I have a group G and subgroup N that is normal to G. As a part of proving the 3rd isomorphism theorm (a version of it) I need to prove that the transformation
from the subgroups of G containing N
that means from H such that $N \le H \le G$
to the subgroups of G/N (that means H/N)
$\varphi : H \to H /N$
is one-to-one correspondence.
What i did is I showed that ker($\varphi$) = N = $e_{H/N}$
is that ok ? in all the proves i see people are proving it differently..
any help will be appriciated.
It seems to me like the correspondence $\varphi$ should not be going from a subgroup $H$ to a subgroup $H/N$.
The way you have it written $\varphi$ acts on the elements of $H$, so $\varphi(h)=(hN)$, but I think what the problem was asking you was to justify the injectivity of a map from the set of subgroups of $G$ containing $N$ to the set of subgroups of $G/N$, so $$ \varphi : \{H \leq G : N \leq H\} \to G/N \\ \varphi(H)=H/N$$
In general, this is not a group homomorphism, and proving that a function is injective by showing its kernel is trivial only works for functions that are homomorphisms. The statement of the theorem you used goes something like this:
If $\psi$ is a group homomorphism, then $\ker\psi =\{1\} \iff \psi$ is injective.
Since $\varphi$ is not a group homomorphism, you can't use that theorem, so to show $\varphi$ is injective, you need to use the definition: assume that $\psi(H_1/N)=\psi(H_2/N)$ and show this implies that $H_1=H_2$.