Proving a certain lemma about subgroups of $A_n$ In proving $A_n$ is simple for $n\neq4$, my teacher established the cases 1, 2, 3 as obvious, then proved the case 5, and proceded by induction on the rest. In the midst of that induction, he stated that if $N$ is a (normal) subgroup of $A_n$ such that it acts transitively on the set $\{1,\dotsc,n\}$, then $NA_{n-1}=A_n$. How do I prove this? 

Frattini Argument;
Let $G$ be a group acting on a set $\Omega$ and $N$ be a subgroup of $G$. If action of $N$ is transitive then $$G=NStab(x)$$
You question can be directly answered by Frattini argument as $A_{n-1}=Stab(n)$. 
Proof: Let $g\in G$ and $x\in \Omega$. Clearly, $gx\in \Omega$. Since the action of $N$ is transitive, we have $n$ such that
$$gx=nx$$
$$n^{-1}gx=x$$
$n^{-1}g\in Stab(x)\implies g\in nStab(x)$. Since $g$ is arbitrary, we have $G=NStab(x)$.
Corollary: If $N$ is a normal subgroup of $G$ with Sylow-$p$ subgroup $P$ then $G=NN_G(P)$.
Proof: Set $\Omega$ as all Sylow-$p$ subgroup of $N$ and notice that you can apply Frattini argument. Then the result follows. 
