Prove that all normal subgroup definitions are equivalent. given $ N<G $
I need to prove that all of the below are equivalent: 
1) for each $g \in G$ , $n \in N$ $gng^{-1} \in N $
2) for each $g \in G$  $gNg^{-1} = N $
3) for each $g \in G$  $gN = Ng$
4) G/N = N\G
5) There exists a Group G' and homomorphism $f: G$$\to$ $G'$ such that $ker(f) = N$
What I did - I managed to start from number 2 and prove 
2 $\to$ 3 $\to$ 4
about 4 $\to$ 5 : I know that 4 means that N is Normal with G and that means that G' = G/N satisfies 5 but I can't explain it is like that. 
also I can't get to and from number 1 (best would be 5 to one and 1 to 2..)
 A: Consider the induced operation $$\begin{align} \dot\, : \frac{G}{N} \times \frac{G}{N} &\to \frac{G}{N}\\ (xN, yN) &\mapsto x\dot\,y N\end{align}$$
If $(xN , yN) = (\tilde xN , \tilde y N)$ then $xN = \tilde xN \iff \tilde x \in xN \iff \tilde x = x n$, similarly $\tilde y = y n'$, then $$\begin{align}x\dot \,y N = \tilde x \dot\,\tilde y N &\iff x\dot\,y N = x\dot\,n\, y\dot\,n' N , \forall x, y \in G, \forall n,n' \in N\\&\iff y^{-1}\dot\,x^{-1}\dot\, x\dot\, y N = y^{-1}\dot\,x^{-1}\dot\, x\dot\, n\dot \, y N\\&\iff N = y^{-1}ny N, \, \forall y \in G \,\, \text{and} \, \, \forall n \in N \\&\iff y^{-1} n y \in N, \, \forall y \in G \,\, \text{and} \, \, \forall n \in N \\&\iff yN = Ny, \, \forall y \in G\end{align}$$
Therefore you may consider the group $\frac{G}{N}$, where $N$ is the identity element, and the inverse element of $gN$ is $g^{-1}N$.
Edit: Your notation is quite different. If by $G/N = N/G$  you mean $GN = NG$, then $GN = NG$ if , and only if, $GN$ is a  subgroup of $G$.
Take $G' = GN$ and $$\begin{align}\psi : G &\to GN \\g &\mapsto gn\end{align}$$ where $\ker \psi = N$.
