Possible Duplicate:
A, B subgroups of G, B/A abelian. Show that BN/AN is abelian.
Let A, B, N be subgroups of a group G such that A $\triangleleft$ B and B/A is Abelian. Also suppose N $\triangleleft$ G. Prove that AN $\triangleleft$ BN and that BN/AN is abelian.
I'm awful at these kinds of questions and never know what I need to do!
I know $AN \subseteq BN$ as $A \subseteq B$ and AN and BN are subgroups as N $\triangleleft$ G. Now I take $x=an \in AN$ and $y=bm \in BN$ $m,n \in N$ . Then $yxy^{-1} = bmanm^{-1}b^{-1}$ I change this to $bmab^{-1}bnm^{-1}b^{-1}=bmab^{-1}n'$ with B/A abelian I can change this to $bmb^{-1}an' = m'an'$. I'm missing some step to get it in the form $an$ but don't know what? For the second part I'm completely stumped so any help appreciated!