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,

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

  • $\begingroup$ @Bungo yeah im just confusing myself at this point trying to write this... $\endgroup$ – Derek Luna Jan 13 at 20:39
  • $\begingroup$ 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... $\endgroup$ – Derek Luna Jan 13 at 21:01
  • $\begingroup$ 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 $$m^{-1}b^{-1}anbm = m^{-1}b^{-1}abb^{-1}nbm = m^{-1}a'n'm = a'n''n'm \in AN$$ $\endgroup$ – Matthew Buck Jan 13 at 21:04
  • $\begingroup$ Can the downvoters explain? By normality, swap the order and adding a subscript each time. $\endgroup$ – Derek Luna Jan 13 at 21:05
  • $\begingroup$ Great, unfortunately I have a sensitive guy (people?) on this site downvoting my posts so it will stay at $-1$ haha. $\endgroup$ – Derek Luna Jan 13 at 21:10

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