Let $G$ and $G'$ be two groups, $f:G \to G'$ a surjective morphism, $H'$ normal subgroup of $G'$ and $H = f^{-1}(H')$.
How to show these isomorphisms ? $\require{AMScd}$ \begin{CD} \frac{G'}{H'} @>\sim>> \frac{G}{H} @>\sim>> \frac{G \over \ker{f}}{H \over \ker{f}} \end{CD}
But before may I need to find what is the third quotient ? And is it possible to show the existence generally or I have to show that a specific morphism is bijective ? Clearly it seem obvious by the hypothesis that a bijection exist between $H$ and $H'$ but after I don't know what to do.
In algebra, a no-brainer that works more than often is to throw the most natural attempt of a map, and then check that it is well defined.
eg for the isomorphism $G/H \cong G'/H'$ :
For $g\in G$, define $\varphi(g + H) := f(g) + H'$.
$\varphi$ is well defined : We first have to show that the value of $\varphi(g + H)$ doesn't depend on the choice of $g$, so consider $g'$ such that $g + H = g' + H'$, ie $g-g' \in H$.
One has $$\varphi(g + H) - \varphi(g' + H) = (f(g) + H') - (f(g') + H') = f(g-g') + H'$$ But $g-g' \in H$, hence $f(g-g') \in H'$, and we get $\varphi(g + H) - \varphi(g' + H) = 0$. Hence, $\boxed{\varphi(g + H) = \varphi(g' + H)}$.
Clearly, $\varphi$ is a surjective homomorphism. Let's show that it is also injective (and hence an isomorphism) :
Consider $g \in G$ such that $\varphi(g + H) = H'$. Then we have $f(g) + H' = H'$ hence $f(g) \in H'$ hence $g \in H$. The only preimage of $0_{G'/H'}$ is $0_{G/H}$ : $\varphi$ is injective.