I am solving this problem:
Let $H,~K$, and $L$ be normal subgroups of $G$ with $H<K<L$. Let $A=G/H,~B=K/H$, and $C=L/H$
Show that $B$ and $C$ are normal subgroups of $A$, and $B<C$
My solution is:
Let us use the third isomorphism theorem: $$(G/H)/(K/H)\simeq G/K$$ Replacing: $$ A/B\simeq G/K$$ As the isomorphism preserves the structure of the group, then $B$ is normal to $A$. Also, $A/B$ would not be a group if $B$ was not normal in $A$.
An Analogous argument shows that $C$ is normal to $A$.
It follows from $H<K<L$ that $K/H<L/H$ so $B<C$.
Is my first argument correct? I have found another solution that uses the restricted natural homomorphisms, but I think it is unnecessarily confusing.