Homomorphism and normal subgroups 
Suppose that $\phi : G \to G'$ is a homomorphism between the groups $G$ and $G'$. Let $N'$ be a normal subgroup of $G'.$ Prove that the inverse image of $N'$ is a normal subgroup of $G$. 

How can I prove this using the defintions? 
A proof that was given confused me. The proof states:
Consider the homomorphism $\phi': G' \to G'/N'$ then $N'$ is the kernal of $\phi'$ (why?). Now consider the composition of $\phi$ and $\phi'$ then the inverse image $N$ is the kernal of $\phi' \circ \phi$ (again why?).
 A: Alternative proof:
Define $N=\left\{ x\in G\mid\phi\left(x\right)\in N'\right\} $. Then
$x,y\in N\Rightarrow xy^{-1}\in N$ since $\phi\left(xy^{-1}\right)=\phi\left(x\right)\phi\left(y\right)^{-1}\in N'$.
This proves that $N$ is a subgroup of $G$. Let $n\in N$ and $g\in G$.
Then $\phi\left(gng^{-1}\right)=\phi\left(g\right)\phi\left(n\right)\phi\left(g\right)^{-1}\in N'$
because $N'$ is normal. This shows that $gng^{-1}\in N$ and proved
is now that $N$ is a normal subgroup of $G$.
When $\nu:G'\rightarrow G'/N'$ is the natural map then $\psi=\nu\circ\phi:G\rightarrow G'/N'$.
Notice that $N'$ is the kernel of $\nu$ so the set $N=\left\{ x\in G\mid\phi\left(x\right)\in N'\right\} $
is the kernel of $\psi$. 
Every kernel is a normal subgroup, so this also proves that $N$ is a normal subgroup.
In your question $\nu$ is denoted as $\phi'$. 
A: If you want to prove this by definition: Let $y\in \phi^{-1}(N')$ and $g\in G$. Then you want to know if $gyg^{-1}\in \phi^{-1}(N')$. If so, $\phi^{-1}(N')$ is normal. To check that, we calculate 
$$\phi(gyg^{-1}) = \phi(g) \phi(y) \phi(g^{-1}) = \phi(g) \phi(y) (\phi(g))^{-1} \in N'$$
as $\phi(y)\in N'$. Thus $\phi^{-1}(N')$ is normal.
A: An element of $G'$ is in the kernel of $\phi'$ precisely when it lies in the coset $e N' = N'$. The kernel of the composition $\phi' \circ \phi$ is the set of elements $x$ such that $\phi'(\phi(x)) = 0$. These are precisely the elements $x$ for which $\phi(x) \in $ ker$(\phi')$, which is the inverse image of $N' = $ ker$(\phi')$ under $\phi$.
