then $AN \lhd BN$? The content of the question is in the title. I'm asking this in relation to a proof that (assuming the above conditions): $$BN/AN \cong B/A(B \cap N)$$
Which I believe follows directly from an application of the second isomorphism theorem for groups.
$AN \subseteq BN$ is clear since $A \subseteq B$.
Then we must show that $xANx^{-1} \subseteq AN$ for $x \in BN$.
$x=bn$ so show that $bn(an_{1})n^{-1}b^{-1} \in AN.$
Using the normality of $N$ in $G$ and of $A$ in $B$ as well the extensive use of the generalized associative law for groups,
$bn(an_{1})n^{-1}b^{-1} = b(na)n_{1}n^{-1}b^{-1}=b(an_{2})(n_{3}b^{-1})= (ba)n_{2}(n_{3}b^{-1})=(a_{2}b)n_{2}(b^{-1}n_{4})=a_{2}(bn_{2})(b^{-1}n_{4})=a_{2}(n_{5}b)(b^{-1}n_{4})=(a_{2}n_{5})(b^{-1}b)n_{4}=(a_{2}n_{5})n_{4}=a_{2}(n_{5}n_{4})=a_{2}n_{6} \in AN$.
$AN \subseteq BN$ is clear since $A \subseteq B$.
Then we must show that $xANx^{-1} \subseteq AN$ for $x \in BN$.
$x=bn$ so show that $bn(an_{1})n^{-1}b^{-1} \in AN.$
Using the normality of $N$ in $G$ and of $A$ in $B$ as well the extensive use of the generalized associative law for groups,
$bn(an_{1})n^{-1}b^{-1} = b(na)n_{1}n^{-1}b^{-1}=b(an_{2})(n_{3}b^{-1})= (ba)n_{2}(n_{3}b^{-1})=(a_{2}b)n_{2}(b^{-1}n_{4})=a_{2}(bn_{2})(b^{-1}n_{4})=a_{2}(n_{5}b)(b^{-1}n_{4})=(a_{2}n_{5})(b^{-1}b)n_{4}=(a_{2}n_{5})n_{4}=a_{2}(n_{5}n_{4})=a_{2}n_{6} \in AN$.