Normal subgroup of a group acting transitively on a set with $p$ elements 
Let $G$ be a group acting transitively on $X=\{1,2,\dots,p\}$, $p$ prime. If $N$ is a normal subgroup of $G$, prove that either $N$ acts transitively on $X$ or $N$ fixes every element in $X$. 

Let $\phi: G\longrightarrow S_p$. I only get the proof for the case $A_p\leq\phi(G)$. 
 A: The key point to prove is that all orbits under $N$ on the set of $p$ points have the same size; then by primality their sizes are either all$~1$ or there is just one orbit of order$~p$. Now the stabilisers in$~G$ of all points are conjugate, since these points form a single $G$-orbit. The stabilisers of the points in $N$ are obtained by intersecting those conjugate subgroups with the normal subgroup$~N$; you should have no difficulty showing that these intersections therefore all have the same size. Now the orbit-stabiliser theorem for $N$ will finish the proof.
A: I'll ginve another answer just paraphrasing Marc van Leeuwen's answer.
Firstly, we show that all of the $N-$ orbits have the same size:
Let $x,y\in X$. Then since the $G-$ action is transitive there is some $g\in G$ s.t. $y=gx$. So we have the bijection $Nx\to Ny=Ngx=gNx$ with $q\mapsto gq$.
So let $s$ the common size of $N-$orbits. Then we have that:
$$p=|X|=(\# N-\text{orbits})(\text{size of each } N- \text{orbit})=(\# N-\text{orbits})s$$

*

*If $s>1$ then $s=p$ which means that there is only one $N-$orbit hence the $N-$action is transitive

*If $s=1$ then $N$ fixes every element (which means that $N=1$).

