# 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,

$$bn(an_{1})n^{-1}b^{-1} = b(na)n_{1}n^{-1}b^{-1}=b(an_{2})(n_{3}b^{-1})= (ba)n_{2}(n_{3}b^{-1})=(a_{2}b)n_{2}(b^{-1}n_{4})=a_{2}(bn_{2})(b^{-1}n_{4})=a_{2}(n_{5}b)(b^{-1}n_{4})=(a_{2}n_{5})(b^{-1}b)n_{4}=(a_{2}n_{5})n_{4}=a_{2}(n_{5}n_{4})=a_{2}n_{6} \in AN$$.

• @Bungo yeah im just confusing myself at this point trying to write this... – Derek Luna Jan 13 at 20:39
• This was very ugly before, so I hope the person that downvoted reconsiders! It might be easier/less confusing just to show the "free" general movement of elements in normal subgroups by induction than doing this... – Derek Luna Jan 13 at 21:01
• Thank you Derek. I've accepted your answer, but I have found a cleaner way to do the algebraic manipulation: for $a \in A$, $b \in B$, $n,m \in N$ we have $$m^{-1}b^{-1}anbm = m^{-1}b^{-1}abb^{-1}nbm = m^{-1}a'n'm = a'n''n'm \in AN$$ – Matthew Buck Jan 13 at 21:04
• Can the downvoters explain? By normality, swap the order and adding a subscript each time. – Derek Luna Jan 13 at 21:05
• Great, unfortunately I have a sensitive guy (people?) on this site downvoting my posts so it will stay at $-1$ haha. – Derek Luna Jan 13 at 21:10