$AB\leq N_G(B)$ implies $B\trianglelefteq AB$? I am reading the proof of the Second Isomorphism Theorem on Dummit and Foote's Abstract Algebra, 3rd edition. 
Could anyone tell why they mention $AB\leq N_G(B)$ on the highlighted part? Does $AB\leq N_G(B)$ imply $B\trianglelefteq AB$?
If you just wanted to show $B\trianglelefteq AB$, then you could say for any $ab\in AB$, 
$$abB(ab)^{-1}=abBb^{-1}a^{-1}=aBa^{-1}=B,$$
where the last equality follows from the assumption $A\leq N_G(B)$. 
So I am wondering what they mean by $AB\leq N_G(B)$, i.e., $B$ is normal subgroup of the subgroup $AB$.

 A: 
If $K\le H \le N_G(K),$ then $K\lhd H$.  

Proof.  
Let $h\in H$. Then $h\in N_G(K)$. Hence $h^{-1}Kh=K$. So $K\lhd H$. 
So in your case, $B\le AB\le N_G(B)$, hence $B\lhd AB$.
Your way to show that $B\lhd AB$ is also correct. Just that the author applied the tool of normalizer to get the result faster.
A: $N_G(B)$ is the largest subgroup of $G$ in which $B$ is normal.  We always have $B\le N_G(B)\le G$ and $B$ is normal in $H\ge B$ if and only if $H\le N_G(B)$.
This definition makes sense because if $H$ and $K$ both normalize $B$, then so does the subgroup generated by $H$ and $K$, which guarantees that there is a unique subgroup maximal with the property that it normalizes $B$.  It contains $B$ since $B$ always normalizes itself.  It also happens to comprise precisely the elements of $G$ that normalize $B$, since that subset is in fact a subgroup.  (It's the stabilizer of $B$ in an action of $G$ on its subgroups by conjugation, and stabilizers are always subgroups, for any group action.)
Thus not only does $AB\le N_G(B)$ imply $B\trianglelefteq AB$ for a subgroup $A$, but the converse is also true.  Also note, that either statement implies that $AB=<A,B>$, which need not be the case if $A$ does not normalize $B$.
"$H$ normalizes $B$" means here that for every $h\in H$, $h^{-1}Bh=B$.  
If Dummit & Foote didn't define $N_G(B)$ before using the notation, their bad.  I hope this helps.
There is also another useful notation $Z_G(B)$ for the centralizer of $B$...i.e., all elements that commute with every element of $B$.  Note that while $B\le N_G(B)$ it will not be the case that $B\le Z_G(B)$ unless $B$ is abelian.
