Let $N \unlhd G$ and let $N \leqslant H \leqslant G$. Show that $N \unlhd H$. 
Let $N \unlhd G$ and let $N \leqslant H \leqslant G$. Show that $N \unlhd H$.

$N \unlhd G$  implies that there is some homomorphism $f$ on $G$ for which $N = ker(f)$. We want to show that there is a homormorphism $f'$ on H such that $N = ker(f')$.
Edit:
Proof. The function $f: G \to G/N$ is a surjective group homorphism with $ker(f) = N$. For $a,b \in G$, we have $f(ab) = abN$ and $f(a)f(b) = (aN)(bN) = abN$, demonstrating that $f$ is a homomorphism. For $a \in G$, we have $$a \in ker(f) \Longleftrightarrow aN = N \Longleftrightarrow a \in N.$$ This shows $ker(f) = N$. Finally, $f$ is surjective, since every left coset is of the form $aN = f(a)$ for some $a \in G$. Since $H \leqslant G$, we can define a map $f':H \hookrightarrow G \to G/N$, $h \mapsto aN$ for $h \in H \leqslant G$. It follows that $f'$ is also a surjective homomorphism and that $ker(f') = N$. Thus $N \unlhd H$. $\text{ } \Box$
Is this proof correct?

Consider $H \hookrightarrow G \to G/N$

Edit: (Concerning your edit)
Your proof looks good, just a few things to mention:


*

*you don't need surjectivity of any map (it is given but not necessary). Also if a map $G \to M$ is not surjective, it will factor through a surjective map, since the image $M'$ of the map is a subgroup. Hence we have $G\to M' \to M$

*probably a typo, but the map $f':H \hookrightarrow G \to G/N$ is $h \mapsto hN$ 

*hence $f'$ is not necessarily surjective! (but as mentioned above we don't need that)

*if you already go into detail you could mention why $ker(f')=N$ (which is easy)


As a last comment, let me mention that the properties of the quotient map are usually assumed as given, because they are very basic. But if you use this the first time it is good to prove everything.
