Using induction to show $A_n$ is simple $A_n$ is simple for $n\geq 5$
I want to proceed via induction.$A_5$ is simple I have successfully proved it.
Now assuming $A_n$ is simple it is required to show that $A_{n+1}$ is also simple
Is this the correct approach ?How to proceed here if this approach is correct?
 A: OK, here is a sketch. Let $N$ be a nontrivial normal subgroup of $A_{n+1}$ with $n \ge 5$. Since $A_{n+1}$ is $2$-transitive, $N$ must be transitive.
In more detail: A group acts $k$-transitively on a set $X$ if for any two sequences of distinct points $a_1,\ldots,a_k$ and $b_1,\ldots,b_k$ in $X$, there exists $g \in G$ with $g.a_i=b_i$ for $1 \le i \le k$. So, for example, $S_n$ acts $n$-transitively and $A_n$ acts $(n-2)$-transitively in tis natural action on $n$ points.
Theorem. If $N$ is a normal subgroup of a group $G$ acting $2$-transitively on $X$ and $N$ acts nontrivially on $X$, then $N$ is transitive on $X$.
Proof. By assumption, there exists $g \in N$ and $x,y\in X$ with $g.x=y$ and $x \ne y$. Let $a,b \in X$ $a \ne b$. Then, by $2$-transitivity of $G$, there exists $h \in G$ with $h.x=a$, $h.y=b$. Then $hgh^{-1}.a=b$ with $hgh^{-1} \in N$, QED.
If the stabilizer $N_1$ of a point is trivial, then $N_1$ is a nontrivial normal subgroup of $A_n$, and so by the inductive hypothesis $N_1=A_n$, and then, since $N$ is transitive, $N= A_{n+1}$, and we are done.
So $N_1 = 1$, and $N$ is a regular normal subgroup of $A_{n+1}$. Note that the point stabilizer $A_n$ of $A_{n+1}$ acts by conjugation on the elements of $N$ and, since $n \ge 5$, it is acting $3$-transitively. The transitive action implies that all elements of $N$ have the same order, which must be a prime $p$. So $N$ is a $p$-group. Hence $Z(N) \ne 1$ and since all automorphisms fix $Z(N)$, $Z(N)=N$, so $N$ is elementary abelian.
So we can think of $A_n$ as a subgroup of ${\rm Aut}(N) = {\rm GL}(k,p)$, where $|N|=p^k$. But ${\rm GL}(k,p)$ acts $2$-transitively only when $p=2$, and $3$-transitively only where $p=2$, $k=2$, which is not the case here, so we have a contradiction.
