A, B subgroups of G, B/A abelian. Show that BN/AN is abelian. I'm going through a past paper on Group theory and I wondered if someone could help with a solution to this.
Let $A$, $B$ be subgroups of $G$ such that $A\triangleleft B$ and $B/A$ is abelian. Also $N\triangleleft G$. I proved that $AN\triangleleft BN$ and now I need to show that $BN/AN$ is abelian.
Thanks

All cosets in $BN/AN$ are of the form $bAN$, where $b \in B$. Then for any $b, b' \in B$ we have 
$bANb'AN = bAb'NAN = bAb'AN = b'AbAN = b'AbNAN = b'ANbAN$ .
