Exercise
Let $f:G \to G'$ be an isomorphism and let $H\unlhd G$. If $H'=f(H)$, prove that $G/H \cong G'/H'$.
As I've shown that $H'\unlhd G'$, I thought of defining $$\phi(Ha)=H'f(a)$$I was trying to show that this function is well defined and that it is an isomorphism.
I'll work with right cosets (which, since $H$ and $H'$ are normal, it's the same as working with left cosets). I need to know if what I did is correct and I would appreciate some help to show injectivity (maybe the $\phi$ I've defined is not the correct one).
So, what I mean by well-defined is that if $Ha=Ha'$, then $H'f(a)=H'f(a')$. It will be sufficient to show that $f(a)f(a')^{-1} \in H'$; by hypothesis, we have $aa'^{-1} \in H$, which means $f(aa'^{-1}) \in H'$. But then $f(aa'^{-1})=f(a)f(a'^{-1})=f(a)f(a')^{-1} \in H'$. From here one deduces the well definition.
To check $\phi$ is a morphism, we have to show $\phi((Ha)(Hb))=\phi(Ha)\phi(Hb)$. But $(Ha)(Hb)=H(ab)$, so $\phi(((Ha)(Hb))=\phi(H(ab))=H'f(ab)=H'f(a)f(b)=(H'f(a))(H'f(b))=\phi(Ha)\phi(Hb)$.
Surjectivity is almost immediate, take a right coset $H'y$, since $f$ is isomorphic, there is $g \in G$ such that $f(g)=y$, so $\phi(Hg)=H'f(g)=H'y$
Now for injectivity, suppose $\phi(Ha)=\phi(Ha')$, then $H'f(a)=H'f(a')$, I got stuck there.
Any suggestions would be appreciated. Thanks in advance.
For well-defined, it's slightly simpler to apply the function $f$ to the equation $Ha=Ha'$ to deduce $f(H)f(a)=f(H)f(a')$ which is $H'f=H'f(a')$.
For injectivity, apply the inverse isomorphism $f^{-1}$ to the equation $H' f(a) = H' f(a')$.