The Set Product Is A Binary Operation On Cosets Of Normal Subgroups

You've got to make use of the fact that $Na=aN$ for every $a\in G$. So we have $Nb=bN$, $Nc=cN$, $Nx=xN$, etc. Why is this useful? Any time you see $na$ in your work for some $n\in N$, you know that $na\in Na$ which means $na\in aN$ which means there exists some $n'\in N$ with $na=an'$. You don't have the ability to commute, but we get ALMOST commutativity. We can write $na=an'$ for some different $n'\in N$. The $'$ symbol just means that it's a different element. It's not a derivative.

So in your work if you ever see $bn_1a$, you can instead either write $bn=n_2b$ or you could write $na=an_3$ for some $n_2,n_3\in N$, and then you have $bn_1a = n_2ba=ban_3$. Similarly, if you see $n_1an_2b$, then you should be able to write this, after some work, in either the form $n_3ab$ or the form $abn_4$ (you'll need to use the closure of the operation in $N$ to combine elements in $N$).

You can't use the commutative law, but you have something pretty close.