In the book Permutation Groups by Dixon & Mortimer, page 78, the well-known fact that for $|\Omega| \ge 5$ the alternating group $Alt(\Omega)$ is simple is proven.
It uses a Theorem that if a primitive group $G$ contains a $3$-cycle, then it contains the alternating group of the set acted on. And in the book it is presented as an implication of this Theorem, but somehow at the end of the proof it is shown that an arbitary normal subgroup $\ne 1$ contains every $3$-cycle, and somehow earlier in the book (as exercise 1.6.9) it is stated that the $3$-cycles generate the alternating group. So this would already yield that the subgroup equals the full alternating group. My question is why invoke this somewhat deep theorem about primitive groups, when it is actually not needed (or maybe I have just overlooked something)?
I will give the proof as written in the book and mark the relevant part (and added explanatory footnotes):
Corollary: If $|\Omega| \ge 5$, then $Alt(\Omega)$ is simple.
Proof: Put $A := Alt(\Omega)$ with $|\Omega| \ge 5$. Suppose we have $N \unlhd A$ and $N \ne 1$; then $N$ is transitive$^{(1)}$. Since $N \le A$, the minimal degree$^{(2)}$ $m$ of $N$ is at least $3$. Our first step is to prove that $m = 3$. Every element of finite support$^{(3)}$ has finite order, and so we can choose an element $u \in N$ of prime order, say $p$, and all of the nontrivial cycles of $u$ have length $p$. We observe that if $x \in A$, then the commutator $y := [u^{-1}x^{-1}u, x] = u^{-1}(xux^{-1})u^{-1}(x^{-1}ux) \in N$. In particular if$^{(4)}$ $|\operatorname{supp}(u^{-1}x^{-1}u)\cap \operatorname{supp}(x)| = 1$, then $N$ contains a $3$-cycle$^{(5)}$. Consider three special cases:
(i) If $p > 3$, then $u$ has a cycle $(\alpha \beta \gamma \delta \epsilon\ldots)$ of length at least $5$. Take $x = (\alpha \beta \delta)$, and then $y = (\beta\delta\gamma) \in N$.
(ii) If $p = 3$, and $u$ is not a $3$-cycle, then $u$ has at least two $3$-cycles $(\alpha\beta\gamma)(\delta\epsilon\theta)\ldots$. Take $x = (\alpha\beta\delta)$, and then $y = (\beta\delta\epsilon) \in N$.
(iii) If $p = 2$, then $u$ has at least two $2$-cycles $(\alpha\beta)(\gamma\delta)\ldots$. Take $x = (\alpha\beta\gamma)$, and then $y = (\alpha\beta)(\gamma\delta) \in N$.
It follows from (i) and (ii) that $m = 3$ if $p > 3$. In the case $p = 2$, (iii) shows that $u$ can be chosen in the form $(\alpha\beta)(\gamma\delta)(\epsilon)\ldots$; taking $x = (\alpha\gamma\epsilon)$ in this case gives $y = (\alpha\beta\epsilon) \in N$. Thus in all cases $m = 3$ as asserted, and $N$ contains a $3$-cycle $z$.
Finally, since $A$ is $3$-transitive, $x^{-1}zx$ runs over the sets of all $3$-cycles of $A$ as $x$ runs over $A$; hence $N$ contains all $3$-cycles. This implies that $N$ has no nontrivial blocks, and so $N$ is primitive. Thus $N = A$ by the theorem (Remark: The Theorem mentioned in the introduction about containment of $3$-cycles in primitive groups). This shows that $A$ has no proper nontrivial subgroups, and so $A$ is simple. $\square$
So why taking this additonal step and invoking the theorem, when I guess it could be finished at the marked spot?
(1) This follows by a previous theorem that the orbits of a normal subgroup form blocks, and because the alternating group is primitive.
(2) The minimal degree is the smallest number of points moved by any non-identity element.
(3) An element has finite support iff it just moves a finite number of points.
(4) $\operatorname{supp}(x) := \{ \alpha \in \Omega : \alpha^x \ne \alpha \}$
(5) This is an exercise in the book, namely that if $x,y \in Sym(\Omega)$ and $\Gamma = \operatorname{supp}(x) \cap \operatorname{supp}(y)$, then $\operatorname{supp}([x,y]) \subseteq \Gamma \cup \Gamma^x \cup \Gamma^y$, in particular if $\Gamma = \{\alpha\}$, then $[x,y]$ is a $3$-cycle.