Prove that if $H\triangleleft G$, then $H/N$ $\triangleleft$ $G/N$. And vice versa. Let $G$ be a group, $H$ and $N$ subgroups of $G$, with $N\triangleleft G$ and $N$ a subgroup of $H$.
a.) Prove that if $H\triangleleft G$, then $H/N$ $\triangleleft$ $G/N$.
b.) Prove that if $H/N$ $\triangleleft$ $G/N$, then $H\triangleleft G$.
Which definition of normal subgroups should I use to help me with these? Need a good amount of help getting these started. Any explanation would be appreciated!
 A: Direct approach (I have not given a complete argument so you can try and solve it)
We will use:


*

*$N \leq G$ is normal if and only if for each $x$ in $G$ and for each $n\in N$, $xnx^{-1} \in N$;

*$G/N = \{xN\,|\,x \in G\}$ where $xN = \{xn\,|\,n\in N\}$ and is a group since $N$ is normal in $G$. It has identity element $eN=N$ and multiplication $(xN) (yN) = (xy)N$.

*For $xN, yN \in G/N$, $xN = yN$ if and only if $y^{-1}x \in N$


Since $H\leq G$ we know that $H/N = \{hN\,|\,h \in H\} \leq \{xN\,|\,x \in G\}=G/N$.
For (a) we must prove: if $H$ normal in $G$ , then $H/N$ normal in $G/N$. Let $xN$ and $hN$ be elements in $H/N$ and $G/N$ respectively and show that $(xN) (hN) (xN)^{-1}$ is in $H/N$ (use the second item in the list above and the fact that $H$ normal in $G$).
For (b) we must prove: if $H/N$ normal in $G/N$, then $H$ normal in $G$. Let $x$ and $h$ be elements of $H$ and $G$ respectively and show that $xhx^{-1}$ is in $H$. To do so note that $xhx^{-1} N = (xN)(hN)(xN)^{-1} \in H/N$ and hence $xhx^{-1} N = h'N$ where $h'$ is in $H$ (use the third item in the list and the fact that $N \leq H$).
